This classic work has been a unique resource for thousands of mathematicians, scientists and engineers since its first a

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A C O U R S E O F M O D E R N A NA LY S I S

This classic work has been a unique resource for thousands of mathematicians, scientists and engineers since its first appearance in 1902. Never out of print, its continuing value lies in its thorough and exhaustive treatment of special functions of mathematical physics and the analysis of differential equations from which they emerge. The book also is of historical value as it was the first book in English to introduce the then modern methods of complex analysis. This fifth edition preserves the style and content of the original, but it has been supplemented with more recent results and references where appropriate. All the formulas have been checked and many corrections made. A complete bibliographical search has been conducted to present the references in modern form for ease of use. A new foreword by Professor S. J. Patterson sketches the circumstances of the book’s genesis and explains the reasons for its longevity. A welcome addition to any mathematician’s bookshelf, this will allow a whole new generation to experience the beauty contained in this text. e . t. w h i t ta k e r was Professor of Mathematics at the University of Edinburgh. He was awarded the Copley Medal in 1954, ‘for his distinguished contributions to both pure and applied mathematics and to theoretical physics’. g . n. wat s o n was Professor of Pure Mathematics at the University of Birmingham. He is known, amongst other things, for the 1918 result now known as Watson’s lemma and was awarded the De Morgan Medal in 1947. v i c t o r h . m o l l is Professor in the Department of Mathematics at Tulane University. He co-authored Elliptic Curves (Cambridge, 1997) and was awarded the Weiss Presidential Award in 2017 for his Graduate Teaching. He first received a copy of Whittaker and Watson during his own undergraduate studies at the Universidad Santa Maria in Chile.

(Left): Edmund Taylor Whittaker (1873–1956); (Right): George Neville Watson (1886–1965): Universal History Archive/Contributor/Getty Images.

A COURSE OF MODERN ANALYSIS Fi f t h E d i t i o n

An introduction to the general theory of infinite processes and of analytic functions with an account of the principal transcendental functions E . T. W H I T TA K E R A N D G . N . WAT S O N Fifth edition edited and prepared for publication by Victor H. Moll Tulane University, Louisiana

University Printing House, Cambridge CB2 8BS, United Kingdom One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia 314–321, 3rd Floor, Plot 3, Splendor Forum, Jasola District Centre, New Delhi – 110025, India 103 Penang Road, #05–06/07, Visioncrest Commercial, Singapore 238467 Cambridge University Press is part of the University of Cambridge. It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning, and research at the highest international levels of excellence. www.cambridge.org Information on this title: www.cambridge.org/9781316518939 DOI: 10.1017/9781009004091 © Cambridge University Press 1902, 1915, 1920, 1927, 2021 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First edition 1902 Second edition 1915 Third edition 1920 Fourth edition 1927 Reprinted 1935, 1940, 1946, 1950, 1952, 1958, 1962, 1963 Reissued in the Cambridge Mathematical Library Series 1996 Sixth printing 2006 Fifth edition 2021 Printed in the United Kingdom by TJ Books Limited, Padstow Cornwall A catalogue record for this publication is available from the British Library. ISBN 978-1-316-51893-9 Hardback Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

Contents

Foreword by S.J. Patterson

xvii

Preface to the Fifth Edition

xxi

Preface to the Fourth Edition

xxiii

Preface to the Third Edition

xxiv

Preface to the Second Edition

xxv

Preface to the First Edition

xxvi

Introduction

xxvii

Part I

The Process of Analysis

1

1 1.1 1.2 1.3 1.4 1.5 1.6

Complex Numbers Rational numbers Dedekind’s theory of irrational numbers Complex numbers The modulus of a complex number The Argand diagram Miscellaneous examples

3 3 4 6 7 8 9

2 2.1

The Theory of Convergence The definition of the limit of a sequence

10 10 10 10 11

2.11

2.2

2.21 2.22

2.3

Definition of the phrase ‘of the order of’

The limit of an increasing sequence Limit-points and the Bolzano–Weierstrass theorem Cauchy’s theorem on the necessary and sufficient condition for the existence of a limit

Convergence of an infinite series 2.31 2.32 2.33 2.34

Dirichlet’s test for convergence Absolute and conditional convergence Í 1 The geometric series, and the series ∞ n=1 n s The comparison theorem

v

12 13 16 17 17 18

Contents

vi

2.4

2.35 2.36 2.37

Cauchy’s test for absolute convergence D’Alembert’s ratio test for absolute convergence A general theorem on series for which lim |un+1 /un | = 1

2.38

Convergence of the hypergeometric series

Effect of changing the order of the terms in a series 2.41

2.5

Convergence of series derived from a power series

Infinite products 2.71

2.8

Methods of summing a double series Absolutely convergent double series Cauchy’s theorem on the multiplication of absolutely convergent series

Power series 2.61

2.7

The fundamental property of absolutely convergent series

Double series 2.51 2.52 2.53

2.6

n→∞

Some examples of infinite products

Infinite determinants 2.81 2.82

Convergence of an infinite determinant The rearrangement theorem for convergent infinite determinants

2.9

Miscellaneous examples

3 3.1 3.2

Continuous Functions and Uniform Convergence The dependence of one complex number on another Continuity of functions of real variables 3.21 3.22

3.3

Series of variable terms. Uniformity of convergence 3.31 3.32 3.33 3.34 3.35

3.4 3.5 3.6

On the condition for uniformity of convergence Connexion of discontinuity with non-uniform convergence The distinction between absolute and uniform convergence A condition, due to Weierstrass, for uniform convergence Hardy’s tests for uniform convergence

Discussion of a particular double series The concept of uniformity The modified Heine–Borel theorem 3.61 3.62 3.63 3.64

3.7

Simple curves. Continua Continuous functions of complex variables

Uniformity of continuity A real function, of a real variable, continuous in a closed interval, attains its upper bound A real function, of a real variable, continuous in a closed interval, attains all values between its upper and lower bounds The fluctuation of a function of a real variable

Uniformity of convergence of power series 3.71 3.72 3.73

Abel’s theorem Abel’s theorem on multiplication of convergent series Power series which vanish identically

3.8

Miscellaneous examples

4 4.1

The Theory of Riemann Integration The concept of integration 4.11 4.12

Upper and lower integrals Riemann’s condition of integrability

20 20 21 22 23 24 24 25 26 27 28 29 30 31 34 34 35 36 40 40 40 41 42 43 44 45 46 47 48 49 51 51 52 53 54 54 55 55 55 56 56 58 58 58 59

Contents 4.13 4.14

4.2 4.3 4.4

Differentiation of integrals containing a parameter Double integrals and repeated integrals Infinite integrals 4.41 4.42 4.43 4.44

4.5

Infinite integrals of continuous functions. Conditions for convergence Uniformity of convergence of an infinite integral Tests for the convergence of an infinite integral Theorems concerning uniformly convergent infinite integrals

Improper integrals. Principal values 4.51

4.6

A general theorem on integration Mean-value theorems

The inversion of the order of integration of a certain repeated integral

Complex integration 4.61 4.62

The fundamental theorem of complex integration An upper limit to the value of a complex integral

4.7 4.8

Integration of infinite series Miscellaneous examples

5

The Fundamental Properties of Analytic Functions; Taylor’s, Laurent’s and Liouville’s Theorems Property of the elementary functions

5.1

5.11 5.12 5.13

5.2

Occasional failure of the property Cauchy’s definition of an analytic function of a complex variable An application of the modified Heine–Borel theorem

Cauchy’s theorem on the integral of a function round a contour 5.21

The value of an analytic function at a point, expressed as an integral taken round a contour enclosing the point The derivatives of an analytic function f (z) Cauchy’s inequality for f (n) (a)

vii 60 62 64 65 67 67 68 68 71 72 73 75 76 76 77 79

81 81 82 82 83 83

5.7 5.8

Many-valued functions Miscellaneous examples

86 88 89 89 90 91 91 94 95 97 98 100 101 103 104 105 106

6 6.1 6.2

The Theory of Residues; Application to the Evaluation of Definite Integrals Residues The evaluation of definite integrals

110 110 111

5.22 5.23

5.3

Analytic functions represented by uniformly convergent series 5.31 5.32

5.4

Taylor’s theorem 5.41

5.5

Forms of the remainder in Taylor’s series

The process of continuation 5.51

5.6

Analytic functions represented by integrals Analytic functions represented by infinite integrals

The identity of two functions

Laurent’s theorem 5.61 5.62 5.63 5.64

6.21 6.22

The nature of the singularities of one-valued functions The ‘point at infinity’ Liouvillle’s theorem Functions with no essential singularities

The evaluation of the integrals of certain periodic functions taken between the limits 0 and 2π The evaluation of certain types of integrals taken between the limits −∞ and +∞

111 112

Contents

viii 6.23 6.24

6.3

Principal values of integrals ∫∞ Evaluation of integrals of the form 0 x a−1 Q (x) dx

Cauchy’s integral 6.31

The number of roots of an equation contained within a contour

6.4 6.5

Connexion between the zeros of a function and the zeros of its derivative Miscellaneous examples

7 7.1 7.2

The Expansion of Functions in Infinite Series A formula due to Darboux The Bernoullian numbers and the Bernoullian polynomials 7.21

7.3

7.31 7.32

7.4 7.5 7.6 7.7 7.8

The Euler–Maclaurin expansion

Bürmann’s theorem Teixeira’s extended form of Bürmann’s theorem Lagrange’s theorem

The expansion of a class of functions in rational fractions The expansion of a class of functions as infinite products The factor theorem of Weierstrass Expansion in a series of cotangents Borel’s theorem 7.81 7.82

Borel’s integral and analytic continuation Expansions in series of inverse factorials

7.9

Miscellaneous examples

8 8.1 8.2

Asymptotic Expansions and Summable Series Simple example of an asymptotic expansion Definition of an asymptotic expansion 8.21

8.3

8.31 8.32

8.4

Another example of an asymptotic expansion

Multiplication of asymptotic expansions Integration of asymptotic expansions Uniqueness of an asymptotic expansion

Methods of summing series 8.41 8.42 8.43 8.44

Borel’s method of summation Euler’s method of summation Cesàro’s method of summation The method of summation of Riesz

8.5 8.6

Hardy’s convergence theorem Miscellaneous examples

9 9.1

Fourier Series and Trigonometric Series Definition of Fourier series 9.11 9.12

9.2

On Dirichlet’s conditions and Fourier’s theorem 9.21 9.22

9.3

The representation of a function by Fourier series for ranges other than (−π, π) The cosine series and the sine series

The nature of the coefficients in a Fourier series 9.31 9.32

9.4

Nature of the region within which a trigonometrical series converges Values of the coefficients in terms of the sum of a trigonometrical series

Differentiation of Fourier series Determination of points of discontinuity

Fejér’s theorem

116 117 118 119 120 121 125 125 125 127 129 131 133 134 137 138 140 141 142 143 145 153 153 154 154 156 156 157 157 158 158 158 159 159 161 163 163 164 167 167 168 169 171 172 173 174

Contents 9.41 9.42 9.43 9.44

9.5 9.6

The Riemann–Lebesgue lemmas The proof of Fourier’s theorem The Dirichlet–Bonnet proof of Fourier’s theorem The uniformity of the convergence of Fourier series

The Hurwitz–Liapounoff theorem concerning Fourier constants Riemann’s theory of trigonometrical series 9.61 9.62 9.63

Riemann’s associated function Properties of Riemann’s associated function; Riemann’s first lemma Riemann’s theorem on trigonometrical series

9.7 9.8

Fourier’s representation of a function by an integral Miscellaneous examples

10 10.1 10.2

Linear Differential Equations Linear differential equations Solutions in the vicinity of an ordinary point 10.21

10.3

10.31 10.32

10.4 10.5 10.6 10.7

Uniqueness of the solution

Points which are regular for a differential equation Convergence of the expansion of §10.3 Derivation of a second solution in the case when the difference of the exponents is an integer or zero

Solutions valid for large values of |z| Irregular singularities and confluence The differential equations of mathematical physics Linear differential equations with three singularities 10.71 10.72

Transformations of Riemann’s P-equation The connexion of Riemann’s P-equation with the hypergeometric equation

10.8 10.9

Linear differential equations with two singularities Miscellaneous examples

11 11.1

Integral Equations Definition of an integral equation 11.11

11.2

11.21 11.22 11.23

11.3

The solution of Fredholm’s equation by a series

Solution of Abel’s integral equation 11.81

11.9

The connexion of orthogonal functions with homogeneous integral equations

The development of a symmetric nucleus 11.71

11.8

Schmidt’s theorem that, if the nucleus is symmetric, the equation D(λ) = 0 has at least one root

Orthogonal functions 11.61

11.7

Volterra’s equation

The Liouville–Neumann method of successive substitutions Symmetric nuclei 11.51

11.6

Investigation of Fredholm’s solution Volterra’s reciprocal functions Homogeneous integral equations

Integral equations of the first and second kinds 11.31

11.4 11.5

An algebraical lemma

Fredholm’s equation and its tentative solution

Schlömilch’s integral equation

Miscellaneous examples

ix 177 179 181 183 185 187 188 189 191 193 195 201 201 201 203 204 206 207 209 210 210 214 215 215 216 216 219 219 220 221 223 226 228 229 229 230 231 232 233 234 236 237 238 238 239

Contents

x

Part II 12 12.1

The Gamma-Function Definitions of the Gamma-function 12.11 12.12 12.13 12.14 12.15 12.16

12.2

The Transcendental Functions

Euler’s formula for the Gamma-function The difference equation satisfied by the Gamma-function The evaluation of a general class of infinite products Connexion between the Gamma-function and the circular functions The multiplication-theorem of Gauss and Legendre Expansion for the logarithmic derivates of the Gamma-function

Euler’s expression of Γ(z) as an infinite integral 12.21

241 243 243 245 245 246 248 248 249 250

Extension of the infinite integral to the case in which the argument of the Gamma-function is negative Hankel’s expression of Γ(z) as a contour integral

12.5 12.6

252 253 Gauss’ infinite integral for Γ 0(z)/Γ(z) 255 12.31 Binet’s first expression for log Γ(z) in terms of an infinite integral 257 12.32 Binet’s second expression for log Γ(z) in terms of an infinite integral 259 12.33 The asymptotic expansion of the logarithms of the Gamma-function 261 The Eulerian integral of the first kind 263 12.41 Expression of the Eulerian integral of the first kind in terms of the Gamma-function 264 12.42 Evaluation of trigonometrical integrals in terms of the Gamma-function 265 12.43 Pochhammer’s extension of the Eulerian integral of the first kind 266 Dirichlet’s integral 267 Miscellaneous examples 268

13 13.1

The Zeta-Function of Riemann Definition of the zeta-function

12.22

12.3

12.4

13.2 13.3 13.4 13.5

13.11 13.12 13.13 13.14 13.15

The generalised zeta-function The expression of ζ(s, a) as an infinite integral The expression of ζ(s, a) as a contour integral Values of ζ(s, a) for special values of s The formula of Hurwitz for ζ(s, a) when σ < 0

13.21

Deductions from Hermite’s formula

13.31

Riemann’s hypothesis concerning the zeros of ζ(s)

13.51

Inequalities satisfied by ζ(s, a) when σ ≤ 0

Hermite’s formula for ζ(s, a) Euler’s product for ζ(s)

Riemann’s integral for ζ(s) Inequalities satisfied by ζ(s, a) when σ > 0

13.6 13.7

The asymptotic expansion of log Γ(z + a) Miscellaneous examples

14 14.1

The Hypergeometric Function The hypergeometric series 14.11

14.2 14.3 14.4 14.5

The value of F(a, b; c; 1) when Re(c − a − b) > 0

The differential equation satisfied by F(a, b; c; z) Solutions of Riemann’s P-equation Relations between particular solutions Barnes’ contour integrals

276 276 276 276 277 278 279 280 282 282 283 283 285 286 288 290 293 293 293 295 295 298 299

Contents 14.51 14.52 14.53

14.6

The continuation of the hypergeometric series Barnes’ lemma The connexion between hypergeometric functions of z and of 1 − z

Solution of Riemann’s equation by a contour integral 14.61

Determination of an integral which represents P(α)

14.7 14.8

Relations between contiguous hypergeometric functions Miscellaneous examples

15 15.1

Legendre Functions Definition of Legendre polynomials 15.11 15.12 15.13 15.14

15.2

Legendre functions 15.21 15.22 15.23

15.3

The integral properties of the associated Legendre functions

Hobson’s definition of the associated Legendre functions 15.61

15.7

Neumann’s expansion of an arbitrary function in a series of Legendre polynomials

Ferrers’ associated Legendre functions Pnm (z) and Q m n (z) 15.51

15.6

Expansion of Q n (z) as a power series The recurrence formulae for Q n (z) The Laplacian integral for Legendre functions of the second kind Neumann’s formula for Q n (z), when n is an integer

Heine’s development of (t − z)−1 15.41

15.5

The recurrence formulae Murphy’s expression of Pn (z) as a hypergeometric function Laplace’s integrals for Pn (z)

Legendre functions of the second kind 15.31 15.32 15.33 15.34

15.4

Rodrigues’ formula for the Legendre polynomials Schläfli’s integral for Pn (z) Legendre’s differential equation The integral properties of the Legendre polynomials

Expression of Pnm (z) as an integral of Laplace’s type

The addition-theorem for the Legendre polynomials 15.71

The addition theorem for the Legendre functions

15.8 15.9

The function Cnν (z) Miscellaneous examples

16 16.1

The Confluent Hypergeometric Function The confluence of two singularities of Riemann’s equation 16.11 16.12

16.2 16.3

Expression of various functions by functions of the type Wk,m (z) The asymptotic expansion of Wk,m (z), when |z| is large 16.31

16.4

The second solution of Weber’s equation The general asymptotic expansion of Dn (z)

A contour integral for Dn (z) 16.61

16.7

Relations between Wk,m (z) and Mk,±m (z)

The parabolic cylinder functions. Weber’s equation 16.51 16.52

16.6

The second solution of the equation for Wk,m (z)

Contour integrals of the Mellin–Barnes type for Wk,m (z) 16.41

16.5

Kummer’s formulae Definition of the function Wk,m (z)

Recurrence formulae for Dn (z)

Properties of Dn (z) when n is an integer

xi 300 301 303 303 306 307 309 316 316 317 317 318 319 320 322 326 327 331 331 333 334 335 337 338 339 340 341 342 342 344 346 347 355 355 356 357 358 360 361 361 363 364 365 366 366 367 367

Contents

xii 16.8

Miscellaneous examples

369

17 17.1

Bessel Functions The Bessel coefficients

373 373 375 376 377 379 380 382 383 385 386 388 390 391 392 393 394

17.11

17.2

17.21 17.22 17.23 17.24

17.3 17.4 17.5 17.6

The ascending series for Yn (z)

Bessel functions with purely imaginary argument 17.71

17.8

The recurrence formulae for the Bessel functions The zeros of Bessel functions whose order n is real Bessel’s integral for the Bessel coefficients Bessel functions whose order is half an odd integer

Hankel’s contour integral for Jn (z) Connexion between Bessel coefficients and Legendre functions Asymptotic series for Jn (z) when |z| is large The second solution of Bessel’s equation 17.61

17.7

Bessel’s differential equation

Bessel’s equation when n is not necessarily an integer

Modified Bessel functions of the second kind

Neumann’s expansions 17.81 17.82

Proof of Neumann’s expansion Schlömilch’s expansion of an arbitrary function in a series of Bessel coefficients of order zero

17.9 Tabulation of Bessel functions 17.10 Miscellaneous examples 18 18.1 18.2 18.3

The Equations of Mathematical Physics The differential equations of mathematical physics Boundary conditions A general solution of Laplace’s equation 18.31

18.4 18.5

18.51

18.6

Solutions of Laplace’s equation involving Legendre functions

The solution of Laplace’s equation Laplace’s equation and Bessel coefficients The periods of vibration of a uniform membrane

A general solution of the equation of wave motions 18.61

Solutions of the equation of wave motions which involve Bessel functions

18.7

Miscellaneous examples

19 19.1

Mathieu Functions The differential equation of Mathieu 19.11 19.12

19.2

Periodic solutions of Mathieu’s equation 19.21 19.22

19.3

The integral formulae for the Mathieu functions

Floquet’s theory 19.41 19.42

19.5

An integral equation satisfied by even Mathieu functions Proof that the even Mathieu functions satisfy the integral equation

The construction of Mathieu functions 19.31

19.4

The form of the solution of Mathieu’s equation Hill’s equation

Hill’s method of solution The evaluation of Hill’s determinant

The Lindemann–Stieltjes theory of Mathieu’s general equation

396 397 397 407 407 408 409 412 414 417 417 418 418 420 426 426 428 428 428 429 430 431 433 434 435 437 438

Contents 19.51 19.52 19.53

19.6

xiii

Lindemann’s form of Floquet’s theorem The determination of the integral function associated with Mathieu’s equation The solution of Mathieu’s equation in terms of F(ζ)

A second method of constructing the Mathieu function 19.61

The convergence of the series defining Mathieu functions

19.7 19.8 19.9

The method of change of parameter The asymptotic solution of Mathieu’s equation Miscellaneous examples

20 20.1

Elliptic Functions. General Theorems and the Weierstrassian Functions Doubly-periodic functions

20.2

20.3

20.4

20.5

20.11 20.12 20.13 20.14

Period-parallelograms Simple properties of elliptic functions The order of an elliptic function Relation between the zeros and poles of an elliptic function

20.21 20.22

Periodicity and other properties of ℘(z) The differential equation satisfied by ℘(z)

20.31 20.32 20.33

Another form of the addition-theorem The constants e1 , e2 , e3 The addition of a half-period to the argument of ℘(z)

20.41 20.42

The quasi-periodicity of the function ζ(z) The function σ(z)

The construction of an elliptic function. Definition of ℘(z) The addition-theorem for the function ℘(z)

Quasi-periodic functions. The function ζ(z)

Formulae in terms of Weierstrassian functions 20.51 20.52 20.53 20.54

The expression of any elliptic function in terms of ℘(z) and ℘0 (z) The expression of any elliptic function as a linear combination of zeta-functions and their derivatives The expression of any elliptic function as a quotient of sigma-functions The connexion between any two elliptic functions with the same periods

20.6 20.7 20.8

On the integration of a0 x 4 + 4a1 x 3 + 6a2 x 2 + 4a3 x + a4 The uniformisation of curves of genus unity Miscellaneous examples

21 21.1

The Theta-Functions The definition of a theta-function 21.11 21.12

21.2

The addition-formulae for the theta-functions Jacobi’s fundamental formulae

Theta-functions as infinite products The differential equation satisfied by the theta-functions 21.41 21.42 21.43

21.5

The four types of theta-functions The zeros of the theta-functions

The relations between the squares of the theta-functions 21.21 21.22

21.3 21.4

−1/2

A relation between theta-functions of zero argument The value of the constant G Connexion of the sigma-function with the theta-functions

Elliptic functions in terms of theta-functions 21.51

Jacobi’s imaginary transformation

439 439 441 442 444 446 447 448 451 451 452 452 453 454 455 456 458 462 462 465 466 467 468 469 471 471 472 473 474 475 477 478 486 486 487 489 490 491 491 493 494 495 496 498 498 499

Contents

xiv 21.52

21.6

Differential equations of theta quotients 21.61 21.62

21.7

Landen’s type of transformation The genesis of the Jacobian elliptic function sn u Jacobi’s earlier notation. The theta-function Θ(u) and the eta-function H(u)

The problem of inversion 21.71 21.72 21.73

The problem of inversion for complex values of c. The modular functions f (τ), g(τ), h(τ) The periods, regarded as functions of the modulus The inversion-problem associated with Weierstrassian elliptic functions

21.8 The numerical computation of elliptic functions 21.9 The notations employed for the theta-functions 21.10 Miscellaneous examples 22 22.1

The Jacobian Elliptic Functions Elliptic functions with two simple poles 22.11 22.12

22.2

The addition-theorem for the function sn u 22.21

22.3

Fourier series for reciprocals of Jacobian elliptic functions

Elliptic integrals 22.71 22.72 22.73 22.74

22.8

Proof of Jacobi’s imaginary transformation by the aid of theta-functions Landen’s transformation

Infinite products for the Jacobian elliptic functions Fourier series for the Jacobian elliptic functions 22.61

22.7

The periodic properties (associated with K) of the Jacobian elliptic functions The constant K 0 The periodic properties (associated with K + iK 0 ) of the Jacobian elliptic functions The periodic properties (associated with iK 0 ) of the Jacobian elliptic functions General description of the functions sn u, cn u, dn u

Jacobi’s imaginary transformation 22.41 22.42

22.5 22.6

The addition-theorems for cn u and dn u

The constant K 22.31 22.32 22.33 22.34 22.35

22.4

The Jacobian elliptic functions, sn u, cn u, dn u Simple properties of sn u, cn u, dn u

The expression of a quartic as the product of sums of squares The three kinds of elliptic integrals The elliptic integral of the second kind. The function E(u) The elliptic integral of the third kind

The lemniscate functions 22.81 22.82

The values of K and K 0 for special values of k A geometrical illustration of the functions sn u, cn u, dn u

22.9

Miscellaneous examples

23 23.1 23.2

Ellipsoidal Harmonics and Lamé’s Equation The definition of ellipsoidal harmonics The four species of ellipsoidal harmonics 23.21 23.22 23.23 23.24 23.25

The construction of ellipsoidal harmonics of the first species Ellipsoidal harmonics of the second species Ellipsoidal harmonics of the third species Ellipsoidal harmonics of the fourth species Niven’s expressions for ellipsoidal harmonics in terms of homogeneous harmonics

501 502 503 504 505 506 510 510 511 512 513 517 517 517 519 521 523 525 526 527 529 530 531 532 533 534 535 537 539 540 541 542 545 551 552 554 556 557 567 567 568 568 571 572 573 574

Contents 23.26

23.3

23.31 23.32 23.33

23.4

The integral equation satisfied by Lamé functions of the third and fourth species Integral formulae for ellipsoidal harmonics Integral formulae for ellipsoidal harmonics of the third and fourth species

Generalisations of Lamé’s equation 23.71

23.8

Solutions in series of Lamé’s equation The definition of Lamé functions The non-repetition of factors in Lamé functions The linear independence of Lamé functions The linear independence of ellipsoidal harmonics Stieltjes’ theorem on the zeros of Lamé functions Lamé functions of the second kind

Lamé’s equation in association with Jacobian elliptic functions The integral equation for Lamé functions 23.61 23.62 23.63

23.7

Uniformising variables associated with confocal coordinates Laplace’s equation referred to confocal coordinates Ellipsoidal harmonics referred to confocal coordinates

Various forms of Lamé’s differential equation 23.41 23.42 23.43 23.44 23.45 23.46 23.47

23.5 23.6

Ellipsoidal harmonics of degree n

Confocal coordinates

The Jacobian form of the generalised Lamé equation

Miscellaneous examples

Appendix. The Elementary Transcendental Functions A.1 On certain results assumed in Chapters 1 to 4 A.11 A.12

A.2

The exponential function exp z A.21 A.22

A.3

The fundamental properties of sin z and cos z The addition-theorems for sin z and cos z

The periodicity of the exponential function A.51 A.52

A.6 A.7

The continuity of the Logarithm Differentiation of the Logarithm The expansion of Log(1 + a) in powers of a

The definition of the sine and cosine A.41 A.42

A.5

The addition-theorem for the exponential function, and its consequences Various properties of the exponential function

Logarithms of positive numbers A.31 A.32 A.33

A.4

Summary of the Appendix A logical order of development of the elements of analysis

The solution of the equation exp γ = 1 The solution of a pair of trigonometrical equations

Logarithms of complex numbers The analytical definition of an angle

xv 577 578 580 582 584 585 587 589 590 590 591 591 593 594 595 597 598 600 601 604 607 611 611 612 612 613 613 614 615 616 616 616 617 618 618 619 619 621 623 623

References

625

Author index

648

Subject index

652

Foreword S.J. Patterson There are few books which remain in print and in constant use for over a century; “Whittaker and Watson” belongs to this select group. In fact there were two books with the title “A Course in Modern Analysis”, the first in 1902 by Edmund Whittaker alone, a textbook with a very specific agenda, and then the joint work, first published in 1915 as a second edition. It is an extension of the first edition but in such a fashion that it becomes a handbook for those working in analysis. As late as 1966 J.T. Whittaker, the son of E.T. Whittaker, wrote in his Biographical Memoir of Fellows of the Royal Society (i.e. obituary) of G.N. Watson that there were still those who preferred the first edition but added that for most readers the later edition was to be preferred. Indeed the joint work is superior in many different ways. The first edition was written at a time when there was a movement for reform in mathematics at Cambridge. Edmund Whittaker’s mentor Andrew Forsyth was one of the driving forces in this movement and had himself written a Theory of Functions (1893) which was, in its time, very influential but is now scarcely remembered. In the course of the nineteenth century the mathematics education had become centered around the Mathematical Tripos, an intensely competitive examination. Competitions and sports were salient features of Victorian Britain, a move away from the older system of patronage and towards a meritocracy. The reader familiar with Gilbert and Sullivan operettas will think of the Modern Major-General in The Pirates of Penzance. The Tripos had become not only a sport but a spectator sport, followed extensively in middle-class England 1 . The result of this system was that the colleges were in competition with one another and employed coaches to prepare the talented students for the Tripos. They developed the skills needed to answer difficult questions quickly and accurately – many Tripos questions can be found in the exercises in Whittaker and Watson. The Tripos system did not encourage the students to become mathematicians and separated them from the professors who were generally very well informed about the developments on the Continent. It was a very inward-looking, self-reproducing system. The system on the Continent, especially in the German universities, was quite different. The professors there sought contact with the students, either as note-takers for lectures or in seminar talks, and actively supported those by whom they were most impressed. The students vied with one another for the attention of the professor, a different and more fruitful form of competition. This 1

Some idea of this may be gleaned from G.B. Shaw’s play Mrs Warren’s Profession, written in 1893 but held back by censorship until 1902. In this play Mrs Warren’s daughter Vivie has distinguished herself in Cambridge – she tied with the third Wrangler, described as a “magnificent achievement” by a character who has no mathematical background. She herself could not be ranked as a Wrangler as she was female. She would have been a contemporary of Grace Chisholm, later Grace Chisholm Young, whose family background was by no means as colourful as that of the fictional Vivie Warren.

xvii

Foreword

xviii

system allowed the likes of Weierstrass and Klein to build up groups of talented and highly motivated students. It had become evident to Andrew Forsyth and others that Cambridge was missing out on the developments abroad because of the concentration on the Tripos system 2 . It is interesting to read what Whittaker himself wrote about the situation at the end of the nineteenth century in Cambridge and so of the conditions under which Whittaker and Watson was written. We quote from his Royal Society Obituary Notice (1942) of Andrew Russell Forsyth: He had for some time past realized, as no one else did, the most serious deficiency of the Cambridge school, namely its ignorance of what had been and was being done on the continent of Europe. The college lecturers could not read German, and did not read French. .. . The schools of Göttingen and Berlin to a great extent ignored each other (Berlin said that Göttingen proved nothing, and Göttingen retorted that Berlin had no ideas) and both of them ignored French work. But Cambridge had hitherto ignored them all: and the time was ripe for Forsyth’s book. The younger men, even undergraduates, had heard in his lectures of the extraordinary riches and beauty of the domain beyond Tripos mathematics, and were eager to enter into it. From the day of its publication in 1893, the face of Cambridge was changed: the majority of the pure mathematicians who took their degrees in the next twenty years became function-theorists. and further As head of the Cambridge school of mathematics he was conspicuously successful. British mathematicians were already indebted to him for the first introduction of the symbolic invariant-theory, the Weierstrassian elliptic functions, the Cauchy–Hermite applications of contour-integration, the Riemannian treatment of algebraic functions, the theory of entire functions, and the theory of automorphic functions: and the importation of novelties continued to occupy his attention. A great traveller and a good linguist, he loved to meet eminent foreigners and invite them to enjoy Trinity hospitality: and in this way his post-graduate students had opportunities of becoming known personally to such men as Felix Klein (who came frequently), Mittag-Leffler, Darboux and Poincaré. To the students themselves, he was devoted: young men fresh from the narrow examination routine of the Tripos were invited to his rooms and told of the latest research papers: and under his fostering care, many of the wranglers of the period 1894–1910 became original workers of distinction. The two authors were very different people. Edmund Whittaker (1874–1956) went on from Cambridge in 1906 to become the Royal Astronomer in Ireland (then still a part of the 2

For his arguments see A. Forsyth: Old Tripos Days at Cambridge, Math. Gazette 19 162–179 (1935). For a dissenting opinion see K. Pearson: Old Tripos Days at Cambridge, as seen from another viewpoint, Math. Gazette 20 27–36 (1936).

Foreword

xix

United Kingdom) and Director of Dunsink Observatory, thereby following in the footsteps of William Rowan Hamilton. In 1985, on the occasion of the bicentenary of Dunsink, the then Director, Patrick A. Wayman, singled out Whittaker as the greatest director aside from Hamilton and one who, despite his relatively short tenure of office, 1906–1912, had achieved most for the Observatory 3 . This appointment brought out his skills as an administrator. Following this he moved to Edinburgh where he exerted his influence to guide mathematics there into the new century. Some indication of the success is given by the fact that it was W.V.D. Hodge, a student of his, who, at the International Congress of Mathematicians in 1954, invited the International Mathematical Union to hold the next Congress in Edinburgh. Whittaker himself did not live to experience the event which reflected the status in which Edinburgh was held at the end of his life. George Neville Watson (1886–1965) on the other hand was a retiring scholar who, after leaving Cambridge, at least in the flesh, spent four years (1914–1918) in London, and then became professor in Birmingham where he remained for the rest of his life 4 , living a relatively withdrawn life devoted to his mathematical work and with stamp-collecting and the study of the history of railways as hobbies. His early work was very much in the direction of E.W. Barnes and A.G. Greenhill. After Ramanujan’s death he took over from Hardy the analysis of many of Ramanujan’s unpublished papers, especially those connected with the theory of modular forms and functions, and of complex multiplication. It is worth remarking that Greenhill, a student and ardent admirer of James Clerk Maxwell and primarily an applied mathematician, concerned himself with the computation of singular moduli, and it was probably he who aroused Ramanujan’s interest in this topic. Watson’s work in this area is, besides his books, that for which he is best remembered today. Both authors wrote other books that are still used today. In Whittaker’s case these are his A Treatise on the Analytical Dynamics of Particles & Rigid Bodies, reprinted in 1999, with a foreword by Sir William McCrea in the CUP series “Cambridge Mathematical Library”, a source of much mathematics which is difficult to find elsewhere, and his History of Theories of the Aether and Electricity which, despite some unconventional views, is an invaluable source on the history of these parts of physics and the associated mathematics. Watson, on the other hand, wrote his A Treatise on the Theory of Bessel Functions, published in 1922, which like Whittaker and Watson has not been out of print since its appearance. On coming across it for the first time as a student I was taken aback by such a thick book being devoted to what seemed to be a very circumscribed subject. One of the Fellows of my college, a physicist, replying to a fellow student who had made a similar observation, declared that it was a work of genius and he would have been proud to have written something like it. In the course of the years I have had recourse to it over and over again and would now concur with this opinion. Watson’s Bessel Functions, like Whittaker and Watson, despite being somewhat oldfashioned, has retained a freshness and relevance that has made both of them classics. Unlike many other books of this period the terminology, although not the style, is that of today. It is less a Cours d’Analyse and more of a Handbuch der Funktionentheorie. Perhaps my own experiences can illuminate this. My copy was given to me in 1967 by my mathematics teacher, 3 4

Irish Astronomical Journal 17 177–178 (1986). It is worth noting that from 1924 on E.W. Barnes was a disputative Bishop of Birmingham.

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Foreword

Mr Cecil Hawe, after I had been awarded a place to study mathematics in Cambridge. He had bought it 20 years earlier as a student. During my student years the textbook on second year analysis was J. Dieudonné’s Foundations of Modern Analysis. People then were prone to be a bit supercilious at least about the “modern” in the title of Whittaker and Watson. 5 At that time it lay on my bookshelf unused. Five years later I was coming to terms with the theory of non-analytic automorphic forms, especially with Selberg’s theory of Eisenstein series. At this point I discovered how useful a book it was, both for the treatment of Bessel functions and for that of the hypergeometric function. It also has a very useful chapter on Fredholm’s theory of integral equations which Selberg had used. In the years since then several other chapters have proved useful, and ones I thought I knew became useful in novel ways. It became a constant companion. This was mainly in connection with doing mathematics but it also proved its worth in teaching – for example the chapter on Fourier series gives very useful results which can be obtained by relatively elementary methods and are suitable for undergraduate lectures. Dieudonné’s book is tremendous for the university teacher; it gives the fundamentals of analysis in a concentrated form, something very useful when one has an overloaded syllabus and a limited number of hours to teach it in. On the other hand it is much less useful as a “Handbuch” for the working analyst, at least in my experience. Nor was it written for this purpose. Whittaker and Watson started, in the first edition, as such a book for teaching but in the second and later editions became that book which has remained on the bookshelves of generations of working mathematicians, be they formally mathematicians, natural scientists or engineers. One aspect that probably contributed to the long popularity of Whittaker and Watson is the fact that it is not overloaded with many of the topics that are within range of the text. Thus, for example, the authors do not go into the arithmetic theory of the Riemann zetafunction beyond the Euler product over primes. Whereas they discuss the 24 solutions to the hypergeometric equation in terms of the hypergeometric series from Riemann’s point of view they do not go into H.A. Schwarz’ beautiful solution of Gauss’ problem as to which of these functions is algebraic. Schwarz’ theory is covered in Forsyth’s Function Theory. The decision to leave this out must have been difficult for Whittaker for it is a topic close to his early research. Finally they touch on the theory of Hilbert spaces only very lightly, just enough for their purposes. On the other hand Fredholm’s theory, well treated here, has often been pushed aside by the theory of Hilbert spaces in other texts and it is a topic about which an analyst should be aware. So, gentle reader, you have in your hands a book which has been useful and instructive to those working in mathematics for well over a hundred years. The language is perhaps a little quaint but it is a pleasure to peruse. May you too profit from this new edition.

5

B.L. v.d. Waerden’s Moderne Algebra became simply Algebra from the 1955 edition on; with either name it remains a great text on algebra.

Preface to the Fifth Edition

In 1896 Edmund Whittaker was elected to a Fellowship at Trinity College, Cambridge. Amongst other duties, he was employed to teach students, many of whom would later become distinguished figures in science and mathematics. These included G.H. Hardy, Arthur Eddington, James Jeans, J.E. Littlewood and a certain G. Neville Watson. His course on mathematical analysis changed the way the subject was taught, and he decided to write a book. So was born A Course of Modern Analysis, which was first published in 1902. It introduced students to functions of a complex variable, to the ‘methods and processes of higher mathematical analysis’, much of which was then fairly modern, and above all to special functions associated with equations that were used to describe physical phenomena. It was one of the first books in English to describe material developed on the continent, mostly in France and Germany. Its breadth and depth of coverage were unparalleled at the time and it became an instant classic. A second edition was called for, but in 1906 Whittaker had left Cambridge, moving first to Dublin, and then in 1912 to Edinburgh. His various duties, and no doubt, the moves themselves, impeded work on the new edition, and Whittaker gratefully accepted the offer from Watson to help him. A greatly expanded second edition duly appeared in 1915. The third edition, published five years later, was also enlarged by the addition of chapters, but the fourth edition was not much more than a corrected reprint with added references. I do not know if a fifth edition was ever planned. Both authors remained active for many years (Watson wrote, amongst other publications, the definitive Treatise on Bessel Functions), but perhaps they had nothing more to say to warrant a new edition. Nevertheless, the book remained a classic, being continually in print and reissued in paperback, first in 1963, and again, in 1996, as a volume of the Cambridge Mathematical Library. It never lost its appeal and occupied a unique place in the heart and work of many mathematicians (in particular, me) as an indispensable reference. The original editions were typeset using ‘hot metal’, and over the years successive reprintings led to the degrading of the original plates. Photographic printing methods slowed this decline, but David Tranah at Cambridge University Press had the idea to halt, indeed reverse, the degradation, by rekeying the book and at the same time updating it with new references and commentary. He spoke to me about this, and we agreed that if he arranged for the rekeying into LaTeX, I would do the updating. I did not need much persuading: it has been a labor of love. So much so that I have preserved the archaic spelling of the original, along with the Peano decimal system of numbering paragraphs, as described by Watson in the Preface to the fourth edition! This will make it straightforward for users of this fifth edition to refer to the previous one. I have however decided to create a complete reference list and to refer readers to that rather than to items in footnotes, items that were often hard to identify. Many xxi

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Preface to the Fifth Edition

of these items are now available in digital libraries and so for many people will be easier to access than they were in the authors’ time. I have made no substantial changes to the text: in particular, the original idea of adding commentaries on the text was abandoned. I have checked and rechecked the mathematics, and I have added some additional references. I have also written an introduction that describes what’s in the book and how it may be used in contemporary teaching of analysis. I have also provided summaries of each chapter, and, within them, make mention of more recent work where appropriate. As I said, preparing this edition has been a labor of love. I have also learned a lot of mathematics, evidence of the enduring quality and value of the original work. It has been a rewarding experience to edit A Course of Modern Analysis: I hope that it will be equally rewarding for readers. Victor H. Moll 2020, New Orleans

Preface to the Fourth Edition

Advantage has been taken of the preparation of the fourth edition of this work to add a few additional references and to make a number of corrections of minor errors. Our thanks are due to a number of our readers for pointing out errors and misprints, and in particular we are grateful to Mr E. T. Copson, Lecturer in Mathematics in the University of Edinburgh, for the trouble which he has taken in supplying us with a somewhat lengthy list. E. T. W. G. N. W. June 18, 1927 The decimal system of paragraphing, introduced by Peano, is adopted in this work. The integral part of the decimal represents the number of the chapter and the fractional parts are arranged in each chapter in order of magnitude. Thus, e.g., on pp. 187, 188 6 , §9.632 precedes §9.7 [because 9.632 < 9.7.] G.N.W. July 1920

6

in the fourth edition

xxiii

Preface to the Third Edition

Advantage has been taken of the preparation of the third edition of this work to add a chapter on Ellipsoidal Harmonics and Lamé’s Equation and to rearrange the chapter on Trigonometric Series so that the parts which are used in Applied Mathematics come at the beginning of the chapter. A number of minor errors have been corrected and we have endeavoured to make the references more complete. Our thanks are due to Miss Wrinch for reading the greater part of the proofs and to the staff of the University Press for much courtesy and consideration during the progress of the printing. E. T. W. G. N. W. July, 1920

xxiv

Preface to the Second Edition

When the first edition of my Course of Modern Analysis became exhausted, and the Syndics of the Press invited me to prepare a second edition, I determined to introduce many new features into the work. The pressure of other duties prevented me for some time from carrying out this plan, and it seemed as if the appearance of the new edition might be indefinitely postponed. At this juncture, my friend and former pupil, Mr G. N. Watson, offered to share the work of preparation; and, with his cooperation, it has now been completed. The appearance of several treatises on the Theory of Convergence, such as Mr Hardy’s Course of Pure Mathematics and, more particularly, Dr Bromwich’s Theory of Infinite Series, led us to consider the desirability of omitting the first four chapters of this work; but we finally decided to retain all that was necessary for subsequent developments in order to make the book complete in itself. The concise account which will be found in these chapters is by no means exhaustive, although we believe it to be fairly complete. For the discussion of Infinite Series on their own merits, we may refer to the work of Dr Bromwich. The new chapters of Riemann Integration, on Integral Equations, and on the Riemann Zeta-Function, are entirely due to Mr Watson: he has revised and improved the new chapters which I had myself drafted and he has enlarged or partly rewritten much of the matter which appeared in the original work. It is therefore fitting that our names should stand together on the title-page. Grateful acknowledgement must be made to Mr W. H. A. Lawrence, B.A., and Mr C. E. Winn, B.A., Scholars of Trinity College, who with great kindness and care have read the proof-sheets, to Miss Wrinch, Scholar of Girton College, who assisted in preparing the index, and to Mr Littlewood, who read the early chapters in manuscript and made helpful criticisms. Thanks are due also to many readers of the first edition who supplied corrections to it; and to the staff of the University Press for much courtesy and consideration during the progress of the printing. E.T. Whittaker July 1915

xxv

Preface to the First Edition

The first half of this book contains an account of those methods and processes of higher mathematical analysis, which seem to be of greatest importance at the present time; as will be seen by a glance at the table of contents, it is chiefly concerned with the properties of infinite series and complex integrals and their applications to the analytical expression of functions. A discussion of infinite determinants and of asymptotic expansions has been included, as it seemed to be called for by the value of these theories in connexion with linear differential equations and astronomy. In the second half of the book, the methods of the earlier part are applied in order to furnish the theory of the principal functions of analysis – the Gamma, Legendre, Bessel, Hypergeometric, and Elliptic Functions. An account has also been given of those solutions of the partial differential equations of mathematical physics which can be constructed by the help of these functions. My grateful thanks are due to two members of Trinity College, Rev. E. M. Radford, M.A. (now of St John’s School, Leatherhead), and Mr J. E. Wright, B.A., who with great kindness and care have read the proof-sheets; and to Professor Forsyth, for many helpful consultations during the progress of the work. My great indebtedness to Dr Hobson’s memoirs on Legendre functions must be specially mentioned here; and I must thank the staff of the University Press for their excellent cooperation in the production of the volume. E. T. WHITTAKER Cambridge 1902 August 5

xxvi

Introduction

The book is divided into two distinct parts. Part I. The Processes of Analysis discusses topics that have become standard in beginning courses. Of course the emphasis is in concrete examples and regrettably, this is different nowadays. Moreover the quality and level of the problems presented in this part is higher than what appears in more modern texts. During the second part of the last century, the tendency in introductory Analysis texts was to emphasize the topological aspects of the material. For obvious reasons, this is absent in the present text. There are 11 chapters in Part I. For a student in an American university, the material presented here is roughly distributed along the following lines: • • • •

Chapter 1 (Complex Numbers) Chapter 2 (The Theory of Convergence) Chapter 3 (Continuous Functions and Uniform Convergence) Chapter 4 (The Theory of Riemann Integration)

are covered in Real Analysis courses. • Chapter 5 (The Fundamental Properties of Analytic Functions; Taylor’s, Laurent’s and Liouville’s Theorems) • Chapter 6 (The Theory of Residues, Applications to the Evaluations of Definite Integrals) • Chapter 7 (The Expansion of Functions in Infinite Series) are covered in Complex Analysis. These courses usually cover the more elementary aspects of • Chapter 12 (The Gamma-Function) appearing in Part II. Most undergraduate programs also include basic parts of • Chapter 9 (Fourier Series and Trigonometric Series) • Chapter 10 (Linear Differential Equations) and some of them will expose the student to the elementary parts of • Chapter 8 ( Asymptotic Expansions and Summable Series) • Chapter 11 (Integral Equations) The material covered in Part II is mostly absent from a generic graduate program. Students interested in Number Theory will be exposed to some parts of the contents in xxvii

xxviii

Introduction

• Chapter 12 (The Gamma-Function) • Chapter 13 (The Zeta-Function of Riemann) • Chapter 14 (The Hypergeometric Function) and a glimpse of • • • •

Chapter 17 (Bessel Functions) Chapter 20 (Elliptic Functions. General Theorems and the Weierstrassian Functions) Chapter 21 (The Theta-Functions) Chapter 22 (The Jacobian Elliptic Functions).

Students interested in Applied Mathematics will be exposed to • Chapter 15 (Legendre Functions) • Chapter 16 (The Confluent Hypergeometric Function) • Chapter 18 (The Equations of Mathematical Physics) and some parts of • Chapter 19 (Mathieu Functions) • Chapter 23 (Ellipsoidal Harmonics and Lamé’s Equation) It is perfectly possible to complete a graduate education without touching upon the topics in Part II. For instance, in the most commonly used textbooks for Analysis, such as Royden [565] and Wheeden and Zygmund [666] there is no mention of special functions. On the complex variables side, in Ahlfors [13] and Greene–Krantz [260] one finds some discussion on the Gamma function, but not much more. This is not a new phenomenon. Fleix Klein [377] in 1928 (quoted in [91, p. 209]) writes ‘When I was a student, Abelian functions were, as an effect of the Jacobian tradition, considered the uncontested summit of mathematics, and each of us was ambitious to make progress in this field. And now? The younger generation hardly knows Abelian functions. During the last two decades, the trend towards the abstraction is being complemented by a group of researchers who emphasize concrete examples as developed by Whittaker and Watson. Among the factors influencing this return to the classics one should include 7 the appearance of symbolic languages and algorithms producing automatic proofs of identities. The work initiated by Wilf and Zeilberger, described in [518], shows that many identities have automatic proofs. A second influential factor is the monumental work by B. Berndt, G. Andrews and collaborators to provide context and proofs of all results appearing in S. Ramanujan’s work. This has produced a collection of books, starting with [60] and currently at [25]. The third example in this list is the work developed by J. M. Borwein and his collaborators in the propagation of Experimental Mathematics. In the volumes [88, 89] the authors present their ideas on how to transform mathematics into a subject, similar in flavor to other experimental sciences. The point of view expressed in the three examples mentioned above has attracted a new generation of researchers to get involved in this point of view type of mathematics. This is just one direction in which Whittaker and Watson has been a profound influence in modern authors. 7

This list is clearly a subjective one.

Introduction

xxix

The remainder of this chapter outlines the content of the book and a comparison with modern practices. The first part is named The Processes of Analysis. It consists of 11 chapters. A brief description of each chapter is provided next. Chapter 1: Complex Numbers. The authors begin with an informal description of positive integers and move on to rational numbers. Stating that from the logical standpoint it is improper to introduce geometrical intuition to supply deficiencies in arithmetical arguments, they adopt Dedekind’s point of view on the construction of real numbers as classes of rational numbers, later called Dedekind’s cuts. An example is given to show that there is no rational number whose square is 2. The arithmetic of real numbers is defined in terms of these cuts. Complex numbers are then introduced with a short description of Argand diagrams. The current treatment offers two alternatives: some authors present the real number from a collection of axioms (as an ordered infinite field) and other approach them from Cauchy’s theory of sequences: a real number is an equivalence class of Cauchy sequences of rational numbers. The reader will find the first point of view in [304] and the second one is presented in [599]. Chapter 2. The Theory of Convergence. This chapter introduces the notion of convergence of sequences of real or complex numbers starting with the definition of lim xn = L currently n→∞ given in introductory texts. The authors then consider monotone sequences of real numbers and show that, for bounded sequences, there is a natural Dedekind cut (that is, a real number) associated to them. A presentation of Bolzano’s theorem a bounded sequence of real numbers contains a limit point and Cauchy’s formulation of the completeness of real numbers; that is, the existence of the limit of a sequence in terms of elements being arbitrarily close, is discussed. These ideas are then illustrated in the analysis of convergence of series. The discussion begins with Dirichlet’s test for convergence: Assume an is a sequence of complex p Í an are uniformly numbers and fn is a sequence of positive real numbers. If the partial sums bounded and fn is decreasing and converges to 0, then

∞ Í n=1

n=1

an fn converges. This is used to give

examples of convergence of Fourier series (discussed in detail in Chapter 9). The convergence ∞ ∞ Í Í 1 , for real s, are presented in detail. This of the geometric series x n and the series ns n=1

n=1

last series defines the Riemann zeta function ζ(s), discussed in Chapter 13. The elementary ∞ Í an converges if lim |an+1 /an | < 1 and diverges if the limit is strictly ratio test states that n=1

n→∞

above 1. A discussion of the case when the limit is 1 is presented and illustrated with the convergence analysis of the hypergeometric series (presented in detail in Chapter 14). The chapter contains some standard material on the convergence of power series as well as some topics not usually found in modern textbooks: discussion on double series, convergence of infinite products and infinite determinants. The final exercise 8 in this chapter presents the evaluation of an infinite determinant considered by Hill in his analysis of the Schrödinger 8

In this book, Examples are often what are normally known as Exercises and are numbered by section, i.e., ‘Example a.b.c’. At the end of most chapters are Miscellaneous Examples, all of which are Exercises, and which are numbered by chapter: thus ‘Example a.b’. This is how to distinguish them.

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Introduction

equation with periodic potential (this is now called the Hill equation). The reader will find in [451] and [536] information about this equation. Chapter 3. Continuous Functions and Uniform Convergence. This chapter also discusses functions f (x, y) of two real variables as well as functions of one complex variable g(z). The notion of uniform convergence of a series is discussed in the context of the limiting function of a series of functions. This is normally covered in every introductory course in Analysis. The classical M-test of Weierstrass is presented. The reader will also find a test for uniform convergence, due to Hardy, and its application to the convergence of Fourier series. The chapter also contains a discussion of the series Õ 1 g(z) = α (z + 2mω 1 + 2nω2 ) m,n which will be used to analyze the Weierstrass ℘-function: one of the fundamental elliptic functions (discussed in Chapter 20). The chapter contains a discussion on the fact that a continuous function defined of a compact set (in the modern terminology) attains its maximum/minimum value. This is nowadays a standard result in elementary analysis courses. Chapter 4. The Theory of Riemann Integration. The authors present the notion of the Riemann integral on a finite interval [a, b], as it is currently done: as limiting values of upper and lower sums. The fact that a continuous function is integrable is presented. The case with finite number of discontinuities is given as an exercise. Basic results, such as integration by parts, differentiation with respect to the limits of integration, differentiation with respect to a parameter, the mean value theorem for integrals and the representation of a double integral as iterated integral are presented. This material has become standard. The chapter also contains a discussion on integrals defined on an infinite interval. There is a variety of tests to determine convergence and criteria that can be used to evaluate the integrals. Two examples of integral representations of the beta integral (discussed in Chapter 12) are presented. A basic introduction to complex integration is given at the end of the chapter; the reader is referred to Watson [650] for more details. This material is included in basic textbooks in Complex Analysis (for instance, see [13, 26, 155, 260, 455, 552]). Chapter 5. The Fundamental Properties of Analytic Functions; Taylor’s, Laurent’s and Liouville’s Theorems. This chapter presents the basic properties of analytic functions that have become standard in elementary books in complex analysis. These include the Cauchy– Riemann equations and Cauchy’s theorem on the vanishing of the integral of an analytic function taken over a closed contour. This is used to provide an integral representation as ∫ f (ξ) 1 dξ f (z) = 2πi Γ z − ξ where Γ is a closed contour containing ξ in its interior. This is then used to establish classical results on analytic functions such as bounds on the derivatives and Taylor theorem. There is also a small discussion on the process of analytic continuation and many-valued functions. This chapter contains also basic properties on functions having poles as isolated singularities: Laurent’s theorem on expansions and Liouville’s theorem on the fact that every entire function that is bounded must be constant (a result that plays an important role in the presentation of elliptic functions in Chapter 20). The Bessel function Jn , defined by its

Introduction

xxxi

integral representation Jn (x) =

1 2π

∫

2π

cos(nθ − x sin θ) dθ

0

makes its appearance in an exercise. This function is discussed in detail in Chapter 17. The chapter also contains a proof of the following fact: any function that is analytic, including at ∞, except for a number of non-essential singularities, must be a rational function. This has become a standard result. It represents the most elementary example of characterizing functions of rational character on a Riemann surface. This is the case of P1 , the Riemann sphere. The next example corresponds to the torus C/L, where L is a lattice. This is the class of elliptic functions described in Chapters 20, 21 and 22. The reader is referred to [461, 553, 600, 665] for more details. Chapter 6. The Theory of Residues: Application to the Evaluation of Definite Integrals. This chapter presents application of Cauchy’s integral representation of functions analytic except for a certain number of poles. Most of the material discussed here has become standard. One of the central concepts is that of the residue of a function at a pole z = zk , defined as the coefficient of (z − zk )−1 in the expansion of f near z = zk . As a first sign of the importance of these residues is the statement that the integral of f (z) over the boundary of a domain Ω is given by the sum of the residues of f inside Ω, the so-called argument principle which gives the difference between zeros and poles of a function as a contour integral. This chapter also presents methods based on residues to evaluate a variety of definite integrals including rational functions of cos θ, sin θ over [0, 2π], integrals over the whole real line via deformation of a semicircle, integrals involving some of the kernels such as 1/(e2πz − 1) (coming from the Fermi–Dirac distribution in Statistical Mechanics) and 1/(1−2a cos x +a2 ) related to Legendre polynomials (discussed in Chapter 15). An important function makes its appearance as Exercise 17: ψ(t) =

∞ Õ

e−n

2

πt

,

n=−∞

introduced by Poisson in 1823. The exercise outlines a proof of the transformation rule ψ(t) = t −1/2 ψ(1/t) known as Poisson summation formula. It plays a fundamental role in many problems in Number Theory, including the proof of the prime number theorem. This states that, for x > 0, the number of primes up to x, denoted by π(x), has the asymptotic behavior π(x) ∼ x/log x as x → ∞. The reader will find in [492] how to use contour integration and the function ψ(t) to provide a proof of the asymptotic behavior of ψ(t). This function reappears in Chapter 21 in the study of theta functions. Chapter 7. The Expansion of Functions in Infinite Series. This chapter begins with a result of Darboux on the expansion of an analytic function defined on a region Ω. For points a, x,

Introduction

xxxii

with the segment from a to x contained in Ω, one has the expansion φ(n) (0)[ f (z) − f (0)] =

n Õ

(−1)k−1 (z − a)k φ(n−k) (1) f (m) (z) − φn−k (0) f (k) (a)

k=1

+ (−1) (z − a) n

n+1

∫

1

φ(t) f (n+1) (a + t(z − a)) dt,

0

for any polynomial φ. The formula is then applied to the Bernoulli polynomials currently defined by the generating function ∞

Õ Bn (z) tezt = tn. t e − 1 n=0 n! (The text employs the notation φn (t) without giving the value for n = 0.) Darboux’s theorem then becomes the classical Euler–MacLaurin summation formula n Õ j=0

f ( j) =

∫ 0

n

bp/2c f (n) + f (0) Õ B2k (2k−1) f (x) dx + + f (n) − f (2k−1) (0) 2 (2k)! k=1 ∫ n Bp (x − bxc) dx. + (−1) p−1 f (p) (x) p! 0

The quantity x − bxc is the fractional part of x, denoted by {x}. This formula is used to estimate partial sums of series of values of an analytic function in terms of the corresponding integrals. The important example of the Riemann zeta function ζ(s) is presented in Chapter 13. The chapter contains a couple of examples of expansions of one function in terms of another one. The first one, due to Bürmann, starts with an analytic function φ(z) defined on a region and φ(a) = b with φ 0(a) , 0. Define ψ(z) = (z − a)/(φ(z) − a), then one obtains the expansion k−1 n−1 Õ 0 [φ(z) − b]k d f (a)ψ k (a) + Rn f (z) = f (a) + k! da k=1 where the error term has the integral representation n−1 0 ∫ z∫ 1 φ(z) − b f (t)φ 0(z) Rn = dt dz, 2πi a γ φ(t) − b φ(t) − φ(z) where γ is a contour in the t-plane, enclosing a and t and such that, for any µ interior to γ, the equation φ(t) = φ(µ) has a unique solution t = µ. The discussion also contains results of Teixeira on conditions for the convergence of the series for f (z) obtained by letting n → ∞. This type of result also contains an expansion of Lagrange for solutions of the equation µ = a + tφ(µ), for analytic function φ satisfying |tφ(z)| < |z − a|. The theorem states that any analytic function f of the solution µ can be expanded as n−1 ∞ Õ tn d [ f 0(a)φn (a)] . f (µ) = f (a) + n! da n=1 This expansion has interesting applications in Combinatorics; see [681] for details. The

Introduction

xxxiii

last type of series expansion described here corresponds to the classical partial fraction expansions of a rational function and its extensions to trigonometric functions. The results of this chapter are then used to prove representations of an entire function f in the form mn ∞ Ö z gn (z) G(z) 1− e f (z) = f (0)e an n=1 where an is a zero of f of multiplicity mn and G(z) is an entire function. The function gn (z) is a polynomial, introduced by Weierstrass, which makes the product converge. An application to 1/Γ(z) is discussed in Chapter 12. Chapter 8. Asymptotic Expansions and Summable Series. This chapter presents an introduction ∫to the basic concepts behind asymptotic expansion. The initial example considers ∞ ∞ Í (−1)k k! f (x) = t −1 e x−t dt. A direct integration by parts shows that the sum Sn (x) = x k+1 k=0

x

satisfies, for fixed x, the inequality | f (x) − Sn (x)| ≤ n!/x n+1 . Therefore, for x ≥ 2n, one obtains | f (x) − Sn (x)| < 1/n2 2n+1 . It follows that the integral f can be evaluated with great accuracy for large values of x by computing the partial sum of the divergent series Sn (x). ∞ Í This type of behavior is written as f (x) ∼ An x −n and the series is called the asymptotic n=0

expansion of f . The chapter covers the basic properties of asymptotic series: such expansions can be multiplied and integrated but not differentiated. Examples of asymptotic expansions of special functions appear in later chapters: for the Gamma function in Chapter 12 and for the Bessel function in Chapter 17. The final part of the chapter deals with summation methods, concentrating on methods assigning a value to a function given by a power series outside its circle of convergence D. The first example, due to Borel, starts with the identity ∫ ∞ ∞ ∞ Õ Õ an n n u valid for z ∈ D. an z = e−t φ(tz) dt where φ(u) = n! 0 n=0 n=0 The series

∞ Í n=0

an z n is said to be Borel summable if the integral on the right converges for z

outside D. For such z, the Borel sum of the series is assigned to be the value of the integral. The discussion continues with Cesàro summability, a notion to be discussed in the context of Fourier series in Chapter 9. Extensions by Riesz and Hardy are mentioned. More details on asymptotic expansions can be found in [468, 508]. Chapter 9. Fourier Series and Trigonometric Series. The authors discuss trigonometrical series defined as series of the form ∞ Õ 1 (an cos nx + bn sin nx) a0 + 2 n=1 for two sequences of real numbers {an } and {bn }. Such series are named Fourier series if there is a function f , with finite integral over (−π, π), such that the coefficients are given by ∫ ∫ 1 π 1 π an = f (t) cos nt dt and bn = f (t) sin nt dt. π −π π −π

Introduction

xxxiv

The chapter contains a variety of results dealing with conditions under which the Fourier series associated to a function f converges to f . These include Dirichlet’s theorem stating that, under some technical conditions, the Fourier series converges to 21 [ f (x + 0) + f (x − 0)]. This is followed by Fejer’s theorem that the Fourier series is Césaro summable at all points where the limits f (x ±0) exist. The proofs are based on the analysis of the so-called Dirichlet– Féjer kernel. Examples are provided where there is not a single analytic expression for the Fourier series. The notion of orthogonality of the sequence of trigonometric functions makes an implicit appearance in all the proofs. The so-called Riemann–Lebesgue theorem, on the behavior of Fourier coefficients, is established. This result states that if ψ(θ) is integrable ∫ b on the interval (a, b), then lim ψ(θ) sin(λθ) dθ = 0. The chapter contains results on the n→∞

a

function f which imply pointwise convergence of the Fourier series. The results of Dini and Jordan, with conditions on the expressions f (x ± 2θ) − f (x ± 0) near θ = 0, are presented. The reader will find more information about convergence of Fourier series in [368] and in the treatise [690]. The results of Kolmogorov [381, 382] on an integrable function with a Fourier series diverging everywhere, as well as the theorem of Carleson [118] on the almosteverywhere convergence of the Fourier series of a continuous function, are some of the high points of this difficult subject. The chapter also includes a discussion on the uniqueness of the representation of a Fourier series for a function f and also of the Gibbs phenomenon on the behavior of a Fourier series in a neighborhood of a point of discontinuity of f . Chapter 10. Linear Differential Equations. This chapter discusses properties of solutions of second order linear differential equations du d2u + p(z) + q(z)u = 0, dz 2 dz where p, q are analytic functions of z except for a finite number of points. The discussion is local; that is, in a neighborhood of a point c ∈ C. The points c are classified as ordinary, where the functions p, q are assumed to be analytic at c and otherwise singular. The question of existence and uniqueness of solutions of the equation is discussed. The d2v + J(z)v = 0, by an elementary change of equation is transformed first into the form dz 2 variables. Existence of solutions is obtained from an integral equation equivalent to the original problem. An iteration process is used to produce a sequence of analytic functions {vn }. Then it is shown that, in a neighborhood of an ordinary point, this sequence converges uniformly to a solution of the equation. Uniqueness of the solution comes also from this process. The solutions near an ordinary point are presented in the case of an ordinary singular point. These are points c ∈ C where p or q have a pole, but (z − c)p(z) and (z − c)2 q(z) are analytic functions in a deleted neighborhood of z = c. The so-called method of Frobenius is then used to seek formal series solutions in the form " # ∞ Õ α n u(z) = (z − c) 1 + an (z − c) . n=1

The so-called indicial equation α2 + (p0 − 1)α + q0 = 0 and its roots α1 , α2 , control the

Introduction

xxxv

properties of these formal power series. The numbers p0 , q0 are the leading terms of (z−c)p(z) and (z − c)2 q(z), respectively. It is shown that if α1 , α2 do not differ by an integer, there are two formal solutions and these series actually converge and thus represent actual solutions. Otherwise one of the formal series is an actual solution and there is a procedure to obtain a second solution containing a logarithmic term. The reader will find in [151] all the details. It is a remarkable fact that the behavior of the singularities determines the equation itself. For example, the most general differential equation of second order which has every point except a1 , a2 , a3 , a4 and ∞ as ordinary points and these five points as regular points, must be of the form ( 4 ) Õ 1 − αr − βr du d2u + dz 2 z − ar dz r=1 ) ( 4 Õ αr βr Az2 + Bz + C + u = 0, + (z − ar )2 (z − a1 )(z − a2 )(z − a3 )(z − a4 ) r=1 for some constants αr , βr , A, B, C. F. Klein [376] describes how all the classical equations of Mathematical Physics appear in this class. Six classes, carrying the names of their discoverers (Lamé, Mathieu, Legendre, Bessel, Weber–Hermite and Stokes) are discussed in later chapters. The chapter finally discusses the so-called Riemann P-function. This is a mechanism used to write a solution of an equation with three singular points and the corresponding roots of the indicial equation. Some examples of formal rules on P, which allow to transform a solution with expansion at one singularity to another are presented. The chapter concludes showing that a second order equation with three regular singular points may be converted to the hypergeometric equation. This is the subject of Chapter 14. The modern theory of this program, to classify differential equations by their singularities, is its extension to nonlinear equations. A singularity of an ordinary differential equation is called movable if its location depends on the initial condition. An equation is called a Painlevé equation if its only movable singularities are poles. Poincaré and Fuchs proved that any first-order equation with this property may be transformed into the Ricatti equation or it may be solved in terms of the Weierstrass elliptic function. Painlevé considered the case of second order, transformed them into the form u 00 = R(u, u 0, z), where R is a rational function. Then he put them into 50 canonical forms and showed that all but six may be solved in terms of previously known functions. The six remaining cases gave rise to the six Painlevé functions PI, . . . , PVI . See [261, 310, 336] for details. It is a remarkable fact that these functions, created for an analytic study, have recently appeared in a large variety of problems. See [37] and [562] for their appearance in combinatorial questions, [76, 636] for their relations to classical functions, [640] for connections to orthogonal polynomials, [632] for their appearance in Statistical Physics. The reader will find in [212] detailed information about their asymptotic behavior. Chapter 11. Integral Equations. Given a function f , continuous on an interval [a, b] and a kernel K(x, y), say continuous on both variables or in the region a ≤ y ≤ x ≤ b and

Introduction

xxxvi

vanishing for y > x, the equation φ(x) = f (x) + λ

∫

b

K(x, y)φ(y) dy a

for the unknown function φ, is called the Fredholm integral equation of the second kind. The solution presented in this chapter is based on the construction of functions D(x, y, λ) and D(λ), both entire in λ, as a series in which the nth-term consists of determinants of order n × n based on the function K(x, y). The solution is then expressed as ∫ b 1 D(x, ξ, λ) f (ξ) dξ. φ(x) = f (x) + D(λ) a In particular, in the homogeneous case f ≡ 0, there is a unique solution φ ≡ 0 for those values of λ with D(λ) , 0. A process to obtain a solution for those values of λ with D(λ) = 0 is also described. Volterra introduced the concept of reciprocal functions for a pair of functions K(x, y) and k(x, y; λ) satisfying the relation ∫ b K(x, y) + k(x, y; λ) = λ k(x, ξ; λ)K(ξ, y)dξ. a

Then the solution to the Fredholm equation is given by ∫ b f (x) = φ(x) + λ k(x, ξ; λ)φ(ξ)dξ. a

The last part of the chapter discusses the equation ∫ b K(x, ξ)Φ(ξ) dξ Φ(x) = f (x) + λ a

and the solution is expressed as a series in terms of a sequence of orthonormal functions and Í the sequence {λn } of eigenvalues of the kernel K(x, y). In detail, if f (x) = bn φn (x), then Í bn λn the solution Φ is given by Φ(x) = λ−λn φn (x). The Fredholm equation is written formally as Φ = f +KΦ and this gives Φ = f +K f +K 2 Φ. ∞ Í K n f , expressing the Iteration of this process gives the so-called Neumann series Φ = n=0

unknown Φ in terms of iterations of the functional defined by the kernel K. The study of Fredholm integral equations is one of the beginnings of modern Functional Analysis. The reader will find more details in P. Lax [415]. The ideas of Fredholm have many applications: the reader will find in H. P. McKean [460] a down-to-earth explanation of Fredholm’s work and applications to integrable systems (such as the Korteweg–de Vries equation ut = u xxx + 6uu x and some special solutions called solitons), to the calculations of some integrals involving Brownian paths (such as P. Lévy’s formula for the area generated by a two-dimensional Brownian path) and finally to explain the appearance of the so-called sine kernel in the limiting distribution of eigenvalues of random unitary matrices. This subject has some mysterious connections to the Riemann hypothesis as described by B. Conrey [154]. The second part of the book is called The Transcendental Functions and it consists of 12 chapters. A brief description of them is provided next.

Introduction

xxxvii

Chapter 12. The Gamma-Function. This function, introduced by Euler, represents an extension of factorials n! from positive integers to complex values of n. The presentation begins with the infinite product ∞ Ö z −z/n 1+ e P(z) = zeγz n n=1 where γ = lim 1 + 21 + · · · + n1 − log n is nowadays called the Euler–Mascheroni constant. n→∞

The product is an entire function of z ∈ C and the Gamma function is defined by Γ(z) = 1/P(z). Therefore Γ(z) is an analytic function except for simple poles at z = 0, −1, 2, . . . . The constant γ is identified as −Γ 0(1). The fact that Γ is a transcendental function is reflected by the fact, mentioned in this chapter, that Γ does not satisfy a differential equation with coefficients being rational functions of z. The chapter contains proofs of a couple of representations by Euler z ∞ 1Ö z −1 1 Γ(z) = 1+ 1+ z n=1 z n = lim

n→∞

(n − 1)! nz . z(z + 1) · · · (z + n − 1)

The functional equation Γ(z+1) = zΓ(z) follows directly from here. Using the value Γ(1) = 1, this leads to Γ(n) = (n − 1)! for n ∈ N, showing that Γ interpolates factorials. The chapter also presents proofs of the reflection formula π Γ(z)Γ(1 − z) = sin πz √ 1 leading to the special value Γ( 2 ) = π. There is also a discussion of the multiplication formula due to Gauss n−1 Ö k −(n−1)/2 −1/2+nz Γ z+ Γ(nz) = (2π) n n k=0 and the special duplication formula of Legendre 1 Γ(2z) = √ 22z−1 Γ(z)Γ(z + 12 ). π This may be used to derive the relation Γ( 13 )Γ( 23 ) = √2π3 . Arithmetical properties of these values are difficult to establish. The reader is referred to [92] and [166] for an elementary presentation of the Gamma function, and to [106] for an introduction to issues of transcendence. There are several integral representations of the Gamma function established in this chapter. Most of them appear in the collection of integrals by Gradshetyn and Ryzhik [258]. The first one, due to Euler, is ∫ ∞ Γ(z) = t z−1 e−t dt, 0

valid for Re z > 0. This may be transformed to the logarithmic scale z−1 ∫ 1 1 dx. Γ(z) = log x 0

Introduction

xxxviii

There is also a presentation of Hankel’s contour integral ∫ 1 (−t)z−1 e−t dt, Γ(z) = − 2i sin πz C

z 0, the series defines an analytic function of s on the half-plane σ = Re s ≥ 1 + δ. The function admits the integral representation ∫ ∞ s−1 −x x e 1 dx. ζ(s) = Γ(s) 0 1 − e−x Euler produced the infinite product ζ(s) =

Ö

(1 − p−s )−1

p

where the product extends over all prime numbers. This formula shows that ζ(s) has no zeros in the open half-plane Re s > 1. The auxiliary function s 1 ξ(s) = π −s/2 s(s − 1)Γ ζ(s) 2 2 is analytic and satisfies the identity ξ(s) = ξ(1 − s). This function now shows that ζ(s) has no zeros for Re s < 0, aside for the so-called trivial zeros at s = −2, −4, −6, . . . coming from the poles of Γ(s/2). Thus all the non-trivial zeros lie on the strip 0 ≤ Re s ≤ 1. The Riemann hypothesis states that all the roots of ζ(s) = 0 are on the critical line Re s = 12 . At the end of §13.3 the authors state that: It was conjectured by Riemann, but it has not yet been proved, that all the zeros of ζ(s) in this strip lie on the line σ = 21 ; while it has quite recently been proved by Hardy [279] that an infinity of zeros of ζ(s) actually lie on σ = 21 . It is highly probable that Riemann’s conjecture is correct, and the proof of it would have far-reaching consequences in the theory of Prime Numbers. The reader will find in [93, 153, 458] more information about the Riemann hypothesis. In a remarkable new connection, it seems that the distribution of the zeros of ζ(s) is related to the eigenvalues of random matrices [367, 369]. This chapter establishes the identity ζ(2n) =

(−1)n−1 B2n (2π)2n 2 (2n)!

where n ∈ N and B2n is the Bernoulli number. This is a generalization of the so-called Basel

Introduction

xl

problem ζ(2) = π 2 /6. The solution of this problem won the young Euler instant fame. It follows that ζ(2n) is a rational multiple of π 2n , therefore this is a transcendental number. The arithmetic properties of the odd zeta values are more difficult to obtain. Apéry proved in 1979 that ζ(3) is not a rational number; see [27, 72, 689]. It is still unknown whether ζ(5) is irrational, but Zudilin [688] proved that one of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational. It is conjectured that all odd zeta values are irrational. The literature contains a large variety of extensions of the Riemann zeta function. The chapter contains information about some of them: the Hurwitz zeta function ζ(s, a) =

∞ Õ n=0

1 , (n + a)s

with 0 < a ≤ 1

with integral representation ζ(s, a) =

1 Γ(s)

∫

∞

0

x s−1 e−ax dx. 1 − e−x

The chapter establishes the values of ζ(−m, a) in terms of derivatives of the Bernoulli polynomials and presents a proof of Lerch’s theorem Γ(a) d ζ(s, a) = log √ . s=0 ds 2π The chapter mentions two further generalizations: one introduced by Lerch (see [414] for details) ∞ Õ e2πinx φ(x, a; s) = , (n + a)s n=0 and another one by Barnes [43, 44, 45, 46] Õ ζN (s, w | a1, . . . , a N ) = n1 ,...,n N

1 . (w + n1 a1 + · · · + n N a N )s

The reader will find in [566] more recent information on this function. Chapter 14. The Hypergeometric Function. This function is defined by the series F(a, b; c, z) =

∞ Õ (a)n (b)n n=0

(c)n n!

z n,

provided c is not a negative integer. Here (u)n = Γ(u + n)/Γ(u) is the Pochhammer symbol. The series converges for |z| < 1 and on the unit circle |z| = 1 if Re (c − a − b) > 0. Many elementary functions can be expressed in hypergeometric form, for instance 1 z F(1, 1; 1; z) = and ez = lim F 1, b; 1; . b→∞ 1−z b The chapter begins with Gauss’ evaluation F(a, b; c; 1) in the form F(a, b; c; 1) =

Γ(c)Γ(c − a − b) . Γ(c − a)Γ(c − b)

Introduction

xli

The function F satisfies the differential equation z(1 − z)

du d2u + [c − (a + b + 1)z] − abu = 0. 2 dz dz

This equation has 0, 1, ∞ as regular singular points and every other point is ordinary. The generalization to singular points at a, b, c with exponents given by {α, α 0 }, {β, β 0 }, {γ, γ 0 }, respectively, is the Riemann differential equation d2w 1 − α − α 0 1 − β − β 0 1 − γ − γ 0 dw + + + dz 2 z−a z−b z−c dz 0 0 0 αα (a − b)(a − c) ββ (b − c)(b − a) γγ (c − a)(c − b) w + + = 0. + z−a z−b z−c (z − a)(z − b)(z − c) It is shown that α γ z−a z−c (z − a)(c − b) F(α + β + γ, α + β 0 + γ; 1 + α − α 0); z−b z−b (z − b)(c − a) solves the Riemann differential equation. Using the invariance of this equation with respect to some permutations of the parameters (for example, the exchange of α and α 0) produces from F(a, b; c; z) Kummer’s 24 new solutions of Riemann’s equation, for example z z −a −b (1 − z) F a, c − b; c; and (1 − z) F c − a, b; c; . z−1 z−1 Since the solutions of a second-order differential equation form a two dimensional vector space, this type of transformation can be used to generate identities among hypergeometric series. The reader will find in [20, 339, 534, 641] more details on these ideas. The corresponding equation with four regular singular points at 0, 1, ∞, a is called the Heun equation γ du δ ε αβx − q d2u + u = 0. + + + dx 2 x x − 1 x − a dx x(x − 1)(x − a) The corresponding process on the symmetries of the equation now gives 192 solutions. These are described in [452]. The reader will find in [190] an example of the appearance of Heun’s equation in integrable systems. The chapter also contains a presentation of Barnes’ integral representation ∫ i∞ Γ(a + s)Γ(b + s)Γ(−s) 1 F(a, b; c; z) = (−z)s ds 2πi −i∞ Γ(c + s) and its use in producing an analytic continuation of the hypergeometric series. Finally, the identities of Clausen 2 F(a, b; a + b + 21 ; x) = 3 F2 (2a, a + b, 2b; a + b + 12 , 2a + 2b; x) where 3 F2 is the analog of the hypergeometric series, now with three Pochhammer symbols on top and two in the bottom of the summand and Kummer’s quadratic transformation F(2a, 2b; a + b + 12 ; x) = F(a, b; a + b + 21 ; 4x(1 − x))

Introduction

xlii

appear as exercises in this chapter. The reader will find in [20] a detailed analysis of these topics. Chapter 15. Legendre Functions. This chapter discusses Legendre polynomials Pn (z) and some of their extensions. These days, the usual starting point for these functions is defining them as orthogonal polynomials on the interval (−1, 1); that is, ∫ 1 Pn (z)Pm (z) dz = 0 if n , m, −1

plus some normalization in the case n = m. The starting point in this chapter is the generating function ∞ Õ (1 − 2zh + h2 )−1/2 = Pn (z)h n . n=0

It is established from here that j k n

Pn (z) =

2 Õ

(−1)r

r=0

(2n − 2r)! z n−2r − r)!(n − 2r)!

2n r!(n

showing that Pn (z) is a polynomial of degree n with leading coefficient 2−n The properties of these polynomials established in this chapter include

2n . n

Rodriguez formula n d 1 (z 2 − 1)n . Pn (z) = n 2 n! dz Legendre’s differential equation The polynomials Pn (z) are solutions of the differential equation (1 − z2 )

du d2u − 2z + n(n + 1)u = 0. dz 2 dz

In the new scale x = z2 , this equation takes its hypergeometric form x(1 − x)

d2 y 1 du 1 + (1 − 3x) + n(n + 1)u = 0. dx 2 2 dx 4

The (more convenient) hypergeometric form Pn (z) = 2 F1 (n + 1, −n; 1; 12 (1 − z) is also established. Recurrences The chapter presents proofs of the recurrences (n + 1)Pn+1 (z) − (2n + 1)zPn (z)zPn (z) − nzPn (z) = 0 and 0 Pn+1 (z) − zPn0 (z) − (n + 1)Pn (z) = 0.

Integral representations A variety of integral representations for the Legendre polynomials are presented:

Introduction

xliii

• Schläfli: Pn (z) =

1 2πi

∮ C

(t 2 − 1)n dt − z)n+1

2n (t

where C is a contour enclosing z. This is then used to prove the orthogonality relation ( ∫ 1 0 if n , m Pn (z)Pm (z) dz = 2/(2n + 1) if n = m. −1 • Laplace: 1 Pn (z) = π

∫

Pn (cos θ) =

1 π

π

z + (z2 − 1)1/2 cos θ

n

dθ

0

• Mehler–Dirichlet: ∫

θ

−θ

1

e(n+ 2 )iϕ dϕ. (2 cos ϕ − 2 cos θ)1/2

The formula of Schläfli given above is then used to extend the definition of Pn (z) for n < N. In order to obtain a single-valued function, the authors introduce a cut from −1 to −∞ in the domain of integration. Since the differential equation for the Legendre polynomials is of second order, it has a second solution independent of Pn (z). This is called the Legendre function of degree n of the second type. It is denoted by Q n (z). The chapter discusses integral representations and other properties similar to those described for Pn (z). For example, one has the hypergeometric expression √ πΓ(n + 1) 1 n+1 n 3 −2 F Q n (z) = n+1 , + 1; n + ; z . 2 2 2 2 Γ n + 32 z n+1 One obtains from here 1 z+1 log 2 z−1 1 z+1 Q1 (z) = z log − 1. 2 z−1

Q0 (z) =

z+1 In general Q n (z) = An (z) + Bn (z) log z−1 for polynomials An , Bn .

The chapter also includes further generalizations of the Legendre functions introduced by Ferrer and Hobson. These are called associated Legendre functions. Some of their properties are presented. There is also a discussion of the addition theorem for Legendre polynomial, as well as a short section on the Gegenbauer function. The reader will find in the Digital Library of Mathematical Functions developed at NIST [443] more information about these functions. Chapter 16. The Confluent Hypergeometric Function. This chapter discusses the second-order differential equation with singularities at {0, ∞, c} and corresponding exponents 1 1 { 2 + m, 2 − m}, {−c, 0}, {c − k, k} in the limiting situation c → ∞. This is the case of confluent singularities (the limiting equation now has only two singularities: 0 and ∞,

Introduction

xliv

with 0 remaining regular and ∞ becomes an irregular singularity). After a change of variables to eliminate the first derivative term, the limiting equation becomes ! 1 1 k 4 − m2 d 2W + − + + W = 0. dz 2 4 z z2 This is called Whittaker equation. The authors introduce the functions Mk,m (z) = z1/2+m e−z/2

+m−k ( 1 + m − k)( 32 + m − k) 2 1+ z+ 2 z +··· 1! (2m + 1) 2! (2m + 1)(2m + 2) 1 2

!

and show that, when 2m < N, the functions Mk,m (z) and Mk,−m (z) form a fundamental set of solutions. It turns out that it is more convenient to work with the functions Wk,m (z) defined by the integral representation k−1/2+m ∫ ∞ t z k e−k/2 −k−1/2+m t 1+ e−t dt. Wk,m (z) = 1 z Γ( 2 − k + m) 0 The reader is referred to [59, Chapter 6] for a readable description of the basic properties of these functions, called Whittaker functions in the literature. The chapter also presents a selection of special functions that can be expressed in terms of Wk,m (z). This includes the incomplete gamma function ∫ x γ(a, x) = t a−1 e−t dt 0

that can be expressed as γ(a, x) = Γ(a) − x (a−1)/2 e−x/2W 1 (a−1), 1 a (x), 2

2

as well as the logarithmic integral function, defined by ∫ x dt li(z) = = −(− log z)−1/2 z1/2W− 1 ,0 (− log z). 2 0 log t This function appears in the description of the asymptotic behavior of the function π(x) = number of primes p ≤ x. The prime number theorem may be written as π(x) ∼ li(x) as x → ∞. See [191] for details. The final example is the function 2 z n/2+1/4 −1/2 Dn (z) = 2 z W n + 1 ,− 1 , 2 2 4 4 related in a simple manner to the Hermite polynomials, defined by n d 2 n z 2 /2 Hn (z) = (−1) e e−z /2 . dz See [20] for information on this class of orthogonal polynomials.

Introduction

xlv

Chapter 17. Bessel Functions. This chapter discusses the Bessel functions defined, for n ∈ Z, by the expansion ∞ Õ z 1 exp t− = Jn (z)t n . 2 t n=−∞ Some elementary properties of Jn (z) are derived directly from this definition, such as J−n (z) = (−1)n Jn (z), the series ∞ Õ (−1)r z n+2r , Jn (z) = r! (n + r)! 2 r=0 and the addition theorem Jn (y + z) =

∞ Õ

Jm (y)Jn−m (z).

m=−∞

The Cauchy integral formula is then used to produce the representation ∮ 1 z n 2 t −n−1 et−z /4t dt, Jn (z) = 2πi 2 C where C is a closed contour enclosing the origin. From here it is possible to extend the definition of Jn (z) to values n < Z and produce the series representation Jn (z) =

∞ Õ r=0

(−1)r z n+2r . 2n+2r r!Γ(n + r + 1)

This function is called the Bessel function of the first kind of order n. The integral representation of Jn (z) is then used to show that y(z) = Jn (z) is a solution of the differential equation n2 d 2 y 1 dy + + 1 − 2 y = 0, dz 2 z dz z called the Bessel equation. In the case n < Z, the functions Jn (z) and J−n (z) form a basis for the space of solutions. In the case n ∈ Z a second solution, independent of Jn (z), is given by πi(n+ε) Jn+ε (z) cos(π(n + ε)) − J−(n+ε) (z) Yn (z) = lim 2πe . ε→0 sin(2π(n + ε)) The functions Yn (z) are called the Bessel functions of the second kind. This chapter also contains some information on some variations of the Bessel function such as π In (z) = i −n Jn (iz) and Kn (z) = [I−n (z) − In (z)] cot(πn). 2 Among the results presented here one finds Recurrences such as 2n Jn (z), z n Jn0 (z) = Jn (z) − Jn+1 (z) z

Jn−1 (z) + Jn+1 (z) =

Introduction

xlvi

and z−n−1 Jn+1 (z) = −

1 d −n [z Jn (z)] z dz

which produces relations of Bessel functions of consecutive indices. Zeros of Bessel functions it is shown that between any two non-zero consecutive roots of Jn (z) = 0 there is a unique root of Jn+1 (z) = 0. Integral representations such as ∫ ∫ 1 π sin πn ∞ −nθ−z sinh θ Jn (z) = cos(nθ − z sin θ) dθ − e dθ, π 0 π 0 where, for n ∈ Z, the second term vanishes. Hankel representation in the form ∫ Γ( 1 − n) z n Jn (z) = 2 √ (t 2 − 1)n−1/2 cos(zt) dt 2πi π 2 C where C is a semi-infinite contour on the real line. Evaluation of definite integrals such as one due to Mahler ∫ ∞ t J0 (t x) dt K0 (x) = 1 + t2 0 and an example due to Sonine giving an expression for ∫ ∞ x 1−m Jm (ax)Jm (bx)Jm (cx) dx. 0

A large selection of integrals involving Bessel functions may be found in [105], [258] and [544]. Series expansion The chapter also contains information about expansions of a function f (z) in a series of the form f (z) =

∞ Õ n=0

an Jn (z) or

f (z) =

∞ Õ

an J0 (nz).

n=0

The reader will find in [20] and [59] more information on these functions at the level discussed in this chapter. Much more appears in the volume [653]. There are many problems whose solutions involve the Bessel functions. As a current problem of interest, consider the symmetric group SN of permutations π of N symbols. An increasing sequence of length k is a collection of indices 1 ≤ i1 < · · · < ik ≤ N such that π(i1 ) < π(i2 ) < · · · < π(ik ). Define on SN a uniform probability distribution; that is, P(π) = 1/N! for each permutation π. Then the maximal length of an increasing subsequence of π is a random variable, denoted by `N (π), and its distribution is of interest. This is the Ulam problem. Introduce the centered and scaled function √ `N (π) − 2 N χN (π) = . N 1/6

Introduction

xlvii

Baik–Deift–Johannson [36] proved that lim P( χN (π) ≤ x) = F(x), where F(x), the so-called N →∞ Tracy–Widom distribution, is given by ∫ ∞ 2 F(x) = exp − (y − x)u (y) dy . x

Here u(x) is the solution of the Painlevé PII equation u 00(x) = 2u3 (x) + xu(x), with asymptotic behavior Ai(x) as x → ∞. The Airy function Ai(x) is defined by ∼ u(x) √ √ 2 3/2 /π 3. The reader will find in [37] an introduction to this fasciAi(x) = xK1/3 3 x nating problem. Chapter 18. The Equations of Mathematical Physics. This chapter contains a brief description of methods of solutions for the basic equations encountered in Mathematical Physics. The results are given for Laplace’s equation ∆V =

∂ 2V ∂ 2V ∂ 2V + + 2 ∂ x2 ∂ y2 ∂z

on a domain Ω ⊂ R3 . The chapter has a presentation on the physical problems modeled by this equation. The results include the integral representation of the solution ∫ π V(x, y, z) = f (z + ix cos u + iy sin u, u) du −π

as the 3-dimensional analog of the form V(x, y) = f (x + iy) + g(x − iy) valid in the 2dimensional case as well as an expression for V(x, y, z) as a series with terms of the form ∫ π cos mu (z + ix cos u + iy sin u)n du. sin mu −π This series is then converted into one of the form ( ) ∞ ∞ Õ Õ (m) m r n An Pn (cos θ) + A(m) V= n cos mφ + Bn sin mφ Pn (cos θ) n=0

m=1

Pnm

where is Ferrer’s version of the associated Legendre function. The chapter also contains similar results for Laplace’s equation on a sphere. For this type of domain, the authors obtain the formula ∫ ∫ a(a2 − r 2 ) π π f (θ 0, φ 0) sin θ 0 dθ 0 dφ 0 V(r, θ, φ) = , 2 0 0 0 2 3/2 4π −π 0 [r − 2ar {cos θ cos θ + sin θ sin θ cos(φ − φ )} + a ] and refer to Thompson and Tait [628] for further discussions on the theory of Green’s functions. A similar analysis for an equation on a cylinder also appears in this chapter. In that case the Legendre functions are replaced by Bessel functions. Some of the material discussed in this chapter has become standard in basic textbooks in Mathematical Physics; see for instance [476]. Chapter 19. Mathieu Functions. This chapter discusses the wave equation Vtt = c2 ∆V and assuming a special form V(x, y, t) = u(x, y) cos(pt + ε) of the unknown V in a special system

xlviii

Introduction

of coordinates (ξ, η) (introduced by Lamé) yields an equation for u(x, y). Using the classical method of separation of variables (u ∗ x, y) = F(ξ)G(η) produces the equation d2 y + (a + 16q cos(2z)) y = 0. dz 2 This is called Mathieu’s equation. The value of a is determined by the periodicity condition G(η + 2π) = G(η) and q is determined by a vanishing condition at the boundary. This type of equation is now called Hill’s equation, considered by Hill [306] in a study on lunar motion. Details about this equation appear in [451] and connections to integrable systems appear in [459, 462, 463]. The authors show that G(η) satisfies the integral equation ∫ π G(η) = λ ek cos η cos θ G(θ) dθ −π

and this λ must be a characteristic value as described in Chapter 11. A sequence of functions, named Mathieu functions, are introduced from the study of Mathieu’s equation. In the case q = 0, the solutions are {1, cos nz, sin nz}n∈N, and via Fourier series the authors introduce functions {ce0 (z, q), cen (z, q), sen (z, q)} n∈N , reducing to the previous set as q → 0. Some expressions for the first coefficients in the Fourier series of these functions are produced (it looks complicated to obtain exact expressions for them). The authors present basic aspects of Floquet theory (more details appear in [451]). One looks for solution of Mathieu’s equation in the form y(z) = eµz φ(z), with φ periodic. The values of µ producing such solutions are obtained in terms of a determinant (called the Hill determinant). The modern theory yields these values in terms of a discriminant attached to the equation. The chapter also discusses results of Lindemann, transforming Mathieu’s equation into the form 4ξ(1 − ξ)u 00 + 2(1 − 2ξ)u 0 + (a − 16q + 32qξ)u = 0. This equation is not of hypergeometric type: the points 0, 1 are regular, but ∞ is an irregular singular point. Finally, the chapter includes some description of the asymptotic behavior of Mathieu functions. More details appear in [34] and [509]. Chapter 20. Elliptic Functions. General Theorems and the Weierstrassian Functions. Consider two complex numbers ω1 , ω2 with non-real ratio. An elliptic function is a doublyperiodic functions: f (z + 2ω1 ) = f (z + 2ω2 ) = f (z) where its singularities are at worst poles. The chapter discusses basic properties of the class E of elliptic functions. It is simple to verify that E is closed under differentiation and that the values of f ∈ E are determined by its values on the parallelogram with vertices 0, 2ω1 , 2ω1 + 2ω2 , 2ω2 . (Observe the factor of 2 in the periods.) This is called a fundamental cell and is denoted by L. One may always assume that there are no poles of the function on the boundary of the cell. The first type of results deal with basic properties of an elliptic function: (1) the number of poles is always finite; the same is true for the number of solutions of f (z) = c. This is independent of c ∈ C and is called the degree of the function f . (2) any elliptic function without poles must be constant.

Introduction

xlix

This result is used throughout the chapter to establish a large number of identities. The fundamental example 1 1 1 Õ − ℘(z) = 2 + z (z − ω)2 ω2 where the sum runs over all non-zero ω = 2nω1 + 2mω2 , was introduced by Weierstrass. It is an elliptic function of order 2. It has a double pole at z = 0. It is an even function, so its zeros in the fundamental cell are of the form ±z0 mod L. A remarkably recent formula for z0 is given by Eichler and Zagier [192]. The ℘ function satisfies a differential equation 2 d℘(z) = 4℘(z)3 − g2 ℘(z) − g3, dz where g2 , g3 are the so-called invariants of the lattice L. This function is then used to parametrize the algebraic curve y 2 = 4x 3 + ax + b, for a, b ∈ C with a3 + 27b2 , 0. The subject is also connected to differential equation by showing that if y = ℘(z), then the inverse z = ℘−1 (y) (given by an elliptic integral) can be written as the quotient of two solutions of ! 3 3 d2v 3 Ö 3 Õ −2 −1 + (y − er ) − y (y − er ) v = 0. dy 2 16 r=1 8 r=1 Here er are the roots of the cubic polynomial appearing in the differential equation for ℘(z). The addition theorem 2 1 ℘0(z) − ℘0(y) ℘(z + y) = − ℘(z) − ℘(y) 4 ℘(z) − ℘(y) is established by a variety of methods. One presented by Abel deals with the intersection of the cubic curve y 2 = 4x 3 + ax + b and a line and it is the basis for an addition on the elliptic curve, as the modern language states. Take two points a, b on the curve and compute the line joining them. This line intersects the cubic at three points: the third is declared −a ⊕ b. The points on the curve now form an abelian group: this is expected since the cubic may be identified with a torus C/L. The remarkable fact is that the addition of points preserves points with rational coordinates, so this set is also an abelian group. A theorem of Mordell and Weil states that this group is finitely generated. More information about the arithmetic of elliptic curves may be found in [331, 461, 592, 593]. The chapter also contains some information about two additional functions: the Weierstrass ζ-function, defined by ζ 0(z) = −℘(z) with lim ζ(z) − 1/z = 0 and the Weierstrass σ-function, defined z→0

by (log σ(z))0 = ζ(z) with lim σ(z)/z = 1. These are the elliptic analogs of the cotangent z→0

and sine functions. The chapter contains some identities for them, for instance one due to Stickelberger: if x + y + z = 0, then [ζ(x) + ζ(y) + ζ(z)]2 + ζ 0(x) + ζ 0(y) + ζ 0(z) = 0, as well as the identity ℘(z) − ℘(y) = −

σ(z + y) σ(z − y) , σ 2 (z)σ 2 (y)

Introduction

l

just to cite two of many. Among the many important results established in this chapter, we select three: (1) any elliptic function f can be written in the form R1 (℘) + R2 (℘)℘0(z), with R1 , R2 rational functions; (2) every elliptic function f satisfies an algebraic differential equation; (3) any curve of genus 1 can be parametrized by elliptic functions. The chapter contains a brief discussion on the uniformization of curves of higher genus. This problem is discussed in detail in [7, 12, 477]. Chapter 21. The Theta-Functions. The study of the function ϑ(z, q) =

∞ Õ

2

(−1)n q n e2niz

n=−∞

with q = exp(πiτ) and Im τ > 0 was initiated by Euler and perfected by Jacobi in [349]. This is an example of a theta function. It is a non-constant analytic function of z ∈ C, so it cannot be elliptic, but it has a simple transformation rule under z 7→ z + τ. This chapter considers ϑ, relabelled as ϑ1 as well as three other companion functions ϑ2 , ϑ3 and ϑ4 . These functions have a single zero in the fundamental cell L and since they transform in a predictable manner under the elements of L, it is easy to produce elliptic functions from them. This leads to a remarkable series of identities such as ϑ3 (z, q) = ϑ3 (2z, q4 ) + ϑ2 (2z, q4 ) and ϑ24 (0, q) + ϑ44 (0, q) = ϑ34 (0, q) that represents a parametrization of the Fermat projective curve x 4 + y 4 = z 4 . The chapter also discusses the addition theorem ϑ3 (z + y)ϑ3 (z − y)ϑ32 (0) = ϑ32 (y)ϑ32 (z) + ϑ12 (y)ϑ12 (z) (where the second variable q has been omitted) as well as an identity of Jacobi ϑ10 (0) = ϑ2 (0)ϑ3 (0)ϑ4 (0). This corresponds to the triple product identity, written as ∞ ∞ Ö Õ 2 (1 − q2n )(1 + q2n−1 p2 )(1 + q2m−1 p−2 ) = q n p2n, n=1

n=−∞

using the representation of theta functions as infinite products. The literature contains a variety of proofs of this fundamental identity; see Andrews [19] for a relatively simple one, Lewis [433] and Wright [683] for enumerative proofs and [311] for more general information on the so-called q-series. The chapter also shows that a quotient of theta functions ξ satisfies the differential equation 2 dξ = ϑ22 (0) − ξ 2 ϑ32 (0) ϑ32 (0) − ξ 2 ϑ22 (0) . dτ

Introduction

li

This is Jacobi’s version of the differential equation satisfied by the Weierstrass ℘-function. The properties of solutions of this equation form the subject of the next chapter. The reader will find in Baker [39, 40] a large amount of information on these functions from the point of view of the 19th century, Mumford [478, 479, 480] for a more modern point of view and [208, 209] for their connections to Riemann surfaces. Theta functions appeared scattered in the magnificent collection by Berndt [60, 61, 62, 63, 64] and Andrews–Berndt [21, 22, 23, 24, 25] on the formulas stated by Ramanujan. Chapter 22. The Jacobian Elliptic Functions. Each elliptic function f has a degree attached to it. This is defined as the number of solutions to f (z) = c in a fundamental cell. Constants have degree 0 and there are no functions of degree 1. A function of degree 2 either has a double pole (say at the origin) or two simple poles. The first case corresponds to the Weierstrass ℘ function described in Chapter 20. The second case is discussed in this chapter. The starting point is to show that any such function y = y(u) may be written as a quotient of theta functions. From here the authors show that y must satisfy the equation 2 dy = 1 − y2 1 − k 2 y2 du where k ∈ C is the modulus. An expression for k as a ratio of null-values of theta functions is provided. Then y = y(u) is seen to come from the inversion of the relation ∫ y −1/2 −1/2 u= 1 − t2 1 − k 2t 2 dt 0

and, following Jacobi, the function y is called the sinus amplitudinus and is denoted by y = sn(u, k). This function becomes the trigonometrical y = sin u when k → 0. Two companion functions cn(u, k) and dn(u, k) are also introduced. These functions satisfy a system of nonlinear differential equations XÛ = Y Z,

YÛ = −Z X,

ZÛ = −k 2 XY,

and they are shown to parametrize the curve ξ 2 = (1 − η2 )(1 − k 2 η2 ). The chapter also contains an addition theorem for these functions, such as sn(u + v) =

sn u cn v dn v + sn v cn u dnu 1 − k 2 sn2 u sn2 v

and other similar expressions. The complete elliptic integral of the first kind K(k) (and the complementary one K 0(k)) appears here from sn(K(k), k) = 1. The authors establish an expression for K(k) in terms of theta values, prove Legendre’s identity d 0 2 dK k(k ) = kK, dk dk and present a discussion of the periods of the (Jacobian elliptic) functions sn, cn, dn in terms of elliptic integrals. The reader will find details of these properties in [90, 461]. Other results appearing here include product representations of Jacobi functions, the Landen transformation and several definite integrals involving these functions. There is also a discussion on the so-called singular values: these are special values of the modulus k such that the ratio

Introduction √ √ K 0(k)/K(k) has the form (a + b n)/(c + d n) with a, b, c, d and n ∈ Z. These values of k satisfy polynomial equations with integer coefficients. The authors state that the study of these equation lies beyond the scope of this book. The reader will find information about these equations in [90]. lii

Chapter 23. Ellipsoidal Harmonics and Lamé’s Equation. This chapter presents the basic theory of ellipsoidal harmonics. It begins with the expression Θp =

y2 z2 x2 + + −1 a2 + θ p b2 + θ p c2 + θ p

where a > b > c are the semi-axis of the ellipsoid Θ p = 0. A function of the form Πm (Θ) = Θ1 · · · Θm is called an ellipsoidal harmonic of the first species. The chapter describes harmonic functions (that is, one satisfying ∆u = 0) of this form. It turns out that every such function (with n even) has the form n/2 Ö x2 y2 z2 + + −1 a2 + θ p b2 + θ p c2 + θ p p=1 where θ 1, . . . , θ n/2 are zeros of a polynomial Λ(θ) of degree n/2. This polynomial solves the Lamé equation p d p 2 dΛ 2 2 2 2 2 = [n(n + 1)θ + C] Λ(θ). 4 (a + θ)(b + θ)(c + θ) (a + θ)(b + θ)(c + θ) dθ dθ The value C is constant and it is shown that there are 12 n + 1 possible choices. There are three other types of ellipsoidal harmonics with a similar theory behind them. The chapter contains many versions of Lamé’s equation: the algebraic form [n(n + 1)λ + C] Λ 1 dΛ d2Λ 1 1 1 + + 2 + 2 = 2 2 2 dλ 2 a + λ b + λ c + λ dλ 4(a + λ)(b2 + λ)(c2 + λ) as well as the Weierstrass elliptic form d2Λ = [n(n + 1)℘(u) + B] Λ du2 and finally the Jacobi elliptic form d2Λ = n(n + 1)k 2 sn2 α + A Λ. 2 dα These equations are used to introduce Lamé functions. These are used to show that there are 2n + 1 ellipsoidal harmonics that form a fundamental system of the harmonic functions of degree n. The chapter contains a brief comment on work by Heun [300, 301] mentioning the study of an equation with four singular points. The reader will find in Ronveaux [563] more information about this equation.

Part I The Process of Analysis

1

1 Complex Numbers

1.1 Rational numbers The idea of a set of numbers is derived in the first instance from the consideration of the set of positive integral numbers, or positive integers; that is to say, the numbers 1, 2, 3, 4, . . .. (Strictly speaking, a more appropriate epithet would be, not positive, but signless.) Positive integers have many properties, which will be found in treatises on the Theory of Integral Numbers; but at a very early stage in the development of Mathematics it was found that the operations of Subtraction and Division could only be performed among them subject to inconvenient restrictions; and consequently, in elementary Arithmetic, classes of numbers are constructed such that the operations of subtraction and division can always be performed among them. To obtain a class of numbers among which the operation of subtraction can be performed without restraint we construct the class of integers, which consists of the class of positive integers (in the strict sense) (+1, +2, +3, . . .) and of the class of negative integers (−1, −2, −3, . . .) and the number 0. To obtain a class of numbers among which the operations both of subtraction and of division can be performed freely, with the exception of division by the rational number 0, we construct the class of rational numbers. Symbols which denote members of this class are 1 , 3, 0, − 15 . We have thus introduced three classes of numbers, (i) the signless integers, (ii) 2 7 the integers, (iii) the rational numbers. It is not part of the scheme of this work to discuss the construction of the class of integers or the logical foundations of the theory of rational numbers. Such a discussion, defining a rational number as an ordered number-pair of integers in a similar manner to that in which a complex number is defined in §1.3 as an ordered number-pair of real numbers, will be found in Hobson [315, §1-12]. The extension of the idea of number, which has just been described, was not effected without some opposition from the more conservative mathematicians. In the latter half of the eighteenth century, Maseres (1731–1824) and Frend (1757–1841) published works on Algebra, Trigonometry, etc., in which the use of negative numbers was disallowed, although Descartes had used them unrestrictedly more than a hundred years before. A rational number x may be represented to the eye in the following manner: If, on a straight line, we take an origin O and a fixed segment OP1 (P1 being on the right of O), we can measure from O a length OPx such that the ratio OPx /OP1 is equal to x; the point Px is taken on the right or left of O according as the number x is positive or negative. We may regard 3

4

Complex Numbers

either the point Px or the displacement OPx (which will be written OPx ) as representing the number x. All the rational numbers can thus be represented by points on the line, but the converse is not true. For if we measure off on the line a length OQ equal to the diagonal of a square of which OP1 is one side, it can be proved that Q does not correspond to any rational number. Points on the line which do not represent rational numbers may be said √ to represent irrational numbers; thus the point Q is said to represent the irrational number 2 = 1.414213 · · · . But while such an explanation of the existence of irrational numbers satisfied the mathematicians of the eighteenth century and may still be sufficient for those whose interest lies in the applications of mathematics rather than in the logical upbuilding of the theory, yet from the logical standpoint it is improper to introduce geometrical intuitions to supply deficiencies in arithmetical arguments; and it was shewn by Dedekind [169] in 1858 that the theory of irrational numbers can be established on a purely arithmetical basis without any appeal to geometry.

1.2 Dedekind’s theory of irrational numbers The geometrical property of points on a line which suggested the starting point of the arithmetical theory of irrationals was that, if all points of a line are separated into two classes such that every point of the first class is on the right of every point of the second class, there exists one and only one point at which the line is thus severed. The theory, though elaborated in 1858, was not published before the appearance of Dedekind’s tract [169]. Other theories are due to Weierstrass (see [642]) and Cantor [116]. Following up this idea, Dedekind considered rules by which a separation or section of all rational numbers into two classes can be made. This procedure formed the basis of the treatment of irrational numbers by the Greek mathematicians in the sixth and fifth centuries b.c. The advance made by Dedekind consisted in observing that a purely arithmetical theory could be built up on it. These classes, which will be called the L-class and the R-class, or the left class and the right class, being such that they possess the following properties: (i) At least one member of each class exists. (ii) Every member of the L-class is less than every member of the R-class. It is obvious that such a section is made by any rational number x; and x is either the greatest number of the L-class or the least number of the R-class. But sections can be made in which no rational number x plays this part. Thus, since there is no rational number 1 whose square is 2, it is easy to see that we may form a section in which the R-class consists of the positive rational numbers whose squares exceed 2, and the L-class consists of all other rational numbers. Then this section is such that the R-class has no least member and the L-class has no greatest 2 2 ) +6) , then y − x = 2x(2−x and member; for, if x be any positive rational fraction, and y = x(x 3x 2 +2 3x 2 +2 (x 2 −2)3 2 2 2 y − 2 = (3x 2 +2)2 , so x , y and 2 are in order of magnitude; and therefore given any member 1

For if p/q be such a number, this fraction being in its lowest terms, it may be seen that (2q − p)/(p − q) is another such number, and 0 < p − q < q, so that p/q is not in its lowest terms. The contradiction implies that such a rational number does not exist.

1.2 Dedekind’s theory of irrational numbers

5

x of the L-class, we can always find a greater member of the L-class, or given any member x 0 of the R-class, we can always find a smaller member of the R-class, such numbers being, for instance, y and y 0, where y 0 is the same function of x 0 as y of x. If a section is made in which the R-class has a least member A2 , or if the L-class has a greatest member A1 , the section determines a rational-real number; which it is convenient to denote by the same symbol A2 or A1 . This causes no confusion in practice. If a section is made, such that the R-class has no least member and the L-class has no greatest member, the section determines an irrational-real number. Note B. A. W. Russell [567] defines the class of real numbers as actually being the class of all L-classes; the class of real numbers whose L-classes have a greatest member corresponds to the class of rational numbers, and though the rational-real number x which corresponds to a rational number x is conceptually distinct from it, no confusion arises from denoting both by the same symbol. If x, y are real numbers (defined by sections) we say that x is greater than y if the L-class defining x contains at least two members of the R-class defining y. If the classes had only one member in common, that member might be the greatest member of the L-class of x and the least member of the R-class of y. Let α, β, . . . be real numbers and let A1, B1, . . . be any members of the corresponding L-classes while A2, B2, . . . are any members of the corresponding R-classes. The classes of which A1, A2, . . . are respectively members will be denoted by the symbols (A1 ), (A2 ), . . . . Then the sum (written α + β) of two real numbers α and β is defined as the real number (rational or irrational) which is determined by the L-class (A1 + B1 ) and the R-class (A2 + B2 ). It is, of course, necessary to prove that these classes determine a section of the rational numbers. It is evident that A1 + B1 < A2 + B2 and that at least one member of each of the classes (A1 + B1 ), (A2 + B2 ) exists. It remains to prove that there is, at most, one rational number which is greater than every A1 + B1 and less than every A2 + B2 ; suppose, if possible, that there are two, x and y, (y > x). Let α1 be a member of (A1 ) and let α2 be a member of (A2 ); and let N be the integer next greater than (α2 − α1 )/{ 12 (y − x)}. Take the last of the numbers α1 + m (α2 − α1 ), (where m = 0, 1, . . . , N), which belongs to (A1 ) and the first of N them which belongs to (A2 ); let these two numbers be c1 , c2 . Then c2 − c1 =

1 1 (α2 − α1 ) < (y − x). N 2

Choose d1, d2 in a similar manner from the classes defining β; then c2 + d2 − c1 − d1 < y − x. But c2 + d2 ≥ y, c1 + d1 ≤ x, and therefore c2 + d2 − c1 − d1 ≥ y − x; we have therefore arrived at a contradiction by supposing that two rational numbers x, y exist belonging neither to (A1 + B2 ) nor to (A2 + B2 ). If every rational number belongs either to the class (A1 + B1 ) or to the class (A2 + B2 ), then the classes (A1 + B1 ), (A2 + B2 ) define an irrational number. If one rational number x exists belonging to neither class, then the L-class formed by x and (A1 + B1 ) and the R-class (A2 + B2 ) define the rational number-real x. In either case, the number defined is called the sum α + β.

Complex Numbers

6

The difference α − β of two real numbers is defined by the L-class (A1 − B2 ) and the R-class (A2 − B1 ). The product of two positive real numbers α, β is defined by the R-class (A2 B2 ) and the L-class of all other rational numbers. The reader will see without difficulty how to define the product of negative real numbers and the quotient of two real numbers; and further, it may be shewn that real numbers may be combined in accordance with the associative, distributive and commutative laws. The aggregate of rational-real and irrational-real numbers is called the aggregate of real numbers; for brevity, rational-real numbers and irrational-real numbers are called rational and irrational numbers respectively.

1.3 Complex numbers We have seen that a real number may be visualised as a displacement along a definite straight line. If, however, P and Q are any two points in a plane, the displacement PQ needs two real numbers for its specification; for instance, the differences of the coordinates of P and Q referred to fixed rectangular axes. If the coordinates of P be (ξ, η) and those of Q(ξ + x, η + y), the displacement PQ may be described by the symbol [x, y]. We are thus led to consider the association of real numbers in ordered pairs. The order of the two terms distinguishes the ordered number-pair [x, y] from the ordered number-pair [y, x]. The natural definition of the sum of two displacements [x, y], [x 0, y 0] is the displacement which is the result of the successive applications of the two displacements; it is therefore convenient to define the sum of two number-pairs by the equation [x, y] + [x 0, y 0] = [x + x 0, y + y 0]. The product of a number-pair and a real number x 0 is then naturally defined by the equation x 0 × [x, y] = [x 0 x, x 0 y]. We are at liberty to define the product of two number-pairs in any convenient manner; but the only definition, which does not give rise to results that are merely trivial, is that symbolised by the equation [x, y] × [x 0, y 0] = [x x 0 − yy 0, xy 0 + x 0 y]. It is then evident that [x, 0] × [x 0, y 0] = [x x 0, xy 0] = x × [x 0, y 0] and [0, y] × [x 0, y 0] = [−yy 0, x 0 y] = y × [−y 0, x 0]. The geometrical interpretation of these results is that the effect of multiplying by the displacement [x, 0] is the same as that of multiplying by the real number x; but the effect of multiplying a displacement by [0, y] is to multiply it by a real number y and turn it through a right angle. It is convenient to denote the number-pair [x, y] by the compound symbol x + iy; and a number-pair is now conveniently called (after Gauss) a complex number; in the fundamental operations of Arithmetic, the complex number x + i0 may be replaced by the real number x

1.4 The modulus of a complex number

7

and, defining i to mean [0, 1], we have i 2 = [0, 1] × [0, 1] = [−1, 0]; and so i 2 may be replaced by −1. The reader will easily convince himself that the definitions of addition and multiplication of number-pairs have been so framed that we may perform the ordinary operations of algebra with complex numbers in exactly the same way as with real numbers, treating the symbol i as a number and replacing the product ii by −1 wherever it occurs. Thus he will verify that, if a, b, c are complex numbers, we have a + b = b + a, ab = ba, (a + b) + c = a + (b + c), (ab)c = a(bc), a(b + c) = ab + ac, and if ab is zero, then either a or b is zero. It is found that algebraical operations, direct or inverse, when applied to complex numbers, do not suggest numbers of any fresh type; the complex number will therefore for our purposes be taken as the most general type of number. The introduction of the complex number has led to many important developments in mathematics. Functions which, when real variables only are considered, appear as essentially distinct, are seen to be connected when complex variables are introduced: thus the circular functions are found to be expressible in terms of exponential functions of a complex argument, by the equations 1 ix 1 (e − e−ix ). cos x = (eix + e−ix ), sin x = 2 2i Again, many of the most important theorems of modern analysis are not true if the numbers concerned are restricted to be real; thus, the theorem that every algebraic equation of degree n has n roots is true in general only when regarded as a theorem concerning complex numbers. Hamilton’s quaternions furnish an example of a still further extension of the idea of number. A quaternion w + xi + y j + zk is formed from four real numbers w, x, y, z, and four number-units 1, i, j, k, in the same way that the ordinary complex number x + iy might be regarded as being formed from two real numbers x, y, and two number-units 1, i. Quaternions however do not obey the commutative law of multiplication.

1.4 The modulus of a complex number Let x + iy be a complex number, x and y being real numbers. Then the positive square root of x 2 + y 2 is called the modulus of (x + iy), and is written | x + iy |. Let us consider the complex number which is the sum of two given complex numbers, x + iy and u + iv. We have (x + iy) + (u + iv) = (x + u) + i(y + v).

Complex Numbers

8

The modulus of the sum of the two numbers is therefore {(x + u)2 + (y + v)2 }1/2 = {(x 2 + y 2 ) + (u2 + v 2 ) + 2(xu + yv)}1/2 . But {| x + iy | + | u + iv |}2 = {(x 2 + y 2 )1/2 + (u2 + v 2 )1/2 }2 = (x 2 + y 2 ) + (u2 + v 2 ) + 2(x 2 + y 2 )1/2 (u2 + v 2 )1/2 = (x 2 + y 2 ) + (u2 + v 2 ) + 2 {(xu + yv)2 + (xv − yu)2 }1/2, and this latter expression is greater than (or at least equal to) (x 2 + y 2 ) + (u2 + v 2 ) + 2(xu + yv). We have therefore |x + iy| + |u + iv| ≥ |(x + iy) + (u + iv)|, i.e. the modulus of the sum of two complex numbers cannot be greater than the sum of their moduli; and it follows by induction that the modulus of the sum of any number of complex numbers cannot be greater than the sum of their moduli. Let us consider next the complex number which is the product of two given complex numbers, x + iy and u + iv, we have (x + iy)(u + iv) = (xu − yv) + i(xv + yu), and so |(x + iy)(u + iv)| = {(xu − yv)2 + (xv + yu)2 }1/2 = {(x 2 + y 2 )(u2 + v 2 )}1/2 = |x + iy||u + iv|. The modulus of the product of two complex numbers (and hence, by induction, of any number of complex numbers) is therefore equal to the product of their moduli.

1.5 The Argand diagram We have seen that complex numbers may be represented in a geometrical diagram by taking rectangular axes Ox, Oy in a plane. Then a point P whose coordinates referred to these axes are x, y may be regarded as representing the complex number x + iy. In this way, to every point of the plane there corresponds some one complex number; and, conversely, to every possible complex number there corresponds one, and only one, point of the plane. The complex number x + iy may be denoted by a single letter z. It is convenient to call x and y the real and imaginary parts of z respectively. We frequently write x = Re z, y = Im z. The point P is then called the representative point of the number z; we shall also speak of the number z as being the affix of the point P. If we denote (x 2 + y 2 )1/2 by r and choose θ so that r cos θ = x, r sin θ = y, then r and θ are clearly the radius vector and vectorial angle of the point P, referred to the origin O and axis Ox. The representation of complex numbers thus afforded is often called the Argand diagram.

1.6 Miscellaneous examples

9

It was published by J. R. Argand [33]; it had however previously been used by Gauss [235] in his Helmstedt dissertation in 1799, who had discovered it in Oct. 1797 [375]; and Caspar Wessel had discussed it in a memoir presented to the Danish Academy in 1797 and published by that Society in 1798–9 [664]. The phrase complex number first occurs in [237, p. 102]. By the definition already given, it is evident that r is the modulus of z. The angle θ is called the argument or phase, of z. We write θ = arg z. From geometrical considerations, it appears that (although the modulus of a complex number is unique) the argument is not unique (see the Appendix, §A.521) if θ be a value of the argument, the other values of the argument of a complex number are comprised in the expression 2nπ + θ where n is any integer, not zero. The principal value of the argument of a complex number value of arg z is that which satisfies the inequality −π < arg z ≤ π. If P1 and P2 are the representative points corresponding to values z1 and z2 respectively of z, then the point which represents the value z1 + z2 is clearly the terminus of a line drawn from P1 , equal and parallel to that which joins the origin to P2 . To find the point which represents the complex number z1 z2 , where z1 and z2 are two given complex numbers, we notice that if z1 = r1 (cos θ 1 + i sin θ 1 ), z2 = r2 (cos θ 2 + i sin θ 2 ) then, by multiplication, z1 z2 = r1 r2 {cos(θ 1 + θ 2 ) + i sin(θ 1 + θ 2 )}. The point which represents the number z1 z2 has therefore a radius vector measured by the product of the radii vectors of P1 and P2 and a vectorial angle equal to the sum of the vectorial angles of P1 and P2 .

1.6 Miscellaneous examples Example 1.1 Shew that the representative points of the complex numbers 1 + 4i, 2 + 7i, 3 + 10i, are collinear. Example 1.2 Shew that a parabola can be drawn to pass through the representative points of the complex numbers 2 + i, 4 + 4i, 6 + 9i, 8 + 16i, 10 + 25i. Example 1.3 (Math. Trip. 1895). Determine the nth roots of unity by aid of the Argand diagram; and shew that the number of primitive roots (roots the powers of each of which give all the roots) is the number of integers (including unity) less than n and prime to it. Í Prove that if θ 1, θ 2, θ 3, . . . be the arguments of the primitive roots, cos pθ = 0 when p is n , where a, b, c, . . . , k are the different constituent primes a positive integer less than abc · · · k of n; and that, when Õ n (−1)µ n p= , then cos pθ = , abc · · · k abc · · · k where µ is the number of constituent primes.

2 The Theory of Convergence

2.1 The definition of the limit of a sequence Let z1, z2, z3, . . . be an unending sequence of numbers, real or complex. Then, if a number ` exists such that, corresponding to every positive 1 number ε, no matter how small, a number n0 can be found, such that | zn − ` | < ε for all values of n greater than n0 , the sequence (zn ) is said to tend to the limit ` as n tends to infinity. A definition equivalent to this was first given by John Wallis in 1655 [645, p. 382]. Symbolic forms of the statement,‘the limit of the sequence (zn ), as n tends to infinity, is `’ are: lim zn = `,

n→∞

lim zn = `,

zn → `

as

n → ∞.

The arrow notation is due to Leathem (see [416]). If the sequence be such that, given an arbitrary number N (no matter how large), we can find n0 such that |zn | > N for all values of n greater than n0 , we say that ‘ |zn | tends to infinity as n tends to infinity’, and we write |zn | → ∞. In the corresponding case when −xn > N when n > n0 we say that xn → −∞. If a sequence of real numbers does not tend to a limit or to ∞ or to −∞, the sequence is said to oscillate.

2.11 Definition of the phrase ‘of the order of’ If (ζn ) and (zn ) are two sequences such that a number n0 exists such that | (ζn /zn ) | < K whenever n > n0 , where K is independent of n, we say that ζn is ‘of the order of’ zn , and we 1 = O . This notation is due to Bachmann [35, p. 401] and write ζn = O(zn ); thus 15n+19 1+n3 n2 Landau [405, p. 61]. Note If lim

ζn zn

= 0, we write ζn = o(zn ).

2.2 The limit of an increasing sequence Let (xn ) be a sequence of real numbers such that xn+1 ≥ xn for all values of n; then the sequence tends to a limit or else tends to infinity (and so it does not oscillate). Let x be any rational-real number; then either: (i) xn ≥ x for all values of n greater than some number n0 depending on the value of x; or (ii) xn < x for every value of n. 1

The number zero is excluded from the class of positive numbers.

10

2.2 The limit of an increasing sequence

11

If (ii) is not the case for any value of x (no matter how large), then xn → ∞. But if values of x exist for which (ii) holds, we can divide the rational numbers into two classes, the L-class consisting of those rational numbers x for which (i) holds and the R-class of those rational numbers x for which (ii) holds. This section defines a real number α, rational or irrational. And if ε be an arbitrary positive number, α − 21 ε belongs to the L-class which defines α, and so we can find n1 such that xn ≥ α − 21 ε whenever n > n1 ; and α + 12 ε is a member of the R-class and so xn < α + 12 ε. Therefore, whenever n > n1 , | α − xn | < ε. Therefore xn → α. Corollary 2.2.1 A decreasing sequence tends to a limit or to −∞. 0 0 Example 2.2.1 If lim zm = `, lim zm = ` 0, then lim(zm + zm ) = ` + ` 0.

For, given ε, we can find n and n 0 such that (i) when m > n, |zm − `| < 12 ε; 0 (ii) when m > n 0, |zm − ` 0 | < 21 ε. Let n1 be the greater of n and n 0; then, when m > n1 , (zm + z 0 ) − (` + ` 0) ≤ |(zm − `)| + (z 0 − ` 0) , m m < ε; 0 ) = ` + ` 0. and this is the condition that lim(zm + zm 0 0 Example 2.2.2 Prove similarly that lim(zm − zm ) = ` − ` 0, lim(zm zm ) = `` 0, and, if ` 0 , 0, 0 0 lim(zm /zm ) = `/` .

Example 2.2.3 If 0 < x < 1, x n → 0. For if x = (1 + α)−1 , α > 0, and 0 < xn =

1 1 < , n (1 + a) 1 + na

by the binomial theorem for a positive integral index. And it is obvious that, given a positive number ε, we can choose n0 such that (1 + na)−1 < ε when n > n0 ; and so x n → 0.

2.21 Limit-points and the Bolzano–Weierstrass theorem This theorem, frequently ascribed to Weierstrass, was proved by Bolzano [81]. It seems to have been known to Cauchy. Let (xn ) be a sequence of real numbers. If any number G exists such that, for every positive value of ε, no matter how small, an unlimited number of terms of the sequence can be found such that G − ε < xn < G + ε, then G is called a limit-point, or cluster-point of the sequence. Theorem 2.2.2 (Bolzano) If λ ≤ xn ≤ ρ, where λ, ρ are independent of n, then the sequence (xn ) has at least one limit-point.

12

The Theory of Convergence

To prove the theorem, choose a section in which (i) the R-class consists of all the rational numbers which are such that, if A be any one of them, there are only a limited number of terms xn satisfying xn > A; and (ii) the L-class is such that there are an unlimited number of terms xn such that xn ≥ α for all members α of the L-class. This section defines a real number G; and, if ε be an arbitrary positive number, G − 21 ε and G + 12 ε are members of the L- and R-classes respectively, and so there are an unlimited number of terms of the sequence satisfying 1 1 G − ε < G − ε ≤ xn ≤ G + ε < G + ε, 2 2 and so G satisfies the condition that it should be a limit-point.

2.211 Definition of ‘the greatest and the least of the limits’ The number G obtained in §2.21 is called ‘the greatest of the limits of the sequence (xn ).’ The sequence (xn ) cannot have a limit-point greater than G; for if G 0 were such a limit-point, and ε = 12 (G 0 − G), G 0 − ε is a member of the R-class defining G, so that there are only a limited number of terms of the sequence which satisfy xn > G 0 − ε. This condition is inconsistent with G 0 being a limit-point. We write G = lim xn . n→∞

The ‘least of the limits’ L, of the sequence (written lim xn ) is defined to be n→∞

− lim (−xn ). n→∞

2.22 Cauchy’s theorem on the necessary and sufficient condition for the existence of a limit [120, p. 125] We shall now shew that the necessary and sufficient condition for the existence of a limiting value of a sequence of numbers z1, z2, z3, . . . is that, corresponding to any given positive number ε, however small, it shall be possible to find a number n such that zn+p − zn < ε for all positive integral values of p. This result is one of the most important and fundamental theorems of analysis. It is sometimes called the Principle of Convergence. First, we have to shew that this condition is necessary, i.e. that it is satisfied whenever a limit exists. Suppose then that a limit ` exists; then (§2.1) corresponding to any positive number ε, however small, an integer n can be chosen such that zn+p − ` < 1 ε, |zn − `| < 21 ε, 2 for all positive values of p; therefore zn+p − zn = zn+p − ` − (zn − `) ≤ zn+p − ` + |zn − `| < ε,

2.3 Convergence of an infinite series

13

which shews the necessity of the condition zn+p − zn < ε, and thus establishes the first half of the theorem. Second, we have to prove that this condition is sufficient, i.e. that if it is satisfied, then a limit exists. This proof is given in Stolz–Gmeiner [613, p. 144]. (I) Suppose that the sequence of real numbers (xn ) satisfies Cauchy’s condition; that is to say that, corresponding to any positive number ε, an integer n can be chosen such that xn+p − xn < ε for all positive integral values of p. Let the value of n, corresponding to the value 1 of ε, be m. Let λ1, ρ1 be the least and greatest of x1, x2, . . . , xm ; then λ1 − 1 < xn < ρ1 + 1, for all values of n; write λ1 − 1 = λ, ρ1 + 1 = ρ. Then, for all values of n, λ < xn < ρ. Therefore by Theorem 2.2.2, the sequence (xn ) has at least one limit-point G. Further, there cannot be more than one limit-point; for if there were two, G and H (H < G), take ε < 41 (G − H). Then, by hypothesis, a number n exists such that xn+p − xn < ε for every positive value of p. But since G and H are limit-points, positive numbers q and r exist such that G − xn+q < ε, | H − xn+r | < ε. Then G − xn+q + xn+q − xn + | xn − xn+r | + | xn+r − H | < 4ε. But, by §1.4, the sum on the left is greater than or equal to | G − H |. Therefore G − H < 4ε, which is contrary to hypothesis; so there is only one limit-point. Hence there are only a finite number of terms of the sequence outside the interval (G − δ, G + δ), where δ is an arbitrary positive number; for, if there were an unlimited number of such terms, these would have a limit-point which would be a limit-point of the given sequence and which would not coincide with G; and therefore G is the limit of (xn ). (II) Now let the sequence (zn ) of real or complex numbers satisfy Cauchy’s condition; and let zn = xn + iyn , where xn and yn are real; then for all values of n and p xn+p − xn ≤ zn+p − zn , yn+p − yn ≤ zn+p − zn . Therefore the sequences of real numbers (xn ) and (yn ) satisfy Cauchy’s condition; and so, by (I), the limits of (xn ) and (yn ) exist. Therefore, by Example 2.2.1, the limit of (zn ) exists. The result is therefore established.

2.3 Convergence of an infinite series Let u1, u2, u3, . . . , un, . . . be a sequence of numbers, real or complex. Let the sum u1 + u2 + · · · + u n be denoted by Sn .

The Theory of Convergence

14

Then, if Sn tends to a limit S as n tends to infinity, the infinite series u1 + u2 + u3 + u4 + · · · is said to be convergent, or to converge to the sum S. In other cases, the infinite series is said to be divergent. When the series converges, the expression S − Sn , which is the sum of the series un+1 + un+2 + un+3 + · · · , is called the remainder after n terms, and is frequently denoted by the symbol Rn . The sum un+1 + un+2 + · · · + un+p will be denoted by Sn,p . It follows at once, by combining the above definition with the results of the last paragraph, that the necessary and sufficient condition for the convergence of an infinite series is that, given an arbitrary positive number ε, we can find n such that Sn,p < ε for every positive value of p. Since un+1 = Sn,1 , it follows as a particular case that lim un+1 = 0, in other words, the nth term of a convergent series must tend to zero as n tends to infinity. But this last condition, though necessary, is not sufficient in itself to ensure the convergence of the series, as appears from a study of the series 1 1 1 1 1 + + + + +··· . 1 2 3 4 5 1 1 1 1 In this series, Sn,n = + + + ··· + . The expression on the right is n+1 n+2 n+3 2n −1 diminished by writing (2n) in place of each term, and so Sn,n > 12 . Therefore S2n+1 = 1 + S1, 1 + S2, 2 + S4, 4 + S8, 8 + S16, 16 + · · · + S2n , 2n 1 > (n + 3) → ∞; 2 so the series is divergent; this result was noticed by Leibniz in 1673. There are two general classes of problems which we are called upon to investigate in connexion with the convergence of series: 1. We may arrive at a series by some formal process, e.g. that of solving a linear differential equation by a series, and then to justify the process it will usually have to be proved that the series thus formally obtained is convergent. Simple conditions for establishing convergence in such circumstances are obtained in §§2.31–2.61. n Í 2. Given an expression S, it may be possible to obtain a development S = um + Rn , valid m=1

for all values of n; and, from the definition of a limit, it follows that, if we can prove that ∞ Í Rn → 0, then the series um converges and its sum is S. An example of this problem m=1

occurs in §5.4. Infinite series were used by Lord Brouncker in [103, pp. 645–649], and the term convergent was introduced by James Gregory, Professor of Mathematics at Edinburgh, in the same year; the term divergent was introduced by N. Bernoulli in 1713. Infinite series were used systematically by Newton [494, pp. 206–247], and he investigated the convergence of

2.3 Convergence of an infinite series

15

hypergeometric series (§14.1) in 1704. (See also the convergence of products in §2.7.) But the great mathematicians of the eighteenth century used infinite series freely without, for the most part, examining their convergence. Thus Euler gave the sum of the series ···+

1 1 1 + 2 + + 1 + z + z2 + z3 + · · · 3 z z z

(a)

as zero, on the ground that z 1−z

(b)

1 1 z + 2 +··· = z z z−1

(c)

z + z2 + z3 + · · · = and 1+

The error of course arises from the fact that the series (b) converges only when |z| < 1, and the series (c) converges only when |z| > 1, so the series (a) never converges. For the history of researches on convergence, see Pringsheim and Molk [543] and Reiff [551].

2.301 Abel’s inequality. This appears in [1, pp. 311–339]. A particular case of Corollary 2.3.3 also appears in that memoir. Theorem 2.3.1 Let fn ≥ fn+1 > 0 for all integer values of n. Then m Õ a f ≤ A f1, n=1 n n where A is the greatest of the sums | a1 | , | a1 + a2 | , | a1 + a2 + a3 | , . . . , | a1 + a2 + · · · + am | . For, writing a1 + a2 + · · · + an = sn, we have m Õ

an fn = s1 f1 + (s2 − s1 ) f2 + (s3 − s2 ) f3 + · · · + (sm − sm−1 ) fm

n=1

= s1 ( f1 − f2 ) + s2 ( f2 − f3 ) + · · · + sm−1 ( fm−1 − fm ) + sm fm . Since f1 − f2, f2 − f3, . . . are not negative, we have, when n = 2, 3, . . . , m, |sn−1 | ( fn−1 − fn ) ≤ A( fn−1 − fn ) also |sm | fm ≤ A fm , and so, summing and using §1.4, we get m Õ a f ≤ A f1 . n=1 n n Corollary 2.3.2 (Hardy) If a1, a2, . . . , w1, w2, . . . are any numbers, real or complex, ( m−1 ) m Õ Õ |wn+1 − wn | + |wm | , a w ≤A n=1 n n n=1 Í p where A is the greatest of the sums n=1 an , (p = 1, 2, . . . , m).

The Theory of Convergence

16

2.31 Dirichlet’s test for convergence This appears in [177, pp. 253–255]. Before the publication of the Second edition of Jordan’s Cours d’Analyse [361], Dirichlet’s test and Abel’s test were frequently jointly described as the Dirichlet–Abel test, see e.g. Pringsheim [537, p. 423]. p Í Let an < K, where K is independent of p. Then, if fn ≥ fn+1 > 0 and lim fn = 0, n=1 ∞ Í an fn converges. n=1

In these circumstances, we say fn → 0 steadily. Proof For, since lim fn = 0, given an arbitrary positive number ε, we can find m such that fm+1 < ε/2K. Then m+q m+q m Õ Õ Õ an ≤ an + an < 2K, n=m+1 n=1 n=1 for all positive values of q; so that, by Abel’s inequality, we have, for all positive values of p, m+p Õ an fn ≤ A fm+1, n=m+1 where A < 2K. ∞ m+p Í Í Therefore an fn < 2K fm+1 < ε; and so, by §2.3, an fn converges. n=m+1 n=1 Corollary 2.3.3

∞ Í

Abel’s test for convergence. If

an converges and the sequence (un )

n=1

is monotonic (i.e. un ≥ un+1 always or else un ≤ un+1 always) and |un | < κ, where κ is ∞ Í independent of n, then an un converges. n=1 ∞ Í

For, by §2.2, un tends to a limit u; let | u − un | = fn . Then fn → 0 steadily; and therefore ∞ Í an fn converges. But, if (un ) is an increasing sequence, fn = u − un , and so (u − un )an

n=1

converges; therefore since

∞ Í n=1

uan converges,

∞ Í n=1

n=1

un an converges. If (un ) is a decreasing

sequence fn = un − u, and a similar proof holds.

Corollary 2.3.4 Taking an = (−1)n−1 in Dirichlet’s test, it follows that, if fn ≥ fn+1 and lim fn = 0, f1 − f2 + f3 − f4 + · · · converges. p Í Example 2.3.1 Shew that if 0 < θ < 2π, sin nθ < cosec 12 θ; and deduce that, if fn → 0 n=1 ∞ ∞ Í Í steadily, fn sin nθ converges for all real values of θ, and that fn cos nθ converges if θ n=1

n=1

is not an even multiple of π. Example 2.3.2 Shew that, if fn → 0 steadily, not an odd multiple of π and π + θ for θ in Example 2.3.1.

∞ Í n=1

∞ Í

(−1)n fn cos nθ converges if θ is real and

n=1

(−1)n fn sin nθ converges for all real values of θ. Hint. Write

2.3 Convergence of an infinite series

17

2.32 Absolute and conditional convergence ∞ Í In order that a series un of real or complex terms may converge, it is sufficient (but not n=1

∞ Í |un | should converge. For, if σn,p = |un+1 | + |un+2 | + necessary) that the series of moduli n=1 ∞ Í |un | converges, we can find n, corresponding to a given number ε, such · · · + un+p and if n=1 ∞ Í un converges. that σn,p < ε for all values of p. But Sn,p ≤ σn,p < ε, and so n=1

The condition is not necessary; for, writing fn = 1/n in Corollary 2.3.4, we see that 1 1 − + 31 − 41 + · · · converges, though (§2.3) the series of moduli 11 + 12 + 31 + 14 + · · · is known 1 2 to diverge. In this case, therefore, the divergence of the series of moduli does not entail the divergence of the series itself. Series which are such that the series formed by the moduli of their terms are convergent, possess special properties of great importance, and are called absolutely convergent series. Series which though convergent are not absolutely convergent (i.e. the series themselves converge, but the series of moduli diverge) are said to be conditionally convergent.

2.33 The geometric series, and the series

Í∞

1 n=1 n s

The convergence of a particular series is in most cases investigated, not by the direct consideration of the sum Sn,p , but (as will appear from the following articles) by a comparison of the given series with some other series which is known to be convergent or divergent. We shall now investigate the convergence of two of the series which are most frequently used as standards for comparison. (I) The geometric series. The geometric series is defined to be the series 1 + z + z2 + z3 + z4 + · · · . Consider the series of moduli 1 + |z| + |z| 2 + |z| 3 + · · · ;

(2.1)

for this series Sn,p = |z| n+1 + |z| n+2 + · · · + |z| n+p 1 − |z| p = |z| n+1 . 1 − |z| n+1

| Hence, if |z| < 1, then Sn,p < |z1−|z for all values of p, and, by Example 2.2.3, given any | positive number ε, we can find n such that

|z| n+1 {1 − |z|}−1 < ε. Thus, given ε, we can find n such that, for all values of p, Sn,p < ε. Hence, by §2.22, the series (2.1) is convergent so long as |z| < 1, and therefore the geometric series is absolutely convergent if |z| < 1.

The Theory of Convergence

18

When |z| ≥ 1, the terms of the geometric series do not tend to zero as n tends to infinity, and the series is therefore divergent. (II) The series

+

1 1s

1 2s

+

1 3s

+

1 4s

+

1 5s

+ · · · . Consider now the series Sn =

n Í m=1

1 , ms

where s

is greater than 1. We have 1 1 2 1 + < s = s−1 , 2s 3s 2 2 1 1 1 4 1 1 + s + s + s < s = s−1 , s 4 5 6 7 4 4 and so on. Thus the sum of 2 p − 1 terms of the series is less than 1 1 1 1 1 1 + + + + · · · + (p−1)(s−1) < , 1s−1 2s−1 4s−1 8s−1 1 − 21−s 2 and so the sum of any number of terms is less than (1 − 21−s )−1 . Therefore the increasing n ∞ Í Í 1 sequence m−s cannot tend to infinity; therefore, by §2.2, the series is convergent if ns m=1

n=1

s > 1; and since its terms are all real and positive, they are equal to their own moduli, and so the series of moduli of the terms is convergent; that is, the convergence is absolute. If s = 1, the series becomes 1 1 1 1 + + + +··· , 1 2 3 4 which we have already shewn to be divergent; and when s < 1, it is a fortiori divergent, since ∞ Í 1 the effect of diminishing s is to increase the terms of the series. The series is therefore ns n=1

divergent if s ≤ 1.

2.34 The comparison theorem We shall now shew that a series u1 + u2 + u3 + · · · is absolutely convergent, provided that |un | is always less than C|vn |, where C is some number independent of n, and vn is the nth term of another series which is known to be absolutely convergent. For, under these conditions, we have |un+1 | + |un+2 | + · · · + un+p < C |vn+1 | + |vn+2 | + · · · + vn+p , Í where n and p are any integers. But since the series vn is absolutely convergent, the series Í |vn | is convergent, and so, given ε, we can find n such that | vn+1 | + | vn+2 | + · · · + vn+p < ε/C, for all values of p. It follows therefore that we can find n such that | un+1 | + | un+2 | + · · · + un+p < ε, Í Í for all values of p, i.e. the series |un |is convergent. The series un is therefore absolutely convergent.

2.3 Convergence of an infinite series

19

Corollary 2.3.5 A series is absolutely convergent if the ratio of its nth term to the nth term of a series which is known to be absolutely convergent is less than some number independent of n. Example 2.3.3 Shew that the series 1 1 1 cos 2z + 2 cos 3z + 2 cos 4z + · · · 2 2 3 4 is absolutely convergent for all real values of z. When z is real, we have |cos nz| ≤ 1, and therefore cosn2nz ≤ n12 . The moduli of the terms of the given series are therefore less than, or at most equal to, the corresponding terms of the series 1 1 1 1 + 2 + 2 + 2 +··· , 1 2 3 4 which by §2.33 is absolutely convergent. The given series is therefore absolutely convergent. cos z +

Example 2.3.4 Shew that the series 1 1 1 1 + + + +··· , 12 (z − z1 ) 22 (z − z2 ) 32 (z − z3 ) 42 (z − z4 ) where zn = ein , (n = 1, 2, 3, . . .) is convergent for all values of z that are not on the circle |z| = 1. The geometric representation of complex numbers is helpful in discussing a question of this kind. Let values of the complex number z be represented on a plane; then the numbers z1, z2, z3, . . . will give a sequence of points which lie on the circumference of the circle whose centre is the origin and whose radius is unity; and it can be shewn that every point on the circle is a limit-point (§2.21) of the points zn . For these special values zn of z, the given series does not exist, since the denominator of the nth term vanishes when z = zn . For simplicity we do not discuss the series for any point z situated on the circumference of the circle of radius unity. Suppose now that |z| , 1. Then for all values of n, |z − zn | ≥ |{1 − |z|}| > c−1 , for some value of c; so the moduli of the terms of the given series are less than the corresponding terms of the series c c c c + + + +··· , 12 22 32 42 which is known to be absolutely convergent. The given series is therefore absolutely convergent for all values of z, except those which are on the circle |z| = 1. It is interesting to notice that the area in the z-plane over which the series converges is divided into two parts, between which there is no intercommunication, by the circle |z| = 1. Example 2.3.5 Shew that the series z z z z 2 sin + 4 sin + 8 sin + · · · + 2n sin n + · · · 3 9 27 3 converges absolutely for all values of z. Since lim 3n sin(z/3n ) = z (this is evident from results proved in the Appendix), we can n→∞

The Theory of Convergence

20

find a number k, independent of n (but depending on z), such that |3n sin(z/3n )| < k; and therefore n 2 z n . 2 sin n < k 3 3 ∞ n Í Since k 23 converges, the given series converges absolutely. n=1

2.35 Cauchy’s test for absolute convergence This appears in [120, p. 132–135]. ∞ Í If lim |un | 1/n < 1, then un converges absolutely. n→∞

n=1

For we can find m such that, when n ≥ m, |un | 1/n ≤ ρ < 1, where ρ is independent of ∞ Í n. Then, when n > m, |un | < ρn ; and since ρn converges, it follows from §2.34 that ∞ Í n=m+1

un (and therefore

n=m+1

∞ Í n=1

un ) converges absolutely.

Note If lim |un | 1/n > 1, un does not tend to zero, and, by §2.3,

∞ Í n=1

un does not converge.

2.36 D’Alembert’s ratio test for absolute convergence This appears in [159, pp. 171–182]. We shall now shew that a series u1 + u2 + u3 + u4 + · · · is absolutely convergent, provided that for all values of n greater than some fixed value r, the ratio | uun+1 | is less than ρ, where ρ is a positive number independent of n and less than unity. n For the terms of the series | ur+1 | + | ur+2 | + | ur+3 | + · · · are respectively less than the corresponding terms of the series | ur+1 | (1 + ρ + ρ2 + ρ3 + · · · ), which is absolutely convergent when ρ < 1; therefore

∞ Í n=r+1

un (and hence the given series)

is absolutely convergent. A particular case of this theorem is that if lim |un+1 /un | = ` < 1, the series is absolutely n→∞ convergent. For, by the definition of a limit, we can find r such that un+1 − ` < 1 (1 − `), when n > r, 2 un and then un+1 1 un < 2 (1 + `) < 1,

when

n > r.

2.3 Convergence of an infinite series

21

Note If lim |un+1 /un | > 1, then un does not tend to zero, and, by §2.3, converge.

∞ Í n=1

un does not

Example 2.3.6 If |c| < 1, shew that the series ∞ Õ

2

c n enz

n=1

converges absolutely for all values of z. Hint. For un+1 /un = c(n+1) −n ez = c2n+1 ez → 0, as n → ∞, if |c| < 1. 2

2

Example 2.3.7 Shew that the series z+

(a − b) 2 (a − b)(a − 2b) 3 (a − b)(a − 2b)(a − 3b) 4 z + z + z +··· 2! 3! 4!

converges absolutely if |z| < |b| −1 . Hint. For uun+1 z → −bz, as n → ∞; so the = a−nb n+1 n condition for absolute convergence is |bz| < 1, i.e. |z| < |b| −1 . ∞ Í nz n−1 Shew that the series converges absolutely if |z| < 1. Hint. z n −(1+n−1 ) n n=1 n For, when | z | < 1, z − (1 + n−1 )n ≥ (1 + n−1 )n − | z n | ≥ 1 + 1 + n−1 + · · · − 1 > 1, 2n so the moduli of the terms of the series are less than the corresponding terms of the series ∞ Í n z n−1 ; but this latter series is absolutely convergent, and so the given series converges

Example 2.3.8

n=1

absolutely.

un+1 =1 2.37 A general theorem on series for which lim n→∞ un It is obvious that if, for all values of n greater than some fixed value r, |un+1 | is greater than |un |, then the terms of the series do not tend to zero as n → ∞, and the series is therefore divergent. On the other hand, if uun+1 is less than some number which is itself less than unity n and independent of n (when n > r), we have shewn in §2.36 that the series is absolutely un+1 convergent. The critical case is that in which, as n increases, un tends to the value unity. In this case a further investigation is necessary. un+1 = 1 will be We shall now shew that a series u1 + u2 + u3 + · · · , in which lim n→∞ un absolutely convergent if a positive number c exists such that un+1 − 1 = −1 − c. lim n n→∞ un This is the second (D’Alembert’s theorem given in §2.36 being the first) of a hierarchy of theorems due to De Morgan. See Chrystal [146, p. xxvi] for an historical account of these theorems. Í Í For, compare the series |un | with the convergent series vn , where vn = An−1− 2 c 1

The Theory of Convergence

22

and A is a constant; we have −(1+ 12 c) 1 + 12 c 1 vn+1 n 1+ 12 c 1 = =1− +O 2 . = 1+ vn n+1 n n n o n − 1 → −1 − 12 c, and hence we can find m such that, when n > m, As n → ∞, n vvn+1 n un+1 vn+1 u n ≤ vn . By a suitable choice of the constant A, we can therefore secure that for all values of n we shall have |un | < vn . Í vn is convergent, |un | is also convergent, and so un is absolutely convergent. Corollary 2.3.6 If uun+1 = 1 + An1 + O n12 , where A1 is independent of n, then the series n is absolutely convergent if A1 < −1. ∞ n Í Í 1 r Example 2.3.9 Investigate the convergence of n exp −k , when r > k and when m As

Í

Í

n=1

r < k.

m=1

2.38 Convergence of the hypergeometric series The theorems which have been given may be illustrated by a discussion of the convergence of the hypergeometric series 1+

a(a + 1)b(b + 1) 2 a(a + 1)(a + 2)b(b + 1)(b + 2) 3 a·b z+ z + z +··· , 1·c 1 · 2 · c(c + 1) 1 · 2 · 3 · c(c + 1)(c + 2)

which is generally denoted (see Chapter 14) by F(a, b; c; z). If c is a negative integer, all the terms after the (1 − c)th have zero denominators; and if either a or b is a negative integer the series will terminate at the (1 − a)th or (1 − b)th term as the case may be. We shall suppose these cases set aside, so that a, b, and c are assumed not to be negative integers. In this series un+1 (a + n − 1)(b + n − 1) = z → |z| , un n(c + n − 1) as n → ∞. We see therefore, by §2.36, that the series is absolutely convergent when |z| < 1, and divergent when |z| > 1. When |z| = 1, we have 2 un+1 = 1 + a − 1 1 + b− 1 1 − c − 1 +O 1 un 2 n n n n a+b−c−1 1 = 1 + + O 2 . n n 2

The symbol O(1/n2 ) does not denote the same function of n throughout. See §2.11.

2.4 Effect of changing the order of the terms in a series

23

Let a, b, c be complex numbers, and let them be given in terms of their real and imaginary parts by the equations a = a 0 + ia 00, Then we have un+1 un

b = b0 + ib00,

c = c 0 + ic 00 .

0 0 0 00 00 00 = 1 + a + b − c − 1 + i(a + b − c ) + O 1 n n2 ( 2 00 2 ) 1/2 a 0 + b0 − c 0 − 1 a + b00 − c 00 1 = 1+ + +O 2 n n n a 0 + b0 − c 0 − 1 1 =1+ +O . n n2

By Corollary 2.3.6, a condition for absolute convergence is a 0 + b0 − c 0 < 0. Hence when |z| = 1, a sufficient condition for the absolute convergence of the hypergeometric series is that the real part of a + b − c shall be negative. The condition is also necessary. See Bromwich [102, pp. 202–204].

2.4 Effect of changing the order of the terms in a series In an ordinary sum the order of the terms is of no importance, for it can be varied without affecting the result of the addition. In an infinite series, however, this is no longer the case 3 , as will appear from the following example. Let 1 1 1 1 1 1 1 1 1 1 1 1 1 − + · · · and S = 1 − + − + − + · · · , T =1+ − + + − + + 3 2 5 7 4 9 11 6 2 3 4 5 6 and let Tn and Sn denote the sums of their first n terms. These infinite series are formed of the same terms, but the order of the terms is different, and so Tn and Sn are quite distinct functions of n. Let σn = 11 + 12 + · · · + n1 , so that S2n = σ2n − σn . Then 1 1 1 1 1 1 + +···+ − − −···− 1 3 4n − 1 2 4 2n = σ4n − 12 σ2n − 12 σn

T3n =

= (σ4n − σ2n ) + 21 (σ2n − σn )

= S4n + 21 S2n .

Making n → ∞, we see that T = S + 21 S; and so the derangement of the terms of S has altered its sum. 3

We say that the series

∞ Í n=1

vn consists of the terms of

∞ Í n=1

un in a different order if a law is given by which

corresponding to each positive integer p we can find one (and only one) integer q and vice versa, and vq is taken equal to u p . The result of this section was noticed by Dirichlet [173, p. 48]. See also Cauchy [125, p. 57].

24

The Theory of Convergence

Example 2.4.1 (Manning) If in the series 1 1 1 + − +··· 2 3 4 the order of the terms be altered, so that the ratio of the number of positive terms to the number of negative terms in the first n terms is ultimately a2 , shew that the sum of the series will become log(2a). 1−

2.41 The fundamental property of absolutely convergent series We shall shew that the sum of an absolutely convergent series is not affected by changing the order in which the terms occur. Let S = u1 + u2 + u3 + · · · be an absolutely convergent series, and let S 0 be a series formed by the same terms in a different order. Let ε be an arbitrary positive number, and let n be chosen so that 1 |un+1 | + |un+2 | + · · · + un+p < ε 2 for all values of p. Suppose that in order to obtain the first n terms of S we have to take m terms of S 0; then if k > m, Sk 0 − Sn + terms of S with suffices greater than n, so that Sk 0 − S = Sn − S + terms of S with suffices greater than n. Now the modulus of the sum of any number of terms of S with suffices greater than n does not exceed the sum of their moduli, and therefore is less than 21 ε. Therefore Sk0 − S < |Sn − S| + 21 ε. But 1 |Sn − S| ≤ lim |un+1 | + |un+2 | + · · · + un+p ≤ ε. p→∞ 2 0 Therefore given ε we can find m such that Sk − S < ε when k > m; therefore Sm0 → S, which is the required result. If a series of real terms converges, but not absolutely, and if Sp be the sum of the first p positive terms, and if σn be the sum of the first n negative terms, then Sp → ∞, σn → −∞; and lim(Sp + σn ) does not exist unless we are given some relation between p and n. It has, in fact, been shewn by Riemann [558, p. 221], that it is possible, by choosing a suitable relation, to make lim(Sp + σn ) equal to any given real number.

2.5 Double series A complete theory of double series, on which this account is based, is given by Pringsheim [541]. See further memoirs by that writer [542] and by London [442], and also Bromwich [102], which, in addition to an account of Pringsheim’s theory, contains many developments of the subject. Other important theorems are given by Bromwich [101].

2.5 Double series

25

Let um,n be a number determinate for all positive integral values of m and n; consider the array u1,1 u1,2 u1,3 · · · u2,1 u2,2 u2,3 · · · u3,1 u3,2 u3,3 · · · .. .. .. .. . . . . Let the sum of the terms inside the rectangle, formed by the first m rows of the first n columns of this array of terms, be denoted by Sm,n . If a number S exists such that, given any arbitrary positive number ε, it is possible to find integers m and n such that Sµ,ν − S < ε whenever both µ > m and ν > n, we say 4 that the double series of which the general element is uµ,ν converges to the sum S, and we write lim

µ→∞, ν→∞

Sµ,ν = S.

If the double series, of which the general element is |uµ,ν |, is convergent, we say that the given double series is absolutely convergent. Since uµ,ν = (Sµ,ν − Sµ,ν−1 ) − (Sµ−1,ν − Sµ−1,ν−1 ), it is easily seen that, if the double series is convergent, then lim

µ→∞,ν→∞

uµ,ν = 0.

Stolz’ necessary and sufficient condition for convergence. This condition, stated by Stolz [612], appears to have been first proved by Pringsheim. A condition for convergence which is obviously necessary (see §2.22) is that, given ε, we can find m and n such that Sµ+ρ,ν+σ − Sµ,ν < ε whenever µ > m and v > n and ρ, σ may take any of the values 0, 1, 2, . . .. The condition is also sufficient; for, suppose it satisfied; then, when µ > m + n, Sµ+ρ,µ+ρ − Sµ,µ < ε. Therefore, by §2.22, Sµ,µ has a limit S; and then making ρ and σ tend to infinity in such a way that µ + ρ = ν + σ, we see that S − Sµ,ν ≤ ε whenever µ > m and ν > n; that is to say, the double series converges. Corollary 2.5.1 An absolutely convergent double series is convergent. For if the double series converges absolutely and if tm,n be the sum of m rows of n columns of the series of moduli, then, given ε, we can find µ such that, when ρ > m > µ and q > n > µ, t p,q −tm,n < ε. But Sp,q − Sm,n ≤ t p,q − tm,n and so Sp,q − Sm,n < ε when p > m > µ, q > n > µ; and this is the condition that the double series should converge.

2.51 Methods of summing a double series ∞ Í These methods are due to Cauchy. Let us suppose that uµ,ν converges to the sum Sµ . ∞ Í

ν=1

Then Sµ is called the sum by rows of the double series; that is to say, the sum by rows is µ=1 ∞ ∞ ∞ ∞ Í Í Í Í uµ,ν . Similarly, the sum by columns is defined as uµ,ν . That these two sums µ=1 ν=1 4

This definition is practically due to Cauchy [120, p. 540].

ν=1 µ=1

The Theory of Convergence

26

are not necessarily the same is shewn by the example Sµ,ν = rows is −1, the sum by columns is +1; and S does not exist.

µ−ν , in which the sum by µ+ν

Theorem 2.5.2 (Pringsheim’s theorem) [541, p. 117]. If S exists and the sums by rows and columns exist, then each of these sums is equal to S. For since S exists, then we can find m such that Sµ,ν − S < ε, if µ > m, ν > m. And µ Í therefore, since lim Sµ,ν exists, lim Sµ,ν − S ≤ ε; that is to say, Sp − S ≤ ε when ν→∞

ν→∞

p=1

µ > m, and so (§2.22) the sum by rows converges to S. In like manner the sum by columns converges to S.

2.52 Absolutely convergent double series We can prove the analogue of §2.41 for double series, namely that if the terms of an absolutely convergent double series are taken in any order as a simple series, their sum tends to the same limit, provided that every term occurs in the summation. Let σµ,ν be the sum of the rectangle of µ rows and ν columns of the double series whose general element is uµ,ν ; and let the sum of this double series be σ. Then given ε we can find m and n such that σ − σµ,ν < ε whenever both µ > m and ν > n. Now suppose that it is necessary to take N terms of the deranged series (in the order in which the terms are taken) in order to include all the terms of SM+1,M+1 , and let the sum of these terms be t N . Then t N − SM+1,M+1 consists of a sum of terms of the type u p,q in which p > m, q > n whenever M > m and M > n; and therefore t N − SM+1,M+1 ≤ σ − σM+1,M+1 < 1 ε. 2 Also, S−SM+1, M+1 consists of terms u p,q in which p > m, q > n; therefore S − SM+1,M+1 ≤ σ − σM+1, M+1 < 21 ε; therefore |S − t N | < ε; and, corresponding to any given number ε, we can find N; and therefore t N → S. Example 2.5.1

Prove that in an absolutely convergent double series,

∞ Í n=1

um,n exists, and

thence that the sums by rows and columns respectively converge to S. Hint. Let the sum of µ rows of ν columns of the series of moduli be tµ,ν , and let t be the sum of the series of moduli. ∞ ∞ Í uµ,ν < t, and so Í uµ,ν converges; let its sum be bµ ; then Then ν=1

ν=1

|b1 | + |b2 | + · · · + bµ ≤ lim tµ,ν ≤ t, ν→∞

and so

∞ Í µ=1

bµ converges absolutely. Therefore the sum by rows of the double series exists, and

similarly the sum by columns exists; and the required result then follows from Pringsheim’s theorem.

2.5 Double series

27

Example 2.5.2 Shew from first principles that if the terms of an absolutely convergent double series be arranged in the order u1,1 + (u2,1 + u1,2 ) + (u3,1 + u2,2 + u1,3 ) + (u4,1 + · · · + u1,4 ) + · · · , this series converges to S.

2.53 Cauchy’s theorem on the multiplication of absolutely convergent series This appears in [120, Note VII]. We shall now shew that if two series S = u1 + u2 + u3 + · · ·

and T = v1 + v2 + v3 + · · ·

are absolutely convergent, then the series P = u1 v1 + u2 v1 + u1 v2 + · · · , formed by the products of their terms, written in any order, is absolutely convergent, and has for sum ST. Let Sn = u1 + u2 + · · · + un, Tn = v1 + v2 + · · · + vn . Then ST = lim Sn lim Tn = lim(SnTn ) by Example 2.2.2. Now SnTn =

u1 v1 + u1 v2 .. + .

+ u1 v n

+ u2 v1 + u2 v2 .. .. . . + u2 v n

+ ··· + ··· .. .. . . + ···

+ + .. .

un v1 un v2 .. .

+ u n vn .

But this double series is absolutely convergent; for if these terms are replaced by their moduli, the result is σn τn, where σn = |u1 | + |u2 | + · · · + |un | , τn = |v1 | + |v2 | + · · · + |vn | , and σn τn is known to have a limit. Therefore, by §2.52, if the elements of the double series, of which the general term is um vn , be taken in any order, their sum converges to ST. Example 2.5.3 Shew that the series obtained by multiplying the two series 1+

z z2 z3 z4 + + + +··· 2 22 23 24

and

1+

1 1 1 + + +··· , z z2 z3

and rearranging according to powers of z, converges so long as the representative point of z lies in the ring-shaped region bounded by the circles |z| = 1 and |z| = 2.

The Theory of Convergence

28

2.6 Power series The results of this section are due to Cauchy [120, Ch. IX]. A series of the type a0 + a1 z + a2 z2 + a3 z3 + · · · , in which the coefficients a0, a1, a2, a3, . . . are independent of z, is called a series proceeding according to ascending powers of z, or briefly a power series. We shall now shew that if a power series converges for any value z0 of z, it will be absolutely convergent for all values of z whose representative points are within a circle which passes through z0 and has its centre at the origin. ∞ Í For, if z be such a point, we have |z| < |z0 |. Now, since an z0n converges, an z0n must tend n=0

to zero as n → ∞, and so we can find M (independent of n) such that |an z0n | < M. Thus n z n |an z | < M . z0 Therefore every term in the series convergent geometric series

∞ Í n=0

∞ Í n=0

an z n is less than the corresponding term in the

M |z/z0 | n ; the series is therefore convergent; and so the

power series is absolutely convergent, as the series of moduli of its terms is a convergent series; the result stated is therefore established. ∞ Í Let lim |an | −1/n = r; then, from §2.35, an z n converges absolutely when |z| < r; if n=0

|z| > r, an z does not tend to zero and so n

∞ Í n=0

an z n diverges (§2.3). The circle |z| = r, which

includes all the values of z for which the power series a0 + a1 z + a2 z2 + a3 z3 + · · · converges, is called the circle of convergence of the series. The radius of the circle is called the radius of convergence. In practice there is usually a simpler way of finding r, derived from d’Alembert’s test (§2.36); r is lim(an /an+1 ) if this limit exists. A power series may converge for all values of the variable, as happens, for instance, in the case of the series 5 z3 z5 + −··· , z− 3! 5! which represents the function sin z; in this case the series converges over the whole z-plane. On the other hand, the radius of convergence of a power series may be zero; thus in the case of the series 1 + 1!z + 2!z2 + 3!z3 + 4!z4 + · · · we have |un+1 /un | = n|z|, which, for all values of n after some fixed value, is greater than 5

The series for e z , sin z, cos z and the fundamental properties of these functions and of log z will be assumed throughout. A brief account of the theory of the functions is given in the Appendix.

2.6 Power series

29

unity when z has any value different from zero. The series converges therefore only at the point z = 0, and the radius of its circle of convergence vanishes. A power series may or may not converge for points which are actually on the periphery of the circle; thus the series z z2 z3 z4 1+ s + s + s + s +··· , 1 2 3 4 whose radius of convergence is unity, converges or diverges at the point z = 1 according as s is greater or not greater than unity, as was seen in §2.33. Corollary 2.6.1 If (an ) be a sequence of positive terms such that lim(an+1 /an ) exists, this limit is equal to lim an1/n .

2.61 Convergence of series derived from a power series Let a0 + a1 z + a2 z2 + a3 z3 + a4 z4 + · · · be a power series, and consider the series a1 + 2a2 z + 3a3 z2 + 4a4 z3 + · · · , which is obtained by differentiating the power series term by term. We shall now shew that the derived series has the same circle of convergence as the original series. For let z be a point within the circle of convergence of the power series; and choose a positive number r1 , intermediate in value between |z| and r the radius of convergence. ∞ Í an r1n converges absolutely, its terms must tend to zero as n → ∞; Then, since the series n=0

and it must therefore be possible to find a positive number M, independent of n, such that |an | < Mr1−n for all values of n. ∞ Í n|an | |z| n−1 are less than the corresponding terms of the Then the terms of the series n=1

series ∞ M Õ n|z| n−1 . r1 n=1 r1n−1

But this series converges, by §2.36, since |z| < r1 . Therefore, by §2.34, the series ∞ Õ

n|an ||z| n−1

n=1

converges; that is, the series

∞ Í n=1

nan z n−1 converges absolutely for all points z situated within

the circle of convergence of the original series

∞ Í n=0

an z n . When |z| > r, an z n does not tend to

zero, and a fortiori nan z n does not tend to zero; and so the two series have the same circle of convergence. Corollary 2.6.2 The series

∞ Í n=0

a n n+1 z , n+1

obtained by integrating the original power series

term by term, has the same circle of convergence as

∞ Í n=0

an z n .

The Theory of Convergence

30

2.7 Infinite products We next consider a class of limits, known as infinite products. Let 1 + a1, 1 + a2, 1 + a3, . . . be a sequence such that none of its members vanish. If, as n → ∞, the product (1 + a1 )(1 + a2 )(1 + a3 ) · · · (1 + an ) Î

(which we denote by n ) tends to a definite limit other than zero, this limit is called the value of the infinite product Ö = (1 + a1 )(1 + a2 )(1 + a3 ) · · · , and the product is said to be convergent. (The convergence of the product in which an−1 = −1/n2 was investigated by Wallis as early as 1655.) It is almost obvious that a necessary Î Î condition for convergence is that lim an = 0, since lim n−1 = lim n , 0. The limit of the ∞ Î product is written (1 + an ). n=1

Now ! m m Õ Ö log(1 + an ) , (1 + an ) = exp n=1

(2.2)

n=1

and (see Appendix §A.2), exp lim um = lim (exp um ) if the former limit exists; hence m→∞ m→∞ ∞ Í log(1 + an ) should cona sufficient condition that the product should converge is that n=1

verge when the logarithms have their principal values. If this series of logarithms converges absolutely, the convergence of the product is said to be absolute. The condition for absolute convergence is given by the following theorem: in order that the infinite product (1 + a1 )(1 + a2 )(1 + a3 ) · · · may be absolutely convergent, it is necessary and sufficient that the series a1 + a2 + a3 + · · · should be absolutely convergent. Î For, by definition, is absolutely convergent or not according as the series log(1 + a1 ) + log(1 + a2 ) + log(1 + a3 ) + · · · is absolutely convergent or not. Now, since lim an = 0, we can find m such that, when n > m, |an | < 12 ; and then 2 3 −1 an log(1 + an ) − 1 = − an + an − an + · · · 2 3 4 1 1 1 < 2 + 3 +··· = . 2 2 2 And thence, when n > m, 1 log(1 + an ) 3 ≤ ≤ 2; 2 an

2.7 Infinite products

31

Í therefore, by the comparison theorem, the absolute convergence of log(1 + an ) entails that Í of an and conversely, provided that an , −1 for any value of n. This establishes the result. Note A discussion of the convergence of infinite products, in which the results are obtained without making use of the logarithmic function, is given by Pringsheim [539], and also by Bromwich [102, Ch. VI]. If, in a product, a finite number of factors vanish, and if, when these are suppressed, the resulting product converges, the original product is said to converge to zero. But such a ∞ Î product as (1 − n−1 ) is said to diverge to zero. n=2

Corollary 2.7.1 Since, if Sn → `, exp(Sn ) → exp `, it follows from §2.41 that the factors of an absolutely convergent product can be deranged without affecting the value of the product. Example 2.7.1 Shew that if

∞ Î

(1 + an ) converges, so does

n=1

∞ Í n=1

log (1 + an ), if the logarithms

have their principal values. Example 2.7.2 Shew that the infinite product 1 1 1 sin z sin 2 z sin 3 z sin 4 z · 1 · 1 · 1 ··· z z z z 2 3 4

is absolutely convergent for all values of z. Hint. For sin nz / form 1 −

λn , n2

z n

can be written in the ∞ Í λn where |λn | < k and k is independent of n; and the series is absolutely n2

convergent, as is seen on comparing it with

∞ Í n=1

n=1

1 . n2

The infinite product is therefore absolutely

convergent.

2.71 Some examples of infinite products Consider the infinite product z2 z2 z2 1− 2 1− 2 2 1− 2 2 ··· , π 2 π 3 π sin z . z ∞ Í In order to find whether it is absolutely convergent, we must consider the series

which, as will be proved later (§7.5), represents the function

n=1

z2 , n2 π 2

∞ 1 z2 Í or 2 ; this series is absolutely convergent, and so the product is absolutely convergent π n=1 n2 for all values of z. Now let the product be written in the form z z z z 1+ 1− 1+ ··· . 1− π π 2π 2π

The Theory of Convergence

32

The absolute convergence of this product depends on that of the series z z z z + −··· . − + − π π 2π 2π But this series is only conditionally convergent, since its series of moduli |z| |z| |z| |z| + + + +··· π π 2π 2π is divergent. In this form thereforethe infinite product is not absolutely convergent, and so, z if the order of the factors 1 ± nπ is deranged, there is a risk of altering the value of the product. Lastly, let the same product be written in the form n z πz o n z − πz o n z 2πz o n z − 2πz o 1− 1+ 1− 1+ ··· , e e e e π π 2π 2π in which each of the expressions z z ∓ mπ 1± e mπ is counted as a single factor of the infinite product. The absolute convergence of this product depends on that of the series of which the (2m − 1)th and (2m)th terms are z z ± mπ 1∓ e − 1. mπ But it is easy to verify that z 1 z ± mπ , e = 1+O 1∓ mπ m2 and so the absolute convergence of the series in question follows by comparison with the series 1 1 1 1 1 1 1+1+ 2 + 2 + 2 + 2 + 2 + 2 +··· . 2 2 3 3 4 4 The infinite product in this last form is therefore again absolutely convergent, the adjunction z of the factors e± n π having changed the convergence from conditional to absolute. This result is a particular case of the first part of the factor theorem of Weierstrass (§7.6). ∞ z Î z 1 − c+n Example 2.7.3 Prove that e n is absolutely convergent for all values of z, if c n=1

is a constant other than a negative integer. Hint. For the infinite product converges absolutely with the series ∞ n o Õ z nz 1− e −1 . c+n n=1 Now the general term of this series is zc − 21 z2 z z z2 1 1 1 1− 1+ + 2 +O 3 −1 = +O 3 =O 2 . 2 c+n n 2n n n n n ∞ ∞ z Í 1 Í z But converges, and so, by §2.34, 1 − c+n e n − 1 converges absolutely, and n2 n=1

n=1

therefore the product converges absolutely.

2.7 Infinite products

Example 2.7.4 Shew that

∞ Î n=2

1− 1−

1 −n n

33

z−n converges for all points z situated outside

a circle whose centre is the origin and radius unity. Hint. For the infinite product is absolutely convergent provided that the series −n ∞ Õ 1 1− z−n n n=2 −n = e, so the limit of the ratio of the (n + 1)th term is absolutely convergent. But lim 1 − n1 n→∞

of the series to the nth term is 1/z; there is therefore absolute convergence when |1/z| < 1, i.e. when |z| > 1. Example 2.7.5 Shew that 1 · 2 · 3 · · · (m − 1) mz (z + 1)(z + 2) · · · (z + m − 1) tends to a finite limit as m → ∞, unless z is a negative integer. Hint. For the expression can be written as a product of which the nth factor is s s n n+1 1 z −1 1 z(z − 1) = 1+ 1+ + O . = 1+ 2 z+n n n n 2n n3 This product is therefore absolutely convergent, provided the series ∞ Õ z(z − 1) 1 +O 3 2 2n n n=1 is absolutely convergent; and a comparison with the convergent series

∞ Í n=1

1 n2

shews that this

is the case. When z is a negative integer the expression does not exist because one of the factors in the denominator vanishes. Example 2.7.6 Prove that z z z z z z z 1− 1− 1+ 1− 1− 1+ · · · = e−z log 2/π sin z. π 2π π 3π 4π 2π For the given product z z z z z z lim z 1 − 1− 1+ ··· 1− 1− 1+ k→∞ π 2π π (2k − 1)π 2kπ kπ # " z 1 1 1 1 1 1 1 e π −1 − 2 + 1 − 3 − 4 + 2 − · · · − 2k−1 − 2k + k = lim z z z −z k→∞ z z z × z 1 − πz e π · 1 − 2π e 2π · · · 1 − 2kπ e 2k π · 1 + kπ e kπ 1 1 − πz (1− 12 + 13 −···+ 2k−1 − 2k ) = lim e k→∞ z πz z − πz z 2πz z − 2πz × z 1− e 1+ e 1− e 1+ e ··· , π π 2π 2π z since the product whose factors are 1 − er π is absolutely convergent, and so the order rπ of its factors can be altered. Since log 2 = 1 − 21 + 13 − 41 + 15 − · · ·, this shews that the given product is equal to e−z log 2/π sin z.

The Theory of Convergence

34

2.8 Infinite determinants Infinite series and infinite products are not by any means the only known cases of limiting processes which can lead to intelligible results. The researches of G. W. Hill in the Lunar Theory, reprinted in [306], brought into notice the possibilities of infinite determinants. Infinite determinants had previously occurred in the researches of Fürstenau [230] on the algebraic equation of the nth degree. Special types of infinite determinants (known as continuants) occur in the theory of infinite continued fractions; see Sylvester [617, p. 504] and [618, p. 249]. The actual investigation of the convergence is due not to Hill but to Poincaré [528]. We shall follow the exposition given by H. von Koch [643, p. 217]. Let Aik be defined for all integer values (positive and negative) of i, k, and denote by Dm = [Aik ]i,k=−m,...,+m the determinant formed of the numbers Aik (i, k = −m, . . . , +m); then if, as m → ∞, the expression Dm tends to a determinate limit D, we shall say that the infinite determinant [Aik ]i,k=−∞,...,+∞ is convergent and has the value D. If the limit D does not exist, the determinant in question will be said to be divergent. The elements Aii (where i takes all values) are said to form the principal diagonal of the determinant D; the elements Aik (where i is fixed and k takes all values) are said to form the row i; and the elements Aik (where k is fixed and i takes all values) are said to form the column k. Any element Aik is called a diagonal or a non-diagonal element, according as i = k or i , k. The element A0,0 is called the origin of the determinant.

2.81 Convergence of an infinite determinant We shall now shew that an infinite determinant converges, provided the product of the diagonal elements converges absolutely, and the sum of the non-diagonal elements converges absolutely. For let the diagonal elements of an infinite determinant D be denoted by 1+ aii , and let the non-diagonal elements be denoted by aik , (i , k), so that the determinant is .. .. .. .. .. . . . . . · · · 1 + a−1−1 a−10 a−11 · · · · · · a0−1 1 + a00 a01 · · · · · · a1−1 a10 1 + a11 · · · . . . .. .. .. .. .. . . Then, since the series

∞ Í

|aik | is convergent, the product

t,k=−∞

P=

∞ Ö i=−∞

is convergent.

1+

∞ Õ k=−∞

! |aik |

2.8 Infinite determinants

35

Now form the products Pm =

m Ö i=−m

1+

m Õ

! aik ,

Pm =

m Ö

1+

i=−m

k=−m

m Õ

! |aik | ;

k=−m

then if, in the expansion of Pm , certain terms are replaced by zero and certain other terms have their signs changed, we shall obtain Dm ; thus, to each term in the expansion of Dm there corresponds, in the expansion of P m , a term of equal or greater modulus. Now Dm+p − Dm represents the sum of those terms in the determinant Dm+p which vanish when the numbers aik {i, k = ± (m + 1) · · · ± (m + p)} are replaced by zero; and to each of these terms there corresponds a term of equal or greater modulus in p¯m+p − p¯m . Hence Dm+p − Dm ≤ P m+p − P m . Therefore, since Pm tends to a limit as m → ∞, so also Dm tends to a limit. This establishes the proposition.

2.82 The rearrangement theorem for convergent infinite determinants We shall now shew that a determinant, of the convergent form already considered, remains convergent when the elements of any row are replaced by any set of elements whose moduli are all less than some fixed positive number. Replace, for example, the elements . . . , A0,−m, . . . , A0, . . . , A0,m, . . . of the row through the origin by the elements . . . , µ−m, . . . , µ0, . . . , µm, . . . which satisfy the inequality | µr | < µ, where µ is a positive number; and let the new values of Dm and D be denoted by Dm0 , and D 0. Moreover, denote by P¯m0 and P¯ 0 the products obtained by suppressing in P¯m and P¯ the factor corresponding to the index zero; we see that no terms of Dm0 can have a greater modulus than the corresponding term in the expansion of µP¯m0 ; and consequently, reasoning as in the last article, we have 0 D ¯ ¯0 m+p − Dm0 < µ P m+p − µ Pm, which is sufficient to establish the result stated. Example 2.8.1 (von Koch) Shew that the necessary and sufficient condition for the absolute convergence of the infinite determinant 1 α1 0 0 · · · 0 β1 1 α2 0 · · · 0 lim 0 β2 1 α3 · · · 0 m→∞ . . . . . . .. .. .. .. .. .. 0 · · · 0 βm 1 is that the series α1 β1 + α2 β2 + α3 β3 + · · · shall be absolutely convergent.

The Theory of Convergence

36

2.9 Miscellaneous examples Example 2.1 Evaluate lim e−na nb , lim n−a log n when a > 0, b > 0. n→∞

n→∞

Example 2.2 (Trinity, 1904) Investigate the convergence of ∞ Õ 2n + 1 1 − n log . 2n − 1 n=1 Example 2.3 (Peterhouse, 1906) Investigate the convergence of 2 ∞ Õ 1 · 3 · · · 2n + 1 4n + 3 . . 2 · 4 · · · 2n 2n + 2 n=1 Example 2.4 Find the range of values of z for which the series 2 sin2 z − 4 sin4 z + 8 sin6 z − · · · + (−1)n+1 2n sin2n z + · · · is convergent. Example 2.5 (Simon) Shew that the series 1 1 1 1 − + − +··· z z+1 z+2 z+3 is conditionally convergent, except for certain exceptional values of z; but that the series 1 1 1 1 1 + +···+ − − −··· z z+1 z+p−1 z+p z+p+1 −

1 1 + +··· , z + 2p + q − 1 z + 2p + q

in which (p + q)-negative terms always follow p positive terms, is divergent. Example 2.6 (Trinity, 1908) Shew that 1 1 1 1 1 1 1 − + − − + − · · · = log 2. 2 4 3 6 8 5 2 Example 2.7 (Cesàro) Shew that the series 1−

1 1 1 1 + β + α + β +··· α 1 2 3 4 is convergent, although u2n+1 /u2n → ∞.

(1 < α < β)

Example 2.8 (Cesàro) Shew that the series α + β2 + α3 + β4 + · · · (with 0 < α < β < 1) is convergent although u2n /u2n−1 → ∞. Example 2.9 Shew that the series ∞ Õ n=1

nz n−1 (1 + n−1 )n − 1 (z n − 1) {z n − (1 + n−1 )n }

converges absolutely for all values of z, except the values a 2kπi/m e (a = 0, 1; k = 0, 1, . . . , m − 1; m = 1, 2, 3, . . .). z = 1+ m

2.9 Miscellaneous examples

37

Example 2.10 (de la Vallée Poussin [638]) Shew that, when δ > 1, ∞ ∞ Õ Õ 1 1 1 1 1 1 = + + − , nδ δ − 1 n=1 nδ δ − 1 (n + 1)δ−1 nδ−1 n=1 and shew that the series on the right converges when 0 < δ < 1. Example 2.11 In the series whose general term is un = q n−ν x 2 ν(ν+1) , (0 < q < 1 < x) where ν denotes the number of digits in the expression of n in the ordinary decimal scale of notation, shew that 1

lim u1/n n = q,

n→∞

and that the series is convergent, although lim un+1 /un = ∞. n→∞

Example 2.12 (Cesàro) Shew that the series q1 + q12 + q23 + q14 + q25 + q36 + q17 + · · · , where qn = q1+4/n , (0 < q < 1) is convergent, although the ratio of the (n + 1)th term to the nth is greater than unity when n is not a triangular number. Example 2.13 Shew that the series ∞ Õ e2nπix , (w + n)s n=0

where w is real, and where (w + n)s is understood to mean es log(w+n) , the logarithm being taken in its arithmetic sense, is convergent for all values of s, when Im x is positive, and is convergent for values of s whose real part is positive, when x is real and not an integer. Í Example 2.14 If un > 0, shew that if un converges, then lim (nun ) = 0, and that, if in n→∞

addition un ≥ un+1 , then lim nun = 0. n→∞

Example 2.15 (Trinity, 1904) If

am,n =

m, n > 0, n = 0, m , 0, m = 0, n , 0, n = m = 0.

m−n (m+n−1)! 2 m+n m! n! 2−m −2−n 0

shew that ∞ ∞ Õ Õ m=0

n=0

! am,n = −1,

∞ ∞ Õ Õ n=0

! am,n = 1.

m=0

Example 2.16 (Jacobi) By converting the series 1+

8q 16q2 24q3 + + +··· , 1 − q 1 + q2 1 − q3

The Theory of Convergence

38

(in which |q| < 1), into a double series, shew that it is equal to 8q2 8q3 8q + + +··· . 2 2 2 (1 − q) (1 + q ) (1 − q3 )2 ∞ Î z2 Example 2.17 (Math. Trip. 1904) Assuming that sin z = z 1 − 2 2 , shew that if r π r=1 m → ∞ and n → ∞ in such a way that lim m/n = k, where k is finite, then 1+

m Ö 0

lim

1+

r=−n

z sin z = k z/π , rπ z

the prime indicating that the factor for which r = 0 is omitted. Example 2.18 (Math. Trip. 1906) If u0 = u1 = u2 = 0, and if, when n > 1, 1 u2n−1 = − √ , n then

∞ Î n=0

(1 + un ) converges, though

∞ Í n=0

1 1 1 u2n = √ + + √ , n n n n

un and

∞ Í n=0

u2n are divergent.

Example 2.19 Prove that ( ∞ Ö n=1

k+1 k−m m Õ n z z nk 1− exp n m m=1

!) ,

where k is any positive integer, converges absolutely for all values of z. Example 2.20 (Cauchy) ∞ Î

If

∞ Í n=1

an be a conditionally convergent series of real terms, then

(1 + an ) converges (but not absolutely) or diverges to zero according as

n=1

∞ Í n=1

an2 converges

or diverges. Example 2.21 (Hill; see §19.42) Let the infinite determinant . . . · · · · · · ∆(c) = · · · · · · · · · . ..

∞ Í n=1

θ n be an absolutely convergent series. Shew that

.. .

.. .

.. .

.. .

.. .

(c−4)2 −θ0 42 −θ0

−θ1 42 −θ0

−θ2 42 −θ0

−θ3 42 −θ0

−θ3 42 −θ0

−θ1 22 −θ0

(c−2)2 −θ0 22 −θ0

−θ1 22 −θ0

−θ2 22 −θ0

−θ3 22 −θ0

−θ2 02 −θ0

−θ1 02 −θ0

c 2 −θ0 02 −θ0

−θ1 02 −θ0

−θ2 02 −θ0

−θ3 22 −θ0 −θ4 42 −θ0

−θ2 22 −θ0 −θ3 42 −θ0

−θ1 22 −θ0 −θ2 42 −θ0

(c+2)2 −θ0 22 −θ0 −θ1 42 −θ0

−θ1 22 −θ0 (c+4)2 −θ0 42 −θ0

.. .

.. .

.. .

.. .

.. .

.. . · · · · · · · · · · · · · · · .. .

2.9 Miscellaneous examples

converges; and shew that the equation ∆(c) = 0 is equivalent to the equation p sin2 (πc/2) = ∆(0) sin2 π θ 0 /2 .

39

(2.3)

3 Continuous Functions and Uniform Convergence

3.1 The dependence of one complex number on another The problems with which Analysis is mainly occupied relate to the dependence of one complex number on another. If z and ζ are two complex numbers, so connected that, if z is given any one of a certain set of values, corresponding values of ζ can be determined, e.g. if ζ is the square of z, or if ζ = 1 when z is real and ζ = 0 for all other values of z, then ζ is said to be a function of z. This dependence must not be confused with the most important case of it, which will be explained later under the title of analytic functionality. If ζ is a real function of a real variable z, then the relation between ζ and z, which may be written ζ = f (z), can be visualised by a curve in a plane, namely the locus of a point whose coordinates referred to rectangular axes in the plane are (z, ζ). No such simple and convenient geometrical method can be found for visualising an equation ζ = f (z), considered as defining the dependence of one complex number ζ = ξ + iη on another complex number z = x + iy. A representation strictly analogous to the one already given for real variables would require four-dimensional space, since the number of variables ξ, η, x, y is now four. One suggestion (made by Lie and Weierstrass) is to use a doubly-manifold system of lines in the quadruply-manifold totality of lines in three-dimensional space. Another suggestion is to represent ξ and η separately by means of surfaces ξ = ξ(x, y), η = η(x, y). A third suggestion, due to Heffter [284], is to write ζ = reiθ , then draw the surface r = r(x, y), which may be called the modular-surface of the function, and on it to express the values of θ by surface-markings. It might be possible to modify this suggestion in various ways by representing θ by curves drawn on the surface r = r(x, y).

3.2 Continuity of functions of real variables The reader will have a general idea (derived from the graphical representation of functions of a real variable) as to what is meant by continuity. We now have to give a precise definition which shall embody this vague idea. Let f (x) be a function of x defined when a ≤ x ≤ b. Let x1 be such that a ≤ x1 ≤ b. If there exists a number ` such that, corresponding to an arbitrary positive number ε, we can find a positive number η such that | f (x) − `| < ε, whenever |x − x1 | < η, x , x1 , and a ≤ x ≤ b, then ` is called the limit of f (x) as x → x1 . It may happen that we can find a number `+ (even when ` does not exist) such that 40

3.2 Continuity of functions of real variables

41

| f (x) − `+ | < ε when x1 < x < x1 + η. We call `+ the limit of f (x) when x approaches x1 from the right and denote it by f (x1 + 0); in a similar manner we define f (x1 − 0) if it exists. If f (x1 + 0), f (x1 ), f (x1 − 0) all exist and are equal, we say that f (x) is continuous at x1 ; so that if f (x) is continuous at x1 , then, given ε, we can find η such that | f (x) − f (x1 )| < ε, whenever |x − x1 | < η, and a ≤ x ≤ b. If `+ and `− exist but are unequal, f (x) is said to have an ordinary discontinuity 1 at x1 ; and if `+ = `− , f (x1 ), f (x) is said to have a removable discontinuity at x1 . If f (x) is a complex function of a real variable, and if f (x) = g(x) + ih(x) where g(x) and h(x) are real, the continuity of f (x) at x1 implies the continuity of g(x) and of h(x). For when | f (x) − f (x1 )| < ε, then |g(x) − g(x1 )| < ε and |h(x) − h(x1 )| < ε; and the result stated is obvious. Example 3.2.1 From Examples 2.2.1 and 2.2.2 deduce that if f (x) and φ(x) are continuous at x1 , so are f (x) ± φ(x), f (x) × φ(x) and, if φ(x1 ) , 0, f (x)/φ(x). The popular idea of continuity, so far as it relates to a real variable f (x) depending on another real variable x, is somewhat different from that just considered, and may perhaps best be expressed by the statement “The function f (x) is said to depend continuously on x if, as x passes through the set of all values intermediate between any two adjacent values x1 and x2, f (x) passes through the set of all values intermediate between the corresponding values f (x1 ) and f (x2 ).” The question thus arises, how far this popular definition is equivalent to the precise definition given above. Cauchy shewed that if a real function f (x), of a real variable x, satisfies the precise definition, then it also satisfies what we have called the popular definition; this result will be proved in §3.63. But the converse is not true, as was shewn by Darboux. This fact may be illustrated by the following example due to Mansion [454]. π ; and let f (0) = 0. It can Between x = −1 and x = +1 (except at x = 0), let f (x) = sin 2x then be proved that f (x) depends continuously on x near x = 0, in the sense of the popular definition, but is not continuous at x = 0 in the sense of the precise definition. Example 3.2.2 If f (x) be defined and be an increasing function in the range (a, b), the limits f (x ± 0) exist at all points in the interior of the range. Hint. If f (x) be an increasing function, a section of rational numbers can be found such that, if a, A be any members of its L-class and its R-class, a < f (x + h) for every positive value of h and A ≥ f (x + h) for some positive value of h. The number defined by this section is f (x + 0).

3.21 Simple curves. Continua Let x and y be two real functions of a real variable t which are continuous for every value of t such that a ≤ t ≤ b. We denote the dependence of x and y on t by writing x = x(t),

y = y(t)

(a ≤ t ≤ b)

The functions x(t), y(t) are supposed to be such that they do not assume the same pair of 1

If a function is said to have ordinary discontinuities at certain points of an interval it is implied that it is continuous at all other points of the interval.

42

Continuous Functions and Uniform Convergence

values for any two different values of t in the range a < t < b. Then the set of points with coordinates (x, y) corresponding to these values of t is called a simple curve. If x(a) = x(b),

y(a) = y(b),

the simple curve is said to be closed. Example 3.2.3 The circle x 2 + y 2 = 1 is a simple closed curve; for we may write 2 x = cos t,

y = sin t,

(0 ≤ t ≤ 2π).

A two-dimensional continuum is a set of points in a plane possessing the following two properties: (i) If (x, y) be the Cartesian coordinates of any point of it, a positive number δ (depending on x and y) can be found such that every point whose distance from (x, y) is less than δ belongs to the set. (ii) Any two points of the set can be joined by a simple curve consisting entirely of points of the set. Example 3.2.4 The points for which x 2 + y 2 < 1 form a continuum. For if P be any point inside the unit circle such that OP = r < 1, we may take δ = 1 − r; and any two points inside the circle may be joined by a straight line lying wholly inside the circle. The following two theorems will be assumed in this work; simple cases of them appear obvious from geometrical intuitions and, generally, theorems of a similar nature will be taken for granted, as formal proofs are usually extremely long and difficult. Formal proofs will be found in Watson [650]. (I) A simple closed curve divides the plane into two continua (the interior and the exterior). (II) If P be a point on the curve and Q be a point not on the curve, the angle between QP and Ox increases by ±2π or by zero, as P describes the curve, according as Q is an interior point or an exterior point. If the increase is +2π, P is said to describe the curve counter-clockwise. A continuum formed by the interior of a simple curve is sometimes called an open twodimensional region, or briefly an open region, and the curve is called its boundary; such a continuum with its boundary is then called a closed two-dimensional region, or briefly a closed region or domain. A simple curve is sometimes called a closed one-dimensional region; a simple curve with its end-points omitted is then called an open one-dimensional region.

3.22 Continuous functions of complex variables Let f (z) be a function of z defined at all points of a closed region (one- or two-dimensional) in the Argand diagram, and let z1 be a point of the region. 2

For a proof that the sine and cosine are continuous functions, see the Appendix, §A.41.

3.3 Series of variable terms. Uniformity of convergence

43

Then f (z) is said to be continuous at z1 , if given any positive number ε, we can find a corresponding positive number η such that | f (z) − f (z1 )| < ε, whenever |z − z1 | < η and z is a point of the region.

3.3 Series of variable terms. Uniformity of convergence Consider the series x2 +

x2 x2 x2 + + · · · + +··· . 1 + x 2 (1 + x 2 )2 (1 + x 2 )n

This series converges absolutely (§2.33) for all real values of x. If Sn (x) be the sum of n terms, then 1 Sn (x) = 1 + x 2 − ; (1 + x 2 )n−1 and so lim Sn (x) = 1 + x 2 ; (x , 0), but Sn (0) = 0, and therefore lim Sn (0) = 0. n→∞ n→∞ Consequently, although the series is an absolutely convergent series of continuous functions of x, the sum is a discontinuous function of x. We naturally enquire the reason of this rather remarkable phenomenon, which was investigated in 1841–1848 by Stokes [608], Seidel [590] and Weierstrass [662, pp. 67, 75], who shewed that it cannot occur except in connexion with another phenomenon, that of non-uniform convergence, which will now be explained. Let the functions u1 (z), u2 (z), . . . be defined at all points of a closed region of the Argand diagram. Let Sn (z) = u1 (z) + u2 (z) + · · · + un (z). The condition that the series

∞ Í n=1

un (z) should converge for any particular value of z is that,

given ε, a number n should exist such that |Sn+p (z) − Sn (z)| < ε for all positive values of p, the value of n of course depending on ε. Let n have the smallest integer value for which the condition is satisfied. This integer will in general depend on the particular value of z which has been selected for consideration. We denote this dependence by writing n(z) in place of n. Now it may happen that we can find a number N, independent of z, such that n(z) < N for all values of z in the region under consideration. If this number N exists, the series is said to converge uniformly throughout the region. If no such number N exists, the convergence is said to be non-uniform. The reader who is unacquainted with the concept of uniformity of convergence will find it made much clearer by consulting Bromwich [102], where an illuminating account of Osgood’s graphical investigation is given. Uniformity of convergence is thus a property depending on a whole set of values of z, whereas previously we have considered the convergence of a series for various particular

Continuous Functions and Uniform Convergence

44

values of z, the convergence for each value being considered without reference to the other values. We define the phrase ‘uniformity of convergence near a point z’ to mean that there is a definite positive number δ such that the series converges uniformly in the domain common to the circle |z − z1 | ≤ δ and the region in which the series converges.

3.31 On the condition for uniformity of convergence This section shews that it is indifferent whether uniformity of convergence is defined by means of the partial remainder Rn,p (z) or by Rn (z). Writers differ in the definition taken as fundamental. If Rn,p (z) = un+1 (z) + un+2 (z) + · · · + un+p (z), we have seen that the necessary and sufficient ∞ Í condition that un (z) should converge uniformly in a region is that, given any positive n=1

number ε, it should be possible to choose N independent of z (but depending on ε) such that |RN ,p (z)| < ε for all positive integral values of p. If the condition is satisfied, by §2.22, Sn (z) tends to a limit, S(z), say for each value of z under consideration; and then, since ε is independent of p, | lim RN ,p (z)| ≤ ε, p→∞

and therefore, when n > N, S(z) − Sn (z) = lim RN ,p (z) − RN ,n−N (z), p→∞

and so |S(z) − Sn (z)| < 2ε. Thus (writing ε/2 for ε) a necessary condition for uniformity of convergence is that |S(z) − Sn (z)| < ε, whenever n > N and N is independent of z; the condition is also sufficient; for if it is satisfied it follows as in §2.22 (I) that |RN ,p (z)| < 2ε, which, by definition, is the condition for uniformity. Example 3.3.1 Shew that, if x be real, the sum of the series x x x + +···+ +··· 1(x + 1) (x + 1)(2x + 1) {(n − 1)x + 1}{nx + 1} is discontinuous at x = 0 and the series is non-uniformly convergent near x = 0. 1 Solution The sum of the first n terms is easily seen to be 1 − nx+1 ; so when x = 0 the sum 1 is 0; when x , 0, the sum is 1. The value of Rn (x) = S(x) − Sn (x) is nx+1 if x , 0; so when x is small, say x = one-hundred-millionth, the remainder after a million terms is 1 1+1 or 100

1 1 − 101 , so the first million terms of the series do not contribute one per cent of the sum. And 1 in general, to make nx+1 < ε, it is necessary to take n > x1 ε1 − 1 . Corresponding to a given ε, no number N exists, independent of x, such that n < N for all values of x in any interval including x = 0; for by taking x sufficiently small we can make n greater than any number N which is independent of x. There is therefore non-uniform convergence near x = 0.

3.3 Series of variable terms. Uniformity of convergence

45

Example 3.3.2 Discuss the series ∞ Õ n=1

in which x is real. The nth term can be written

x{n(n + 1)x 2 − 1} , {1 + n2 x 2 }{1 + (n + 1)2 x 2 }

nx 1+n2 x 2

−

(n+1)x , 1+(n+1)2 x 2

Rn (x) =

so S(x) =

x , 1+x 2

and

(n + 1)x . 1 + (n + 1)2 x 2

Note In this example the sum of the series is not discontinuous at x = 0. But (taking ε < 21 , and x , 0), |Rn (x)| < ε if ε −1 (n + 1)|x| < 1 + (n + 1)2 x 2 ; i.e. if √ √ 1 1 n + 1 > {ε −1 + ε −2 − 4}|x| −1 , or n + 1 < {ε −1 − ε −2 − 4}|x| −1 . 2 2 Now it is not the case that the second inequality is satisfied for all values of n greater than a certain value and for all values of x; and the first inequality gives a value of n(x) which tends to infinity as x → 0; so that, corresponding to any interval containing the point x = 0, there is no number N independent of x. The series, therefore, is non-uniformly convergent near x = 0. The reader will observe that n(x) is discontinuous at x = 0; for n(x) → ∞ as x → 0, but n(0) = 0.

3.32 Connexion of discontinuity with non-uniform convergence We shall now shew that if a series of continuous functions of z is uniformly convergent for all values of z in a given closed domain, the sum is a continuous function of z at all points of the domain. For let the series be f (z) = u1 (z) + u2 (z) + · · · + un (z) + · · · = Sn (z) + Rn (z), where Rn (z) is the remainder after n terms. Since the series is uniformly convergent, given any positive number ε, we can find a corresponding integer n independent of z, such that |Rn (z)| < 31 ε for all values of z within the domain. Now n and ε being thus fixed, we can, on account of the continuity of Sn (z), find a positive number η such that |Sn (z) − Sn (z 0)| < 13 ε, whenever |z − z 0 | < η. We have then | f (z) − f (z 0)| = |Sn (z) − Sn (z 0) + Rn (z) − Rn (z 0)| < |Sn (z) − Sn (z 0)| + |Rn (z)| + |Rn (z 0)| < ε, which is the condition for continuity at z. Example 3.3.3 Shew that near x = 0 the series u1 (x) + u2 (x) + u3 (x) + · · · , where u1 (x) = x,

un (x) = x 2n−1 − x 2n−3 , 1

1

and real values of x are concerned, is discontinuous and non-uniformly convergent.

Continuous Functions and Uniform Convergence

46

In this example it is convenient to take a slightly different form of the test; we shall shew that, given an arbitrarily small number ε, it is possible to choose values of x, as small as we please, depending on n in such a way that |Rn (x)| is not less than ε for any value of n, no matter how large. The reader will easily see that the existence of such values of x is inconsistent with the condition for uniformity of convergence. The value of Sn (x) is x 1/2n−1 ; as n tends to infinity, Sn (x) tends to +1, 0, or −1, according as x is positive, zero, or negative. The series is therefore absolutely convergent for all values of x, and has a discontinuity at x = 0. In this series Rn (x) = 1 − x 1/(2n−1) , (x > 0); however great n may be, by taking x = e−(2n−1) (this value of x satisfies the condition |x| < δ whenever 2n − 1 > log δ−1 ), we can cause this remainder to take the value 1 − e−1, which is not arbitrarily small. The series is therefore non-uniformly convergent near x = 0. Example 3.3.4 Shew that near z = 0 the series ∞ Õ n=1

−2z(1 + z)n−1 {1 + (1 + z)n−1 }{1 + (1 + z)n }

is non-uniformly convergent and its sum is discontinuous. The nth term can be written 1 − (1 + z)n 1 − (1 + z)n−1 − , 1 + (1 + z)n 1 + (1 + z)n−1 1 − (1 + z)n . Thus, considering real values of z greater 1 + (1 + z)n than −1, it is seen that the sum to infinity is +1, 0, or −1, according as z is negative, zero, or positive. There is thus a discontinuity at z = 0. This discontinuity is explained by the fact that the series is non-uniformly convergent near z = 0; for the remainder after n terms in 1 −2 , and, however great n may be, by taking z = , the series when z is positive is 1 + (1 + z)n n 2 this can be made numerically greater than , which is not arbitrarily small. The series is 1+e therefore non-uniformly convergent near z = 0. so the sum of the first n terms is

3.33 The distinction between absolute and uniform convergence The uniform convergence of a series in a domain does not necessitate its absolute convergence ∞ Í z2 converges absolutely, at any points of the domain, nor conversely. Thus the series (1+z 2 ) n n=1 ∞ Í

but (near z = 0) not uniformly; while in the case of the series is

∞ Í n=1

1 , |z 2 +n |

n=1

(−1) n−1 , z 2 +n

the series of moduli

which is divergent, so the series is only conditionally convergent; but for all

real values of z, the terms of the series are alternately positive and negative and numerically decreasing, so the sum of the series lies between the sum of its first n terms and of its first (n + 1) terms, and so the remainder after n terms is numerically less than the nth term. Thus we only need take a finite number (independent of z) of terms in order to ensure that for

3.3 Series of variable terms. Uniformity of convergence

47

all real values of z the remainder is less than any assigned number ε, and so the series is uniformly convergent. Absolutely convergent series behave like series with a finite number of terms in that we can multiply them together and transpose their terms. Uniformly convergent series behave like series with a finite number of terms in that they are continuous if each term in the series is continuous and (as we shall see) the series can then be integrated term by term.

3.34 A condition, due to Weierstrass, for uniform convergence This appears in [661, p. 70]. The test given by this condition is usually described (e.g. by Osgood, [512]) as the M-test for uniform convergence. A sufficient, though not necessary, condition for the uniform convergence of a series may be enunciated as follows: If, for all values of z within a domain, the moduli of the terms of a series S = u1 (z) + u2 (z) + u3 (z) + · · · are respectively less than the corresponding terms in a convergent series of positive terms T = M1 + M2 + M3 + · · · , where Mn is independent of z, then the series S is uniformly convergent in this region. This follows from the fact that, the series T being convergent, it is always possible to choose n so that the remainder after the first n terms of T, and therefore the modulus of the remainder after the first n terms of S, is less than an assigned positive number ε; and since the value of n thus found is independent of z, it follows (§3.31) that the series S is uniformly convergent; by §2.34, the series S also converges absolutely. Example 3.3.5 The series 1 1 cos2 z + 2 cos3 z + · · · 22 3 is uniformly convergent for all real values of z, because the moduli of its terms are not greater than the corresponding terms of the convergent series cos z +

1+

1 1 + 2 +··· , 2 2 3

whose terms are positive constants.

3.341 Uniformity of convergence of infinite products The definition is, effectively, that given by Osgood [513, p. 462]. The condition here given for uniformity of convergence is also established in that work. ∞ Î (1 + un (z)) is said to converge uniformly in a domain of values A convergent product n=1

of z if, given ε, we can find m independent of z such that m+p m Ö Ö (1 + un (z)) − (1 + un (z)) < ε n=1 n=1 for all positive integral values of p. The only condition for uniformity of convergence which will be used in this work is that ∞ Í the product converges uniformly if |un (z)| < Mn where Mn is independent of z and Mn n=1 converges.

Continuous Functions and Uniform Convergence

48

To prove the validity of the condition we observe that

∞ Î n=1

(1 + Mn ) converges (§2.7), and

so we can choose m such that m+p Ö

(1 + Mn ) −

m Ö

(1 + Mn ) < ε;

n=1

n=1

and then we have m+p " m+p # m m Ö Ö Ö Ö (1 (1 (1 (1 + u (z)) − + u (z)) = + u (z)) + u (z)) − 1 n n n n n=1 n=1 n=1 n=m+1 " # m+p m Ö Ö (1 + Mn ) (1 + Mn ) − 1 ≤ n=1

n=m+1

< ε, and the choice of m is independent of z.

3.35 Hardy’s tests for uniform convergence These results, which are generalizations of Abel’s theorem (§3.71, below), though well known, do not appear to have been published before 1907 [276]. From their resemblance to the tests of Dirichlet and Abel for convergence, Bromwich proposes to call them Dirichlet’s and Abel’s tests respectively. p Í The reader will see, from §2.31, that if, in a given domain, an (z) ≤ k where an (z) n=1 is real and k is finite and independent of p and z, and if fn (z) ≥ fn+1 (z) and fn (z) → 0 ∞ Í an (z) fn (z) converges uniformly. uniformly as n → ∞, then n=1

Also that if k ≥ un (z) ≥ un+1 (z) ≥ 0, where k is independent of z and uniformly, then

∞ Í n=1

∞ Í n=1

an (z) converges

an (z)un (z) converges uniformly. Hint. To prove the latter, observe that m

can be found such that am+1 (z), am+1 (z) + am+2 (z), . . . , am+1 (z) + am+2 (z) + · · · + am+p (z) are numerically less than ε/k; and therefore (§2.301) m+p Õ an (z)un (z) < εum+1 (z)/k < ε, n=m+1 and the choice of ε and m is independent of z. Example 3.3.6 Shew that, if δ > 0, the series ∞ Õ cos nθ n=1

n

,

∞ Õ sin nθ n=1

n

3.4 Discussion of a particular double series

49

converge uniformly in the range δ ≤ θ ≤ 2π − δ. Obtain the corresponding result for the series ∞ ∞ Õ Õ (−1)n sin nθ (−1)n cos nθ , , n n n=1 n=1 by writing θ + π for θ. Example 3.3.7 (Hardy) If, when a ≤ x ≤ b, |ωn (x)| < k1 and where k1 , k2 are independent of n and x, and if ∞ Í

∞ Í n=1

∞ Í n=1

|ωn+1 (x) − ωn (x)| < k2 ,

an is a convergent series indepen-

an ωn (x) converges uniformly when a ≤ x ≤ b. Hint. For we can m+p Í an < ε, and then, by Corollary 2.3.2, we choose m, independent of x, such that n=m+1 m+p Í have an ωn (x) < (k1 + k 2 )ε. n=m+1

dent of x, then

n=1

3.4 Discussion of a particular double series Let ω1 and ω2 be any constants whose ratio is not purely real; and let α be positive. The Í 1 series , in which the summation extends over all positive and negative α m,n (z + 2mω1 + 2nω2 ) integral and zero values of m and n, is of great importance in the theory of Elliptic Functions. At each of the points z = −2mω1 − 2nω2 the series does not exist. It can be shewn that the series converges absolutely for all other values of z if α > 2, and the convergence is uniform for those values of z such that |z + 2mω1 + 2nω2 | ≥ δ for all integral values of m and n, where δ is an arbitrary positive number. Í Let 0 denote a summation for all integral values of m and n, the term for which m = n = 0 being omitted. Now, if m and n are not both zero, and if |z + 2mω1 + 2nω2 | ≥ δ > 0 for all integral values of m and n, then we can find a positive number C, depending on δ but not on z, such that 1 1 < C (2mω1 + 2nω2 )α . (z + 2mω1 + 2nω2 )α Consequently, by §3.34, the given series is absolutely and uniformly convergent in the Í 1 domain considered if 0 converges. (The reader will easily define uniformity |mω1 + nω2 | α of convergence of double series (see §3.5).) To discuss the convergence of the latter series, let ω1 = α1 + i β1,

ω2 = α2 + i β2,

where α1 , α2 , β1 , β2 , are real. Since ω2 /ω1 is not real, α1 β2 − α2 β1 , 0. Then the series is Õ0 1 . {(α1 m + α2 n)2 + (β1 m + β2 n)2 }α/2

Continuous Functions and Uniform Convergence

50

This converges (Corollary 2.5.1) if the series Õ0 1 S= 2 (m + n2 )α/2 converges; for the quotient of corresponding terms is α/2 (α1 + α2 µ)2 + (β1 + β2 µ)2 , 1 + µ2 where µ = n/m. This expression, qua function of a continuous real variable µ, can be proved to have a positive minimum 3 (not zero) since α1 β2 − α2 β1 , 0; and so the quotient is always greater than a positive number K (independent of µ). We have therefore only to study the convergence of the series S. Let Sp,q =

q p Õ Õ 0 m=−p n=−q

∞ ∞ Õ Õ 0 1 1 ≤ 4 . 2 2 α/2 2 + n2 )α/2 (m + n ) (m m=0 m=0

Separating Sp,q into the terms for which m = n, m > n, and m < n, respectively, we have p m−1

p

q

n−1

ÕÕ ÕÕ Õ 1 1 1 1 + + . Sp,q = 2 α/2 2 2 α/2 2 4 (2m ) (m + n ) (m + n2 )α/2 m=1 n=0 n=1 m=0 m=1 But m−1 Õ n=0

(m2

1 m 1 < 2 α/2 = α−1 ; 2 α/2 m +n ) (m )

(3.1)

therefore ∞ ∞ ∞ Õ Õ Õ 1 1 1 1 + + . S≤ α−1 α−1 α/2 mα 4 m n 2 m=1 n=1 m=1

(3.2)

But these last series are known to be convergent if α − 1 > 1. So the series S is convergent if α > 2. The original series is therefore absolutely and uniformly convergent, when α > 2, for the specified range of values of z. Example 3.4.1 (Eisenstein [193]) Prove that the series Õ 1 (m12 + m22 + · · · + mr2 )µ

,

in which the summation extends over all positive and negative integral values and zero values of m1, m2, . . . , mr , except the set of simultaneous zero values, is absolutely convergent if µ > 12 r. 3

The reader will find no difficulty in verifying this statement; the minimum value in question is given by 1/2 1/2 K 2/α = 12 [α1 2 + α2 2 + β1 2 + β2 2 − (α1 − β2 )3 + (α2 + β1 )2 (α1 + β2 )2 + (α2 − β1 )2 ]. .

3.5 The concept of uniformity

51

3.5 The concept of uniformity There are processes other than that of summing a series in which the idea of uniformity is of importance. Let ε be an arbitrary positive number; and let f (z, ζ) be a function of two variables z and ζ, which for each point z of a closed region, satisfies the inequality | f (z, ζ)| < ε when ζ is given any one of a certain set of values which will be denoted by (ζz ); the particular set of values of course depends on the particular value of z under consideration. If a set (ζ)0 can be found such that every member of the set (ζ)0 is a member of all the sets (ζz ), the function f (z, ζ) is said to satisfy the inequality uniformly for all points z of the region. And if a function φ(z) possesses some property, for every positive value of ε, in virtue of the inequality | f (z, ζ)| < ε, φ(z) is then said to possess the property uniformly. In addition to the uniformity of convergence of series and products, we shall have to consider uniformity of convergence of integrals and also uniformity of continuity; thus a series is uniformly convergent when |Rn (z)| < ε, ζ(= n) assuming integer values independent of z only. Further, a function f (z) is continuous in a closed region if, given ε, we can find a positive number ηs such that | f (z + ζs ) − f (z)| < ε whenever 0 < |ζs | < ηs and z + ζ is a point of the region. The function will be uniformly continuous if we can find a positive number η independent of z, such that η < ηs and | f (z + ζ) − f (z)| < ε whenever 0 < |ζ | < η and z + ζ is a point of the region (in this case the set (ζ)0 is the set of points whose moduli are less than η). We shall find later (§3.61) that continuity involves uniformity of continuity; this is in marked contradistinction to the fact that convergence does not involve uniformity of convergence.

3.6 The modified Heine–Borel theorem The following theorem is of great importance in connexion with properties of uniformity; we give a proof for a one-dimensional closed region. (A formal proof of the theorem for a two-dimensional region will be found in Watson [650].) Given (i) a straight line CD and (ii) a law by which, corresponding to each point P of CD, we can determine a closed interval I(P) of CD, P being an interior point of I(P) (except when P is at C or D, when it is an end point). Examples of such laws associating intervals with points will be found in §3.61 and §5.13. Then the line CD can be divided into a finite number of closed intervals J1, J2, . . . , Jk , such that each interval Jr contains at least one point (not an end point) Pr , such that no point of Jr lies outside the interval I(Pr ) associated (by means of the given law) with that point Pr . This statement of the Heine–Borel theorem (which is sometimes called the Borel–Lebesgue theorem) is due to Baker [41]. Hobson [316] points out that the theorem is practically given in Goursat [254]; the ordinary form of the Heine–Borel theorem will be found in the treatise cited. A closed interval of the nature just described will be called a suitable interval, and will be said to satisfy condition (A). If CD satisfies condition (A), what is required is proved. If not, bisect CD; if either or both

52

Continuous Functions and Uniform Convergence

of the intervals into which CD is divided is not suitable, bisect it or them. A suitable interval is not to be bisected; for one of the parts into which it is divided might not be suitable. This process of bisecting intervals which are not suitable either will terminate or it will not. If it does terminate, the theorem is proved, for CD will have been divided into suitable intervals. Suppose that the process does not terminate; and let an interval, which can be divided into suitable intervals by the process of bisection just described, be said to satisfy condition (B). Then, by hypothesis, CD does not satisfy condition (B); therefore at least one of the bisected portions of CD does not satisfy condition (B). Take that one which does not (if neither satisfies condition (B) take the left-hand one); bisect it and select that bisected part which does not satisfy condition (B). This process of bisection and selection gives an unending sequence of intervals s0, s1, s2, . . . such that: (i) The length of sn is 2−n CD. (ii) No point of sn+1 is outside sn . (iii) The interval sn does not satisfy condition (A). Let the distances of the end points of sn from C be xn, yn ; then xn ≤ xn+1 < yn+1 ≤ yn . Therefore, by §2.2, xn and yn have limits; and, by the condition (i) above, these limits are the same, say ξ; let Q be the point whose distance from C is ξ. But, by hypothesis, there is a number δQ such that every point of CD, whose distance from Q is less than δQ , is a point of the associated interval I(Q). Choose n so large that 2−n CD < δQ ; then Q is an internal point or end point of sn and the distance of every point of sn from Q is less than δQ . And therefore the interval sn satisfies condition (A), which is contrary to condition (iii) above. The hypothesis that the process of bisecting intervals does not terminate therefore involves a contradiction; therefore the process does terminate and the theorem is proved. In the two-dimensional form of the theorem 4 , the interval CD is replaced by a closed two-dimensional region, the interval I(P) by a circle, or the portion of the circle which lies inside the region, with centre P, and the interval Jr by a square with sides parallel to the axes.

3.61 Uniformity of continuity From the theorem just proved, it follows without difficulty that if a function f (x) of a real variable x is continuous when a ≤ x ≤ b, then f (x) is uniformly continuous throughout the range a ≤ x ≤ b. This result is due to Heine [288]. For let ε be an arbitrary positive number; then, in virtue of the continuity of f (x), corresponding to any value of x, we can find a positive number δx , depending on x, such that | f (x 0) − f (x)| < ε/4 for all values of x 0 such that |x 0 − x| < δx . Then by §3.6 we can divide the range (a, b) into a finite number of closed intervals with the property that in each interval there is a number x1 such that | f (x 0) − f (x1 )| < 14 ε, whenever x 0 lies in the interval in which x1 lies. Let δ0 be the length of the smallest of these intervals; and let ξ, ξ 0 be any two numbers 4

The reader will see that a proof may be constructed on similar lines by drawing a square circumscribing the region and carrying out a process of dividing squares into four equal squares.

3.6 The modified Heine–Borel theorem

53

in the closed range (a, b) such that |ξ − ξ 0 | < δ0 . Then ξ, ξ 0 lie in the same or in adjacent intervals; if they lie in adjacent intervals let ξ0 be the common end point. Then we can find numbers x1 , x2 , one in each interval, such that | f (ξ) − f (x1 )| < 14 ε,

| f (ξ0 ) − f (x1 )| < 41 ε,

| f (ξ 0) − f (x2 )| < 14 ε,

| f (ξ0 ) − f (x2 )| < 41 ε,

so that | f (ξ) − f (ξ 0)| = |{ f (ξ) − f (x1 )} − { f (ξ0 ) − f (x1 )} − { f (ξ 0) − f (x2 )} + { f (ξ0 ) − f (x2 )}| < ε. If ξ, ξ 0 lie in the same interval, we can prove similarly that | f (ξ) − f (ξ 0)| < ε/2. In either case we have shewn that, for any number ξ in the range, we have | f (ξ) − f (ξ + ζ)| < ε whenever ξ + ζ is in the range and −δ0 < ζ < δ0 where δ0 is independent of ξ. The uniformity of the continuity is therefore established. Corollary 3.6.1 From the two-dimensional form of the theorem of §3.6 we can prove that a function of a complex variable, continuous at all points of a closed region of the Argand diagram, is uniformly continuous throughout that region. Corollary 3.6.2 A function f (x) which is continuous throughout the range a ≤ x ≤ b is bounded in the range; that is to say we can find a number κ independent of x such that | f (x)| < κ for all points x in the range. Let n be the number of parts into which the range is divided. Let a, ξ1, ξ2, . . . , ξn−1, b be their end points; then if x be any point of the rth interval we can find numbers x1, x2, . . . , xn such that | f (a) − f (x1 )| < 41 ε,

| f (x1 ) − f (ξ1 )| < 41 ε,

| f (ξ1 ) − f (x2 )| < 41 ε,

| f (x2 ) − f (ξ2 )| < 41 ε, . . . , | f (xr−1 ) − f (x)| < 41 ε.

Therefore | f (a) − f (x)| < 12 rε, and so | f (x)| < | f (a)| + 21 nε, which is the required result, since the right-hand side is independent of x. The corresponding theorem for functions of complex variables is left to the reader.

3.62 A real function, of a real variable, continuous in a closed interval, attains its upper bound Let f (x) be a real continuous function of x when a ≤ x ≤ b. Form a section in which the R-class consists of those numbers r such that r > f (x) for all values of x in the range (a, b), and the L-class of all other numbers. This section defines a number α such that f (x) ≤ α, but, if δ be any positive number, values of x in the range exist such that f (x) > α − δ. Then α is called the upper bound of f (x); and the theorem states that a number x 0 in the range can be found such that f (x 0) = α. For, no matter how small δ may be, we can find values of x for which | f (x) − α| −1 > δ−1 ;

54

Continuous Functions and Uniform Convergence

therefore | f (x) − α| −1 is not bounded in the range; therefore (Corollary 3.6.2) it is not continuous at some point or points of the range; but since | f (x) − α| is continuous at all points of the range, its reciprocal is continuous at all points of the range (Example 3.2.1) except those points at which f (x) = α; therefore f (x) = α at some point of the range; the theorem is therefore proved. Corollary 3.6.3 The lower bound of a continuous function may be defined in a similar manner; and a continuous function attains its lower bound. Corollary 3.6.4 If f (z) be a function of a complex variable continuous in a closed region, | f (z)| attains its upper bound.

3.63 A real function, of a real variable, continuous in a closed interval, attains all values between its upper and lower bounds Let M, m be the upper and lower bounds of f (x); then we can find numbers x, ¯ x, by §3.62, such that f ( x) ¯ = M, f (x) = m; let µ be any number such that m < µ < M. Given any positive number ε, we can (by §3.61) divide the range ( x, ¯ x) into a finite number, r, of closed intervals such that | f (x1,r ) − f (x2,r )| < ε, where x1,r , x2,r are any points of the rth interval; take x1,r , x2,r to be the end points of the interval; then there is at least for which f (x1,r ) − µ and f (x2,r ) − µ have one of the intervals opposite signs; and since f (x1,r ) − µ − f (x2,r ) − µ < ε, it follows that | f (x1,r )− µ| < ε. Since we can find a number x1,r to satisfy this inequality for all values of ε, no matter how small, the lower bound of the function | f (x) − µ| is zero; since this is a continuous function of x, it follows from Corollary 3.6.3 that f (x) − µ vanishes for some value of x.

3.64 The fluctuation of a function of a real variable The terminology of this section is partly that of Hobson [316] and partly that of Young [687]. Let f (x) be a real bounded function, defined when a ≤ x ≤ b. Let a ≤ x1 ≤ x2 ≤ · · · ≤ xn ≤ b. Then | f (a) − f (x1 )| + | f (x1 ) − f (x2 )| + · · · + | f (xn ) − f (b)| is called the fluctuation of f (x) in the range (a, b) for the set of subdivisions x1, x2, . . . , xn . If the fluctuation have an upper bound Fab , independent of n, for all choices of x1, x2, . . . , xn , then f (x) is said to have limited total fluctuation in the range (a, b). Fab is called the total fluctuation in the range. Example 3.6.1 If f (x) be monotonic; that is, ( f (x) − f (x 0)) /(x − x 0) is one-signed or zero for all pairs of different values of x and x 0, in the range (a, b), its total fluctuation in the range is | f (a) − f (b)|. Example 3.6.2 A function with limited total fluctuation can be expressed as the difference of two positive increasing monotonic functions. Hint. These functions may be taken to be 1 {Faz + f (x)}, 12 {Faz − f (x)}. 2 Example 3.6.3 If f (x) have limited total fluctuation in the range (a, b), then the limits f (x ± 0) exist at all points in the interior of the range. [See Example 3.2.2].

3.7 Uniformity of convergence of power series

55

Example 3.6.4 If f (x), g(x) have limited total fluctuation in the range (a, b) so has f (x)g(x). Hint. For | f (x 0)g(x 0) − f (x)g(x) ≤ | f (x 0)| · |g(x 0) − g(x)| + |g(x)| · | f (x 0) − f (x)|,

(3.3)

and so the total fluctuation of f (x)g(x) cannot exceed g · Fab + f · G ba , where f , g are the upper bounds of | f (x)|, |g(x)|.

3.7 Uniformity of convergence of power series Let the power series a0 + a1 z + · · · + an z n + · · · converge absolutely when z = z0 . Then, if ∞ ∞ Í Í |z| ≤ |z0 |, |an z n | ≤ |an z0n |. But since |an z0 n | converges, it follows, by §3.34, that an z n n=0

n=0

converges uniformly with regard to the variable z when |z| ≤ |z0 |. Hence, by §3.32, a power series is a continuous function of the variable throughout the closed region formed by the interior and boundary of any circle concentric with the circle of convergence and of smaller radius (§2.6).

3.71 Abel’s theorem Abel’s proof [1] employs directly the arguments by which the theorems of §3.32 and §3.35 Í are proved. In the case when |an | converges, the theorem is obvious from §3.7 on continuity up to the circle of convergence. ∞ Í an z n be a power series, whose radius of convergence is unity, and let it be such that Let n=0

∞ Í n=0

an converges; and let 0 ≤ x ≤ 1; then Abel’s theorem asserts that

lim

x→1

∞ Õ

! an x

n

=

∞ Õ

an .

(3.4)

n=0

n=0

For, with the notation of §3.35, the function x n satisfies the conditions laid on un (x), ∞ Í when 0 ≤ x ≤ 1; consequently f (x) = an x n converges uniformly throughout the range n=0

0 ≤ x ≤ 1; it is therefore, by §3.32, a continuous function of x throughout the range, and so lim− f (x) = f (1), which is the theorem stated. x→1

3.72 Abel’s theorem on multiplication of convergent series This is a modification of the theorem of §2.53 for absolutely convergent series. This is Abel’s original proof [1, Theorem VI]. In some textbooks a more elaborate proof, by the use of Cesàro’s sums (§8.43), is given. ∞ ∞ ∞ Í Í Í Let cn = a0 bn + a1 bn−1 + · · · + an b0 . Then the convergence of an , bn , and cn is a n=0

sufficient condition that ∞ Õ n=0

! an

∞ Õ n=0

! bn =

∞ Õ n=0

cn .

n=0

n=0

Continuous Functions and Uniform Convergence

56

For, let A(x) =

∞ Õ

an x n,

B(x) =

n=0

∞ Õ

bn x n,

C(x) =

n=0

∞ Õ

cn x n .

n=0

Then the series for A(x), B(x), C(x) are absolutely convergent when |x| < 1 (§2.6); and consequently, by §2.53, A(x)B(x) = C(x) when 0 < x < 1; therefore, by Example 2.2.2, lim A(x) · lim− B(x) = lim− C(x)

x→1−

x→1

x→1

provided that these three limits exist; but, by §3.71, these three limits are

∞ Í n=0

an ,

∞ Í n=0

bn ,

∞ Í n=0

cn ;

and the theorem is proved.

3.73 Power series which vanish identically If a convergent power series vanishes for all values of z such that |z| ≤ r1 , where r1 > 0, then all the coefficients in the power series vanish. For, if not, let am be the first coefficient which does not vanish. Then am + am+1 z + am+2 z2 + · · · vanishes for all values of z (zero excepted) and converges absolutely when |z| ≤ r < r1 ; hence, if s = am+1 + am+2 z + · · · , we have ∞ Õ |am+n | r n−1, |s| ≤ n=1

and so we can find a positive number δ ≤ r such that, whenever |z| ≤ δ, am+1 z + am+2 z2 + · · · ≤ 1 |am | ; 2 5

and then |am + s| ≥ |am | − |s| > 12 |am |, and so |am + s| , 0 when |z| < δ. We have therefore arrived at a contradiction by supposing that some coefficient does not vanish. Therefore all the coefficients vanish. Corollary 3.7.1 We may equate corresponding coefficients in two power series whose sums are equal throughout the region |z| < δ, where δ > 0. Corollary 3.7.2 We may also equate coefficients in two power series which are proved equal only when z is real.

3.8 Miscellaneous examples Example 3.1 Shew that the series ∞ Õ n=1

(1 −

1 is equal to (1−z) 2 when |z| < 1 and is equal to the theory of uniform convergence? 5

z n−1 − z n+1 )

z n )(1

1 z(1−z)2

when |z| > 1. Is this fact connected with

It is sufficient to take δ to be the smaller of the numbers r and

1 2

|a m | ÷

∞ Í n=1

|a m+n | r n−1 .

3.8 Miscellaneous examples

57

Example 3.2 Shew that the series 1 1 1 + 4 sin + · · · + 2n sin n + · · · 3z 9z 3 z converges absolutely for all values of z (z = 0 excepted), but does not converge uniformly near z = 0. 2 sin

Example 3.3 (Math. Trip. 1907) If un (x) = −2(n − 1)2 xe−(n−1) ∞ Í un (x) does not converge uniformly near x = 0.

2 2

x

+ 2n2 xe−n

2 2

x

, shew that

n=1

Example 3.4 Shew that the series √11 − √12 + √13 − · · · is convergent, but that its square formed by Abel multiplication, 1 2 2 1 2 2 −√ + √ + − √ + √ −··· , 1 2 3 2 4 6 is divergent. Example 3.5 (Cauchy, Cajori) If the convergent series s = 11r − 21r + 31r − 41r + · · · (with r > 0) be multiplied by itself the terms of the product being arranged as in Abel’s result, shew that the resulting series diverges if r ≤ 21 but converges to the sum s2 if r > 12 . Example 3.6 (Cajori) If the two conditionally convergent series ∞ Õ (−1)n+1 n=1

nr

and

∞ Õ (−1)n+1 n=1

ns

,

where r and s lie between 0 and 1, be multiplied together, and the product arranged as in Abel’s result, shew that the necessary and sufficient condition for the convergence of the resulting series is r + s > 1. Example 3.7 (Cajori) Shew that if the series 1 − 13 + 15 − 71 + · · · be multiplied by itself any number of times, the terms of the product being arranged as in Abel’s result, the resulting series converges. Example 3.8 Shew that the qth power of the series α1 sin θ + α2 sin 2θ + · · · + αn sin nθ + · · · is convergent whenever q(1 − r) < 1, r being the greatest number satisfying the relation αn ≤ n−r for all values of n. Example 3.9 (Math. Trip. 1896) Shew that if θ is not equal to 0 or a multiple of 2π, and Í if u0, u1, u2, . . . be a sequence such that un → 0 steadily, then the series un cos(nθ + a) is convergent. Shew also that, if the limit of un is not zero, but un is still monotonic, the sum of the series is oscillatory if θ/π is rational, but that, if θ/π is irrational, the sum may have any value between certain bounds whose difference is α cosec(θ/2), where α = lim un . n→∞

4 The Theory of Riemann Integration

4.1 The concept of integration The reader is doubtless familiar with the idea of integration as the operation inverse to that of differentiation; and he is equally well aware that the integral (in this sense) of a given elementary function is not always expressible in terms of elementary functions. In order therefore to give a definition of the integral of a function which shall be always available, even though it is not practicable to obtain a function of which the given function is the differential coefficient, we have recourse to the result that the integral of f (x), defined as the (elementary) function whose differential coefficient is f (x), between the limits a and b is the area bounded by the curve y = f (x), the axis of x and the ordinates x = a, x = b. We proceed to frame a formal definition of integration with this idea as the starting-point.

4.11 Upper and lower integrals The following procedure for establishing existence theorems concerning integrals is based on that given by Goursat [255, I, Ch. IV]. The concepts of upper and lower integrals are due to Darboux, [160, p. 64]. Let f (x) be a bounded function of x in the range (a, b). Divide the interval at the points x1, x2, . . . , xn−1 , (a ≤ x1 ≤ x2 ≤ · · · ≤ xn−1 ≤ b). Let U, L be the bounds of f (x) in the range (a, b), and let Ur , Lr , be the bounds of f (x) in the range (xr−1, xr ), where x0 = a, xn = b. The reader will find a figure of great assistance in following the argument of this section. Sn and sn represent the sums of the areas of a number of rectangles which are respectively greater and less than the area bounded by y = f (x), x = a, x = b and y = 0, if this area be assumed to exist. Consider the sums Sn = U1 (x1 − a) + U2 (x2 − x1 ) + · · · + Un (b − xn−1 ), sn = L1 (x1 − a) + L2 (x2 − x1 ) + · · · + Ln (b − xn−1 ). Then U(b − a) ≥ Sn ≥ sn ≥ L(b − a). For a given n, Sn and sn are bounded functions of x1, x2, . . . , xn−1 . Let their lower and upper bounds 1 respectively be S n , s n , so that S n , s n depend only on n and on the form of f (x), and not on the particular way of dividing the interval into n parts. 1

The bounds of a function of n variables are defined in just the same manner as the bounds of a function of a single variable (§3.62).

58

4.1 The concept of integration

59

Let the lower and upper bounds of these functions of n be S, s. Then Sn ≥ S, sn ≤ s. We proceed to shew that s is at most equal to S; i.e. S ≥ s. Let the intervals (a, x1 ), (x1, x2 ), . . . be divided into smaller intervals by new points of subdivision, and let a, y1, y2, . . . , yk−1, yk (= x1 ), yk+1, . . . , yl−1, yl (= x2 ), yl+1, . . . , ym−1, b be the end points of the smaller intervals; let Ur0 , Lr0 be the bounds of f (x) in the interval (yr−1, yr ). Let m m Õ Õ Tm = (yr − yr−1 )Ur0, tm = (yr − yr−1 )Lr0 . r=1

Since

U10 ,

U20, . .

. , Uk0

r=1

do not exceed U1 , it follows without difficulty that Sn ≥ Tm ≥ tm ≥ sn .

Now consider the subdivision of (a, b) into intervals by the points x1, x2, . . . , xn−1 , and also the subdivision by a different set of points x10 , x20 , . . . , xn0 0 −1 . Let Sn0 0 , sn0 0 be the sums for the second kind of subdivision which correspond to the sums Sn , sn for the first kind of subdivision. Take all the points x1, . . . , xn−1 ; x10 , . . . , xn0 0 −1 as the points y1, y2, . . . , ym . Then Sn ≥ Tm ≥ tm ≥ sn,

and

Sn0 0 ≥ Tm ≥ tm ≥ sn0 0 .

Hence every expression of the type Sn exceeds (or at least equals) every expression of the type sn0 0 ; and therefore S cannot be less than s. For if S < s and s − S = 2η we could find an Sn and an sn0 0 such that Sn − S < η, s − sn0 0 < η and so sn0 0 > Sn , which is impossible. ∫ b The bound S is called the upper integral of f (x), and is written U f (x)dx; the bound s a ∫ b is called the lower integral, and written L f (x)dx. If S = s, their common value is called a

the integral of f (x) taken between the limits 2 of integration a and b. The integral is written ∫ b f (x)dx. a ∫ a ∫ b We define f (x)dx, when a < b, to mean − f (x)dx. b

a

∫ Example 4.1.1 Prove that a

b

{ f (x) + φ(x)}dx =

∫ a

b

f (x)dx +

∫

b

φ(x)dx.

a

Example 4.1.2 By means of Example 4.1.1, define the integral of a continuous complex function of a real variable.

4.12 Riemann’s condition of integrability Riemann [558, p. 239] bases his definition of an integral on the limit of the sum occurring in §4.13; but it is then difficult to prove the uniqueness of the limit. A more general definition 2

‘Extreme’ values would be a more appropriate term but ‘limits’ has the sanction of custom. ‘Termini’ has been suggested by Lamb [399, p. 207].

The Theory of Riemann Integration

60

of integration (which is of very great importance in the modern theory of Functions of Real Variables) has been given by Lebesgue [417]. See also [418]. A function is said to be ‘integrable in the sense of Riemann’ if (with the notation of §4.11) Sn and sn have a common limit (called the Riemann integral of the function) when the number of intervals (xr−1, xr ) tends to infinity in such a way that the length of the longest of them tends to zero. The necessary and sufficient condition that a bounded function should be integrable is that Sn − sn should tend to zero when the number of intervals (xr−1, xr ) tends to infinity in such a way that the length of the longest tends to zero. The condition is obviously necessary, for if Sn and sn have a common limit Sn − sn → 0 as n → ∞. And it is sufficient; for, since Sn ≥ S ≥ s ≥ sn , it follows that if lim(Sn − sn ) = 0, then lim Sn = lim sn = S = s. Remark 4.1.1 A continuous function f (x) is integrable. For, given ε, we can find δ such that | f (x 0) − f (x 00)| < ε/(b − a) whenever |x 0 − x 00 | < δ. Take all the intervals (xδ−1, xδ ) less than δ, and then Uδ − Lδ < ε/(b − a) and so Sn − sn < ε; therefore Sn − sn → 0 under the circumstances specified in the condition of integrability. Corollary 4.1.2 If Sn and sn have the same limit S for one mode of subdivision of (a, b) into intervals of the specified kind, the limits of Sn and of sn for any other such mode of subdivision are both S. Example 4.1.3 The product of two integrable functions is an integrable function. Example 4.1.4 A function which is continuous except at a finite number of ordinary discontinuities is integrable. Hint. If f (x) have an ordinary discontinuity at c, enclose c in an interval of length δ1 ; given ε, we can find δ so that | f (x 0) − f (x)| < ε when |x 0 − x| < δ and x, x 0 are not in this interval. Then Sn − sn ≤ ε(b − a − δ1 ) + kδ1 , where k is the greatest value of | f (x 0) − f (x)|, when x, x 0 lie in the interval. When δ1 → 0, k → | f (c + 0) − f (c − 0)|, and hence lim (Sn − sn ) = 0. n→∞

Example 4.1.5 A function with limited total fluctuation and a finite number of ordinary discontinuities is integrable. (See Example 3.6.2.)

4.13 A general theorem on integration Let f (x) be integrable, and let ε be any positive number. Then it is possible to choose δ so that ∫ b n Õ 0 (x p − x p−1 ) f (x p−1 ) − f (x)dx < ε, p=1 a 0 provided that x p − x p−1 ≤ δ, and x p−1 ≤ x p−1 ≤ xp . To prove the theorem we observe that, given ε, we can choose the length of the longest

4.1 The concept of integration

61

interval, δ, so small that Sn − sn < ε. Also Sn ≥

n Õ 0 ) ≥ sn, (x p − x p−1 ) f (x p−1 p=1

∫ Sn ≥

b

f (x)dx ≥ sn .

a

Therefore ∫ b n Õ 0 (x − x p−1 ) f (x p−1 ) − f (x)dx p=1 p a

≤ Sn − sn < ε.

As an example (see Netto [484]) ∫ Xof the evaluation of a definite integral directly from the dx , where X < 1. Take δ = p1 arcsin X and let theorem of this section consider (1 − x 2 )1/2 0 xs = sin sδ, (0 < sδ < 21 π), so that xδ+1 − xδ = 2 sin δ2 cos δ + 21 < δ; also let xδ 0 = sin(δ + 21 ) δ. Then p p Õ Õ sin sδ − sin (δ − 1) δ xδ − xδ−1 = 02 1/2 (1 − xδ−1 ) cos (δ − 21 ) δ δ=1 δ=1

= 2p sin 21 δ ( = arc sin X .

sin 12 δ 1 δ 2

) .

By taking p sufficiently large we can make ∫ p X Õ dx xs − xs−1 − 02 1/2 0 (1 − x 2 )1/2 δ=1 (1 − xs−1 ) n o arbitrarily small. We can also make arcsin X . sin(δ/2) − 1 arbitrarily small. That is, given δ/2 an arbitrary number ε, we can make ∫ X dx − arcsin X < ε 2 1/2 (1 − x ) 0 by taking p sufficiently large. But the expression now under consideration does not depend on p; and therefore it must be zero; for if not we could take ε to be less than it, and we should have a contradiction. That is to say ∫ X dx = arcsin X. (4.1) (1 − x 2 )1/2 0 Example 4.1.6 Shew that 1 x 2x (n − 1) x sin x 1 + cos + cos + · · · + cos = . lim n→∞ n n n n x

The Theory of Riemann Integration

62

Example 4.1.7 If f (x) has ordinary discontinuities at the points a1, a2, . . . , ak , then ∫ a1 −δ1 ∫ a2 −δ2 ∫ b ∫ b f (x)dx = lim + +··· + f (x) dx , a

a1 +ε1

a

ak +εk

where the limit is taken by making δ1, δ2, . . . , δk , ε1, ε2, . . . , εk tend to +0, independently. Example 4.1.8 If f (x) is integrable when a1 ≤ x ≤ b1 and if, when a1 ≤ a < b < b1, we write ∫ b f (x)dx = φ(a, b), a

and if f (b + 0) exists, then φ(a, b + δ) − φ(a, b) = f (b + 0). δ Deduce that, if f (x) is continuous at a and b, ∫ b ∫ b d d f (x)dx = − f (a), f (x)dx = f (b). da a db a lim

δ→+0

dx Example 4.1.9 Prove by differentiation that, if φ(x) is a continuous function of x and dt a continuous function of t, then ∫ t1 ∫ x1 dx φ(x) dt. φ(x)dx = dt t0 x0 Example 4.1.10 If f 0(x) and φ 0(x) are continuous when a ≤ x ≤ b, shew from Example 4.1.8 that ∫ b ∫ b 0 f (x)φ(x)dx + φ 0(x) f (x)dx = f (b)φ(b) − f (a)φ(a). a

a

∫ Example 4.1.11 If f (x) is integrable in the range (a, c) and a ≤ b ≤ c, shew that

b

f (x)dx a

is a continuous function of b.

4.14 Mean-value theorems The two following general theorems are frequently useful. (I) Let U and L be the upper and lower bounds of the integrable function f (x) in the range (a, b). Then from the definition of an integral it is obvious that ∫ b ∫ b (U − f (x)) dx, ( f (x) − L) dx a

a

are not negative; and so ∫ U(b − a) ≥

b

f (x)dx ≥ L(b − a). a

This is known as the First Mean-Value Theorem.

4.1 The concept of integration

63

If f (x) is continuous we can find a number ξ such that a ≤ ξ ≤ b and such that f (ξ) has any given value lying between U and L (§3.63). Therefore we can find ξ such that ∫ b f (x)dx = (b − a) f (ξ). a

If F(x) has a continuous differential coefficient F 0(x) in the range (a, b), we have, on writing F 0(x) for f (x), F(b) − F(a) = (b − a)F 0(ξ) for some value of ξ such that a ≤ ξ ≤ b. Example 4.1.12 If f (x) is continuous and φ(x) ≥ 0, shew that ξ can be found such that ∫ b ∫ b f (x)φ(x) dx = f (ξ) φ(x) dx. a

a

(II) Let f (x) and φ(x) be integrable in the range (a, b) and let φ(x) be a positive decreasing function of x. Then Bonnet’s form of the Second Mean-Value Theorem [83] is that a number ξ exists such that a ≤ ξ ≤ b, and ∫ ξ ∫ b f (x)φ(x) dx = φ(a) f (x) dx. a

a

The proof given is a modified form of an investigation due to Hölder [326]. For, with the notation of §4.1 and §4.13, consider the sum S=

p Õ (xs − xs−1 ) f (xs−1 )φ(xs−1 ). s=1

Writing (xs − xs−1 ) f (xs−1 ) = as−1 , φ(xs−1 ) = φs−1 , a0 + a1 + · · · + as = bs , we have S=

p−1 Õ

bs−1 (φs−1 − φs ) + b p−1 φ p−1 .

s=1

Each term in the summation is increased by writing b¯ for bs−1 and decreased by writing b ¯ and b be the greatest and least of b0, b1, . . . , b p−1 ; and so bφ0 ≤ S ≤ bφ ¯ 0. for bs−1 , if b, m Í Therefore S lies between the greatest and least of the sums φ(x0 ) (xs − xs−1 ) f (xs−1 ) where s=1

m = 1, 2, 3, . . . , p. But, given ε, we can find δ such that, when xs − xs−1 < δ, p ∫ xp Õ f (x)φ(x) dx < ε, (xs − xs−1 ) f (xs−1 )φ(xs−1 ) − s=1 x0 ∫ m xm Õ f (x) dx < ε, φ(x0 ) (xs − xs−1 ) f (xs−1 ) − φ(x0 ) x0 s=1 ∫

b

and so, writing a, b for x0 , x p , we find that f (x)φ(x) dx lies between the upper and a ∫ ξ1 lower bounds of φ(a) f (x) dx ± 2ε, where ξ1 may take all values between a and b. (By a

The Theory of Riemann Integration ∫ ξ1 Example 4.1.11, since f (x) is bounded, f (x) dx is a continuous function of ξ1 .) Let U a ∫ ∫ b ξ1 and L be the upper and lower bounds of φ(a) f (x) dx. Then U +2ε ≥ f (x)φ(x)dx ≥

64

a

a

L − 2ε for all positive values of ε; therefore ∫ b U≥ f (x)φ(x) dx ≥ L. a

Since φ(a)

∫

ξ1

f (x) dx qua function of ξ1 takes all values between its upper and lower ∫ b bounds, there is some value ξ, say, of ξ1 for which it is equal to f (x)φ(x) dx. This proves a

a

the Second Mean-Value Theorem. Example 4.1.13 (Du Bois Reymond) By writing |φ(x) − φ(b)| in place of φ(x) in Bonnet’s form of the mean-value theorem, shew that if φ(x) is a monotonic function, then a number ξ exists such that a ≤ ξ ≤ b and ∫ b ∫ ξ ∫ b f (x)φ(x)dx = φ(a) f (x)dx + φ(b) f (x)dx. a

ξ

a

4.2 Differentiation of integrals containing a parameter The equation d dα

∫

b

f (x, α)dx =

a

∫ a

b

∂f dx ∂α

(4.2)

∂f , is is true if f (x, α) possesses a Riemann integral with respect to x and fα , which equals ∂α a continuous function of the variables x and α. This formula was given by Leibniz, without specifying the restrictions laid on f (x, a). Note φ(x, y) is defined to be a continuous function of both variables if, given ε, we can find δ such that |φ(x 0, y 0) − φ(x, y)| < ε whenever {(x 0 − x)2 + (y 0 − y)2 }1/2 < δ. It can be shewn by §3.6 that if φ(x, y) is a continuous function of both variables at all points of a closed region in a Cartesian diagram, it is uniformly continuous throughout the region (the proof is almost identical with that of §3.61). It should be noticed that, if φ(x, y) is a continuous function of each variable, it is not necessarily a continuous function of both; as an example take φ(x, y) =

(x + y)2 , x2 + y2

φ(0, 0) = 1;

this is a continuous function of x and of y at (0, 0), but not of both x and y. For d dα

∫ a

b

f (x, α) dx = lim

h→0

∫ a

b

f (x, α + h) − f (x, α) dx h

(4.3)

if this limit exists. But, by the first mean-value theorem, since fα is a continuous function of

4.3 Double integrals and repeated integrals

65

α, the second integrand is fa (x, α + θh), where 0 ≤ θ ≤ 1. But, for any given ε, a number δ independent of x exists (since the continuity of fα is uniform with respect to the variable x) such that | fa (x, α 0) − fa (x, α) | < ε/(b − a), whenever | α 0 − α | < δ. It is obvious that it would have been sufficient to assume that fα had a Riemann integral and was a continuous function of a (the continuity being uniform with respect to x), instead of assuming that fa was a continuous function of both variables. This is actually done by Hobson [315, p. 599]. Taking |h| < δ we see that |θh| < δ, and so whenever |h| < δ, ∫ b ∫ b f (x, α + h) − f (x, α) dx − f (x, α) dx α h a a ∫ b ≤ | fα (x, α + θh) − fα (x, α)| dx < ε. a

Therefore by the definition of a limit of a function (§3.2), ∫ b f (x, α + h) − f (x, α) dx lim h→0 a h ∫ b exists and is equal to fα dx. a

Example 4.2.1 If a, b be not constants but functions of α with continuous differential coefficients, shew that ∫ b ∫ b d db da ∂f f (x, α)dx = f (b, α) − f (a, α) + dx. dα a dα dα a ∂α ∫ b Example 4.2.2 If f (x, α) is a continuous function of both variables, f (x, α)dx is a a

continuous functions of α.

4.3 Double integrals and repeated integrals Let f (x, y) be a function which is continuous with regard to both of the variables x and y, when a ≤ x ≤ b, α ≤ y ≤ β. By Example 4.2.2 it is clear that ∫ b ∫ β ∫ β ∫ b f (x, y) dy dx, f (x, y) dx dy a

α

α

a

both exist. These are called repeated integrals. Also, as in §3.62, f (x, y), being a continuous function of both variables, attains its upper and lower bounds. Consider the range of values of x and y to be the points inside and on a rectangle in a Cartesian diagram; divide it into nν rectangles by lines parallel to the axes. Let Um,µ , Lm,µ

The Theory of Riemann Integration

66

be the upper and lower bounds of f (x, y) in one of the smaller rectangles whose area is, say, Am,µ ; and let n Õ ν n Õ ν Õ Õ Um,µ Am,µ = Sn,v , Lm,µ Am,µ = Sn,ν . m=1 µ=1

m=1 µ=1

Then Sn,ν > sn,ν , and, as in §4.11, we can find numbers S n,ν , s¯n,ν , which are the lower and upper bounds of Sn,ν , sn,ν respectively, the values of S n,v , s¯n,v depending only on the number of the rectangles and not on their shapes; and S n,ν ≥ s¯n,ν . We then find the lower and upper bounds (S and s) respectively of S n,ν , s¯n,ν qua functions of n and ν; and Sn,ν ≥ S ≥ s ≥ sn,ν , as in §4.11. Also, from the uniformity of the continuity of f (x, y), given ε, we can find δ such that Um,µ − Lm,µ < ε, (for all values of m and µ) whenever the sides of all the small rectangles are less than the number δ which depends only on the form of the function f (x, y) and on ε. And then Sn,ν − sn,ν < ε(b − a)(β − α), and so S − s < ε(b − a)(β − α). But S and s are independent of ε, and so S = s. The common value of S and s is called the double integral of f (x, y) and is written ∫ b∫ β f (x, y) dy dx. a

α

It is easy to shew that the repeated integrals and the double integral are all equal when f (x, y) is a continuous function of both variables. For let Ym , Λm be the upper and lower bounds of ∫ β f (x, y) dy α

as x varies between xm−1 and xm . Then ∫ n Õ Ym (xm − xm−1 ) ≥

b

a

m=1

β

∫

n Õ Λm (xm − xm−1 ). f (x, y) dy dx ≥

a

m=1

But the upper bound of f (x, y) in the rectangle Am,µ is not less than the upper bound of f (x, y) on that portion of the line x = ξ which lies in the rectangle, therefore ν Õ

Um,µ (yµ − yµ−1 ≥ Ym ≥ Λm ≥

ν Õ

Lm,µ (yµ − yµ−1 ).

µ=1

µ=1

Multiplying these last inequalities by xm − xm−1 , using the preceding inequalities and summing, we get ∫ b ∫ β n Õ ν n Õ ν Õ Õ Um,µ Am,µ ≥ f (x, y)dy dx ≥ Lm,µ Am,µ ; a

m=1 µ=1

a

m=1 µ=1

and so, proceeding to the limit, ∫

b

∫

S≥ a

a

β

f (x, y)dy dx ≥ S.

4.4 Infinite integrals

But S =

b

∫

∫

67

β

f (x, y) dx dy, and so one of the repeated integrals is equal to the double a

a

integral. Similarly the other repeated integral is equal to the double integral. Corollary 4.3.1 If f (x, y) be a continuous function of both variables, ∫ 1 ∫ 1−ν ∫ 1 ∫ 1−x dx f (x, y)dy = dy f (x, y)dx . 0

∫

b

If lim

b→∞

0

0

0

4.4 Infinite integrals ∫ ∞ f (x)dx exists, we denote it by f (x)dx; and the limit in question is called

a

a

an infinite integral. This phrase, due to Hardy [274, p. 16], suggests the analogy between an infinite integral and an infinite series. ∫ ∞ 1 1 1 dx = lim − = . Example 4.4.1 1. 2 b→∞ x a b a a ∫ ∞ x dx 1 1 1 = lim − = 2. 2. 0 + (x 2 + a2 )2 b→∞ 2(b2 + a2 ) 2a2 2a ∫ ∞ 3. (Euler). By integrating by parts, shew that t n e−t dt = n!. 0 b

∫ b f (x) dx to mean lim f (x) dx; if this limit exists; and −∞ ∫ a ∫ ∞ a→−∞ a f (x) dx is defined as f (x) dx + f (x) dx. In this last definition the choice of ∫

Similarly we define ∫

∞

−∞

−∞

a

a is a matter of indifference.

4.41 Infinite integrals of continuous functions. Conditions for convergence ∫ ∞ A necessary and sufficient condition for the convergence of f (x)dx is that, corresponding a ∫ x00 to any positive number ε, a positive number X should exist such that f (x)dx < ε x0

whenever x 00 ≥ x 0 ≥ X. The condition is obviously necessary; to prove that∫it is sufficient, suppose it is satisfied; a+n

then, if n ≥ X − a and n be a positive integer and Sn =

f (x)dx, we have |Sn+p − Sn | < ε.

a

Hence, by §2.22, Sn tends to a limit, S; and then, if ξ > a + n, ∫ ∫ ξ ∫ a+n S− + f (x)dx ≤ S − f (x)dx a

a

< 2ε; ∫ and so lim

ξ→∞

a

ξ

f (x) dx = S; so that the condition is sufficient.

ξ

a+n

f (x)dx

The Theory of Riemann Integration

68

4.42 Uniformity of convergence of an infinite integral ∫ The integral

∞

f (x, α) dx is said to converge uniformly with regard to α in a given domain

a

of values of α if, corresponding to an arbitrary positive number ε, there exists a number X independent of α such that ∫ ∞ 0 f (x, α) dx < ε x

for all values of α in the domain and all values of x 0 ≥ X. The reader will see without difficulty ∫ on comparing §2.22 and §3.31 with §4.41 that a ∞

f (x, α) dx should converge uniformly in a given

necessary and sufficient condition that a

domain is that, corresponding to any positive number ε, there exists a number X independent of α such that ∫ x00 f (x, α) dx < ε x0

for all values of α in the domain whenever x 00 ≥ x 0 ≥ X.

4.43 Tests for the convergence of an infinite integral There are conditions for the convergence of an infinite integral analogous to those given in Chapter 2 for the convergence of an infinite series. The following tests are of special importance. ∫ ∞ (I) Absolutely convergent integrals. It may be shewn that f (x)dx certainly converges if a ∫ ∞ | f (x)|dx does so; and the former integral is then said to be absolutely convergent. The a

proof is similar to that of §2.32. ∫ Example 4.4.2 The comparison test. If | f (x)| ≤ g(x) and ∫ ∞ f (x) dx converges absolutely.

∞

g(x) dx converges, then a

a 2 It was observed by Dirichlet ∫ ∞[175] (with the example f (x) = sin x ) that it is not necessary for the convergence of f (x) dx that f (x) → 0 as x → ∞: the reader may see

Note

a

this by considering the function ( 0 f (x) = (n + 1)4 (n + 1 − x) x − (n + 1) + (n + 1)−2

(n ≤ x ≤ n + 1 − (n + 1)−2 ), (n + 1 − (n + 1)−2 ≤ x ≤ n + 1,

where n takes all integral values. ∫ ξ ∫ n+1 1 For f (x) dx increases with ξ and f (x) dx = (n + 1)−2 ; whence it follows with6 0 n ∫ ∞ out difficulty that f (x) dx converges. But when x = n + 1 − 12 (n + 1)−2 , f (x) = 41 ; and a

so f (x) does not tend to zero.

4.4 Infinite integrals

69

(II) The Maclaurin–Cauchy test. 3 If f (x) > 0 and f (x) → 0 steadily, ∞ Í

∫

∞

f (x) dx and 1

f (n) converge or diverge together. ∫ m+1 For f (m) ≥ f (x) dx ≥ f (m + 1), and so

n=1

m n Õ

∫ f (m) ≥

n+1

f (x) dx ≥ 1

m=1

n+1 Õ

f (m).

m=2

∫ The first inequality shews that, if the series converges, the increasing sequence

n+1

f (x)dx

1

converges (§2.2) when n → ∞ through integral values, and hence it follows without difficulty ∫ x0 that f (x)dx converges when x 0 → ∞; also if the integral diverges, so does the series. 1

The second shews that if the series diverges so does the integral, and if the integral converges so does the series (§2.2). ∫ ∞ λ−1 (III) Bertrand’s test [67, p. 38–39]. If f (x) = O(x ), f (x) dx converges when λ < 0; n ∫ ∞ and if f (x) = O(x −1 {log x}λ−1 ), f (x) dx converges when λ < 0. a

These results are particular cases of the comparison test given in (I). 4 (IV) Chartier’s test [143] for integrals involving functions. ∫ periodic x If f (x) → 0 steadily as x → ∞ and if φ(x) dx is bounded as x → ∞, then a ∫ ∞ f (x)φ(x) dx is convergent. a ∫ x For if the upper bound of φ(x) dx be A, we can choose X such that f (x) < ε/2A a

when x ≥ X; and then by the second mean-value theorem, when x 00 ≥ x 0 ≥ X, we have ∫ x00 ∫ ξ ∫ ξ ∫ x0 0 0 φ(x) dx φ(x) dx − 0 f (x) φ(x) dx = f (x ) 0 φ(x) dx = f (x ) x

x

a

a

≤ 2A f (x 0) < ε, which is the condition for convergence. ∫ ∞ sin x dx converges. Example 4.4.3 x 0 ∫ ∞ sin(x 3 − ax) Example 4.4.4 dx converges. x 0 3

4

Maclaurin [449, vol. I, p. 289–290] makes a verbal statement practically equivalent to this result. Cauchy’s result is given in [129, v. 7, p. 269]. It is remarkable that this test for conditionally convergent integrals should have been given some years before formal definitions of absolutely convergent integrals.

The Theory of Riemann Integration

70

4.431 Tests for uniformity of convergence of an infinite integral. The results of this section and of §4.44 are due to de la Vallée Poussin [637]. (I) De la Vallée Poussin’s test. ∫ This name is due to Osgood. The reader will easily see by using ∞

f (x, α) dx converges uniformly with regard to α in a domain ∫ ∞ of values of α if | f (x, α)| < µ(x), where µ(x) is independent of α and µ(x) dx converges. a∫ x00 ∫ x00 00 0 For, choosing X so that µ(x) dx < ε when x ≥ x ≥ X, we have f (x, α) dx < ε, the reasoning of §3.34 that

a

x0

x0

and the choice of X is independent of α. ∫ ∞ Example 4.4.5 x a−1 e−x dx converges uniformly in any interval (A, B) such that 1 ≤ 0

A ≤ B.

(II) ∫ x The method of change of variable. This may be illustrated by an example. Consider sin ax dx where a is real. We have x 0 ∫ ax00 ∫ x00 sin y sin ax dx = dy. x y 0 0 ax x ∫ ν00 ∫ ∞ sin y sin y Since dy converges we can find Y such that dy < ε when y 00 ≥ y 0 ≥ Y . y y ν0 ∫ x000 sin ax So dx < ε whenever |ax 0 | ≥ Y ; if |a| ≥ δ > 0, we therefore get x x0 ∫ x00 sin ax 0 and when a ≤ −δ < 0. ∫ ∞ ∫ a 2 3 Example 4.4.6 (de la Vallée Poussin) Prove that sin(b x ) db dx is uniformly 1

0

convergent in any range of real values of a. ∫ 2 3 ax Write b2 x 3 = z, and observe that z−1/2 sin z dz does not exceed a constant inde 0 ∫ ∞ pendent of a and x since z−1/2 sin z dz converges. 0

∫ ∫ (III) The method of integration by parts. If f (x, a) dx = φ(x, a) + χ(x, a) dx and if ∫ ∞ φ(x, a) → 0 uniformly as x → ∞ and χ(x, a) dx converges uniformly with regard to a, a ∫ ∞ then obviously f (x, a) dx converges uniformly with regard to a. a

(IV) The method of decomposition.

4.4 Infinite integrals

Example 4.4.7 ∫ ∞ 0

1 cos x sin ax dx = x 2

∫

∞

0

71

1 sin(a + 1)x dx + x 2

∫ 0

∞

sin(a − 1)x dx; x

both of the latter integrals converge uniformly in any closed domain of real values of a from which the points a = ±1 are excluded.

4.44 Theorems concerning uniformly convergent infinite integrals ∫

∞

f (x, α) dx converge uniformly when a lies in a domain S. Then, if f (x, α) is ∫ ∞ a continuous function of both variables when x ≥ a and a lies in S, f (x, a) dx is a

(I) Let

a

a

continuous function of α. This result is due to Stokes. His statement is that the integral is a continuous function of a if it does not ‘converge infinitely slowly’. ∫ ∞ For, given ε, we can find X independent of α, such that f (x, a) dx < ε whenever ξ ξ ≥ X. Also we can find δ independent of x and a, such that | f (x, α) − f (x, α 0)| < ε/(X − a) whenever |α − α 0 | < δ. That is to say, given ε, we can find δ independent of a, such that ∫ ∞ ∫ X ∫ ∞ 0 0 f (x, α ) dx − f (x, α) dx ≤ { f (x, α) − f (x, α )} dx a a a∫ ∞ ∫ ∞ 0 + f (x, α ) dx + f (x, α) dx X

X

< 3ε, whenever |α 0 − α| < δ; and this is the condition for continuity. (II) If f (x, α) satisfies the same conditions as in (I), and if α lies in S when A ≤ α ≤ B, then ∫ ∞ ∫ B ∫ B ∫ ∞ f (x, α) dx dα = f (x, α) dα dx. A

a

a

A

For, by §4.3, B

∫ A

∫

ξ

f (x, α) dx dα =

a

a

ξ

ξ

B

B

∫

a

Therefore ∫ B ∫ ∞ ∫ f (x, α) dx dα − A

∫

f (x, α) dα dx. A

f (x, α) dα dx a A ∫ B ∫ ξ ∫ = f (x, α) dx dα < ∫

A

for all sufficiently large values of ξ.

∞

B

ε dα < ε (B − A) , A

The Theory of Riemann Integration

72

But, from §2.1 and §4.41, this is the condition that ∫ ξ ∫ B lim f (x, α) dα dx ξ→∞

a

A

should exist, and be equal to B

∫

∫

A

∞

f (x, α) dx dα.

a

∫ ∞ d ∂φ dx is true if the integral on φ (x, a) dx = da a ∂a a the right converges uniformly and the integrand is a continuous function of both variables, when x ≥ a and a lies in a domain S, and if the integral on the left is convergent. ∂φ Let A be a point of S, and let = f (x, a), so that, by Example 4.1.8, ∂a ∫ a f (x, a) da = φ (x, a) − φ (x, A) . ∫

Corollary 4.4.1

∞

The equation

A

∫

x

∫

a

∫

∞

{φ (x, a) − φ (x, A)} dx converges, f (x, a) da dx converges, that is a A a∫ ∫ ∞ ∞ and therefore, since φ(x, a) dx converges, so does φ (x, A) dx. Then

a

Then d da

a

∫ a

∞

∫ ∞ d φ(x, a)dx = {φ(x, a) − φ(x, A)}dx da a ∫ ∞ ∫ a d = f (x, a)da dx da a A ∫ a ∫ ∞ d = f (x, a)dx da da A a ∫ ∞ ∫ ∞ ∂φ dx, = f (x, a) dx = ∂a a a

which is the required result; the∫ change of the order of the integrations has been justified a above, and the differentiation of A with regard to a is justified by §4.44 (I) and Example 4.1.8.

4.5 Improper integrals. Principal values ∫ b If | f (x)| → ∞ as x → a + 0, then lim f (x) dx may exist, and is written simply δ→+0 a+δ ∫ b f (x)dx; this limit is called an improper integral. If | f (x)| → ∞ as x → c, where a

a < c < b, then ∫ lim

δ→+0

a

c−δ

f (x)dx + lim 0

δ →+0

∫

b

f (x)dx c+δ 0

4.5 Improper integrals. Principal values 73 ∫ b may exist; this is also written f (x)dx, and is also called an improper integral; it might a

however happen that neither of these limits exists when δ, δ 0 → 0 independently, but ∫ c−δ ∫ b lim f (x)dx + f (x)dx δ→+0

a

c+δ

∫ exists; this is called Cauchy’s principal value of ∫ b P f (x)dx.

b

f (x)dx and is written for brevity a

a

Results similar to those of §4.4–§4.44 may be obtained for improper integrals. But all that is required in practice is (i) the idea of absolute convergence, (ii) the analogue of Bertrand’s test for convergence, (iii) the analogue of de la Vallée Poussin’s test for uniformity of convergence. The construction of these is left to the reader, as is also the consideration of integrals in which the integrand has an infinite limit at more than one point of the range of integration. For a detailed discussion of improper integrals, the reader is referred either to Hobson [315] or to Pierpont [519]. The connexion between infinite integrals and improper integrals is exhibited by Bromwich [102, §164]. ∫ π Example 4.5.1 1. x −1/2 cos x dx is an improper integral. 0 ∫ 1 λ−1 µ−1 2. x (1− x) dx is an improper integral if 0 < λ < 1, 0 < µ < 1. It does not converge 0

for negative values of λ and µ. ∫ 2 α−1 x dx is the principal value of an improper integral when 0 < α < 1. 3. P 1 −x 0

4.51 The inversion of the order of integration of a certain repeated integral General conditions for the legitimacy of inverting the order of integration when the integrand is not continuous are difficult to obtain. The following is a good example of the difficulties to be overcome in inverting the order of integration in a repeated improper integral. Let f (x, y) be a continuous function of both variables, and let 0 < λ, µ, ν ≤ 1; then ∫ 1 ∫ 1−x dx x λ−1 y µ−1 (1 − x − y)ν−1 f (x, y) dy 0 0 ∫ 1 ∫ 1−ν λ−1 µ−1 ν−1 = dy x y (1 − x − y) f (x, y) dx . 0

0

This integral, which was first employed by Dirichlet, is of importance in the theory of integral equations; the investigation which we shall give is due to W. A. Hurwitz [329, p. 183]. Let x λ−1 y µ−1 (1 − x − y)ν−1 f (x, y) = φ(x, y); and let M be the upper bound of | f (x, y)|. Let δ be any positive number less than 1/δ. Draw the triangle whose sides are x = δ, y = δ, x + y = 1 − δ; at all points on and inside this triangle φ(x, y) is continuous, and hence, by

The Theory of Riemann Integration

74

Corollary 4.3.1 ∫

∫

1−2δ

1−x−δ

dx δ

Now ∫ 1−2δ

δ

∫

1−x

dx δ

∫

φ(x, y)dy =

∫ φ(x, y) dy =

1−x−δ

φ(x, y) dy

where I1 =

δ

φ(x, y) dy, and I2 =

0

δ

+ ∫

φ(x, y)dx .

δ

dx

δ

0

1−y−δ

dy δ

∫

1−2δ

∫

1−2δ

∫

1−x

∫

1−2δ

δ

I1 dx +

∫

1−2δ

δ

I2 dx,

φ(x, y) dy. But

1−x−δ

∫

δ

|I1 | ≤

Mx

λ−1 µ−1

y

ν−1

(1 − x − y)

dy ≤ M x

λ−1

ν−1

(1 − x − δ)

0

∫

δ

y µ−1 dy,

0

since (1 − x − y)ν−1 ≤ (1 − x − δ)ν−1 . Therefore, writing x = (1 − δ)x1 , and since ∫ 1 x1λ−1 (1 − x1 )ν−1 dx1 = B(λ, ν) exists if λ > 0, ν > 0 (see (2) of Example 4.5.1), we 0

have ∫

δ

1−2δ

∫ µ −1 I1 dx ≤ Mδ µ

1−δ

x λ−1 (1 − x − δ)ν−1 dx

0

µ −1

≤ Mδ µ (1 − δ)

λ+ν−1

∫

1

x1λ−1 (1 − x1 )ν−1 dx1

0

< Mδ µ µ−1 (1 − δ)λ+ν−1 B(λ, ν) → 0 as δ → 0. The reader will prove similarly that I2 → 0 as δ → 0. Hence 5 ∫ 1−x ∫ 1 ∫ 1−x ∫ 1−2δ dx φ(x, y) dy = lim dx φ(x, y) dy δ→0 δ 0 0 0 ∫ 1−x−δ ∫ 1−2δ = lim dx φ(x, y) dy δ→0 δ δ ∫ 1−y−δ ∫ 1−2δ = lim dy φ(x, y) dx , δ→0

δ

δ

by what has been already proved; but, by a precisely similar piece of work, the last integral is ∫ 1 ∫ 1−v dy φ(x, y) dx . 0 5

0

The repeated integral exists, and is, in fact, absolutely convergent; for ∫ 1−x ∫ |x λ−1 y µ−1 (1 − x − y)ν−1 f (x, y) dy | < M x λ−1 (1 − x)µ+ν−1 0

1

δ µ−1 (1 − δ)ν−1 ds,

0

∫ 1 M x λ−1 (1 − x)µ+ν−1 dx · δ µ−1 (1 − δ)ν−1 dδ exists. And since the integral 0 ∫ 0 ∫ 1−ε 1−2δ exists, its value which is lim may be written lim .

writing y = (1 − x)δ; and

∫

1

δ , ε→0

δ

δ→0

δ

4.6 Complex integration

75

We have consequently proved the theorem in question. Corollary 4.5.1 continuous, ∫

Writing ξ = a + (b − a)x, η = b − (b − a)y, we see that, if φ(ξ, η) is b

∫

λ−1

(ξ − a)

dξ a

=

b

∫

ξ b

∫

η

dη a

(b − η)

µ−1

ν−1

(η − ξ)

φ(ξ, η) dη

(ξ − a)λ−1 (b − η)µ−1 (η − ξ)ν−1 φ(ξ, η) dξ .

α

This is called Dirichlet’s formula. Note What are now called infinite and improper integrals were defined by Cauchy [124], though the idea of infinite integrals seems to date from Maclaurin [449]. The test for convergence was employed by Chartier [143]. Stokes (1847) distinguished between ‘essentially’ (absolutely) and non-essentially convergent integrals though he did not give a formal definition. Such a definition was given by Dirichlet [179] in 1854 and 1858 (see [179, p. 39]). In the early part of the nineteenth century improper integrals received more attention than infinite integrals, probably because it was not fully realised that an infinite integral is really the limit of an integral.

4.6 Complex integration A treatment of complex integration based on a different set of ideas and not making so many assumptions concerning the curve AB will be found in Watson [650]. Integration with regard to a real variable x may be regarded as integration along a particular path (namely part of the real axis) in the Argand diagram. Let f (z) (= P + iQ), be a function of a complex variable z, which is continuous along a simple curve AB in the Argand diagram. Let the equations of the curve be x = x(t), y = y(t) (a ≤ t ≤ b). Let x(a) + iy(a) = z0 , x(b) + iy(b) = Z. 6 ∫ Then if x(t), y(t) have continuous differential coefficients (see Example 4.1.9) we define Z

f (z) dz taken along the simple curve AB to mean z0 b

dy dx +i dt. (P + iQ) dt dt a s ∫ b 2 2 dx dy The ‘length’ of the curve AB will be defined as + dt. It obviously dt dt a dx dy exists if , are continuous; we have thus reduced the discussion of a complex integral dt dt to the discussion of four real integrals, viz. ∫ b ∫ b ∫ b ∫ b dx dy dx dy P dt, P dt, Q dt, Q dt. dt dt dt dt a a a a ∫

6

This assumption will be made throughout the subsequent work.

The Theory of Riemann Integration

76

By Example 4.1.9, this definition is consistent with the definition of an integral when AB happens to be part of the real axis. ∫ Example 4.6.1

Z

f (z) dz = −

z0

z0

∫

f (z) dz, the paths of integration being the same (but Z

in opposite directions) in each integral. ∫

Z

∫

dz = Z − z0,

z0

Z

z dz =

b

dy dy dx dx −y +i x +y dt dt dt dt dt a t=b 1 1 2 1 2 = (Z 2 − z02 ). x − y + ix y = 2 2 2 t=a

∫

z0

x

4.61 The fundamental theorem of complex integration From §4.13, the reader will easily deduce the following theorem: Let a sequence of points be taken on a simple curve z0 Z; and let the first n of them, rearranged in order of magnitude of their parameters, be called z1(n), z2(n), . . . , zn(n)

(z0(n) = z0,

(n) zn+1 = Z);

let their parameters be t1(n), t2(n), . . . , tn(n), and let the sequence be such that, given any number (n) δ, we can find N such that, when n > N, tr+1 − tr(n) < δ, for r = 0, 1, 2, . . . , n; let ζr(n) be any (n) (n) point whose parameter lies between tr and tr+1 ; then we can make ∫ Z n Õ (n) (n) (n) (Z − Z ) f (ζ ) − f (z) dz r r r=0 r+1 z0 arbitrarily small by taking n sufficiently large.

4.62 An upper limit to the value of a complex integral Let M be the upper bound of the continuous function | f (z)|. Then ∫

z0

Z

∫ f (z) dz ≤

b

dx dy | f (z)| +i dt dt dt a ∫ b ( 2 2 ) 1/2 dx dy dt ≤ M + dt dt a ≤ M L,

∫ where L is the ‘length’ of the curve z0 Z. That is to say,

z0

Z

f (z) dz cannot exceed M L.

4.7 Integration of infinite series

77

4.7 Integration of infinite series We shall now shew that if S(z) = u1 (z) + u2 (z) + · · · is a uniformly convergent series of continuous functions of z, for values of z contained within some region, then the series ∫ ∫ u1 (z)dz + u2 (z)dz + · · · , C

C

(where ∫ all the integrals are taken along some path C in the region) is convergent, and has for sum C S(z) dz. For, writing S(z) = u1 (z) + u2 (z) + · · · + un (z) + Rn (z), we have ∫

S(z)dz =

C

∫

u1 (z)dz + · · · +

C

∫

un (z) dz +

∫

C

Rn (z)dz. C

Now since the series is uniformly convergent, to every positive number ε there corresponds a number r independent of z, such that when n > r we have | Rn (z) | < ε, for all values of z in the region considered. Therefore if L be the length of the path of integration, we have (§4.62) ∫ Rn (z) dz < εL. C

∫ S(z) dz and

Therefore the modulus of the difference between C

n Í m=1

∫ um (z) dz can be C

made less than any positive number, by giving n any sufficiently large ∫ ∫ value. This proves ∞ Í um (z) dz is convergent, and that its sum is S(z) dz. both that the series m=1

C

C

Corollary 4.7.1 As in Corollary 4.4.1, it may be shewn that 7 ∞ ∞ Õ d d Õ un (z) = un (z) dz n=0 dz n=0

if the series on the right converges uniformly and the series on the left is convergent. Example 4.7.1 Consider the series ∞ Õ n=1

2x{n(n + 1) sin2 x 2 − 1} cos x 2 , {1 + n2 sin2 x 2 }{1 + (n + 1)2 sin2 x 2 }

in which x is real. The nth term is 2xn cos x 2 2x(n + 1) cos x 2 − , 1 + n2 sin2 x 2 1 + (n + 1)2 sin2 x 2 7

d f (z) z

means lim

h→0

f (z+h)− f (z) h

where h → 0 along a definite simple curve; this definition is modified slightly in

§5.12 in the case when f (z) is an analytic function.

The Theory of Riemann Integration

78

and the sum of n terms is therefore 2x(n + 1) cos x 2 2x cos x 2 − . 1 + sin2 x 2 1 + (n + 1)2 sin2 x 2 √ Hence the series is absolutely convergent for all real values of x except ± mπ where m = 1, 2, . . .; but Rn (x) =

2x (n + 1) cos x 2 , 1 + (n + 1)2 sin2 x 2

and if n be any integer, by taking x = (n + 1)−1 this has the limit 2 as n → ∞. The series is therefore non-uniformly convergent near x = 0. 2x cos x 2 Now the sum to infinity of the series is , and so the integral from 0 to x of the 1 + sin2 x 2 2 sum of the series is arctan sin x . On the other hand, the sum of the integrals from 0 to x of the first n terms of the series is arctan sin x 2 − arctan (n + 1) sin x 2 , and as n → ∞ this tends to arctan sin x 2 − π2 . Therefore the integral of the sum of the series differs from the sum of the integrals of the terms by π2 . Example 4.7.2 Discuss, in a similar manner, the series ∞ Õ 2en x{1 − n(e − 1) + en+1 x 2 } n(n + 1)(1 + en x 2 )(1 + en+1 x 2 ) n=1

for real values of x. Example 4.7.3 Discuss the series u1 + u2 + u3 + · · · , where u1 = ze−z , 2

un = nze−nz − (n − 1)ze−(n−1)z , 2

2

for real values of z. Hint. The sum of the first n terms is nze−nz , so the sum to infinity is 0 for all real values of z. Since the terms un are real and ultimately all of the same sign, the convergence is absolute. In the series ∫ z ∫ z ∫ z u1 dz + u2 dz + u3 dz + · · · , 2

0

0

0

the sum of n terms is 12 (1 − e−ne ), and this tends to the limit 12 as n tends to infinity; this is Í not equal to the integral from 0 to z of the sum of the series un . The explanation of this discrepancy is to be found in the non-uniformity of the convergence 2 near z = 0, for the remainder after n terms in the series u1 + u2 + · · · is −nze−nz ; and by taking z = n−1 we can make this equal to e−1/n , which is not arbitrarily small; the series is therefore non-uniformly convergent near z = 0. 2

4.8 Miscellaneous examples

Example 4.7.4 (Trinity, 1903) Compare the values of ) ∫ z (Õ ∞ ∞ ∫ Õ un dz and 0

n=1

79

z

un dz,

0

n=1

where un =

2(n + 1)2 z 2n2 z − . (1 + n2 z2 ) log(n + 1) {1 + (n + 1)2 z2 } log(n + 2)

4.8 Miscellaneous examples Example 4.1 (Dirichlet, Du Bois Reymond) Shew that the integrals ∫ ∞ ∫ ∞ ∫ ∞ 2 2 sin(x ) dx, cos(x ) dx, x exp(−x 6 sin2 x) dx 0

0

0

converge. Example 4.2 (Stokes) If a be real, the integral ∫ ∞ cos(ax) dx 1 + x2 0 is a continuous function of a. Example 4.3 (de la Vallée Poussin) ∫ ∞ x sin(x 3 − ax) dx. Hint. Use

Discuss the uniformity of the convergence of

0

∫ 1 1 a a 3 3 x sin(x − ax)dx = − cos(x − ax) − + cos(x 3 − ax) dx + x 3x 3 x2 x4 ∫ sin(x 3 − ax) 1 2 + a dx. 3 x3 ∫ ∞ Example 4.4 (Stokes) Shew that exp[−eia (x 3 − nx)] dx converges uniformly in the 0 1 1 range − π, π of values of a. 2 2 ∫ ∞ x µ dx Example 4.5 (Hardy [275]) Discuss the convergence of when µ, ν, p 1 + xν | sin x| p 0 are positive. ∫

3

Example 4.6 (Math. Trip. 1914) Examine the convergence of the integrals ∫ ∞ ∫ ∞ 1 1 −x 1 dx sin(x + x 2 ) − e + , dx. x 2 1 − ex x xn 0 0 ∫ ∞ dx Example 4.7 Shew that exists. 2 (sin x)2/3 x π

The Theory of Riemann Integration ∫ ∞ Example 4.8 (Math. Trip. 1908) Shew that x −n esin x sin 2x dx converges if a > 0,

80

a

n > 0.

If a series g(z) =

∞ Í

(Cν − Cν+1 ) sin(2ν + 1)πz, (in which π is the derivative of the C0 = 0) converges uniformly in an interval, shew that g(z) sin πz ∞ Í Cν sin 2νπz. series f (z) = ν Example 4.9 (Lerch [430])

v=0

ν=1

Example 4.10 (Math. Trip. 1904) Shew that ∫ ∞∫ ∞ ∫ ∞ dx1 dx2 · · · dxn ··· (x12 + x22 + · · · + xn2 )α ∫ and

∞

∫

∞

∫ ···

∞

dx1 dx2 · · · dxn x1α

+ x2β + · · · + xnλ

converge when α > 12 n and α−1 + β−1 + · · · + λ−1 < 1 respectively. Example 4.11 (Bôcher) If f (x, y) be a continuous function of both x and y in the ranges (a ≤ x ≤ b), (a ≤ y ≤ b) except that it has ordinary discontinuities at points on a finite number of curves, with continuously turning tangents, each of which meets any line parallel ∫ b to the coordinate axes only a finite number of times, then f (x, y) dx is a continuous function of y. Hint. Consider ∫ ∫ a1 −δ1 ∫ a2 −δ2 +··· + + a

a1 +ε1

a b

a n +εn

{ f (x, y + h) − f (x, y)} dx,

where the numbers δ1, δ2, . . . , ε1, ε2, . . . are so chosen as to exclude the discontinuities of f (x, y + h) from the range of integration; a1, a2, . . . being the discontinuities of f (x, y).

5 The Fundamental Properties of Analytic Functions; Taylor’s, Laurent’s and Liouville’s Theorems 5.1 Property of the elementary functions The reader will be already familiar with the term elementary function, as used (in textbooks on Algebra, Trigonometry, and the Differential Calculus) to denote certain analytical expressions 1 depending on a variable z, the symbols involved therein being those of elementary algebra together with exponentials, logarithms and the trigonometrical functions; examples of such expressions are z 2,

ez ,

log z,

arcsin z 3/2 .

Such combinations of the elementary functions of analysis have in common a remarkable property, which will now be investigated. Take as an example the function ez . Write ez = f (z). Then, if z be a fixed point and if z 0 be any other point, we have 0

0

(z −z) f (z 0) − f (z) ez − ez −1 z e = = e · 0 0 0 z −z z −z z −z 0 0 z − z (z − z)2 z = e 1+ + + ···+ ; 2! 3!

and since the last series in brackets is uniformly convergent for all values of z 0, it follows (§3.7) that, as z 0 → z, the quotient f (z 0) − f (z) z0 − z tends to the limit ez , uniformly for all values of arg(z 0 − z). This shews that the limit of f (z 0) − f (z) z0 − z is in this case independent of the path by which the point z 0 tends towards coincidence with z. It will be found that this property is shared by many of the well-known elementary functions; namely, that if f (z) be one of these functions and h be any complex number, the limiting value of 1 { f (z + h) − f (z)} h 1

The reader will observe that this is not the sense in which the term function is defined (§3.1) in this work. Thus e.g. x − iy and |z | are functions of z (= x + iy) in the sense of §3.1, but are not elementary functions of the type under consideration.

81

82

The Fundamental Properties of Analytic Functions

exists and is independent of the mode in which h tends to zero. The reader will, however, easily prove that, if f (z) = x − iy, where z = x + iy, then f (z + h) − f (z) is not independent of the mode in which h → 0. lim h

5.11 Occasional failure of the property For each of the elementary functions, however, there will be certain points z at which this property will cease to hold good. Thus it does not hold for the function 1/(z − a) at the point z = a, since 1 1 1 − lim h→0 h z − a + h z−a does not exist when z = a. Similarly it does not hold for the functions log z and z1/2 at the point z = 0. These exceptional points are called singular points or singularities of the function f (z) under consideration; at other points f (z) is said to be analytic. The property does not hold good at any point for the function |z|.

5.12 Cauchy’s definition of an analytic function of a complex variable (See the memoir [121]). The property considered in §5.11 will be taken as the basis of the definition of an analytic function, which may be stated as follows. Let a two-dimensional region in the z-plane be given; and let u be a function of z defined uniquely at all points of the region. Let z, z + δz be values of the variable z at two points, δu and u, u + δu the corresponding values of u. Then, if, at any point z within the area, δz tends to a limit when δx → 0, δy → 0, independently (where δz = δx + iδy), u is said to be a function of z, which is monogenic or analytic at the point. The words ‘regular’ and ‘holomorphic’ are sometimes used. A distinction has been made by Borel [86, p. 137–138], [87] between ‘monogenic’ and ‘analytic’ functions in the case of functions with an infinite number of singularities. See §5.51. If the function is analytic and one-valued at all points of the region, we say that the function is analytic throughout the region. See the footnote after Corollary 5.2.2. We shall frequently use the word ‘function’ alone to denote an analytic function, as the functions studied in this work will be almost exclusively analytic functions. In the foregoing definition, the function u has been defined only within a certain region in the z-plane. As will be seen subsequently, however, the function u can generally be defined for other values of z not included in this region; and (as in the case of the elementary functions already discussed) may have singularities, for which the fundamental property no longer holds, at certain points outside the limits of the region. We shall now state the definition of analytic functionality in a more arithmetical form. Let f (z) be analytic at z, and let ε be an arbitrary positive number; then we can find numbers ` and δ (with δ depending on ε) such that f (z 0) − f (z) 0 unless f (z) = 0 or f 0(z) = 0.

5.3 Analytic functions represented by uniformly convergent series Í∞

Let n=0 fn (z) be a series such that: (i) it converges uniformly along a contour C; (ii) fn (z) is Í analytic throughout C and its interior. Then ∞ n=0 fn (z) converges, and the sum of the series is an analytic function throughout C and its interior. Í For let a be any point inside C; on C, let ∞ n=0 fn (z) = Φ(z). Then ) ∫ ∫ (Õ ∞ 1 Φ(z) 1 dz dz = fn (z) 2πi C z − a 2πi C n=0 z−a ∫ ∞ Õ 1 fn (z) = dz , 2πi C z − a n=0

The Fundamental Properties of Analytic Functions

90

by §4.7. Since |z − a| −1 is bounded when a is fixed and z is on C, the uniformity of the ∞ ∞ Í Í convergence of fn (z)/(z − a) follows from that of fn (z). But this last series, by §5.21, n=0

is

∞ Í n=0

n=0

fn (a); the series under consideration therefore converges at all points inside C; let its

sum inside C (as well as on C) be called Φ(z). Then the function is analytic if it has a unique differential coefficient at all points inside C. But if a and a + h are inside C, ∫ 1 Φ(a + h) − Φ(a) Φ(z) dz = , h 2πi C (z − a)(z − a − h) and hence, as in §5.22, lim {Φ(a + h) − Φ(a)} h−1 exists and is equal to h→0

1 2πi

∫ C

Φ(z) dz ; (z − a)2

and therefore Φ(z) is analytic inside C. Further, by transforming the last integral in the same ∞ ∞ Í Í fn (a) may be way as we transformed the first one, we see that Φ0(a) = fn0(a), so that n=0

n=0

differentiated term by term. If a series of analytic functions converges only at points of a curve which is not closed nothing can be inferred as to the convergence of the derived series. This might have been anticipated as the main theorem of this section deals with uniformity of convergence over a two-dimensional region. ∞ Í cos nx Thus (−1)n converges uniformly for real values of x (§3.34). But the derived n2 n=1 ∞ Í sin nx series (−1)n−1 converges non-uniformly near x = (2m + 1)π, (m any integer); and n n=1 ∞ Í the derived series of this, viz. (−1)n−1 cos nx, does not converge at all. n=1

Corollary 5.3.1 By §3.7, the sum of a power series is analytic inside its circle of convergence.

5.31 Analytic functions represented by integrals Let f (t, z) satisfy the following conditions when t lies on a certain path of integration (a, b) and z is any point of a region S: ∂f are continuous functions of t. ∂z 2. f is an analytic function of z. ∂f 3. The continuity of qua function of z is uniform with respect to the variable t. ∂z ∫ b Then f (t, z) dt is an analytic function of z. For, by §4.2, it has the unique derivative a ∫ b ∂ f (t, z) dt. ∂z a 1. f and

5.4 Taylor’s theorem

91

5.32 Analytic functions represented by infinite integrals ∫ ∞ From Corollary 4.4.1, it follows that f (t, z) dt is an analytic function of z at all points a

of a region S if

(i) the integral converges, (ii) f (t, z) is an analytic function of z when t is on the path of integration and z is on S, ∂ f (t, z) is a continuous function of both variables, (iii) ∂z ∫ ∞ ∂ f (t, z) (iv) dt converges uniformly throughout S. ∂z a ∫ ∞ For if these conditions are satisfied f (t, z)dt has the unique derivative a

∫ a

∞

∂ f (t, z) dt. ∂z ∫

A case of very great importance is afforded by the integral

∞

e−tz f (t) dt, where f (t)

0

is continuous and | f (t)| < Kert where K, r are independent of t; it is obvious from the conditions stated that the integral is an ∫ analytic function of z when R(z) ≥ r1 > r. Condition ∞

te(r−r1 )t dt converges.

(iv) is satisfied, by §4.431 (I), since 0

5.4 Taylor’s theorem Consider a function f (z), which is analytic in the neighborhood of a point z = a. Let C be a circle with a as centre in the z-plane, which does not have any singular point of the function f (z) on or inside it; so that f (z) is analytic at all points on and inside C. Let z = a + h be any point inside the circle C. Then, by §5.21, we have ∫ f (z) dz 1 f (a + h) = 2πi C z − a − h ∫ 1 1 hn h n+1 h = f (z) +···+ + dz + 2πi C z − a (z − a)2 (z − a)n+1 (z − a)n+1 (z − a − h) ∫ h2 00 h n (n) 1 f (z) dz = f (a) + h f 0(a) + f (a) + · · · + f (a) + h n+1 . n+1 2! n! 2πi C (z − a) (z − a − h) But when z is on C, the modulus of f (z)/(z−a−h) is continuous, and so, by Corollary 3.6.2, will not exceed some finite number M. Therefore, by §4.62, ∫ n+1 M · 2πR |h| n+1 1 f (z) dz · h ≤ , 2πi n+1 (z − a − h) 2π R C (z − a) where R is the radius of the circle C, so that 2πR is the length of the path of integration in the last integral, and R = |z − a| for points z on the circumference of C.

The Fundamental Properties of Analytic Functions

92

The right-hand side of the last inequality tends to zero as n → ∞. We have therefore f (a + h) = f (a) + h f 0(a) +

h n (n) h2 00 f (a) + · · · + f (a) + · · · , 2! n!

which we can write (z − a)n (n) (z − a)2 00 f (a) + · · · + f (a) + · · · . 2! n! This result is known as Taylor’s theorem; and the proof given is due to Cauchy. The formal expansion was first published by Dr. Brook Taylor [621]. It follows that the radius of convergence of a power series is always at least so large as only just to exclude from the interior of the circle of convergence the nearest singularity of the function represented by the series. And by Corollary 5.3.1 it follows that the radius of convergence is not larger than the number just specified. Hence the radius of convergence is just such as to exclude from the interior of the circle that singularity of the function which is nearest to a. f (z) = f (a) + (z − a) f 0(a) +

At this stage we may introduce some terms which will be frequently used. If f (a) = 0, the function f (z) is said to have a zero at the point z = a. If at such a point f 0(a) is different from zero, the zero of f (a) is said to be simple; if, however, f 0(a), f 00(a), . . . , f (n−1) (a) are all zero, so that the Taylor’s expansion of f (z) at z = a begins with a term in (z − a)n , then the function f (z) is said to have a zero of the nth order at the point z = a. Example 5.4.1 Find the function f (z), which is analytic throughout the circle C and its interior, whose centre is at the origin and whose radius is unity, and has the value a − cos θ sin θ +i 2 a2 − 2a cos θ + 1 a − 2a cos θ + 1 (where a > 1 and θ is the vectorial angle) at points on the circumference of C. We have ∫ f (z) dz n! f (n) (0) = 2πi C z n+1 ∫ 2π n! a − cos θ + i sin θ = e−niθ · idθ · 2 , (putting z = eiθ ) 2πi 0 a − 2a cos θ + 1 n ∫ 2π −niθ ∫ n! e dθ n! dz d 1 = = = 2π 0 a − eiθ 2πi C z n+1 (a − z) dz n a − z z=0 n! = n+1 . a Therefore by Maclaurin’s theorem 4 , f (z) = 4

∞ Õ zn , a n+1 n=0

The result f (z) = f (0) + z f 0 (0) + z2 f 00 (0) + · · · , obtained by putting a = 0 in Taylor’s theorem, is usually called Maclaurin’s theorem; it was discovered by Stirling in 1717 and published by Maclaurin [449]. 2

5.4 Taylor’s theorem

93

or f (z) = (a − z)−1 for all points within the circle. This example raises the interesting question, will it still be convenient to define f (z) as (a − z)−1 at points outside the circle? This will be discussed in §5.51. ∞ Í Example 5.4.2 Prove that the arithmetic mean of all values of z−n aν zν for points z on ν=0 Í the circumference of the circle |z| = 1, is an ; if aν zν is analytic throughout the circle and ∞ Í f (ν) (0) . Then, writing z = eiθ , and its interior. Solution. Let aν zν = f (z), so that aν = ν! ν=0 calling C the circle |z| = 1,

1 2π

∫

2π

0

1 f (z) dθ = zn 2πi

∫ C

f (n) (0) f (z) dz = = an . z n+1 n!

Example 5.4.3 Let f (z) = zr ; then f (z + h) is an analytic function of h when |h| < |z| for r(r − 1) r−2 2 all values of r; and so (z + h)r = zr + r zr−1 h + z h + · · · , this series converging 2 when |h| < |z|. This is the binomial theorem. Example 5.4.4 Prove that if h is a positive constant, and (1 − 2zh + h2 )−1/2 is expanded in the form 1 + hP1 (z) + h2 P2 (z) + h3 P3 (z) + · · ·

(5.2)

(where Pn (z) is easily seen to be a polynomial of degree n in z), then this series converges so long as z is in the interior of an ellipse whose foci are the points z = 1 and z = −1, and whose semi-major axis is 12 (h + h−1 ). Let the series be first regarded as a function of h. It is a power series in h, and therefore converges so long as the point h lies within a circle in the h-plane. The centre of this circle is the point h = 0, and its circumference will be such as to pass through that singularity of (1 − 2zh + h2 )−1/2 which is nearest to h = 0. But 1 − 2zh + h2 = {h − z + (z2 − 1)1/2 }{h − z − (z2 − 1)1/2 },

(5.3)

so the singularities of (1−2zh+ h2 )−1/2 are the points h = z −(z2 −1)1/2 and h = z +(z2 −1)1/2 . These singularities are branch points (see §5.7). Thus the series (5.2) converges so long as |h| is less than both |z − (z2 − 1)1/2 |

and

|z + (z2 − 1)1/2 |.

Draw an ellipse in the z-plane passing through the point z and having its foci at ±1. Let a be its semi-major axis, and θ the eccentric angle of z on it. Then z = a cos θ +i(a2 −1)1/2 sin θ, which gives z±(z2 −1)1/2 = {a±(a2 −1)1/2 }(cos θ ±i sin θ), so |z±(z2 −1)1/2 | = a±(a2 −1)1/2 . Thus the series (5.2) converges so long as h is less than the smaller of the numbers a+(a2 −1)1/2 and a − (a2 − 1)1/2 , i.e. so long as h is less than a − (a2 − 1)1/2 . But h = a − (a2 − 1)1/2 when a = 21 (h + h−1 ). Therefore the series (5.2) converges so long as z is within an ellipse whose foci are 1 and −1, and whose semi-major axis is 12 (h + h−1 ).

The Fundamental Properties of Analytic Functions

94

5.41 Forms of the remainder in Taylor’s series Let f (x) be a real function of a real variable; and let it have continuous differential coefficients of the first n orders when a ≤ x ≤ a + h. If 0 ≤ t ≤ 1, we have ) ( n−1 h n (1 − t)n−1 (n) d Õ hm (1 − t)m f (m) (a + th) = f (a + th) − h f 0(a + th). dt m=1 m! (n − 1)! Integrating this between the limits 0 and 1, we have ∫ 1 n n−1 m Õ h (m) h (1 − t)n−1 (n) f (a + h) = f (a) + f (a) + f (a + th) dt. m! (n − 1)! 0 m=1 Let Rn =

hn (n − 1)!

∫

1

(1 − t)n−1 f (n) (a + th) dt;

0

and let p be a positive integer such that p ≤ n. Then ∫ 1 hn Rn = (1 − t) p−1 · (1 − t)n−p f (n) (a + th) dt. (n − 1)! 0 Let U, L be the upper and lower bounds of (1 − t)n−p f (n) (a + th). Then ∫ 1 ∫ 1 ∫ 1 L(1 − t) p−1 dt < (1 − t) p−1 · (1 − t)n−p f (n) (a + th) dt < U(1 − t) p−1 dt. 0

0

0

Since (1 − t) f (a + th) is a continuous function it passes through all values between U and L, and hence we can find θ such that 0 ≤ θ ≤ 1, and ∫ 1 (1 − t)n−1 f (n) (a + th) dt = p−1 (1 − θ)n−p f (n) (a + θh). (n)

n−p

0

h n (n) hn (1 − θ)n−p f (n) (a + θh). Writing p = n, we get Rn = f (a + θh), (n − 1)!p n! which is Lagrange’s form for the remainder; and writing p = 1, we get Therefore Rn =

Rn =

hn (1 − θ)n−1 f (n) (a + θh), (n − 1)!

which is Cauchy’s form for the remainder. Taking n = 1 in this result, we get f (a + h) − f (a) = h f 0(a + θh) if f 0(x) is continuous when a ≤ x ≤ a+h; this result is usually known as the First Mean-Value Theorem (see also §4.14). Darboux [162, p. 291] gave a form for the remainder in Taylor’s series, which is applicable to complex variables and resembles the above form given by Lagrange for the case of real variables.

5.5 The process of continuation

95

5.5 The process of continuation Near every point P, z0 , in the neighbourhood of which a function f (z) is analytic, we have seen that an expansion exists for the function as a series of ascending positive integral powers of (z − z0 ), the coefficients in which involve the successive derivatives of the function at z0 . Now let A be the singularity of f (z) which is nearest to P. Then the circle within which this expansion is valid has P for centre and P A for radius. Suppose that we are merely given the values of a function at all points of the circumference of a circle slightly smaller than the circle of convergence and concentric with it together with the condition that the function is to be analytic throughout the interior of the larger circle. Then the preceding theorems enable us to find its value at all points within the smaller circle and to determine the coefficients in the Taylor series proceeding in powers of z − z0 . The question arises, Is it possible to define the function at points outside the circle in such a way that the function is analytic throughout a larger domain than the interior of the circle? In other words, given a power series which converges and represents a function only at points within a circle, to define by means of it the values of the function at points outside the circle. For this purpose choose any point P1 within the circle, not on the line P A. We know the value of the function and all its derivatives at P1 , from the series, and so we can form the Taylor series (for the same function) with P1 as origin, which will define a function analytic throughout some circle of centre P1 . Now this circle will extend as far as the singularity (of the function defined by the new series) which is nearest to P1 , which may or may not be A; but in either case, this new circle will usually 5 lie partly outside the old circle of convergence, and for points in the region which is included in the new circle but not in the old circle, the new series may be used to define the values of the function, although the old series failed to do so. Similarly we can take any other point P2 , in the region for which the values of the function are now known, and form the Taylor series with P2 as origin, which will in general enable us to define the function at other points, at which its values were not previously known; and so on. This process is called continuation. By means of it, starting from a representation of a function by any one power series we can find any number of other power series, which between them define the value of the function at all points of a domain, any point of which can be reached from P without passing through a singularity of the function; and the aggregate 6 of all the power series thus obtained constitutes the analytical expression of the function. Note It is important to know whether continuation by two different paths PBQ, PB 0Q will give the same final power series; it will be seen that this is the case, if the function have no singularity inside the closed curve PBQB 0 P, in the following way: Let P1 be any point on PBQ, inside the circle C with centre P; obtain the continuation of the function with P1 0 as origin, and let it converge inside a circle C1 ; let P1 be any point inside both circles and 0 0 also inside the curve PBQB 0 P; let S, S1, S1 be the power series with P, P1, P1 as origins; then 5

6

The word ‘usually’ must be taken as referring to the cases which are likely to come under the reader’s notice while studying the less advanced parts of the subject. Such an aggregate of power series has been obtained for various functions by M. J. M. Hill [307], by purely algebraical processes.

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The Fundamental Properties of Analytic Functions 0

0

(since each is equal to S), S1 ≡ S1 over a certain domain which will contain P1 , if P1 be taken 0 sufficiently near P1 ; and hence S1 will be the continuation of S1 ; for if T1 were the continuation 0 of S1 , we would have T1 ≡ S1 over a domain containing P1 , and so (§3.73) corresponding coefficients in S1 and T1 are the same. By carrying out such a process a sufficient number of times, we deform the path PBQ into the path PB 0Q if no singular point is inside PBQB 0 P. The reader will convince himself by drawing a figure that the process can be carried out in a finite number of steps. Example 5.5.1 The series z z2 z3 1 + 2 + 3 + 4 +··· a a a a represents the function f (z) =

1 a−z

only for points z within the circle |z| = |a|. But any number of other power series exist, of the type z−b (z − b)2 (z − b)3 1 + + + +··· ; a − b (a − b)2 (a − b)3 (a − b)4 if b/a is not real and positive these converge at points inside a circle which is partly inside and partly outside |z| = |a|; these series represent this same function at points outside this circle.

5.501 On functions to which the continuation-process cannot be applied It is not always possible to carry out the process of continuation. Take as an example the function f (z) defined by the power series n

f (z) = 1 + z2 + z4 + z8 + z16 + · · · + z 2 + · · · , which clearly converges in the interior of a circle whose radius is unity and whose centre is at the origin. Now it is obvious that, as z → 1−, f (z) → +∞; the point +1 is therefore a singularity of f (z). But f (z) = z2 + f (z2 ), and if z2 → 1−, f (z2 ) → ∞ and so f (z) → ∞, and hence the points for which z2 = 1 are singularities of f (z); the point z = −1 is therefore also a singularity of f (z). Similarly since f (z) = z2 + z4 + f (z4 ), we see that if z is such that z4 = 1, then z is a singularity of f (z); and, in general, any root of any of the equations z2 = 1,

z4 = 1,

z8 = 1,

z16 = 1, . . . ,

is a singularity of f (z). But these points all lie on the circle |z| = 1; and in any arc of this circle, however small, there are an unlimited number of them. The attempt to carry out the process of continuation will therefore be frustrated by the existence of this unbroken front of singularities, beyond which it is impossible to pass. In such a case the function f (z) cannot be continued at all to points z situated outside the

5.5 The process of continuation

97

circle |z| = 1; such a function is called a lacunary function, and the circle is said to be a limiting circle for the function.

5.51 The identity of two functions The two series 1 + z + z2 + z3 + · · · and −1 + (z − 2) − (z − 2)2 + (z − 2)3 − (z − 2)4 + · · · do not both converge for any value of z, and are distinct expansions. Nevertheless, we generally say that they represent the same function, on the strength of the fact that they can both be represented by the same rational expression 1/(1 − z). This raises the question of the identity of two functions. When can two different expansions be said to represent the same function? We might define a function (after Weierstrass), by means of the last article, as consisting of one power series together with all the other power series which can be derived from it by the process of continuation. Two different analytical expressions will then define the same function, if they represent power series derivable from each other by continuation. Since if a function is analytic (in the sense of Cauchy §5.12) at and near a point it can be expanded into a Taylor’s series, and since a convergent power series has a unique differential coefficient (§5.3), it follows that the definition of Weierstrass is really equivalent to that of Cauchy. It is important to observe that the limit of a combination of analytic functions can represent different analytic functions in different parts of the plane. This can be seen by considering the series Õ ∞ 1 1 1 1 1 z− − . z+ + 2 z z 1 + z n 1 + z n−1 n=1 The sum of the first n + 1 terms of this series is 1 1 1 . . + z− z z 1 + zn The series therefore converges for all values of z (zero excepted) not on the circle |z| = 1. But, as n → ∞, |z n | → 0 or |z n | → ∞ according as |z| is less or greater than unity; hence we see that the sum to infinity of the series is z when |z| < 1, and 1/z when |z| > 1. This series therefore represents one function at points in the interior of the circle |z| = 1, and an entirely different function at points outside the same circle. The reader will see from §5.3 that this result is connected with the non-uniformity of the convergence of the series near |z| = 1. Note It has been shewn by Borel [86] that if a region C is taken and a set of points S such that points of the set S are arbitrarily near every point of C, it may be possible to define a function which has a unique differential coefficient (i.e. is monogenic) at all points of C which do not belong to S; but the function is not analytic in C in the sense of Weierstrass. The functions are not monogenic strictly in the sense of §5.1 because, in the example quoted,

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98

in working out { f (z + h) − f (z)} /h, it must be supposed that Re(z + h) and Im(z + h) are not both rational fractions. Such a function is ∞ Õ n Õ n Õ exp(− exp n4 ) . f (z) = z − (p + qi)/n n=1 p=0 q=0

5.6 Laurent’s theorem A very important theorem was published in 1843 by Laurent [413]. The theorem is contained in a paper which was written by Weierstrass [662, p. 51–66], but apparently not published before 1894. It relates to expansions of functions to which Taylor’s theorem cannot be applied. Let C and C 0 be two concentric circles of centre a, of which C 0 is the inner; and let f (z) be a function which is analytic at all points on C and C 0 and throughout the annulus between C and C 0. Let a + h be any point in this ring-shaped space. Then we have (Corollary 5.2.4) ∫ ∫ 1 f (z) 1 f (z) f (a + h) = dz − dz, 2πi C z − a − h 2πi C 0 z − a − h where the integrals are supposed taken in the positive or counter-clockwise direction round the circles. This can be written as ∫ 1 1 hn h f (a + h) = f (z) + · · · + + + 2πi C z − a (z − a)2 (z − a)n+1 h n+1 dz + (z − a)n+1 (z − a − h) ∫ 1 1 z−a (z − a)n (z − a)n+1 f (z) − dz. + 2 +···+ 2πi C 0 h h h n+1 h n+1 (z − a − h) We find, as in the proof of Taylor’s theorem, that ∫ ∫ f (z) dz f (z)(z − a)n+1 n+1 dz h and n+1 (z − a − h) n+1 C (z − a) C 0 (z − a − h)h tend to zero as n → ∞; and thus we have b1 b2 f (a + h) = a0 + a1 h + a2 h2 + · · · + + 2 +··· , h h ∫ ∫ 1 1 f (z) dz where an = and bn = (z − a)n−1 f (z) dz. We cannot write an = 2πi C (z − a)n+1 2πi C 0 f (n) (a)/n! as in Taylor’s theorem since f (z) is not necessarily analytic inside C 0. This result is Laurent’s theorem; changing the notation, it can be expressed in the following form: If f (z) be analytic on the concentric circles C and C 0 of centre a, and throughout the annulus between them, then at any point z of the annulus f (z) can be expanded in the form f (z) = a0 + a1 (z − a) + a2 (z − a)2 + · · · +

b2 b1 + +··· , (z − a) (z − a)2

where 1 an = 2πi

∫ C

f (t) dt (t − a)n+1

and

1 bn = 2πi

∫ C0

(t − a)n−1 f (t) dt.

5.6 Laurent’s theorem

99

An important case of Laurent’s theorem arises when there is only one singularity within the inner circle C 0, namely at the centre a. In this case the circle C 0 can be taken as small as we please, and so Laurent’s expansion is valid for all points in the interior of the circle C, except the centre a. Example 5.6.1 Prove that x

e 2 (z−1/z) = J0 (x) + zJ1 (x) + z2 J2 (x) + · · · + z n Jn (x) + · · · 1 1 (−)n − J1 (x) + 2 J2 (x) − · · · + n Jn (x) + · · · , z z z ∫ 2π 1 where Jn (x) = cos(nθ − x sin θ) dθ. 2π 0 For the function of z under consideration is analytic in any domain which does not include the point z = 0; and so by Laurent’s theorem, x

e 2 (z−1/z) = a0 + a1 z + a2 z2 + · · · +

b1 b2 + 2 +··· , z z

where an =

∫

1 2π i

e 2 (z− z ) x

1

C

dz z n+1

and 1 bn = 2π i

∫

e 2 (z− z ) z n−1 dz, x

1

C0

0

and where C and C are any circles with the origin as centre. Taking C to be the circle of radius unity, and writing z = eiθ , we have ∫ 2π ∫ 2π 1 1 ix sin θ −niθ an = e · e idθ = cos (nθ − x sin θ) dθ, 2πi 0 2π 0 ∫ 2π since sin(nθ − x sin θ) dθ vanishes, as may be seen by writing 2π − φ for θ. Thus 0

an = Jn (x), and bn = (−1)n an , since the function expanded is unaltered if −z−1 be written for z, so that bn = (−1)n Jn (x), and the proof is complete. Example 5.6.2 Shew that, in the annulus defined by |a| < |z| < |b|, the function 1/2 bz (z − a) (b − z) can be expanded in the form S0 +

∞ Õ n=1

Sn

an z n + , z n bn

where Sn =

∞ Õ 1 · 3 · · · (2l − 1) · 1 · 3 · · · (2l + 2n − 1) a l . b 22l+n · l ! (l + n) ! l=0

The function is one-valued and analytic in the annulus (see §5.7), for the branch-points 0, a

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neutralise each other, and so, by Laurent’s theorem, if C denotes the circle | z | = r, where | a | < r < | b |, the coefficient of z n in the required expansion is 12 ∫ bz 1 dz . 2π i C z n+1 (z − a) (b − z) Putting z = reiθ , this becomes ∫ 2π −1/2 a 1 r −1/2 1 − e−iθ , e−niθ r −n dθ 1 − eiθ 2π 0 b r or ∫ 2π ∞ ∞ Õ 1 1 · 3 · · · (2k − 1) r k eikθ Õ 1 · 3 · · · (2l − 1) al e−ilθ −inθ −n , e r dθ 2π 0 2k · k ! bk l=0 2l · l! rl k=0 the series being absolutely convergent and uniformly convergent with regard to θ. The only terms which give integrals different from zero are those for which k = l + n. So the coefficient of z n is ∫ 2π Õ ∞ 1 · 3 · · · (2l − 1) 1 · 3 · · · (2l + 2n − 1) al 1 Sn dθ = n. l ·l! l+n (l + n) ! l+n 2π 0 b 2 2 b l=0 Similarly it can be shewn that the coefficient of z−n is Sn a n . Example 5.6.3 Shew that enz+v/z = a0 + a1 z + a2 z2 + · · · +

b1 b2 + 2 +··· , z z

where 1 an = 2π

∫

1 bn = 2π

∫

2π

e(u+v) cos θ cos {(u − v) sin θ − nθ} dθ,

0

and 2π

e(u+v) cos θ cos {(v − u) sin θ − nθ} dθ.

0

5.61 The nature of the singularities of one-valued functions Consider first a function f (z) which is analytic throughout a closed region S, except at a single point a inside the region. Let it be possible to define a function φ(z) such that (i) φ(z) is analytic throughout S, B2 Bn B1 + +···+ . (ii) when z , a, f (z) = φ(z) + z − a (z − a)2 (z − a)n B1 B2 Bn + +· · ·+ 2 z − a (z − a) (z − a)n are called the principal part of f (z) near a. By the definition of a singularity (§5.12) a pole is a singularity. If n = 1, the singularity is called a simple pole. Any singularity of a one-valued function other than a pole is called an essential singularity. Then f (z) is said to have a pole of order n at a; and the terms

5.6 Laurent’s theorem

101

If the essential singularity, a, is isolated (i.e. if a region, of which a is an interior point, can be found containing no singularities other than a), then a Laurent expansion can be found, in ascending and descending powers of (z − a) valid when ∆ > | z − a | > δ, where ∆ depends on the other singularities of the function, and δ is arbitrarily small. Hence the ‘principal part’ of a function near an isolated essential singularity consists of an infinite series. It should be noted that a pole is, by definition, an isolated singularity, so that all singularities which are not isolated (e.g. the limiting point of a sequence of poles) are essential singularities. Note There does not exist, in general, an expansion of a function valid near a non-isolated singularity in the way that Laurent’s expansion is valid near an isolated singularity. Corollary 5.6.1 If f (z) has a pole of order n at a, and ψ(z) = (z − a)n f (z) (z , a),

ψ(a) = lim (z − a)n f (z), z→a

then ψ(z) is analytic at a. Example 5.6.4 A function is not bounded near an isolated essential singularity. Hint. Prove that if the function were bounded near z = a, the coefficients of negative powers of z − a would all vanish. Example 5.6.5 Find the singularities of the function ec/(z−a) . (5.4) ez/a − 1 At z = 0, the numerator is analytic, and the denominator has a simple zero. Hence the function has a simple pole at z = 0. Similarly there is a simple pole at each of the points 2πnia (n = ±1, ±2, ±3, . . .); the denominator is analytic and does not vanish for other values of z. At z = a, the numerator has an isolated singularity, so Laurent’s theorem is applicable, and the coefficients in the Laurent expansion may be obtained from the quotient 1+

c z−a

+

exp 1 +

c2 +··· 2 ! (z−a)2 , z−a +··· −1 a

(5.5)

which gives an expansion involving all positive and negative powers of (z − a). So there is an essential singularity at z = a. Example 5.6.6 (Math. Trip. 1899) Shew that the function defined by the series ∞ Õ nz n−1 (1 + n−1 )n − 1 (z n − 1) {z n − (1 + n−1 )n } n=1 has simple poles at the points z = (1 + n−1 ) e2kiπ/n , (k = 0, 1, 2, . . . , n − 1; n = 1, 2, 3, . . .).

5.62 The ‘point at infinity’ The behaviour of a function f (z) as | z | → ∞ can be treated in a similar way to its behaviour as z tends to a finite limit. If we write z = 1/z 0, so that large values of z are represented by small values of z 0 in the 0 z -plane, there is a one-one correspondence between z and z 0, provided that neither is zero;

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102

and to make the correspondence complete it is sometimes convenient to say that when z 0 is the origin, z is the ‘point at infinity’. But the reader must be careful to observe that this is not a definite point, and any proposition about it is really a proposition concerning the point z 0 = 0. Let f (z) = φ(z 0). Then φ(z 0) is not defined at z 0 = 0, but its behaviour near z 0 = 0 is determined by its Taylor (or Laurent) expansion in powers of z 0 ; and we define φ(0) as lim φ(z 0) if that limit exists. For instance the function φ(z 0) may have a zero of order m at z →0

the point z 0 = 0; in this case the Taylor expansion of φ(z 0) will be of the form A z 0 m + Bz 0 m+1 + Cz 0 m+2 + · · · , and so the expansion of f (z) valid for sufficiently large values of | z | will be of the form f (z) =

B C A + + +··· . z m z m+1 z m+2

In this case, f (z) is said to have a zero of order m at ‘infinity’. Again, the function φ(z 0) may have a pole of order m at the point z 0 = 0; in this case φ(z 0) =

A B C L + 0m−1 + 0m−2 + · · · + 0 + M + N z 0 + Pz 02 + · · · ; 0m z z z z

and so, for sufficiently large values of |z|, f (z) can be expanded in the form f (z) = Az m + Bz m−1 + Cz m−2 + · · · + Lz + M +

N P + 2 +··· . z z

In this case, f (z) is said to have a pole of order m at ‘infinity’. Similarly f (z) is said to have an essential singularity at infinity, if φ(z 0) has an essential singularity at the point z 0 = 0. Thus the function ez has an essential singularity at infinity, 0 since the function e1/z or 1 1 1 + +··· 1+ 0 + 0 2 z 2! z 3! z 0 3 has an essential singularity at z 0 = 0. Example 5.6.7 Discuss the function represented by the series ∞ Õ 1 1 , n ! 1 + a2n z2 n=0

(a > 1).

Hint. The function represented by this series has singularities at z = a−n and z = −ia−n , (n = 1, 2, 3, . . .), since at each of these points the denominator of one of the terms in the series is zero. These singularities are on the imaginary axis, and have z = 0 as a limiting point; so no Taylor or Laurent expansion can be formed for the function valid throughout any region of which the origin is an interior point. For values of z, other than these singularities, the series converges absolutely, since the limit of the ratio of the (n + 1)th term to the nth is lim (n + 1)−1 a−2 = 0. The function is an n→∞

even function of z (i.e. is unchanged if the sign of z be changed), tends to zero as |z| → ∞,

5.6 Laurent’s theorem

103

and is analytic on and outside a circle C of radius greater than unity and centre at the origin. So, for points outside this circle, it can be expanded in the form b2 b4 b6 + + +··· , z2 z4 z6 where, by Laurent’s theorem, b2k

1 = 2πi

∫ z2k−1 C

∞ Õ a−2n 1 dz. n ! a− 2n + z2 n=0

Now ∞ Õ n=0

∞

∞

Õ Õ z2k−3 a−2n a−2n z2k−1 = (−1)m a−2nm z−2m . n ! (a− 2n + z2 ) n=0 m=0 n!

This double series converges absolutely when |z| > 1, and if it be rearranged in powers of z it converges uniformly. ∞ (−1)k−1 a−2kn Í Since the coefficient of z−1 is and the only term which furnishes a nonn! n=0 zero integral is the term in z−1 , we have ∫ Õ ∞ 1 (−1)k−1 a−2kn dz b2k = 2πi C n=0 n! z =

∞ Õ (−1)k−1 n ! a2kn n=0

= (−1)k−1 e1/a . 2k

Therefore, when |z| > 1, the function can be expanded in the form 2

4

6

e1/a e1/a e1/a − 4 + 6 −··· . 2 z z z The function has a zero of the second order at infinity, since the expansion begins with a term in z−2 .

5.63 Liouvillle’s theorem This theorem, which is really due to Cauchy [127], was given this name by Borchardt [84], who heard it in Liouville’s lectures in 1847. Let f (z) be analytic for all values of z and let | f (z)| < K for all values of z, where K is a constant (so that | f (z)| is bounded as |z| → ∞). Then f (z) is a constant. Let z, z 0 be any two points and let C be a contour such that z, z 0 are inside it. Then, by §5.21, ∫ 1 1 1 f (z 0) − f (z) = − f (ζ) dζ; 2πi C ζ − z 0 ζ − z take C to be a circle whose centre is z and whose radius is ρ ≥ 2 | z 0 − z | ; on C write

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ζ = z + ρeiθ ; since | ζ − z 0 | ≥ 12 ρ when ζ is on C it follows from §4.62 that ∫ 0 1 z − z 0 | f (z ) − f (z)| = f (ζ) dζ 0 2π C (ζ − z ) (ζ − z) ∫ 2π 0 1 |z − z| · K dθ < 1 2π 0 ρ 2 = 2|z 0 − z|K ρ−1 . Make ρ → ∞, keeping z and z 0 fixed; then it is obvious that f (z 0) − f (z) = 0; that is to say, f (z) is constant. As will be seen in the next article, and again frequently in the latter half of this volume (Chapters 20, 21 and 22), Liouville’s theorem furnishes short and convenient proofs of some of the most important results in Analysis.

5.64 Functions with no essential singularities We shall now shew that the only one-valued functions which have no singularities, except poles, at any point (including ∞) are rational functions. For let f (z) be such a function; let its singularities in the finite part of the plane be at the points c1, c2, . . . , ck : and let the principal part (§5.61) of its expansion at the pole cr be ar , nr ar , 1 ar , 2 +···+ . + 2 z − cr (z − cr ) (z − cr )nr Let the principal part of its expansion at the pole at infinity be a1 z + a2 z 2 + · · · + an z n ; if there is not a pole at infinity, then all the coefficients in this expansion will be zero. Now the function k Õ ar , nr ar , 1 ar , 2 f (z) − +···+ − a1 z − a2 z 2 − · · · − an z n + 2 nr z − c (z − c ) (z − c ) r r r r=1 has clearly no singularities at the points c1, c2, . . . , ck or at infinity; it is therefore analytic everywhere and is bounded as |z| → ∞, and so, by Liouville’s theorem, is a constant; that is, k Õ ar , nr ar , 1 ar , 2 f (z) = C + a1 z + a2 z2 + · · · + an z n + + + · · · + , z − cr (z − cr )2 (z − cr )nr r=1 where C is constant; f (z) is therefore a rational function, and the theorem is established. It is evident from Liouville’s theorem (combined with Corollary 3.6.2 that a function which is analytic everywhere (including ∞) is merely a constant. Functions which are analytic everywhere except at ∞ are of considerable importance; they are known as integral functions. z Examples of such functions are ez , sin z, ee . From §5.4 it is apparent that there is no finite radius of convergence of a Taylor’s series which represents an integral function; and from the result of this section it is evident that all integral functions (except mere polynomials) have essential singularities at ∞.

5.7 Many-valued functions

105

5.7 Many-valued functions In all the previous work, the functions under consideration have had a unique value (or limit) corresponding to each value (other than singularities) of z. But functions may be defined which have more than one value for each value of z; thus if z = r(cos θ + i sin θ), the function z1/2 has the two values r 1/2 cos 12 θ + i sin 12 θ , r 1/2 cos 12 (θ + 2π) + i sin 12 (θ + 2π) ; and the function arctan x (x real) has an unlimited number of values, viz. Arctan x + nπ, where − π2 < Arctan x < π2 and n is any integer; further examples of many-valued functions are log z, z −5/3 , sin(z1/2 ). Either of the two functions which z1/2 represents is, however, analytic except at z = 0, and we can apply to them the theorems of this chapter; and the two functions are called ‘branches of the many-valued function z1/2 ’. There will be certain points in general at which two or more branches coincide or at which one branch has an infinite limit; these points are called ‘branch-points’. Thus z1/2 has a branch-point at 0; and, if we consider the change in z1/2 as z describes a circle counter-clockwise round 0, we see that θ increases by 2π, r remains unchanged, and either branch of the function passes over into the other branch. This will be found to be a general characteristic of branch-points. It is not the purpose of this book to give a full discussion of the properties of many-valued functions, as we shall always have to consider particular branches of functions in regions not containing branch-points, so that there will be comparatively little difficulty in seeing whether or not Cauchy’s theorem may be applied. Note Thus we cannot apply Cauchy’s theorem to such a function as z3/2 when the path of integration is a circle surrounding the origin; but it is permissible to apply it to one of the branches of z3/2 when the path of integration is like that shewn in §6.24, for throughout the contour and its interior the function has a single definite value. Example 5.7.1 (Math. Trip. 1899) Prove that if the different values of az , corresponding to a given value of z, are represented on an Argand diagram, the representative points will be the vertices of an equiangular polygon inscribed in an equiangular spiral, the angle of the spiral being independent of a. The idea of the different branches of a function helps us to understand such a paradox as the following. Consider the function y = x x , for which dy = x x (1 + log x). dx

(5.6)

dy When x is negative and real, is not real. But if x is negative and of the form p/(2q + 1) dx (where p and q are positive or negative integers), y is real. If therefore we draw the real curve y = x x , we have for negative values of x a set of conjugate points, one point corresponding to each rational value of x with an odd denominator; and then we might think of proceeding to dy , form the tangent as the limit of the chord, just as if the curve were continuous; and thus dx when derived from the inclination of the tangent to the axis of x, would appear to be real. The question thus arises, Why does the ordinary process of differentiation give a non-real

The Fundamental Properties of Analytic Functions

106

dy ? The explanation is, that these conjugate points do not all arise from the same dx branch of the function y = x x . We have in fact y = e x log x+2kπix , where k is any integer. To each value of k corresponds one branch of the function y. Now in order to get a real value of y when x is negative, we have to choose a suitable value for k : and this value of k varies as we go from one conjugate point to an adjacent one. So the conjugate points do not represent values of y arising from the same branch of the function y = x x , and consequently we cannot dy expect the value of when evaluated for a definite branch to be given by the tangent of the dx inclination to the axis of x of the line joining two arbitrarily close members of the series of conjugate points. value for

5.8 Miscellaneous examples Example 5.1 Obtain the expansion f (z) = f (a) + 2

nz − a 2

f

0

z + a 2

(z − a)3 000 z + a (z − a)5 (5) z + a + 3 f + 5 f +··· , 2 ·3 2 2 · 5! 2

and determine the circumstances and range of its validity. Example 5.2 (Corey [156]) Obtain, under suitable circumstances, the expansion z−a 0 z − a 3 (z − a) 0 f (z) = f (a) + f a+ +··· + f a+ m 2m 2m (2m − 1)(z − a) 2 z − a 3 +f0 a + + 2m 3 ! 2m z−a 3(z − a) f 000 a + +··· + f 000 a + 2m 2m (2m − 1)(z − a) 2 z − a 5 h (5) z − a 000 +f a+ + f a+ 2m 5 ! 2m 2m 3(z − a) (2m − 1)(z − a) + f (5) a + + · · · + f (5) a + +··· . 2m 2m Example 5.3 (Weierstrass [660]) Shew that for the series ∞ Õ n=0

1 , z n + z−n

the region of convergence consists of two distinct areas, namely outside and inside a circle of radius unity, and that in each of these the series represents one function and represents it completely. Example 5.4 (Lerch [425]) Shew that the function ∞ Õ n=0

z n!

5.8 Miscellaneous examples

107

tends to infinity as z → exp(2πip/m!) along the radius through the point; where m is any integer and p takes the values 0, 1, 2, . . . , (m ! − 1). Deduce that the function cannot be continued beyond the unit circle. Example 5.5 (Jacobi [348] and Scheibner [574]) Shew that, if z2 − 1 is not a positive real number, then −1/2 1·3 4 1 · 3 · · · (2n − 1) 2n 1 z +···+ z 1 − z2 = 1 + z2 + 2 2·4 ∫ z 2 · 4 · · · 2n 3 · 5 · · · (2n − 1) + (1 − z2 )−1/2 t 2n+1 (1 − t 2 )−1/2 dt. 2 · 4 · · · (2n) 0 Example 5.6 (Jacobi [348] and Scheibner [574]) Shew that, if z − 1 is not a positive real number, then (1 − z)−m = 1 +

m(m + 1) 2 m(m + 1) · · · (m + n − 1) n m z+ z +···+ z 1 2! ∫ zn! m(m + 1) · · · (m + n) + (1 − z)−m t n (1 − t)m−1 dt. n! 0

Example 5.7 (Jacobi [348] and Scheibner [574]) Shew that, if z and 1 − z are not negative real numbers, then ∫ −1/2 z m 1 − z2 t (1 − t 2 )−1/2 dt 0 m+2 2 z m+1 (m + 2) · · · (m + 2n − 2) 2n−2 1+ = z +···+ z m+1 m+3 (m + 3) · · · (m + 2n − 1) ∫ z −1/2 (m + 2)(m + 4) · · · (m + 2n) t m+2n (1 − t 2 )−1/2 dt. + 1 − z2 (m + 1)(m + 3) · · · (m + 2n − 1) 0 Example 5.8 (Scheibner [574]) If, in the expansion of (a0 + a1 z + a2 z2 )m by the multinomial theorem, the remainder after n terms be denoted by Rn (z), so that m a0 + a1 z + a2 z2 = A0 + A1 z + A2 z2 + · · · + An−1 z n−1 + Rn (z), shew that Rn (z) = (a + a1 z + a2 z )

2 m

∫ 0

z

naAn t n−1 + (2m − n + 1)a2 An−1 t n dt. (a + a1 t + a2 t 2 )m+1

Example 5.9 (Scheibner [574]) If (a0 + a1 z + a2 z )

2 −m−1

∫

z

(a0 + a1 t + a2 t 2 )m dt

0

be expanded in ascending powers of z in the form A1 z + A2 z2 + · · · , shew that the remainder after n − 1 terms is ∫ z 2 −m−1 (a0 + a1 z + a2 z ) (a0 + a1 t + a2 t 2 )m {na0 An − (2m + n + 1)a2 An−1 t}t n−1 dt. 0

The results of Examples 5.5, 5.6 and 5.7 are special cases of formulae contained in Jacobi’s dissertation (Berlin, 1825) published in [354, vol. 3, pp. 1-44]. Jacobi’s formulae were generalized by Scheibner [574].

108

The Fundamental Properties of Analytic Functions

Example 5.10 (Pincherle [523]) Shew that the series ∞ Õ

{1 + λn (z)es }

n=0

d n φ(z) , dz n

where λn (z) = −1 + z −

z2 z3 zn + − · · · + (−1)n , 2! 3! n!

and where φ(z) is analytic near z = 0, is convergent near the point z = 0; and shew that if the sum of the series be denoted by f (z), then f (z) satisfies the differential equation f 0(z) = f (z) − φ(z). Example 5.11 (Gutzmer [264]) Shew that the arithmetic mean of the squares of the moduli ∞ Í of all the values of the series an z n on a circle |z| = r, situated within its circle of n−0

convergence, is equal to the sum of the squares of the moduli of the separate terms. Example 5.12 (Lerch [431]) Shew that the series ∞ Õ

e−2(am) z m−1 1/2

m=1

converges when |z| < 1; and that, when a > 0, the function which it represents can also be represented when |z| < 1 by the integral a 1/2 ∫ ∞ e−a/x dx , π e x − z x 3/2 0 and that it has no singularities except at the point z = 1. Example 5.13 (Weierstrass [660]) Shew that the series 2Õ z 2 −1 (z + z )+ + 0 π π (1 − 2ν − 2ν zi)(2ν + 2ν 0 zi)2 z−1 , (1 − 2ν − 2ν 0 z−1i)(2ν + 2ν 0 z−1i)2 in which the summation extends over all integral values of ν, ν 0, except the combination (ν = 0, ν 0 = 0), converges absolutely for all values of z except purely imaginary values; and that its sum is +1 or −1, according as the real part of z is positive or negative. Example 5.14 Shew that sin (u(z + 1/z)) can be expanded in a series of the type a0 + a1 z + a2 z2 + · · · +

b1 b2 + 2 +··· , z z

in which the coefficients, both of z n and of z−n , are ∫ 2π 1 sin(2u cos θ) cos nθ dθ. 2π 0

5.8 Miscellaneous examples

109

∞ Õ

z2 , shew that f (z) is finite and continuous for all + a2 n=1 real values of z, but cannot be expanded as a Maclaurin’s series in ascending powers of z; and explain this apparent anomaly. For other cases of failure of Maclaurin’s theorem, see a posthumous memoir by Cellérier [140]; Lerch [427]; Pringsheim [540]; and Du Bois Reymond [189]. Example 5.15

If f (z) =

n2 z 2

Example 5.16 If f (z) be a continuous one-valued function of z throughout a two-dimensional region, and if ∫ f (z) dz = 0 C

for all closed contours C lying inside the region, then f (z) is an analytic function of z throughout the interior of the region. Hint. Let a be any point of the region and let ∫ z F(z) = f (z) dz. a

It follows from the data that F(z) has the unique derivative f (z). Hence F(z) is analytic (§5.1) and so (§5.22) its derivative f (z) is also analytic. This important converse of Cauchy’s theorem is due to Morera [474].

6 The Theory of Residues; Application to the Evaluation of Definite Integrals 6.1 Residues If the function f (z) has a pole of order m at z = a, then, by the definition of a pole, an equation of the form f (z) =

a−m a−m+1 a−1 + +···+ + φ(z), m m−1 (z − a) (z − a) z−a

where φ(z) is analytic near and at a, is true near a. The coefficient a−1 in this expansion is called the residue of the function f (z) relative to the pole a. ∫ Consider now the value of the integral

α

f (z)dz, where the path of integration is a circle

α, whose centre is the point a and whose radius ρ is so small that φ(z) is analytic inside and on the circle. The existence of such a circle is implied in the definition of a pole as an isolated singularity. We have ∫ ∫ ∫ m Õ dz a−r f (z) dz = + φ(z) dz. r α α (z − a) α r−1 ∫ Now φ(z) dz = 0 by §5.2; and (putting z − a = ρeiθ ) we have, if r , 1, α

∫ α

dz = (z − a)r

∫ 0

2π

ρeiθ i dθ = ρ−r+1 ρr eriθ

∫

2π

e(1−r)iθ i dθ = ρ−r+1

0

e(1−r)iθ 1−r

2π

= 0.

0

But when r = 1, we have ∫ α

dz = z−a

∫

2π

i dθ = 2πi.

0

Hence finally ∫ α

f (z) dz = 2πia−1 .

Now let C be any contour, containing in the region interior to it a number of poles a, b, c, . . . of a function f (z), with residues a−1, b−1, c−1, . . . respectively: and suppose that the function f (z) is analytic throughout C and its interior, except at these poles. Surround the points a, b, c, . . . by circles α, β, γ, . . . so small that their respective centres are the only singularities 110

6.2 The evaluation of definite integrals

111

inside or on each circle; then the function f (z) is analytic in the closed region bounded by C, α, β, γ, . . . . Hence, by Corollary 5.2.3 ∫ ∫ ∫ f (z) dz = f (z) dz + f (z) dz + · · · α

C

β

= 2πia−1 + 2πib−1 + · · · . Thus we have the theorem of residues, namely that if f (z) be analytic throughout a contour C and its interior except at a number of poles inside the contour, then ∫ Õ f (z) dz = 2πi R, C

Í

where R denotes the sum of the residues of the function f (z) at those of its poles which are situated within the contour C. This is an extension of the theorem of §5.21 (giving (5.21)). Note If a is a simple pole of f (z) the residue of f (z) at that pole is lim (z − a) f (z). z→a

6.2 The evaluation of definite integrals We shall now apply the result of §6.1 to evaluating various classes of definite integrals; the methods to be employed in any particular case may usually be seen from the following typical examples.

6.21 The evaluation of the integrals of certain periodic functions taken between the limits 0 and 2π An integral of the type ∫

2π

R(cos θ, sin θ) dθ

0

where the integrand is a rational function of cos θ and sin θ, finite on the range of integration, can be evaluated by writing eiθ = z; since cos θ =

1 (z + z−1 ), 2

sin θ =

1 (z − z−1 ), 2i

∫ S(z) dz, where S(z) is a rational function of z finite on the path

the integral takes the form C

of integration C, the circle of radius unity whose centre is the origin. Therefore, by §6.1, the integral is equal to 2πi times the sum of the residues of S(z) at those of its poles which are inside that circle. Example 6.2.1 If 0 < p < 1, ∫ 2π 0

dθ = 1 − 2p cos θ + p2

∫ C

dz . i(1 − pz)(z − p)

The Theory of Residues; Application to the Evaluation of Definite Integrals

112

The only pole of the integrand inside the circle is a simple pole at p; and the residue there is lim

z→p

1 z−p = . i(1 − pz)(z − p) i(1 − p2 )

Hence ∫

2π

0

2π dθ = . 1 − 2p cos θ + p2 1 − p2

Example 6.2.2 If 0 < p < 1, 2 ∫ 2π ∫ cos2 3θ dz 1 1 3 1 −3 dθ = z + z 2 2 1 − 2p cos 2θ + p 2 (1 − pz )(1 − pz −2 ) 0 c iz 2 Õ = 2π R, (z6 + 1)2 at its poles inside C; R denotes the sum of the residues of 5 4z (1 − pz2 )(z2 − p) √ √ these poles are 0, − p, p; and the residues at them are where

Í

−

1 + p2 + p4 (p3 + 1)2 (p3 + 1)2 , , ; 4p3 8p3 (1 − p2 ) 8p3 (1 − p2 )

and hence the integral is equal to π(1 − p + p2 ) . 1−p Example 6.2.3 If n be a positive integer, ∫ 2π 2π ecos θ cos(nθ − sin θ) dθ = , n! 0 Example 6.2.4 If a > b > 0, ∫ 2π 2πa dθ = 2 , 2 (a + b cos θ) (a − b2 )3/2 0

∫

2π

ecos θ sin(nθ − sin θ) dθ = 0.

0

∫ 0

2π

dθ π(2a + b) = . (a + b cos2 θ)2 a3/2 (a + b)3/2

6.22 The evaluation of certain types of integrals taken between the limits −∞ and +∞ ∫ ∞ We shall now evaluate Q(x) dx, where Q(z) is a function such that: −∞

(i) it is analytic when the imaginary part of z is positive or zero (except at a finite number of poles); (ii) it has no poles on the real axis; (iii) as |z| → ∞, z Q(z) → 0 uniformly for all values of arg z such that 0 ≤ arg∫z ≤ π; ∞

(iv) provided that when x is real, xQ(x) → 0, as x → ±∞, in such a way 1 that ∫ 0 and Q(x) dx both converge.

Q(x) dx 0

−∞

∫ 1

The condition xQ(x) → 0 is not in itself sufficient to secure the convergence of Q(x) = (x log x) . −1

∞

Q(x) dx; consider 0

6.2 The evaluation of definite integrals

113

Given ε, we can choose ρ0 (independent of arg z) such that |z Q(z)| < ε/π whenever ∫ |z| > ρ0 and 0 ≤ arg z ≤ π. Consider Q(z) dz taken round a contour C consisting of the C

part of the real axis joining the points ±ρ (where ρ > ρ0 ) and a semicircle Γ, of radius ρ, ∫ Õ having its centre at the origin, above the real axis. Then, by §6.1, Q(z) dz = 2πi R, C Í where R denotes the sum of the residues of Q(z) at its poles above the real axis (Q(z) has no poles above the real axis outside the contour). Therefore ∫ ρ Õ ∫ . Q(z) dz − 2πi R = Q(z) dz −ρ

Γ

In the last integral write z = ρe , and then ∫ ∫ π ∫ π iθ iθ Q(z) dz = < Q(ρe )ρe i dθ (ε/π)dθ = ε, iθ

Γ

0

0

by §4.62. Hence ∫

ρ

Q(z) dz = 2πi

lim

ρ→∞

∫

Õ

R.

(6.1)

−ρ

∞

∫

σ

∫

σ

But the meaning of Q(x) dx is lim Q(x) dx; and since lim Q(x) dx and ρ,σ→∞ σ→∞ 0 −∞ −ρ ∫ 0 ∫ ρ lim Q(x) dx both exist, this double limit is the same as lim Q(x) dx. Hence we have ρ→∞

ρ→∞

−ρ

−ρ

proved that ∫

∞

Q(x) dx = 2πi

Õ

R.

−∞

This theorem is particularly useful in the special case when Q(x) is a rational function. Note Even if condition (iv) is not satisfied, we still have ∫ ρ ∫ ∞ Õ {Q(x) + Q(−x)} dx = lim Q(x) dx = 2πi R. 0

ρ→∞

(6.2)

−ρ

Example 6.2.5 The only pole of (z2 + 1)−3 in the upper half plane is a pole at z = i with 3 i. Therefore residue there − 16 ∫ ∞ 3π dx = . 2 3 8 −∞ (x + 1) Example 6.2.6 If a > 0, b > 0, shew that ∫ ∞ x 4 dx π = . 2 )4 3/2 b5/2 (a + bx 16a −∞ ∫ 2 Example 6.2.7 By integrating e−λz dz around a parallelogram whose corners are −R, R, R + ai, −R + ai and making R → ∞, shew that, if λ > 0, then ∫ ∫ ∞ ∫ ∞ −λx 2 −λa2 −λx 2 − 12 −λa2 c cos(2λax) dx = e e dx = 2λ e −∞

−∞

0

∞

e−x dx. 2

114

The Theory of Residues; Application to the Evaluation of Definite Integrals

6.221 Certain infinite integrals involving sines and cosines miz If Q(z) satisfies the conditions ∫ ∞ (i), (ii) and (iii) of §6.22, and m > 0, then Q(z)e also satisfies Í Í those conditions. Hence Q(x)emix + Q(−x)e−mix dx is equal to 2πi R 0, where R 0 0

means the sum of the residues of Q(z)emiz at its poles in the upper half plane; and so (i) If Q(x) is an even function, i.e. if Q(−x) = Q(x), ∫ ∞ Õ Q(x) cos(mx) dx = πi R0. 0

(ii) If Q(x) is an odd function, ∫

∞

Q(x) sin(mx) dx = π

Õ

R0.

0

6.222 Jordan’s lemma (Jordan [362, p. 285–286]). The results of §6.221 are true if Q(z) be subject to the less stringent condition Q(z) → 0 uniformly when 0 ≤ arg z ≤ π as |z| → ∞ in place of the condition z Q(z) → 0 uniformly. To prove this we require a theorem known as Jordan’s lemma, viz. If Q(z) → 0 uniformly with regard to arg z as |z| → ∞ when 0 ≤ arg z ≤ π, and if Q(z) is analytic when both |z| > c (a constant) and 0 ≤ arg z ≤ π, then ∫ lim emiz Q(z)dz = 0, ρ→∞

Γ

where Γ is a semicircle of radius ρ above the real axis with centre at the origin. Given ε, choose ρ0 so that |Q(z)| < ε/π when |z| > ρ0 and 0 ≤ arg z ≤ π; then, if ρ > ρ0 , ∫ ∫ π mi(ρ cos θ+iρ sin θ) iθ iθ emiz Q(z) dz = e Q(ρe )ρe i dθ . Γ

0

But emiρ cos θ = 1, and so ∫ emiz Q(z) dz Γ

π

ε ρ e−mρ sin θ dθ 0 π ∫ 2ε π/2 −mρ sin θ = ρe dθ. π 0

∫ <

Now sin θ ≥ 2θ/π, when 2 0 ≤ θ ≤ π/2, and so ∫ ∫ π/2 eimz Q(z) dz < 2ε ρe−2mρθ/π dθ π Γ 0 2ε π −2mρθ/π π/2 = · −e 0 π 2m ε < . m 2

This inequality appears obvious when we draw the graphs y = sin x, y = 2x/π; it may be proved by shewing that (sin θ)/θ decreases as θ increases from 0 to π/2.

6.2 The evaluation of definite integrals

115

Hence ∫ lim

ρ→∞

emiz Q(z) dz = 0.

Γ

This result is Jordan’s lemma. Now ∫

ρ

{emix Q(x) + e−mix Q(−x)} dx = 2πi

Õ

R0 −

∫

emiz Q(z) dz,

Γ

0

and, making ρ → ∞, we see at once that ∫ ∞ Õ {emix Q(x) + e−mix Q(−x)} dx = 2πi R 0, 0

which is the result corresponding to the result of §6.221. Example 6.2.8 Shew that, if a > 0, then ∫ ∞ π −a cos x dx = e . 2 2 x +a 2a 0 Example 6.2.9 Shew that, if a ≥ 0, b ≥ 0, then ∫ ∞ cos 2ax − cos 2bx dx = π(b − a). x2 0 Hint. Take a contour consisting of a large semicircle of radius ρ, a small semicircle of radius δ, both having their centres at the origin, and the parts of the real axis joining their ends; then make ρ → ∞, δ → 0. Example 6.2.10 Shew that, if b > 0, m ≥ 0, then ∫ ∞ 3x 2 − a2 πe−mb cos mx dx = {3b2 − a2 − mb(3b2 + a2 )}. (x 2 + b2 )2 4b3 0 Example 6.2.11 Shew that, if k > 0, a > 0, then ∫ ∞ 1 x sin ax dx = πe−k a . 2 + k2 x 2 0 Example 6.2.12 Shew that, if m ≥ 0, a > 0, then ∫ ∞ sin mx π πe−ma 2 dx = 4 − m+ . x(x 2 + a2 )2 2a 4a3 a 0 (Take the contour of Example 6.2.9). Example 6.2.13 Shew that, if the real part of z be positive, then ∫ ∞ dt (e−t − e−tz ) = log z. t 0

116

The Theory of Residues; Application to the Evaluation of Definite Integrals

Solution. We have ∫ ∞

(e−t − e−tz )

0

ρ

∫ ρ −tz e− t e dt − dt t t δ δ ∫ ρ − ∫ ρz −u e t e = lim dt − du δ→0, ρ→∞ t u δ δz ∫ δz − ∫ ρz −t e t e = lim dt − dt , δ→0, ρ→∞ t t δ ρ

dt = lim δ→0, ρ→∞ t

∫

since t −1 e−t is analytic inside the quadrilateral whose corners are δ, δz, ρz, ρ. Now ∫ ρz t −1 e−t dt → 0 as ρ → ∞ when Re z > 0; and ρ

∫ δ

δz

t −1 e−t dt = log z −

∫ δ

δz

t −1 (1 − e−t ) dt → log z,

since t −1 (1 − e−t ) → 1 as t → 0.

6.23 Principal values of integrals It was assumed in §6.22, §6.221, and §6.222 that the function Q(x) had no poles on the real axis; if the function has a finite number of simple poles on the real axis, we can obtain theorems corresponding to those already obtained, except that the integrals are all principal Í Í Í Í values (§4.5) and R has to be replaced by R + 12 R0 , where R0 means the sum of the residues at the poles on the real axis. To obtain this result we saw that, instead of the former contour, we had to take as contour a circle of radius ρ and the portions of the real axis joining the points −ρ, a − δ1 ; a + δ1, b − δ2 ; b + δ2, c − δ3, . . . and small semicircles above the real axis of radii δ1, δ2, . . . with centres a, b, c, . . ., where a, b, c, . . . are the poles of Q(z) on the real axis; and then we have to make δ1, δ2, . . . → 0; call these semicircles γ1, γ2, . . .. Then instead of the equation ∫ ρ ∫ Õ Q(z) dz + Q(z) dz = 2πi R, −ρ

we get ∫

ρ

P

Q(z) dz +

−ρ

Γ

Õ n

∫ lim

δ n →0

γn

Q(z) dz +

∫

Q(z) dz = 2πi

Õ

R.

Γ

Let a 0 be the residue of Q(z) at a; then writing z = a + δ1 eiθ on γ1 , we get ∫ ∫ 0 Q(z) dz = Q(a + δ1 eiθ ) δ1 eiθ i dθ. γ1

π

But Q(a + δ1 e )δ1 e → a uniformly as δ1 → 0; and therefore ∫ lim Q(z) dz = −πia 0; iθ

iθ

0

δ1 →0

γ1

6.2 The evaluation of definite integrals

we thus obtain ∫

ρ

P

Q(z)dz +

−ρ

∫

Q(z)dz = 2πi

Õ

R + πi

Õ

117

R0,

Γ

and hence, using the arguments of §6.22, we get ∫ ∞ Õ Õ R0 . P Q(x) dx = 2πi R + 12 −∞

The reader will see at once that the theorems of §6.221 and §6.222 have precisely similar generalisations. The process employed above of inserting arcs of small circles so as to diminish the area of the contour is called indenting the contour.

∫ 6.24 Evaluation of integrals of the form

∞

x a−1 Q (x) dx

0

Let Q(x) be a rational function of x such that it has no poles on the positive part of the real axis and x a Q(x) ∫ → 0 both when x → 0 and when x → ∞. Consider

(−z)a−l Q(z) dz taken round the contour C shewn in the figure,

consisting of the arcs of circles of radii ρ, δ and the straight lines joining their end points; (−z)a−1 is to be interpreted as exp{(a − 1) log(−z)} and log(−z) = log |z| + i arg(−z), where −π ≤ arg(−z) ≤ π; with these conventions the integrand is one-valued and analytic on and within the contour save at the poles of Q(z). Í Hence, if r denote the sum of the residues of (−z)a−1 Q(z) at all its poles, ∫ Õ (−z)a−1 Q(z) dz = 2πi r. C

On the small circle write −z = δeiθ , and the integral along it becomes ∫ −π − (−z)a Q(z)i dθ, π

The Theory of Residues; Application to the Evaluation of Definite Integrals

118

which tends to zero as δ → 0. On the large semicircle write −z = ρeiθ , and the integral along it becomes ∫ π

(−z)a Q(z)i dθ,

− −π

which tends to zero as ρ → ∞. On one of the lines we write −z = xe πi , on the other −z = xe−πi and (−z)a−1 becomes x a−1 e±(a−1)πi . Hence ∫ ρ Õ lim {x a−1 e−(a−1)πi Q (x) − x a−1 e a−1 xi Q(x)} dx = 2πi r; δ→0, ρ→∞

δ

and therefore ∫

∞

x a−1 Q (x) dx = π cosec(aπ)

Õ

r.

0

Corollary 6.2.1 If Q(x) have a number of simple poles on the positive part of the real axis, it may be shewn by indenting the contour that ∫ ∞ Õ Õ P x a−1 Q(x) dx = π cosec(aπ) r − π cot(aπ) r 0, 0

where

Í

0

r is the sum of the residues of z a−1 Q(z) at these poles.

Example 6.2.14 If 0 < a < 1, ∫ ∞ a−1 x dx = π cosec(πa), 1 +x 0

∫ P 0

∞

x a−1 dx = π cot(πa). 1−x

Example 6.2.15 (Minding) If 0 < z < 1 and −π < a < π, ∫ ∞ z−1 πei(z−1)a t dt = . t + eia sin πz 0 Example 6.2.16 Shew that, if −1 < z < 3, then ∫ ∞ x z dx π(1 − z) = . 2 2 (1 + x ) 4 cos 12 πz 0 Example 6.2.17 (Euler) Shew that, if −1 < p < 1 and −π < λ < π, then ∫ ∞ π sin pλ x −p dx = . 2 1 + 2x cos λ + x sin pπ sin λ 0

6.3 Cauchy’s integral We shall next discuss a class of contour-integrals which are sometimes found useful in analytical investigations. Let C be a contour in the z-plane, and let f (z) be a function analytic inside and on C. Let φ(z) be another function which is analytic inside and on C except at a finite number of poles; let the zeros of φ(z) in the interior 3 of C be a1, a2, . . . , and let their degrees of multiplicity be r1, r2, . . . ; and let its poles in the interior of C be b1, b2, . . . , and let their degrees of multiplicity be s1, s2, . . . . 3

φ(z) must not have any zeros or poles on C.

6.3 Cauchy’s integral 119 ∫ φ 0(z) 1 f (z) dz is equal to the sum Then, by the fundamental theorem of residues, 2πi C φ(z) 0 0 f (z)φ (z) f (z)φ (z) of the residues of at its poles inside C. Now can have singularities φ(z) φ (z) only at the poles and zeros of φ(z). Near one of the zeros, say a1 , we have φ(z) = A(z − a1 )r1 + B(z − a1 )r1 +1 + · · · . Therefore φ 0(z) = Ar1 (z − a1 )r1 −1 + B(r1 + 1)(z − a1 )r1 + · · · , f (z)φ 0(z) r1 f (a1 ) and f (z) = f (a1 ) + (z − a1 ) f 0(a1 ) + · · · . Therefore − is analytic at a1 . φ(z) z − a1 0 f (z)φ (z) Thus the residue of , at the point z = a1 , is r1 f (a1 ). Similarly the residue at z = b1 φ(z) is −s1 f (b1 ); for near z = b1 , we have φ(z) = C(z − b1 )−s1 + D(z − b1 )−s1 +1 + · · · , and f (z)φ 0(z) s1 f (b1 ) f (z) = f (b1 ) + (z − b1 ) f 0(b1 ) + · · · , so + is analytic at b1 . Hence φ(z) z − b1 ∫ Õ Õ 1 φ 0(z) f (z) dz = r1 f (a1 ) − s1 f (b1 ), 2πi C φ(z)

the summations being extended over all the zeros and poles of φ(z).

6.31 The number of roots of an equation contained within a contour The result of the preceding paragraph can be at once applied to find how many roots of an equation φ(z) = 0 lie within a contour C. For, on putting f (z) = 1 in the preceding result, we obtain the result that ∫ 0 φ (z) 1 dz 2πi C φ(z) is equal to the excess of the number of zeros over the number of poles of φ(z) contained in the interior of C, each pole and zero being reckoned according to its degree of multiplicity. Example 6.3.1 Shew that a polynomial φ(z) of degree m has m roots. Hint. Let φ(z) = a0 z m + a1 z m−1 + · · · + am , (a0 , 0). Then φ 0(z) ma0 z m−1 + · · · + am−1 = . φ(z) a0 z m + · · · + a m Consequently, for large values of |z|, 1 φ 0(z) m = +O 2 . φ(z) z z Thus, if C be a circle of radius ρ whose centre is at the origin, we have ∫ 0 ∫ ∫ ∫ 1 φ (z) m dz 1 1 1 1 dz = + O 2 dz = m + O 2 dz. 2πi C φ(z) 2πi C z 2πi C z 2πi C z But, as in §6.22,

∫ O C

1 dz → 0 as z2

ρ → 0;

(6.3)

120

The Theory of Residues; Application to the Evaluation of Definite Integrals

and hence as φ(z) has no poles in the interior of C, the total number of zeros of φ(z) is ∫ 0 1 φ (z) dz = m. lim ρ→∞ 2πi C φ (z) Example 6.3.2 If at all points of a contour C the inequality |ak z k | > |a0 + a1 z + · · · + ak−1 z k−1 + ak+1 z k+1 + · · · + am z m | is satisfied, then the contour contains k roots of the equation am z m + am−1 z m−1 + · · · + a1 z + a0 = 0. For write f (z) = am z m + am−1 z m−1 + · · · + a1 z + a0 . Then am z m + · · · + ak+1 z k+1 + ak−1 z k−1 + · · · + a0 k f (z) = ak z 1 + ak z k = ak z k (1 + U), where |U| ≤ a < 1 on the contour, a being independent 4 of z. Therefore the number of roots of f (z) contained in C is ∫ 0 ∫ f (z) k 1 1 1 dU dz = + dz. 2πi C f (z) 2πi C z 1 + U dz ∫ dz But = 2πi; and, since |U| < 1, we can expand (1 + U)−1 in the uniformly convergent z C series 1 − U + U 2 − U 3 + · · · , so ∫ 1 dU dz = U − 12 U 2 + 13 U 3 − · · · C = 0. C 1 + U dz Therefore the number of roots contained in C is equal to k. Example 6.3.3 (Clare, 1900) Find how many roots of the equation z6 + 6z + 10 = 0 lie in each quadrant of the Argand diagram.

6.4 Connexion between the zeros of a function and the zeros of its derivative MacDonald [444] has shewn that if f (z) is a function of z analytic throughout the interior of a single closed contour C, defined by the equation | f (z)| = M, where M is a constant, then the number of zeros of f (z) in this region exceeds the number of zeros of the derived function f 0(z) in the same region by unity. On C let f (z) = Meiθ ; then at points on C ( 2) 2 dθ d θ dθ f 0(z) = Meiθ i , f 00(z) = Meiθ i 2 − . dz dz dz 4

|U | is a continuous function of z on C, and so attains its upper bound (§3.62). Hence its upper bound a must be less than 1.

6.5 Miscellaneous examples

121

Hence, by §6.31, the excess of the number of zeros of f (z) over the number of zeros of f (z) inside 5 C is ∫ 0 ∫ 00 ∫ −1 2 dθ f (z) 1 f (z) 1 d θ 1 dz − dz = − dz. 0 2πi C f (z) 2πi C f (z) 2πi C dz dz2 0

Let s be the arc of C measured from a fixed point and let ψ be the angle the tangent to C makes with 0x; then ∫ −1 2 1 dθ 1 d θ dθ − dz = − log 2πi C dz dz 2 2πi dz C dz 1 dθ − log . =− log 2πi ds ds C dθ dz is purely real and its initial value is the same as its final value; and log = iψ; ds ds 0 hence the excess of the number of zeros of f (z) over the number of zeros of f (z) is the change in ψ/2π in describing the curve C; and it is obvious (for a formal proof, see [651]) that if C is any ordinary curve, ψ increases by 2π as the point of contact of the tangent describes the curve C; this gives the required result. Now log

Example 6.4.1 Deduce from Macdonald’s result the theorem that a polynomial of degree n has n zeros. Example 6.4.2 Prove that, if a polynomial f (z) has real coefficients and if its zeros are all real and different, then between two consecutive zeros of f (z) there is one zero and one only of f 0(z). Pólya has pointed out that this result is not necessarily true for functions other than polynomials, as may be seen by considering the function (z2 − 4) exp(z 2 /3).

6.5 Miscellaneous examples Example 6.1 (Trinity, 1898) A function φ(z) is zero when z = 0, and is real when z is real, and is analytic when |z| ≤ 1; if f (x, y) is the coefficient of i in φ(x + iy), prove that if −1 < x < 1, then ∫ 2π x sin θ f (cos θ, sin θ) dθ = πφ(x). 1 − 2x cos θ + x 2 0 e±aiz Example 6.2 (Legendre) By integrating 2πz round a contour formed by the rectangle e −1 whose corners are 0, R, R + i, i (the rectangle being indented at 0 and i) and making R → ∞, shew that ∫ ∞ sin ax 1 ea + 1 1 dx = − . 2πx − 1 a −1 e 4 e 2a 0 5

f 0 (z) does not vanish on C unless C has a node or other singular point; for, if f = φ + iψ, where φ and ψ are ∂f ∂f ∂φ ∂φ ∂ψ ∂ψ real, since i = , it follows that if f 0 (z) = 0 at any point, then , , , all vanish; and these ∂x ∂y ∂x ∂y ∂x ∂y 2 2 2 are sufficient conditions for a singular point on φ + ψ = M .

122

The Theory of Residues; Application to the Evaluation of Definite Integrals

Example 6.3 By integrating log(−z) Q(z) round the contour of §6.24, where Q(z) is a rational function such that zQ(z) → 0 as |z| →∫ 0 and as |z| → ∞, shew that if Q(z) has ∞

Q(x) dx is equal to minus the sum of

no poles on the positive part of the real axis, 0

the residues of log(−z) Q(z) at the poles of Q(z); where the imaginary part of log(−z) lies between ±π. Example 6.4 Shew that, if a > 0, b > 0, ∫ ∞ dx 1 = π(e a − 1). e a cos bx sin(a sin bx) x 2 0 Example 6.5 (Cauchy) Shew that ( ∫ π/2 π log(1 + a), a sin 2x 4 xdx = π 2 1 − 2a cos 2x + a log(1 + a−1 ), 0 4

(−1 < a < 1) (a2 > 1)

Example 6.6 (Störmer [614]) Shew that ∫ ∞ sin φ1 x sin φ2 x sin φn x sin ax π ··· cos a1 x · · · cos am x dx = φ1 φ2 · · · φn, x x x x 2 0 if φ1, φ2, . . . , φn, a1, a2, . . . , am are real and a be positive and a > |φ1 | + |φ2 | + · · · + |φn | + |a1 | + · · · + |am |. Example 6.7 (Amigues [17]) If a point z describes a circle C of centre a, and if f (z) be analytic throughout C and its interior except at a number of poles inside C, then the point u = f (z) will describe a closed curve γ in the u-plane. Shew that if to each element of γ be attributed a mass proportional to the corresponding element of C, the centre of gravity of γ f (z) at its poles in the interior of C. is the point r, where r is the sum of the residues of z−a Example 6.8 Shew that ∫

∞

−∞

Example 6.9 Shew that ∫ ∞ 0

π(2a + b) dx = . (x 2 + b2 )(x 2 + a2 )2 2a3 b(a + b)2

dx π 1 · 3 · · · (2n − 3) 1 = n 1/2 . 2 n (a + bx ) 1 · 2 · · · (n − 1) a n−1/2 2 b

Example 6.10 (Laurent [412]) If Fn (z) = f (z) = −

n−1 Î n−1 Î

∞ Õ n=2

(1 − z mp ), shew that the series

m=1 p=1

Fn (zn−1 ) (z n n−n − 1)nn−1

is an analytic function when z is not a root of any of the equations z n = nn ; and that the sum of the residues of f (z) contained in the ring-shaped space included between two circles whose centres are at the origin, one having a small radius and the other having a radius between n and n + 1, is equal to the number of prime numbers less than n + 1.

6.5 Miscellaneous examples

123

Example 6.11 (Grace [257]) If A and B represent on the Argand diagram two given roots (real or imaginary) of the equation f (z) = 0 of degree n, with real or imaginary coefficients, shew that there is at least one root of the equation f 0(z) = 0 within a circle whose centre is the middle point of AB and whose radius is 12 AB cot(π/n). Example 6.12 (Kronecker [386]) Shew that, if 0 < ν < 1, n Õ e2πikν 1 e2πiνx = lim . 1 − e2πix 2πi n→∞ k=−n k − x

∫ Hint. Consider

e(2ν−1)πiz dz round a circle of radius n + 12 ; and make n → ∞. sin πz z − x

Example 6.13 Shew that, if m > 0, then ∫ ∞ πmn−1 n(n − 1) n sinn mt dt = (n − 4)n−1 nn−1 − (n − 2)n−1 + n n t 2 (n − 1)! 1 2! 0 n(n − 1)(n − 2) n−1 − (n − 6) + · · · . 3! Discuss the discontinuity of the integral at m = 0. Example 6.14 (Wolstenholme) If A + B + C + · · · = 0 and a, b, c, . . . are positive, shew that ∫ ∞ A cos ax + B cos bx + · · · + K cos k x dx x 0 = −A log a − B log b − · · · − K log k. ∫ Example 6.15 By considering shew that, if k > 0, ∫

ρ

i lim

ρ→∞

−ρ

e x(k+ti) dt taken around a rectangle indented at the origin, k + ti

e x(k+ti) dt = πi + lim P ρ→∞ k + ti

∫

ρ

−ρ

e xti dt, t

and thence deduce, by using the contour of Example 6.2.9, or its reflexion in the real axis (according as x ≥ 0 and x < 0), that ∫ 1 ρ e x(k+ti) lim dt = 2, 1 or 0, ρ→∞ π −ρ k + ti according as x > 0, x = 0 or x < 0. This integral is known as Cauchy’s discontinuous factor. Example 6.16 Shew that, if 0 < a < 2, b > 0, r > 0, then ∫ ∞ πa r dx π x a−1 sin − bx 2 = r a−1 e−br . 2 2 x + r 2 0 Example 6.17 (Poisson [531]; Jacobi [352])

Let t > 0 and let

∞ Í n=−∞

e−n

2

πt

= ψ (t). By

The Theory of Residues; Application to the Evaluation of Definite Integrals ∫ 2 e−s πt dz around a rectangle whose corners are ±(N + 12 ) ± i, where N is considering e2πis − 1 an integer, and making N → ∞, shew that ∫ ∞−i −s2 πt ∫ ∞+i −s2 πt e e ψ (t) = dz − dz. 2πiz 2πiz −1 −1 −∞−i e −∞+i e 124

By expanding these integrands in powers of e−2πis , e2πis respectively and integrating termby-term, deduce from Example 6.2.7 that ∫ ∞ 1 2 ψ (t) = ψ (1/t) e−x dx. 1 (πt) 2 −∞ Hence, by putting t = 1 shew that ψ(t) = t − 2 ψ (1/t) . 1

Example 6.18 (Poisson [532], Jacobi [346] and Landsberg [407]) Shew that, if t > 0, ! ∞ ∞ Õ Õ 2 2 2 e−n π/t cos 2nπa . e−n πt−2nπat = t −1/2 e πa t 1 + 2 n=−∞

See also §21.51.

n=1

7 The Expansion of Functions in Infinite Series

7.1 A formula due to Darboux Let f (z) be analytic at all points of the straight line joining a to z, and let φ(t) be any polynomial of degree n in t. Then if 0 ≤ t ≤ 1, we have by differentiation n d Õ (−1)m (z − a)m φ(n−m) (t) f (m) (a + t (z − a)) dt m=1

= − (z − a) φ(n) (t) f 0 (a + t (z − a)) + (−1)n (z − a)n+1 φ (t) f (n+1) (a + t (z − a)) . Noting that φ(n) (t) is constant = φ(n) (0), and integrating between the limits 0 and 1 of t, we get φ(n) (0){ f (z) − f (a)} n Õ (−1)m−1 (z − a)m {φ(n−m) (1) f (m) (z) − φ(n−m) (0) f m (a)} = m=1

+(−1) (z − a) n

n+1

∫

1

φ(t) f (n+1) (a + t (z − a)) dt,

0

which is the formula in question. It appears in Darboux [162]. Taylor’s series may be obtained as a special case of this by writing φ(t) = (t − 1)n and making n → ∞. Example 7.1.1 By substituting 2n for n in the formula of Darboux, and taking φ(t) = t n (t − 1)n , obtain the expansion (supposed convergent) f (z) − f (a) =

∞ Õ (−1)n−1 (z − a)n n=1

2n n !

{ f (n) (z) + (−1)n−1 f (n) (a)},

and find the expression for the remainder after n terms in this series.

7.2 The Bernoullian numbers and the Bernoullian polynomials The function 12 z cot 21 z is analytic when |z| < 2π and, since it is an even function of z, it can be expanded into a Maclaurin series, thus z z z2 z4 z6 cot = 1 − B1 − B2 − B3 −··· ; 2 2 2! 4! 6! 125

126

The Expansion of Functions in Infinite Series

then Bn is called the nth Bernoullian number. These numbers were introduced by Jakob Bernoulli [66]. It is found that 1 1 1 5 1 B1 = , B2 = , B3 = , B4 = , B5 = , . . . . 6 30 42 30 66 The first sixty-two Bernoullian numbers were computed by Adams [11]; the first nine significant figures of the first 250 Bernoullian numbers were subsequently published by Glaisher [246]. These numbers can be expressed as definite integrals as follows: We have, by Chapter 6, Example 6.2, ∫ ∞ sin px dx 1 i = − + cot ip πx e −1 2p 2 0 1 1 (2p)4 (2p)2 =− + − B2 +··· . 1 + B1 2p 2p 2! 4! ∫ ∞ n x sin px + 21 nπ Since dx converges uniformly (by de la Vallée Poussin’s test) near e πx − 1 0 p = 0 we may, by Corollary 4.4.1, differentiate both sides of this equation any number of times and then put p = 0; doing so and writing 2t for x, we obtain ∫ ∞ 2n−1 t dt Bn = 4n . 2πt e −1 0 A proof of this result, depending on contour integration, is given by Carda [117]. Example 7.2.1 Shew that Bn =

2n π 2n (22n − 1)

∫ 0

∞

x 2n−1 dx > 0. sinh x

zt

e −1 Now consider the function t t , which may be expanded into a Maclaurin series in e −1 powers of t valid when |t| < 2π. n The Bernoullian polynomial of order n is defined to be the coefficient of tn! in this expansion. It is denoted by φn (z), so that ∞

ezt − 1 Õ φn (z)t n t t = . e −1 n! n=1 The name was given by Raabe [547]. For a full discussion of their properties, see Nörlund [507]. This polynomial possesses several important properties. Writing z+1 for z in the preceding equation and subtracting, we find that tezt =

∞ Õ

{φn (z + 1) − φn (z)}

n=1 n

tn . n!

On equating coefficients of t on both sides of this equation we obtain nz n−1 = φn (z + 1) − φn (z),

7.2 The Bernoullian numbers and the Bernoullian polynomials

127

which is a difference-equation satisfied by the function φn (z). An explicit expression for the Bernoullian polynomials can be obtained as follows. We have z2 t 2 z3 t 3 + +··· , ezt − 1 = zt + 2! 2! and t t t t t B1 t 2 B2 t 4 = cot − = 1 − + − +··· . et − 1 2i 2i 2 2 2! 4!

(7.1)

Hence ∞ Õ z2 t 2 z3 t 3 t B1 t 2 B2 t 4 φn (z) t n = zt + + +··· 1− + − +··· . n! 2! 3! 2 2! 4! n=1 From this, by equating coefficients of t n (§3.73), we have n n n 1 B1 z n−2 − B2 z n−4 + B3 z n−6 − · · · , φn (z) = z n − nz n−1 + 2 4 6 2 the last term being that in z or z2 and n2 , n4 , . . . being the binomial coefficients; this is the Maclaurin series for the nth Bernoullian polynomial. When z is an integer, it may be seen from the difference-equation that φn (z) = 1n−1 + 2n−1 + · · · + (z − 1)n−1 . n The Maclaurin series for the expression on the right was given by Bernoulli. Example 7.2.2 Shew that, when n > 1, φn (z) = (−1)n φn (1 − z).

7.21 The Euler–Maclaurin expansion In the formula of Darboux (§7.1) write φn (t) for φ(t), where φn (t) is the nth Bernoullian polynomial. Differentiating the equation φn (t + 1) − φn (t) = nt n−1 n − k times, we have φ(n−k) (t + 1) − φ(n−k) (t) = n(n − 1) · · · kt k−1 . n n Putting t = 0 in this, we have φ(n−k) (1) = φ(n−k) (0). Now, from the Maclaurin series for φn (z), n n we have if k > 0 φ(n−2k−1) (0) = 0, n

φ(n−2k) (0) = n

1 φ(n−1) (0) = − n!, n 2

n! (−1) k−1 Bk , (2k)!

φ(n) n (0) = n!.

Substituting these values of φ(n−k) (1) and φ(n−k) (0) in Darboux’s result, we obtain the n n

128

The Expansion of Functions in Infinite Series

Euler–Maclaurin sum formula 1 , z−a 0 { f (z) − f 0(a)} 2 n−1 Õ (−1)m−1 Bm (z − a)2m (2m) + {f (z) − f (2m) (a)} (2m)! m=1 ∫ (z − a)2n+1 1 − φ2n (t) f (2n+1) {a + (z − a)t} dt. (2n)! 0

(z − a) f 0(a) = f (z) − f (a) −

In certain cases the last term tends to zero as n → ∞, and we can thus obtain an infinite series for f (z) − f (a). If we write ω for z − a and F(x) for f 0(x), the last formula becomes ∫ a+ω 1 F(x) dx = ω{F(a) + F(a + ω)} 2 a n−1 Õ (−1)m Bm ω2m (2m−1) {F (a + ω) − F (2m−1) (a)} + (2m)! m=1 ∫ ω2n+1 1 + φ2n (t)F (2n) (a + ωt) dt. (2n)! 0 Writing a + ω, a + 2ω, . . . , a + (r − 1)ω for a in this result and adding up, we get ∫ a+rω 1 1 F(x) dx = ω F(a) + F(a + ω) + F(a + 2ω) + · · · + F(a + rω) 2 2 a n−1 Õ (−1)m Bm ω2m {F (2m−1) (a + rω) − F (2m−1) (a)} + Rn, + (2m)! m=1 where ω2n+1 Rn = (2n)!

∫

1

φ2n (t)

( r−1 Õ

0

) F

(2n)

(a + mω + ωt) dt.

m=0

This last formula is of the utmost importance in connexion with the numerical evaluation of definite integrals. It is valid if F(x) is analytic at all points of the straight line joining a to a + rω. Example 7.2.3 If f (z) be an odd function of z, shew that z f 0(z) = f (z) +

n Õ

(−1)m

m=2 2n 2n+1

− 1

2 2 (2n)!

∫

Bm−1 (2z)2m−2 (2m−2) f (z) (2m − 2)! 1

φ2n (t) f (2n+1) (−z + 2zt) dt.

0

A history of the formula is given by Barnes [47]. It was discovered by Euler (1732), but was not published at the time. Euler communicated it (June 9, 1736) to Stirling who replied (April 16, 1738) that it included his own theorem (see §12.33) as a particular case, and also that the more general theorem had been discovered by Maclaurin; and Euler, in a lengthy reply, waived his claims to priority. The theorem was published by Euler [204] and by Maclaurin [449]. For information concerning the correspondence between Euler and Stirling, we are indebted to Mr. C. Tweedie.

7.3 Bürmann’s theorem

129

Example 7.2.4 (Math. Trip. 1904) Shew, by integrating by parts, that the remainder after n terms of the expansion of 2z cot 2z may be written in the form ∫ (−1)n+1 z2n+1 1 φ2n (t) cos (zt) dt. (2n)! sin z 0

7.3 Bürmann’s theorem We shall next consider several theorems which have for their object the expansion of one function in powers of another function. This appears in [109]; see also Dixon [182]. Let φ(z) be a function of z which is analytic in a closed region S of which a is an interior point; and let φ(a) = b. Suppose also that φ 0(a) , 0. Then Taylor’s theorem furnishes the expansion φ 00(a) (z − a)2 + · · · , 2! and if it is legitimate to revert this series we obtain φ(z) − b = φ 0(a) (z − a) +

z−a=

1 φ 00(a) 1 {φ(z) − b}2 + · · · , {φ(z) − b} − φ 0(a) 2 {φ 0(a)}3

which expresses z as an analytic function of the variable {φ(z) − b}, for sufficiently small values of |z − a|. If then f (z) be analytic near z = a, it follows that f (z) is an analytic function of {φ(z) − b} when |z − a| is sufficiently small, and so there will be an expansion of the form a2 a3 f (z) = f (a) + a1 {φ (z) − b} + {φ (z) − b}2 + {φ (z) − b}3 + · · · . 2! 3! The actual coefficients in the expansion are given by the following theorem, which is generally known as Bürmann’s theorem. Let ψ(z) be a function of z defined by the equation z−a ψ(z) = ; φ(z) − b then an analytic function f (z) can, in a certain domain of values of z, be expanded in the form n−1 Õ {φ(z) − b} m d m−1 0 f (z) = f (a) + [ f (a){ψ(a)} m ] + Rn, m−1 m! da m=1 where Rn =

1 2πi

∫ a

z

∫ γ

φ(z) − b φ(t) − b

n−1

f 0(t)φ 0(z)dt dz , φ(t) − φ(z)

and γ is a contour in the t-plane, enclosing the points a and z and such that, if ζ be any point inside it, the equation φ(t) = φ(ζ) has no roots on or inside the contour except 2 a simple root t = ζ. 2

It is assumed that such a contour can be chosen if |z − a | be sufficiently small; see §7.31.

The Expansion of Functions in Infinite Series

130

To prove this, we have ∫ f (z) − f (a) =

∫ z∫ 0 f (t) φ 0(ζ) dt dζ 1 f (ζ) dζ = 2πi a γ φ(t) − φ(ζ) a ∫ z∫ 0 1 f (t) φ 0(ζ) dt dζ = 2πi a γ φ(t) − b " n−2 # Õ φ(ζ) − b m {φ(ζ) − b} n−1 × + . φ(t) − b {φ(t) − b} n−2 {φ(t) − φ(ζ)} m=0 z

0

But, by §4.3, m 0 ∫ {φ(z) − b} m+1 f 0(t) dt φ(ζ) − b f (t) φ 0(ζ) dt dζ = φ(t) − b 2πi(m + 1) γ {φ(t) − b} m+1 a γ φ(t) − b ∫ 0 {φ(z) − b} m+1 f (t){ψ(t)} m+1 dt {φ(z) − b} m+1 d m 0 = = [ f (a){ψ(a)} m+1 ]. 2πi(m + 1) γ (t − a)m+1 (m + 1)! da m

1 2πi

∫

z

∫

Therefore, writing m − 1 for m, n−1 Õ {φ(z) − b} m d m−1 0 [ f (a){ψ(a)} m ] m−1 m! da m=1 n−1 0 ∫ z∫ φ(ζ) − b 1 f (t) φ 0(ζ) dt dζ + . 2πi a γ φ(t) − b φ(t) − φ(ζ)

f (z) = f (a) +

If the last integral tends to zero as n → ∞, we may write the right-hand side of this equation as an infinite series. Example 7.3.1 Prove that z=a+

2 2 ∞ Õ (−1)n−1 Cn (z − a)n en(z −a )

n=1

n!

,

where Cn = (2na)n−1 −

n(n − 1)(n − 2) n2 (n − 1)(n − 2)(n − 3)(n − 4) (2na)n−3 + (2na)n−5 − · · · . 1! 2!

To obtain this expansion, write f (z) = z,

φ(z) − b = (z − a)e x

2

−a2

,

ψ(z) = e a

2

−z 2

in the above expression of Bürmann’s theorem; we thus have n−1 ∞ Õ 1 d n n(z 2 −a2 ) n(a2 −z 2 ) z=a+ (z − a) e e . n! dz n−1 z=a n=1

,

7.3 Bürmann’s theorem

131

But, putting z = a + t, n−1 n−1 d d n(a2 −z 2 ) −n(2at+t 2 ) = e e dz n−1 dt n−1 z=a t=0 = (n − 1)! × the coefficient of t n−1 = (n − 1)! × the coefficient of t n−1 = (n − 1)! ×

n−1 Õ r=0

in the expansion of e−nt(2a+t) ∞ Õ (−1)r nr t r (2a + t)r in r! r=0

(−1)r nr (2a)2r−n+1 . (n − 1 − r)!(2r − n + 1)!

The highest value of r which gives a term in the summation is r = n − 1. Arranging therefore the summation in descending indices r, beginning with r = n − 1, we have n−1 d n(n − 1)(n − 2) n−1 n−1 n(a2 −z 2 ) n−3 = (−1) (2na) − e (2na) + · · · dz n−1 1! z=a = (−1)n−1 Cn, which gives the required result. Example 7.3.2 Obtain the expansion 2 sin4 z 2 · 4 sin6 z · + · +··· . 3 2 3·5 3 Example 7.3.3 Let a line p be drawn through the origin in the z-plane, perpendicular to the line which joins the origin to any point a. If z be any point on the z-plane which is on the same side of the line p as the point a is, shew that 2m+1 ∞ Õ 1 z−a . log z = log a + 2 2m + 1 z + a m=1 z2 = sin2 z +

7.31 Teixeira’s extended form of Bürmann’s theorem In the last section we have not investigated closely the conditions of convergence of Bürmann’s series, for the reason that a much more general form of the theorem will next be stated; this generalisation bears the same relation to the theorem just given that Laurent’s theorem bears to Taylor’s theorem: viz., in the last paragraph we were concerned only with the expansion of a function in positive powers of another function, whereas we shall now discuss the expansion of a function in positive and negative powers of the second function. The general statement of the theorem is due to Teixeira [622], whose exposition we shall follow in this section. See also Bateman [56]. Suppose (i) that f (z) is a function of z analytic in a ring-shaped region A, bounded by an outer curve C and an inner curve c; (ii) that θ(z) is a function analytic on and inside C, and has only one zero a within this contour, the zero being a simple one;

The Expansion of Functions in Infinite Series

132

(iii) that x is a given point within A; (iv) that for all points z of C we have |θ(x)| < |θ(z)|, and for all points z of c we have |θ(x)| > |θ(z)|. The equation θ(z) − θ(x) = 0 has, in this case, a single root z = x in the interior of C, as is seen from the equation 3 ∫ 0 ∫ ∫ θ 0(z) dz 1 θ (z) θ 0(z) 1 = dz + θ(x) dz + · · · 2 2πi C θ(z) − θ(x) 2πi C θ(z) C {θ(z)} ∫ 0 1 θ (z) dz = , 2πi C θ(z) of which the left-hand and right-hand members represent respectively the number of roots of the equation considered (§6.31) and the number of the roots of the equation θ(z) = 0 contained within C. Cauchy’s theorem therefore gives ∫ ∫ f (z) θ 0(z) dz f (z) θ 0(z) dz 1 . − f (x) = 2πi C θ(z) − θ(x) c θ(z) − θ(x) The integrals in this formula can be expanded, as in Laurent’s theorem, in powers of θ (x), by the formulae ∫ ∫ ∞ f (z) θ 0(z) dz Õ f (z) θ 0(z) dz n {θ(x)} , = {θ(z)} n+1 C θ(z) − θ(x) C n=0 ∫ ∫ ∞ Õ f (z) θ 0(z) dz 1 f (z){θ(z)} n−1 θ 0(z) dz. =− n {θ(x)} c θ(z) − θ(x) c n=1 We thus have the formula f (x) =

∞ Õ

An {θ(x)} n +

n=0

∞ Õ n=1

Bn , {θ(x)} n

where 1 An = 2πi

∫ C

f (z) θ 0(z) dz , {θ(z)} n+1

Integrating by parts, we get, if n , 0, ∫ f 0 (z) 1 dz, An = 2πin C {θ(z)} n

1 Bn = 2πi

Bn = −

∫

f (z){θ(z)} n−1 θ 0(z) dz.

c

1 2πin

∫

{θ(z)} n f 0(z) dz.

c

This gives a development of f (x) in positive and negative powers of θ(x), valid for all points x within the ring-shaped space A. If the zeros and poles of f (z) and θ(z) inside C are known, An and Bn can be evaluated by §5.22 or by §6.1. 3

The expansion is justified by §4.7, since

on ∞ n Í θ (x) n=1

θ (z)

converges uniformly when z is on C.

7.3 Bürmann’s theorem

133

Example 7.3.4 Shew that, if |x| < 1, then 3 5 1·3 1 1 2x 2x 2x + +··· . x= + 2 1 + x2 2 · 4 1 + x2 2 · 4 · 6 1 + x2 Shew that, when |x| > 1, the second member represents x −1 . (m) denote the sum of all combinations of the numbers 22 , 42 , Example 7.3.5 (Teixeira) If S2n 2 2 6 , . . . , (2n − 2) , taken m at a time, shew that ( ) (1) (n) ∞ Õ S2(n+1) (−1)n S2(n+1) 1 1 (−1)n+1 1 = + − +···+ (sin z)2n+1, z sin z n=0 (2n + 2)! 2n + 3 2n + 1 3

the expansion being valid for all values of z represented by points within the oval whose equation is |sin z| = 1 and which contains the point z = 0.

7.32 Lagrange’s theorem Suppose now that the function f (z) of §7.31 is analytic at all points in the interior of C, and let θ(x) = (x − a)θ 1 (x). Then θ 1 (x) is analytic and not zero on or inside C and the contour c can be dispensed with; therefore the formulae which give An and Bn now become, by §5.22 and §6.1, ∫ f 0(z) dz 1 d n−1 f 0(a) 1 = (n ≥ 1), An = 2πin C (z − a)n {θ 1 (z)} n n! da n−1 θ 1n (a) ∫ 1 f (z) θ 0(z) dz A0 = = f (a), 2πi C θ 1 (z) z − a Bn = 0. The theorem of the last section accordingly takes the following form, if we write θ 1 (z) = 1/φ(z): Let f (z) and φ(z) be functions of z analytic on and inside a contour C surrounding a point a, and let t be such that the inequality | tφ(z) | < |z − a| is satisfied at all points z on the perimeter of C; then the equation ζ = a + tφ(ζ), regarded as an equation in ζ, has one root in the interior of C; and further any function of ζ analytic on and inside C can be expanded as a power series in t by the formula f (ζ) = f (a) +

∞ Õ t n d n−1 0 [ f (a){φ(a)} n ]. n−1 n! da n=1

This result was published by Lagrange [395] in 1770.

The Expansion of Functions in Infinite Series

134

Example 7.3.6 Within the contour surrounding b defined by the inequality |z(z − α)| > |α|, where |α| < 12 |b|, the equation b z−α− =0 z has one root ζ, the expansion of which is given by Lagrange’s theorem in the form ζ = b+

∞ Õ (−1)n−1 (2n − 2)! n=1

n!(n − 1)!b2n−1

αn .

Now, from the elementary theory of quadratic equations, we know that the equation b =0 z n n o o p p has two roots, namely α2 1 + 1 + 4b/α2 and α2 1 − 1 + 4b/α2 ; and our expansion represents the former of these only (the latter is outside the given contour) – an example of the need for care in the discussion of these series. z−α−

Example 7.3.7 If y be that one of the roots of the equation y = 1 + zy 2 which tends to 1 when z → 0, shew that n(n + 3) 2 n(n + 4)(n + 5) 3 z + z 2! 3! n(n + 5)(n + 6)(n + 7) 4 n(n + 6)(n + 7)(n + 8)(n + 9) 5 + z + z +··· 4! 5!

y n = 1 + nz +

so long as |z| < 14 . Example 7.3.8 (McClintock) If x be that one of the roots of the equation x = 1 + yx a which tends to 1 when y → 0, shew that log x = y +

2a − 1 2 (3a − 1)(3a − 2) 3 y + y +··· , 2 2.3

the expansion being valid so long as |y| < |(a − 1)a−1 a−a |.

7.4 The expansion of a class of functions in rational fractions This appears in Mittag-Leffler [470], see also [471]. Consider a function f (z), whose only singularities in the finite part of the plane are simple poles a1, a2, a3, . . . , where |a1 | ≤ |a2 | ≤ |a3 | ≤ · · · . Let b1, b2, b3, . . . be the residues at these poles, and let it be possible to choose a sequence of circles Cm (the radius of Cm being Rm ) with centre at O, not passing through any poles, such that | f (z)| is bounded on Cm . (The function cosec z may be cited as an example of the class of functions considered, and we take Rm = (m + 21 )π.) Suppose further that Rm → ∞ as m → ∞ and that the upper bound (which is a function of m) of | f (z)| on Cm is

7.4 The expansion of a class of functions in rational fractions

135

itself bounded as 4 m → ∞; so that, for all points on the circle Cm , | f (z)| < M, where M is independent of m. Then, if x be not a pole of f (z), since the only poles of the integrand are the poles of f (z) and the point z = x, we have, by §6.1, ∫ Õ br 1 f (z) dz = f (x) + , 2πi Cm z − x ar − x r where the summation extends over all poles in the interior of Cm . But ∫ ∫ ∫ 1 x 1 f (z) dz f (z) dz f (z) dz = + 2πi Cm z − x 2πi Cm z 2πi Cm z(z − x) ∫ Õ br x f (z) dz = f (0) + + , ar 2πi Cm z(z − x) r if we suppose the function f (z) to be analytic at the origin. ∫ f (z) dz Now as m → ∞, is O(Rm −1 ), and so tends to zero as m tends to infinity. Cm z(z − x) Therefore, making m → ∞, we have ∫ ∞ Õ 1 1 x f (z) dz bn 0 = f (x) − f (0) + − − lim , m→∞ a − x a 2πi z(z − x) n n Cm n=1

1 1 i.e. f (x) = f (0) + bn + , which is an expansion of f (x) in rational fractions x − an an n=1 of x; and the summation extends over all the poles of f (x). If |an | < |an+1 | this series converges uniformly throughout the region given by | x | ≤ a, where a is any constant (except near the points an ). For if Rm be the radius of the circle which encloses the points |a1 |, . . . , |an |, the modulus of the remainder of the terms of the series after the first n is ∫ x f (z) dz Ma < , 2πi Rm − a Cm z(z − x) ∞ Í

Ma < ε. Rm − a The convergence is obviously still uniform even if |an | ≤ |an+1 | provided the terms of the series are grouped so as to combine the terms corresponding to poles of equal moduli. If, instead of the condition | f (z)| < M, we have the condition |z−p f (z)| < M, where M is independent of m when z is on Cm , and p is a positive integer, then we should have to expand ∫ f (z) dz by writing C z− x

by §4.62; and, given ε, we can choose n independent of x such that

1 1 x x p+1 = + 2 + · · · + p+1 , z−x z z z (z − x) and should obtain a similar but somewhat more complicated expansion. 4

Of course R m need not (and frequently must not) tend to infinity continuously; e.g. in the example taken R m = (m + 21 ) π, where m assumes only integer values.

The Expansion of Functions in Infinite Series

136

Example 7.4.1 Prove that cosec z =

1 1 1 Õ + (−1)n + , z z − nπ nπ

the summation extending to all positive and negative values of n. To obtain this result, let cosec z − 1/z = f (z). The singularities of this function are at the points z = nπ, where n is any positive or negative integer. The residue of f (z) at the singularity nπ is therefore (−1)n , and the reader will easily see that | f (z)| is bounded on the circle |z| = (n + 12 )π as n → ∞. Applying now the general theorem Õ 1 1 f (z) = f (0) + cn + , z − an an where cn is the residue at the singularity an , we have Õ 1 1 n . f (z) = f (0) + + (−1) z − nπ nπ But f (0) = lim

z→0

z − sin z = 0. Therefore z sin z 1 1 Õ 1 n , cosec z = + (−1) + z z − nπ nπ

(7.2)

which is the required result. Example 7.4.2 If 0 < a < 1, shew that ∞

e az 1 Õ 2z cos 2naπ − 4nπ sin 2naπ . = + ez − 1 z n=1 z2 + 4n2 π 2 Example 7.4.3 Prove that 1 2 1 1 1 1 + 2 = − π 1 4 −π −2π 4 4 4 x − cos x) 2πx e − e π + 4x e π − e (2π) + 14 x 4 3 1 − 3π +··· . e − e−3π (3π)4 + 14 x 4

2πx 2 (cosh

The general term of the series on the right is (er π

(−r)r r , − e−r π ) (rπ)4 + 14 x 4

which is the residue at each of the four singularities r, −r, ri, −ri of the function πz (π 4 z4

+

1 4 x )(e πz 4

− e−πz ) sin πz

.

The singularities of this latter function which are not of the type r, −r, ri, −ri are at

7.5 The expansion of a class of functions as infinite products

137

the five points 0, (±1 ± i)x/2π. At z = 0 the residue is 2/πx 4 ; at each of the four points z = (±1 ± i)x/2π, the residue is {2πx 2 (cos x − cosh x)}−1 . Therefore 4

∞ Õ

2 1 2 (−1)r r + 4− 2 1 4 r π − e−r π 4 e πx πx (cosh x − cos x) (rπ) + 4 x r=1 ∫ 1 πz dz = , lim 1 2πi n→∞ C (π 4 z4 + 4 x 4 )(e πz − e−πz ) sin πz

where C is the circle whose radius is n + 21 (here n an integer), and whose centre is the origin. But, at points on C, this integrand is O(|z| −3 ); the limit of the integral round C is therefore zero. From the last equation the required result is now obvious. Example 7.4.4 Prove that

1 3 5 sec x = 4π 2 − + −··· . π − 4x 2 9π 2 − 4x 2 25π 2 − 4x 2

(7.3)

Example 7.4.5 Prove that cosech x =

1 1 1 1 − + − · · · . − 2x 2 x π + x 2 4π 2 + x 2 9π 2 + x 2

(7.4)

Example 7.4.6 Prove that

1 3 1 sech x = 4π 2 − + −··· . π + 4x 2 9π 2 + 4x 2 25π 2 + 4x 2

(7.5)

Example 7.4.7 Prove that 1 1 1 1 coth x = + 2x 2 + + +··· . x π + x 2 4π 2 + x 2 9π 2 + x 2

(7.6)

Example 7.4.8 (Math. Trip. 1899) Prove that ∞ ∞ Õ Õ

π2 1 = coth πa coth πb. (m2 + a2 )(n2 + b2 ) ab m=−∞ n=−∞

(7.7)

7.5 The expansion of a class of functions as infinite products The theorem of the last article can be applied to the expansion of a certain class of functions as infinite products. For let f (z) be a function which has simple zeros at the points 5 a1, a2, a3, . . . , where lim |an | is infinite; and let f (z) be analytic for all values of z. Then f 0(z) is analytic for all n→∞

values of z (§5.22), and so f 0(z)/ f (z) can have singularities only at the points a1, a2, a3, . . . Consequently, by Taylor’s theorem, f (z) = (z − ar ) f 0(ar ) + 5

These being the only zeros of f (z); and a n , 0.

(z − ar )2 00 f (ar ) + · · · 2

138

The Expansion of Functions in Infinite Series

and f 0(z) = f 0(ar ) + (z − ar ) f 00(ar ) + · · · . It follows immediately that at each of the points ar , the function f 0(z)/ f (z) has a simple pole, with residue +1. If then we can find a sequence of circles Cm of the nature described in §7.4, such that f 0(z)/ f (z) is bounded on Cm as m → ∞, it follows, from the expansion given in §7.4, that ∞ f 0(0) Õ 1 1 f 0(z) = + + . f (z) f (0) n=1 z − an an Since this series converges uniformly when the terms are suitably grouped (§7.4), we may integrate term-by-term (§4.7). Doing so, and taking the exponential of each side, we get ∞ Ö z z/a n f 0 (0)z/ f (0) e , f (z) = ce 1− an n=1 where c is independent of z. Putting z = 0, we see that f (0) = c, and thus the general result becomes ∞ Ö z f 0 (0)z/ f (0) z/a n 1− f (z) = f (0)e e . an n=1 This furnishes the expansion, in the form of an infinite product, of any function f (z) which fulfils the conditions stated. sin z Example 7.5.1 Consider the function f (z) = , which has simple zeros at the points z rπ, where r is any positive or negative integer. In this case we have f (0) = 1, f 0(0) = 0, and so the theorem gives immediately ∞ z nzπ o n sin z Ö n z − nzπ o = 1− e e 1+ ; z nπ nπ n=1 for it is easily seen that the condition concerning the behaviour of

f 0 (z) f (z)

as |z| → ∞ is fulfilled.

Example 7.5.2 (Trinity, 1899) Prove that ( 2) ( 2) ( 2) 2) ( k k k k 1+ 1+ 1+ 1+ x 2π − x 2π + x 4π − x ( ) 2 k cosh k − cos x × 1+ ··· = . 4π + x 1 − cos x

7.6 The factor theorem of Weierstrass This appears in [663, pp. 77–124]. The theorem of §7.5 is very similar to a more general theorem in which the character of the function f (z), as |z| → ∞, is not so narrowly restricted. Let f (z) be a function of z with no essential singularities (except at ‘the point infinity’); and let the zeros and poles of f (z) be at a1, a2, a3, . . ., where 0 < |a1 | ≤ |a2 | ≤ |a3 | ≤ · · · .

7.6 The factor theorem of Weierstrass

139

Let the zero at an be of (integer) order mn . (We here regard a pole as being a zero of negative order.) If the number of zeros and poles is unlimited, it is necessary that |an | → ∞, as n → ∞; for, if not, the points an would have a limit point 6 , which would be an essential singularity of f (z). We proceed to shew first of all that it is possible to find polynomials gn (z) such that mn ∞ Ö z gn (z) e 1− an n=1 converges for all finite values of z, provided that z is not at one of the points an for which mn is negative. Let K be any constant, and let |z| < K; then, since |an | → ∞, we can find N such that, when n > N, |an | > 2K. The first N factors of the product do not affect its convergence; consider any value of n greater than N, and let 2 kn −1 z 1 z 1 z gn (z) = + +···+ . an 2 an k n − 1 an Then m m ∞ ∞ Õ Õ 1 z 1 z + gn (z) = − m=1 m an m=k m an n k n Õ z ∞ z m < an m=0 an < 2 (Kan−1 )kn , since zan−1 < 12 . Hence 1−

mn z = eun (z), egn (z) an

(7.8)

where |un (z)| ≤ 2 mn (Kan−1 )kn . Now mn and an are given, but k n is at our disposal; since Kan−1 < 12 , we choose k n to be ∞ Í bn is any convergent series of the smallest number such that 2 mn (Ka−1 )kn < bn , where n

n=1

positive terms (e.g. we might take bn = 2−n ). Hence mn ∞ ∞ Ö Ö z gn (z) e = eun (z), 1− a n n=N +1 n=N +1

(7.9)

where |un (z)| < bn ; and therefore, since bn is independent of z, the product converges absolutely and uniformly when |z| < K, except near the points an . Now let mn ∞ Ö z F(z) = 1− egn (z) . (7.10) an n=1 6

From the two-dimensional analogue of §2.21.

The Expansion of Functions in Infinite Series

140

f (z) = G1 (z), it follows that G1 (z) is an integral function (§5.64) of z and has no F(z) 1 d zeros. It follows that G1 (z) is analytic for all finite values of z; and so, by Taylor’s G1 (z) dz ∞ Í theorem, this function can be expressed as a series nbn z n−1 converging everywhere; Then, if

n=1

integrating, it follows that G1 (z) = ceG(z), where G(z) =

∞ Í n=1

bn z n and c is a constant; this series converges everywhere, and so G(z) is

an integral function. Therefore, finally, f (z) = f (0)e

G(z)

∞ Ö n=1

mn z gn (z) 1− e , an

where G(z) is some integral function such that G(0) = 0. Note The presence of the arbitrary element G(z) which occurs in this formula for f (z) is due to the lack of conditions as to the behaviour of f (z) as |z| → ∞. Corollary 7.6.1 If mn = 1, it is sufficient to take k n = n, by §2.36.

7.7 The expansion of a class of periodic functions in a series of cotangents Let f (z) be a periodic function of z, analytic except at a certain number of simple poles; for convenience, let π be the period of f (z) so that f (z) = f (z + π). Let z = x + iy and let f (z) → ` uniformly with respect to x as y → +∞, when 0 ≤ x ≤ π; similarly let f (z) → ` 0 uniformly as y → −∞. Let the poles of f (z) in the strip 0 < x ≤ π be at a1, a2, . . . , an ; and let the residues at them be c1, c2, . . . , cn . Further, let ABCD be a rectangle whose ∫ corners are 7 −i ρ, π − i ρ, π + i ρ0 and i ρ0 in order. 1 Consider f (t) cot(t − z) dt taken round this rectangle; the residue of the integrand 2πi at ar is cr cot(ar − z), and the residue at z is f (z). Also the integrals along DA and CB cancel on account of the periodicity of the integrand; and as ρ → ∞, the integrand on AB tends uniformly to i` 0, while as ρ0 → ∞ the integrand on CD tends uniformly to −i`; therefore n

Õ 1 (` + ` 0) = f (z) + cr cot(ar − z). 2 r=1 That is to say, we have the expansion n

f (z) = 7

Õ 1 (` + ` 0) + cr cot(z − ar ). 2 r=1

If any of the poles are on x = π, shift the rectangle slightly to the right; ρ, ρ0 are to be taken so large that a1 , a2 , . . . , a n are inside the rectangle.

7.8 Borel’s theorem

141

Example 7.7.1 Prove that cot(x − a1 ) cot(x − a2 ) · · · cot(x − an ) = n Õ cot(ar − a1 ) · · · ∗ · · · cot(ar − an ) × cot (x − ar ) + (−1)n/2, r=1

or

n Õ

cot(ar − a1 ) · · · ∗ · · · cot(ar − an ) cot(x − ar ),

r=1

according as n is even or odd; the ∗ means that the factor cot(ar − ar ) is omitted. Example 7.7.2 Prove that sin(x − b1 ) sin(x − b2 ) · · · sin(x − bn ) sin(a1 − b1 ) · · · sin(a1 − bn ) = cot(x − a1 ) sin(x − a1 ) sin(x − a2 ) · · · sin(x − an ) sin(a1 − a2 ) · · · sin(a1 − an ) sin(a2 − b1 ) · · · sin(a2 − bn ) + cot(x − a2 ) sin(a2 − a1 ) · · · sin(a2 − an ) .. + . + cos(a1 + a2 + · · · + an − b1 − b2 − · · · − bn ).

7.8 Borel’s theorem This appears in [85, p. 94] and the memoirs there cited. Let f (z) =

∞ Í n=0

an z n be analytic

when |z| ≤ r, so that, by §5.23, | an r n | < M, where M is independent of n. Hence, if ∞ a zn Í n φ(z) = , then φ(z) is an integral function, and n=0 n! ∞ Õ M |z n | |φ(z)| < = Me |z |/r , n · n! r n=0

(7.11)

and similarly φ(n) (z) < Me |z |/r /r n . ∫ ∞ Now consider f1 (z) = e−t φ(zt) dt; this integral is an analytic function of z when 0

|z| < r, by §5.32. Also, if we integrate by parts, f1 (z) = −e φ(zt)

−t

∞ 0

+z

∫∞

e−t φ 0(zt) dt

0

=

n Õ m=0

z

m

−e φ −t

(m)

(zt)

∞ 0

+z

n+1

∫∞

e−t φ(n+1) (zt) dt.

0

But lim e−t φ(m) (zt) = am ; and, when |z| < r, lim e−t φ(m) (zt) = 0. Therefore t→0

t→∞

The Expansion of Functions in Infinite Series

142

f1 (z) =

n Í m=0

am z m + Rn, where |Rn | < z n+1

∫∞

e−t × Me |zt |/r r −n−1 dt

0

n+1 = zr −1 M {1 − |z| r −1 }−1 → 0,

as

n → ∞.

Consequently, when |z| < r, f1 (z) =

Õ∞ m=0

and so f (z) =

∫

am z m = f (z);

∞

e−t φ(zt) dt,

0

where φ(z) = If S =

∞ Í n=0

∞ Í n=0

an z n!

n

; φ(z) is called Borel’s function associated with

an and φ(z) =

∞ Í n=0

an z n n!

∞ Í n=0

an z n .

and if we can establish the relation S =

∫∞

e−t φ (t) dt,

0

the series S is said (§8.41) to be ‘summable (B)’; so that the theorem just proved shews that a Taylor’s series representing an analytic function is summable (B).

7.81 Borel’s integral and analytic continuation We next obtain Borel’s result that his integral represents an analytic function in a more extended region than the interior of the circle |z| = r.

This extended region is obtained as follows: take the singularities a, b, c, . . . of f (z) and

7.8 Borel’s theorem

143

through each of them draw a line perpendicular to the line joining that singularity to the origin. The lines so drawn will divide the plane into regions of which one is a polygon with the origin inside it. Then Borel’s integral represents an analytic function (which, by §5.5 and §7.8, is obviously that defined by f (z) and its continuations) throughout the interior of this polygon. The reader will observe that this is the first actual formula obtained for the analytic continuation of a function, except the trivial one of Example 5.5.1. For, take any point P with affix ζ inside the polygon; then the circle on OP as diameter has no singularity on or inside it 8 ; and consequently we can draw a slightly larger concentric circle (the difference of the radii of the circles being, say, δ) C with no singularity on or inside it. Then, by §5.4, ∫ f (z) 1 dz, an = 2πi C z n+1 and so φ (ζt) =

∫ ∞ 1 Õ ζ nt n f (z) dz; 2πi n=0 n! C z n+1

but ∞ Õ ζ n t n f (z) n! z n+1 n=0

converges uniformly (§3.34) on C since f (z) is bounded and |z| ≥ δ > 0, where δ is independent of z; therefore, by §4.7, ∫ 1 φ(ζt) = z−1 f (z) exp(ζtz−1 ) dz, 2πi C and so, when t is real, |φ(ζt)| < F(ζ)eλt , where F(ζ) is bounded in any closed region lying wholly inside the polygon and is independent of t; and λ is the greatest value of the real part of ζ/z on C. If we draw the circle traced out by the point z/ζ, we see that the real part of ζ/z is greatest −1 when z is at the extremity of the diameter through ζ, and so the value ∫ of λ is |ζ |·(|ζ | + δ) < 1. ∞

We can get a similar inequality for φ 0(ζt) and hence, by §5.32,

e−t φ(ζt) dt is analytic at

0

ζ and is obviously a one-valued function of ζ. This is the result stated above.

7.82 Expansions in series of inverse factorials A mode of development of functions, which, after being used by Nicole [498] and Stirling (see [635], and [634]) in the eighteenth century, was systematically investigated by Schlömilch [583] in 1863, is that of expansion in a series of inverse factorials. More recent investigations are due to Kluyver [379], Nielsen [499, 501, 502] and Pincherle [524, 525]. Properties of functions defined by series of inverse factorials have been studied in an important memoir by Nörlund [506]. 8

The reader will see this from the figure; for if there were such a singularity the corresponding side of the polygon would pass between O and P; i.e. P would be outside the polygon.

The Expansion of Functions in Infinite Series

144

To obtain such an expansion of a function∫analytic when |z| > r, we let the function be ∞ ∞ ∞ Í Í f (z) = an z−n, and use the formula f (z) = ze−tz φ (t) dt, where φ(t) = an t n /n!; this n=0

n=0

0

result may be obtained in the same way as that of §7.8. Modify this by writing e−t = 1 − ξ, ∫ 1 φ(t) = F(ξ); then f (z) = z(1 − ξ)z−1 F(ξ) dξ. Now if t = u + iv and if t be confined to 0

the strip −π < v < π, t is a one-valued function of ξ and F(ξ) is an analytic function of ξ; and ξ is restricted so that −π < arg(1 − ξ) < π. Also the interior of the circle |ξ | = 1 corresponds to the interior of the curve traced out by the point t = − log(2 cos 21 θ) + 12 iθ, 1/2 (writing ξ = exp{i(θ + π)}); and inside this curve | t | − R(t) ≤ R(t)2 + π 2 − R(t) → 0, as R (t) → ∞. of t; It follows that, when |ξ | ≤ 1, |F(ξ)| < Mer |t | < M1 |ert | , where M1 is independent and so F(ξ) < M1 (1−ξ)−r . Now suppose that 0 ≤ ξ < 1; then, by §5.23, F (n) (ξ) < M2 n!ρ−n , where M2 is the upper bound of |F(z)| on a circle with centre ξ and radius ρ < 1 − ξ. Taking ρ = n(1 − ξ)/(n + 1) and observing that (1 + n−1 )n < e we find that 9 −n n oi −r n(1 − ξ) n! ξ n+1 n+1

h n (n) F (ξ) < M1 1 − ξ +

< M1 e(n + 1)r n!(1 − ξ)−r−n . ∫

means lim+ ε→0

0

parts,

∫

1

Remembering that, by §4.5,

f (z) = lim+ [−(1 − ξ)z F(ξ)]1−ε + 0 ε→0

= lim+ [−(1 − ξ)z F(ξ)]1−ε + 0 ε→0

+

1 z+1

∫

1−ε

∫

1−ε

, we have, by repeated integrations by 0

1−ε

(1 − ξ)z F 0 (ξ) dξ

0

1−ε 1 −(1 − ξ)z+1 F 0(ξ) 0 z+1

(1 − ξ)z+1 F 00(ξ) dξ

0

=

.. .

= b0 +

b1 b2 bn + +···+ + Rn, z + 1 (z + 1)(z + 2) (z + 1)(z + 2) · · · (z + n)

where 1−ε bn = lim −(1 − ξ)z+n F (n) (ξ) 0 = F (n) (0), ε→0

9

(1 + x −1 ) x increases with x; for y

−1

= 1 + x,

d dx

1 1−y

> e v , when y < 1, and so log

x log(1 + x ) = log(1 + x ) − −1

−1

1 1+x

> 0.

1 1−y

> y. That is to say, putting

7.9 Miscellaneous examples

145

if the real part of z + n − r − n > 0, i.e. if Re z > r; further ∫ 1−ε 1 (1 − ξ)z+n F (n+1) (ξ) dξ |Rn | ≤ lim |(z + 1)(z + 2) · · · (z + n)| ε→0 0 M1 e(n + 2)r n! < |(z + 1)(z + 2) · · · (z + n)| · R(z − r) M1 e(n + 2)r n! < , (r + 1 + δ)(r + 2 + δ) · · · (r + n + δ) · δ where δ = R(z − r). n Î r + δ − r +δ Now 1+ e m tends to a limit (§2.71) as n → ∞, and so |Rn0 | → 0 if m m=1 (n + 2) e r

−(r+δ)

n Í 1

1/m

tends to zero; but ∫ n+1 n Õ 1 dx > = log(n + 1), m x 1 m=1

by §4.43, and (n + 2)r (n + 1)−r−δ → 0 when δ > 0; therefore Rn → 0 as n → ∞, and so, when R(z) > r, we have the convergent expansion f (z) = b0 +

b1 b2 bn + +···+ +··· . z + 1 (z + 1) (z + 2) (z + 1)(z + 2) · · · (z + n)

Example 7.8.1 Obtain the same expansion by using the results ∫ 1 1 1 un (1 − u)z du, = (z + 1) (z + 2) · · · (z + n + 1) n ! 0 ∫ ∫ 1 ∫ f (t) dt dt = f (t)(1 − u)z−t−1 du. C C z−t 0

(7.12)

Example 7.8.2 (Schlömilch) Obtain the expansion 1 1 a1 a2 log 1 + = − − −··· , (7.13) z z z(z + 1) z(z + 1)(z + 2) ∫ 1 where an = t(1 − t)(2 − t) · · · (n − 1 − t) dt, and discuss the region in which it converges. 0

7.9 Miscellaneous examples Example 7.1 (Levi-Cività [432]) argument, obtain the expansion f (y) = f (x) +

If y − x − φ (y) = 0, where φ is a given function of its m ∞ Õ 1 d 1 {φ (x)} m f (x), m! 1 − φ 0 (x) dx m=1

where f denotes any analytic function of its argument, and discuss the range of its validity.

The Expansion of Functions in Infinite Series

146

Example 7.2 Obtain (from the formula of Darboux or otherwise) the expansion f (z) − f (a) =

∞ Õ (−1)n−1 (z − a)n

n!(1 − r)n

n=1

{ f (n) (z) − r n f (n) (a)};

find the remainder after n terms, and discuss the convergence of the series. Example 7.3 Shew that f (x + h) − f (x) =

n Õ m=1

1 · 3 · 5 · · · (2m − 1) h m (m) { f (x + h) − (−1)m f (m) (x)} (m!)2 2m ∫ 1 +(−1)n h n+1 γn (t) f (n+1) (x + ht) dt,

(−1)m−1

0

where 1

1

x n+ 2 (1 − x)n+ 2 d n − 12 1 1 {x (1 − x)− 2 } = γn (x) = 2 n (n!) dx πn!

∫

1

1

1

(x − z)n z− 2 (1 − z)− 2 dz,

0

and shew that γn (x) is the coefficient of n!t n in the expansion of {(1 − t x)(1 + t − t x)}−1/2 in ascending powers of t. Example 7.4 By taking 1 d n (1 − r)e xu φ(x + 1) = n ! dun 1 − re−u u=0 in the formula of Darboux, shew that n Õ

h m (m) 1 (m) am f (x + h) − f (x) = − f (x + h) − f (x) m! r m=1 ∫ 1 +(−1)n h n+1 φ (t) f (n+1) (x + ht) dt, 0

where u u2 u3 1−r = 1 − a1 + a2 − a3 + · · · . −u 1 − re 1! 2! 3! Example 7.5 Shew that f (z) − f (a) =

n Õ

2Bm (22n − 1)(z − a)2m−1 (2m−1) {f (a) + f (2m−1) (z)} (2m)! m=1 ∫ (z − a)2n+1 1 + ψ2n (t) f (2n+1) {a + t (z − a)} dt, (2n)! 0 (−1)m−1

where ψn (t) =

n+1 2 d uetu . n + 1 dun+1 eu + 1 u=0

7.9 Miscellaneous examples

147

Example 7.6 (Trinity, 1899) Prove that f (z2 ) − f (z1 ) = C1 (z2 − z1 ) f 0(z2 ) + C2 (z2 − z1 )2 f 00(z1 ) − C3 (z2 − z1 )3 f 000(z2 ) −C4 (z2 − z1 )4 f (iv) (z1 ) + · · · + (−1)n (z2 − z1 )n+1 ∫ 1 n d tu f (n+1) (z1 + tz2 − tz1 ) dt; (e sech u) × n du 0 u=0 in the series plus signs and minus signs occur in pairs, and the last term before the integral is that involving (z2 − z1 )n ; also Cn is the coefficient of z n in the expansion of cot π4 − 2z in ascending powers of z. Example 7.7 If x1 and x2 are integers, and φ(z) is a function which is analytic and bounded for all values of z such that x1 ≤ R (z) ≤ x2 , shew (by integrating ∫ φ(z) dz e±2πiz − 1 round indented rectangles whose corners are x1 , x2 , x2 ± ∞ i, x1 ± ∞ i) that 1 φ (x1 ) 2

+ φ (x1 + 1) + φ (x1 + 2) + · · · + φ (x2 − 1) + 21 φ (x2 ) ∫ x3 1 ∞ φ(x2 + iy) − φ(x1 + iy) − φ(x2 − iy) + φ(x1 − iy) dy. = φ(z) dz + i 0 e2πy − 1 x1 ∫

Hence, by applying the theorem ∫ 4n 0

∞

y 2n−1 dy = Bn, e2πy − 1

where B1, B2, . . . are Bernoulli’s numbers, shew that φ(1) + φ(2) + · · · + φ(n) ∫ n ∞ Õ (−1)r−1 Br (2r−1) 1 φ(z) dz + φ (n), = C + φ(n) + 2 2r! r=1 (where C is a constant not involving n), provided that the last series converges. (This important formula is due to Plana [526]; a proof by means of contour integration was published by Kronecker [386]. For a detailed history, see Lindelöf [436]. Some applications of the formula are given in Chapter 12.) Example 7.8 Obtain the expansion ∞

u=

x Õ 1 · 3 · · · (2n − 3) x n + (−1)n−1 2 n=2 n! 2n

for one root of the equation x = 2u + u2 , and shew that it converges so long as |x| < 1. (m) Example 7.9 (Teixeira) If S2n+1 denote the sum of all combinations of the numbers

12, 32, 52, . . . , (2n − 1)2,

The Expansion of Functions in Infinite Series

148

taken m at a time, shew that ∞

Õ (−1)n+1 1 cos z = + z sin z n=0 (2n + 2) ! 2(n+1) 2 22n 22 (1) (n) × − S2(n+1) + · · · + (−1)n S2(n+1) sin2n+1 z. 2n + 3 2n + 1 3 Example 7.10 (Teixeira) If the function f (z) is analytic in the interior of that one of the ovals whose equation is | sin z| = C (where C ≤ 1), which includes the origin, shew that f (z) can, for all points z within this oval, be expanded in the form f (z) = f (0) +

(n−1) 00 (1) (2n−2) ∞ Õ f (0) 2n f (0) + · · · + S2n f (2n) (0) + S2n sin z 2n ! n=1

(n) (1) ∞ Õ f 0 (0) 2n+1 f (2n−1) (0) + · · · + S2n+1 f (2n+1) (0) + S2n+1 + sin z, (2n + 1) ! n=0 (m) where S2n is the sum of all combinations of the numbers

22, 42, 62, . . . , (2n − 2)2, (m) taken m at a time, and S2n+1 denotes the sum of all combinations of the numbers

12, 32, 52, . . . , (2n − 1)2, taken m at a time. Example 7.11 (Kapteyn [366]) Shew that the two series 2z +

2z3 2z5 + 2 +··· , 32 5

and 3 6 2z 2 2z 2·4 2z − + −··· , 1 − z 2 1 · 32 1 − z2 3 · 52 1 − z2 represent the same function in a certain region of the z-plane, and can be transformed into each other by Bürmann’s theorem. Example 7.12 If a function f (z) is periodic, of period 2π, and is analytic at all points in the infinite strip of the plane included between the two branches of the curve | sin z| = C (where C > 1), shew that at all points in the strip it can be expanded in an infinite series of the form f (z) =A0 + A1 sin z + · · · + An sinn z + · · · + + cos z (B1 + B2 sin z + · · · + Bn sinn−1 z + · · · ); and find the coefficients An and Bn . Example 7.13 If φ and f are connected by the equation φ(x) + λ f (x) = 0,

7.9 Miscellaneous examples

149

of which one root is a, shew that λ2 1 φ 0 λ 1 f 2 F 0 0 fF + F(x) =F − 1 φ0 1! 2! φ 03 φ 00 ( f 2 F 0)0 φ 0 (φ2 )0 ( f 3 F 0) 3 λ 1 00 2 00 3 0 0 φ (φ ) ( f F ) − +··· , 1! 2! 3! φ 06 φ 000 (φ2 )000 ( f 3 F 0)00 the general term being (−1)m

λm 1

1!2! · · · m! (φ 0) 2 m(m+1)

multiplied by a determinant in which the elements of the first row are φ 0, (φ2 )0, (φ3 )0, . . . , (φm−1 )0, ( f m F 0) and each row is the differential coefficient of the preceding one with respect to a; and F, f , F 0, . . . denote F(a), f (a), F 0(a), . . .. (See Wronski [684]. For proofs of the theorem see Cayley [135], Transon [633] and Ch. Lagrange [391].) Example 7.14 (Ježek) If the function W(a, b, x) be defined by the series a − b 2 (a − b)(a − 2b) 3 x + x +··· , 2! 3! which converges so long as |x| < 1/|b|, shew that W(a, b, x) = x +

d W(a, b, x) = 1 + (a − b)W(a − b, b, x); dx and shew that if y = W(a, b, x), then x = W(b, a, y). Examples of this function are W(1, 0, x) = e x − 1, W(0, 1, x) = log(1 + x), (1 + x)a − 1 W(a, 1, x) = . a Example 7.15 (Mangeot [453]) Prove that ∞ 1 1 Õ (−1)n x n G n, = + ∞ Í a0 n=1 n! a0n+1 an x n n=0

where 2a1 a0 4a 3a 2 1 6a 5a 3 2 Gn = .. .. . . (2n − 2)an−1 · · · nan (n − 1)an−1 1/2 ∞ Í and obtain a similar expression for an x n . n=0

0 0 2a0 0 4a1 3a0 .. .. . . ··· ··· ··· ···

··· ··· ··· .. . ··· ···

, (n − 1)a0 a1 0 0 0 .. .

150

The Expansion of Functions in Infinite Series

Example 7.16 (Gambioli [232]) Shew that 1 ∞ Í

=−

ar x r

∞ Õ r=0

1 ∂Sr+1 r x , r + 1 ∂a1

r=0

where Sr is the sum of the rth powers of the reciprocals of the roots of the equation n Í ar x r = 0.

r=0

Example 7.17 (Guichard) If fn (z) denote the nth derivate of f (z), and if f−n (z) denote that one of the nth integrals of f (z) which has an n-tuple zero at z = 0, shew that if the series ∞ Õ

fn (z)g−n (x)

n=−∞

is convergent it represents a function of z + x; and if the domain of convergence includes the origin in the x-plane, the series is equal to ∞ Õ

f−n (z + x)gn (0).

n=0

Obtain Taylor’s series from this result, by putting g(z) = 1. Example 7.18 (Math. Trip. 1895) Shew that, if x be not an integer, ν ν Õ Õ m=−ν n=−ν

2x + m + n →0 (x + m)2 (x + n)2

as ν → ∞, provided that all terms for which m = n are omitted from the summation. Example 7.19 (Math. Trip. 1896) Sum the series p Õ 1 1 + , (−1)n x − a − n n n=−q where the value n = 0 is omitted, and p, q are positive integers to be increased without limit. ∫ x Example 7.20 (Trinity, 1898) If F(x) = exp 0 xπ cot(xπ) dx , shew that ∞ n Î

F(x) = e x

n=1 ∞ n Î n=1

1−

x n n

e x+ 2n

1+

x n n

e−x+ 2n

x2

x2

o o,

and that the function thus defined satisfies the relations 1 F(−x) = , F(x)F(1 − x) = 2 sin xπ. F(x) Further, if ∫ z z2 z3 dt ψ(z) = z + 2 + 2 + · · · = − log(1 − t) , 2 3 t 0

7.9 Miscellaneous examples

151

shew that

1 1 ψ(1 − e−2πix ) F(x) = exp πix 2 − 2 2πi

when |1 − e−2πix | < 1. Example 7.21 (Mildner) Shew that n n n k k k 1+ 1+ 1+ x 2π − x 2π + x

≤n/2 Î g=1

=

1+

k 4π − x

n

1 − 2e−αg cos(x + βg ) + e−2αg

1+

1/2

k 4π + x

n ···

1 − 2e−αg cos(x − βg ) + e−2αg

1/2

2n/2 (1 − cos x)n/2 e−k cos π/n

,

where αg = k sin 2g−1 π, βg = k cos 2g−1 π, and 0 < x < 2π. n n Example 7.22 (Lerch [428]) If |x| < 1 and a is not a positive integer, shew that ∫ α−1 ∞ Õ x t − x α−1 xn 2πix α + = dt, n − α 1 − e2απi 1 − e2απi C t−x n=1 where C is a contour in the t-plane enclosing the points 0, x. Example 7.23 If φ1 (z), φ2 (z), . . . are any polynomials in z, and if F(z) be any integrable function, and if ψ1 (z), ψ2 (z), . . . be polynomials defined by the equations ∫ b φ1 (z) − φ1 (x) dx = ψ1 (z), F(x) z−x a ∫ b φ2 (z) − φ2 (x) F(x)φ1 (x) dx = ψ2 (z), z−x a ∫ b φm (z) − φm (x) dx = ψm (z), F(x) φ1 (x)φ2 (x) · · · φm−1 (x) z−x a shew that ∫ b a

F(x) dx ψ1 (z) ψ2 (z) ψ3 (z) = + + +··· z−x φ1 (z) φ1 (z) φ2 (z) φ1 (z) φ2 (z) φ3 (z) ψm (z) + φ1 (z)φ2 (z) · · · φm (z) ∫ b 1 dx + F(x)φ1 (x)φ2 (x) · · · φm (x) . φ1 (z)φ2 (z) · · · φm (z) a z−x

Example 7.24 (Pincherle [521]) A system of functions p0 (z), p1 (z), p2 (z), . . . is defined by the equations p0 (z) = 1, pn+1 (z) = (z2 + an z + bn )pn (z),

The Expansion of Functions in Infinite Series

152

where an and bn are given functions of n, which tend respectively to the limits 0 and −1 as n → ∞. Í Shew that the region of convergence of a series of the form en pn (z), where e1, e2, . . . are independent of z, is a Cassini’s oval with the foci +1, −1. Shew that every function f (z), which is analytic on and inside the oval, can, for points inside the oval, be expanded in a series Õ f (z) = (cn + zcn0 )pn (z), where cn =

1 2πi

∫

(an + z)qn (z) f (z) dz, cn0 =

1 2πi

∫ qn (z) f (z) dz,

the integrals being taken round the boundary of the region, and the functions qn (z) being defined by the equations q0 (z) =

1 1 , qn+1 (z) = 2 qn (z). z2 + a0 z + b0 z + an+1 z + bn+1

Example 7.25 Let C be a contour enclosing the point a, and let φ(z) and f (z) be analytic when z is on or inside C. Let |t| be so small that |tφ(z)| < |z − a| when z is on the periphery of C. By expanding ∫ 1 − tφ 0(z) 1 f (z) dz 2πi C z − a − tφ(z) in ascending powers of t, shew that it is equal to f (a) +

∞ Õ t n d n−1 [ f 0(a) {φ(a)} n ] . n−1 n! da n=1

Hence, by using §6.3, §6.31, obtain Lagrange’s theorem.

8 Asymptotic Expansions and Summable Series

8.1 Simple example of an asymptotic expansion ∫ ∞ Consider the function f (x) = t −1 e x−t dt, where x is real and positive, and the path of x

integration is the real axis. By repeated integrations by parts, we obtain ∫ ∞ x−t 1 2! (−1)n−1 (n − 1)! e dt 1 n + (−1) n! . f (x) = − 2 + 3 − · · · + n x x x x t n+1 x In connexion with the function f (x), we therefore consider the expression un−1 =

(−1)n−1 (n − 1)! , xn

and we shall write n Õ

2! (−1)n n! 1 1 = Sn (x). − 2 + 3 −···+ x x x x n+1 m=0 Í Then we have |um /um−1 | = mx −1 → ∞ as m → ∞. The series um is therefore divergent for all values of x. In spite of this, however, the series can be used for the calculation of f (x) ; this can be seen in the following way. Take any fixed value for the number n, and calculate the value of Sn . We have ∫ ∞ x−t e dt n+1 f (x) − Sn (x) = (−1) (n + 1)! , t n+2 x um =

and therefore, since e x−t ≤ 1, | f (x) − Sn (x)| = (n + 1)!

∫ x

∞

e x−t dt < (n + 1)! t n+2

∫ x

∞

dt t n+2

=

n! . x n+1

For values of x which are sufficiently large, the right-hand member of this equation is very small. Thus, if we take x ≥ 2n, we have | f (x) − Sn (x) |

0 and 0 < c < 1. The ratio of the kth term of this series to the (k − 1)th is less than c, and consequently the

8.2 Definition of an asymptotic expansion

155

series converges for all positive values of x. We shall confine our attention to positive values of x. We have, when x > k, 1 k k2 k3 k4 1 = − 2 + 3 − 4 + 5 −··· . x+k x x x x x 1 If, therefore, it were allowable 1 to expand each fraction in this way, and to rearrange x+k the series for f (x) in descending powers of x, we should obtain the formal series A1 A2 An + 2 +···+ n +··· , x x x where An = (−1)n−1

∞ Í

k n−1 ck . But this procedure is not legitimate, and in fact

k=1

diverges. We can, however, shew that it is an asymptotic expansion of f (x). A1 A2 An + 2 + · · · + n+1 . Then For let Sn (x) = x x x ∞ Õ ck kck k 2 ck (−)n k n ck Sn (x) = − 2 + 3 +···+ x x x x n+1 k=1 ( ) n+1 ∞ Õ k ck = 1− − x x+k k=1

∞ Í

An x −n

n=1

(8.1)

n+1 Í ∞ Í ck k ∞ k n ck . so that | f (x) − Sn (x)| = − < x −n−2 k=1 x x + k k=1 ∞ Í k n ck converges for any given value of n and is equal to Cn , say, and hence Now k=1

| f (x) − Sn (x)| < Cn x −n−2 . Consequently f (x) ∼ Example 8.2.1

If f (x) =

∫

∞ Í n=1

An x −n .

∞

ex

2

−t 2

dt, where x is positive and the path of integration is

x

the real axis, prove that f (x) ∼

1 1·3 1·3·5 1 − 2 3 + 3 5 − 4 7 +··· . 2x 2 x 2 x 2 x

In fact, it was shewn by Stokes [610] in 1857 that ∫ x 1 x2 √ 1 1 1·3 1·3·5 x 2 −t 2 e dt ∼ ± e π − − + − 4 7 +··· ; 2 2x 22 x 3 23 x 5 2 x 0 the upper or lower sign is to be taken according as 1 1 − π < arg x < π 2 2 1

or

1 3 π < arg x < π. 2 2

It is not allowable, since k > x for all terms of the series after some definite term.

Asymptotic Expansions and Summable Series

156

8.3 Multiplication of asymptotic expansions We shall now shew that two asymptotic expansions, valid for a common range of values of arg z, can be multiplied together in the same way as ordinary series, the result being a new asymptotic expansion. ∞ ∞ Í Í For let f (z) ∼ Am z−m , φ(z) ∼ Bm z−m, and let Sn (z) and Tn (z) be the sums of their m=0

m=0

first (n + 1) terms; so that, n being fixed, f (z) − Sn (z) = o(z−n ),

φ(z) − Tn (z) = o(z−n ).

Then, if Cm = A0 Bm + A1 Bm−1 + · · · + Am B0 , it is obvious that 2 Sn (z)Tn (z) =

n Õ

Cm z−m + o(z−n ).

m=0

But f (z)φ(z) = (Sn (z) + o(z−n )) (Tn (z) + o(z−n )) = Sn (z)Tn (z) + o(z−n ) n Õ = Cm z−m + o(z−n ). m=0

This result being true for any fixed value of n, we see that f (z)φ(z) ∼

∞ Õ

Cm z−m .

m=0

8.31 Integration of asymptotic expansions We shall now shew that it is permissible to integrate an asymptotic expansion term by term, the resulting series being the asymptotic expansion of the integral of the function represented by the original series. n ∞ Í Í Am x −m . Then, given any positive number Am x −m, and let Sn (x) = For let f (x) ∼ m=2

m=2

ε, we can find x0 such that

| f (x) − Sn (x)| < ε|x| −n when x > x0, and therefore ∫

x

∞

∫ f (x)dx − x

∞

∫ Sn (x)dx ≤

| f (x) − Sn (x)|dx

x

< 2

∞

ε . (n − 1)x n−1

See §2.11; we use o(z −n ) to denote any function ψ(z) such that z n ψ(z) → 0 as |z | → x.

8.4 Methods of summing series

∫ But x

∞

Sn (x)dx =

157

A3 An A2 + 2 +···+ , and therefore x 2x (n − 1)x n−1 ∫ ∞ ∞ Õ Am . f (x)dx ∼ (m − 1)x m−1 x m=2

(8.2)

On the other hand, it is not in general permissible to differentiate an asymptotic expansion; this may be seen by considering e−x sin(e x ). For a theorem concerning differentiation of asymptotic expansions representing analytic functions, see Ritt [560].

8.32 Uniqueness of an asymptotic expansion A question naturally suggests itself, as to whether a given series can be the asymptotic expansion of several distinct functions. The answer to this is in the affirmative. To shew this, we first observe that there are functions L(x) which are represented asymptotically by a series all of whose terms are zero, i.e. functions such that lim x n L(x) = 0 for every fixed value of x→∞

n. The function e−x is such a function when x is positive. The asymptotic expansion 3 of a function J(x) is therefore also the asymptotic expansion of J(x) + L(x). On the other hand, a function cannot be represented by more than one distinct asymptotic expansion over the whole of a given range of values of z; for, if f (z) ∼

∞ Õ

Am z

−m

,

f (z) ∼

∞ Õ

Bm z−m,

m=0

m=0

then lim z

z→∞

n

A1 B1 An Bn A0 + + · · · + n − B0 − − · · · − n = 0, z z z z

which can only be if A0 = B0, A1 = B1, . . .. Important examples of asymptotic expansions will be discussed later, in connexion with the Gamma-function (Chapter 12) and Bessel functions (Chapter 17).

8.4 Methods of summing series We have seen that it is possible to obtain a development of the form f (x) = where Rn (x) → ∞ as n → ∞, and the series

∞ Í m=0

n Í m=0

Am x −m +Rn (x),

Am x −m does not converge. We now consider

what meaning, if any, can be attached to the sum of a non-convergent series. That is to say, given the numbers a0, a1, a2, . . ., we wish to formulate definite rules by which we can obtain ∞ ∞ Í Í from them a number S such that S = an if an converges, and such that S exists when n=0

n=0

this series does not converge. 3

It has been shewn that when the coefficients in the expansion satisfy certain inequalities, there is only one analytic function with that asymptotic expansion. See Watson [648].

Asymptotic Expansions and Summable Series

158

8.41 Borel’s method of summation [85, p. 97–115] ∞ ∞ ∫∞ Í Í an t n z n , the equation We have seen (§7.81) that an z n = 0 e−t φ(tz) dt, where φ(tz) = n! n=0

n=0

certainly being true inside the circle of convergence of z outside this circle, we define the Borel sum of

∞ Í n=0

∞ Í n=0

an z . If the integral exists at points

an z n to mean the integral.

Thus, whenever Re z < 1, the ‘Borel sum’ of the series ∫

n

∞ Í n=0

z n is

∞

e−t etz dt = (1 − z)−1 .

0

If the Borel sum exists we say that the series is summable (B).

8.42 Euler’s method of summation [85, 201] A method, practically due to Euler, is suggested by the theorem of §3.71; the sum of may be defined as lim− x→1

∞ Í n=0

∞ Í

an

n=0

an x n, when this limit exists.

Thus the sum of the series 1 − 1 + 1 − 1 + · · · would be lim (1 − x + x 2 − · · · ) = lim− (1 + x)−1 =

x→1−

x→1

1 . 2

8.43 Cesàro’s method of summation [141] ∞ Í 1 an Let sn = a1 + a2 + · · · + an ; then if S = lim (s1 + s2 + · · · + sn ) exists, we say that n→∞ n n=1 is ‘summable (C1)’, and that its sum (C1) is S. It is necessary to establish the condition of ∞ Í an when this series is convergent. (See the end of §8.4.) consistency, namely that S = n=1

∞ Í

n Í am = s, sm = nSn ; then we have to prove that m=1 m=1 n+p Í Sn → s. Given ε, we can choose n such that am < ε for all values of p, and so

To obtain the required result, let

m=n+1

|s − sn | ≤ ε. Then, if ν > n, we have 1 n−1 Sν = a1 + a2 1 − + · · · + an 1 − ν ν n ν−1 + · · · + aν 1 − +an+1 1 − . ν ν

Since 1, 1−ν −1, 1−2ν −1, . . . is a positive decreasing sequence, it follows from Abel’s inequality (§2.301) that an+1 1 − n + an+2 1 − n + 1 + · · · + aν 1 − ν − 1 < 1 − n ε. ν ν ν ν

8.5 Hardy’s convergence theorem

159

Therefore sν − a1 + a2 1 − 1 + · · · + an 1 − n − 1 < 1 − n ε. ν ν ν Making ν → ∞, we see that, if S be any one of the limit points (§2.21) of Sν , then n Í ≤ ε. Therefore, since |s − sn | ≤ ε, we have |S − s| ≤ 2ε. This inequality being S − a m m=1 true for every positive value of ε we infer, as in §2.21, that S = s; that is to say, Sν has the unique limit s; this is the theorem which had to be proved. Example 8.4.1 terms’.

Frame a definition of ‘uniform summability (C1) of a series of variable

Example 8.4.2 If bn, ν ≥ bn+1, ν ≥ 0 when n < ν, and if, when n is fixed, lim bn, ν = 1 and ν→∞ ∞ ν Í Í if am = S, then lim an bn, ν = S. ν→∞

m=1

A series

∞ Í n=0

n=1

8.431 Cesàro’s general method of summation ν Í an , is said to be ‘ summable (Cr)’ if lim an bn, ν exists, where ν→∞ n=0

b0, ν = 1, bn, ν =

n

1+

r r r o −1 . 1+ ··· 1+ ν+1−n ν+2−n ν−1

It follows from Example 8.4.2 that the condition of consistency is satisfied; in fact it can be proved [102, §122] that if a series is summable (Cr 0) it is also summable (Cr) when r > r 0; the condition of consistency is the particular case of this result when r = 0.

8.44 The method of summation of Riesz [559] A more extended method of summing a series than the preceding is by means of r ν Õ λn 1− lim an, ν→∞ λν n=1 in which λn is any real function of n which tends to infinity with n. A series for which this limit exists is said to be ‘ summable (Rr) with sum-function λn ’.

8.5 Hardy’s convergence theorem This appears in Hardy [278]. For the proof here given, we are indebted to Mr. Littlewood. ∞ ∞ Í Í Let an be a series which is summable (C1). Then if an = O(1/n), the series an n=1 n=1 converges. ∞ Í Let sn = a1 + a2 + · · · + an ; then since an is summable (C1), we have s1 + s2 + · · · + sn = n=1 ∞ Í

n (s + o(1)) , where s is the sum (C1) of

n=1

an . Let sm − s = tm , (m = 1, 2, . . . , n), and let

160

Asymptotic Expansions and Summable Series

t1 + t2 + · · · + tn = σn . With this notation, it is sufficient to shew that, if |an | < Kn−1 , where K is independent of n, and if σn = n · o(1), then tn → 0 as n → ∞. Suppose first that a1, a2, . . . , are real. Then, if tn does not tend to zero, there is some positive number h such that there are an unlimited number of the numbers tn which satisfy either (i) tn > h or (ii) tn < −h. We shall shew that either of these hypotheses implies a contradiction. Take the former 4 , and choose n so that tn > h. Then, when r = 0, 1, 2, . . ., |an+r | < K/n.

Now plot the points Pr whose coordinates are (r, tn+r ) in a Cartesian diagram. Since tn+r+1 − tn+r = an+r+1 , the slope of the line Pr Pr+1 is less than θ = arctan(K/n). Therefore the points P0, P1, P2, . . . lie above the line y = h − x tan θ. Let Pk be the last of the points P0, P1, . . . which lie on the left of x = h cot θ, so that k ≤ h cot θ. Draw rectangles as shewn in the figure. The area of these rectangles exceeds the area of the triangle bounded by y = h − x tan θ and the axes; that is to say σn+k − σn−1 = tn + tn+1 + · · · + tn+k 1 1 > h2 cot θ = h2 K −1 n. 2 2 But |σn+k − σn−1 | ≤ |σn+k | + |σn−1 | = (n + k) · o(1) + (n − 1) · o(1) = n · o(1), since k ≤ hnK −1, and h, K are independent of n. Therefore, for a set of values of n tending to infinity, 21 h2 K −1 n < n · o(1), which is impossible since 21 h2 K −1 is not o(1) as n → ∞. This is the contradiction obtained on the hypothesis that lim tn ≥ h > 0; therefore lim tn ≤ 0. Similarly, by taking the corresponding case in which tn ≤ −h, we arrive at the result lim tn ≥ 0. Therefore since lim tn ≥ lim tn , we have lim tn = lim tn = 0, and so tn → 0. ∞ Í That is to say sn → s, and so an is convergent and its sum is s. n=1

If an be complex, we consider Re an and Im an separately, and find that 4

∞ Í n=1

Re an and

The reader will see that the latter hypothesis involves a contradiction by using arguments of a precisely similar character to those which will be employed in dealing with the former hypothesis.

8.6 Miscellaneous examples ∞ Í n=1

Im an converge by the theorem just proved, and so

∞ Í n=1

161

an converges. The reader will see

in Chapter 9 that this result is of great importance in the modern theory of Fourier series. Corollary 8.5.1 If an (ξ) be a function of ξ such that

∞ Í

an (ξ) n=1 n−1

throughout a domain of values of ξ, and if |an (ξ)| < K ∞ Í an (ξ) converges uniformly throughout the domain.

is uniformly summable (C1)

, where K is independent of ξ,

n=1

For, retaining the notation of the preceding section, if tn (ξ) does not tend to zero uniformly, we can find a positive number h independent of n and ξ such that an infinite sequence of values of n can be found for which tn (ξn ) > h or tn (ξn ) < −h for some point ξn of the domain 5 ; the value of ξn depends on the value of n under consideration. We then find, as in the original theorem, 12 h2 K −1 n < n · o(1) for a set of values of n tending to infinity. The contradiction implied in the inequality shews 6 that h does not exist, and so tn (ξ) → 0 uniformly.

8.6 Miscellaneous examples 1 2! 4! e−xt dt ∼ − 3 + 6 − · · · when x is real and positive. 2 1+t x x x

∞

∫ Example 8.1 Shew that 0

Example 8.2 Discuss the representation of the function ∫ 0 f (x) = φ(t)et x dt −∞

(where x is supposed real and positive, and φ is a function subject to certain general conditions) by means of the series f (x) =

φ(0) φ 0(0) φ 00(0) −··· . − 2 + x x x3

Shew that in certain cases (e.g. φ(t) = e at ) the series is absolutely convergent, and represents f (x) for large positive values of x; but that in certain other cases the series is the asymptotic expansion of f (x). Example 8.3 (Legendre [421, p. 340]) Shew that ∫ ∞ 1 a − 1 (a − 1)(a − 2) e3 z−a e−x x a−1 dx ∼ + 2 + +··· z z z3 z for large positive values of z. 5

6

It is assumed that a n (ξ) is real; the extension to complex variables can be made as in the former theorem. If no such number h existed, tn (ξ) would tend to zero uniformly. It is essential to observe that the constants involved in the inequality do not depend on ξn . For if, say, K depended on ξn , K −1 would really be a function of n and might be o(1) qua function of n, and the inequality would not imply a contradiction.

162

Asymptotic Expansions and Summable Series

Example 8.4 (Schlömilch) Shew that if, when x > 0, ∫ ∞ du 1 e−xu , f (x) = log u + log −u 1−e u 0 1 B1 B2 B3 then f (x) ∼ − + − + · · · . Shew also that f (x) can be expanded into an 2x 22 x 2 42 x 4 62 x 6 absolutely convergent series of the form f (x) =

∞ Õ k−1

ck . (x + 1)(x + 2) · · · (x + k)

Example 8.5 (Euler, Borel) Shew that if the series 1 + 0 + 0 − 1 + 0 + 1 + 0 + 0 − 1 + · · · , in which two zeros precede each −1 and one zero precedes each +1, be summed by Cesàro’s method, its sum is 35 . Example 8.6 Shew that the series 1 − 2! + 4! − · · · cannot be summed by Borel’s method, but the series 1 + 0 − 2! + 0 + 4! + · · · can be so summed.

9 Fourier Series and Trigonometric Series

9.1 Definition of Fourier series Series of the type 1 a 2 0

+ (a1 cos x + b1 sin x) + (a2 cos 2x + b2 sin 2x) + · · · = 21 a0 +

∞ Õ

(an cos nx + bn sin nx),

n=1

where an , bn are independent of x, are of great importance in many investigations. They are called trigonometrical series. (Throughout this chapter, except in §9.11, it is supposed that all the numbers involved are real.) ∫ π

If there is a function f (t) such that

f (t) dt exists as a Riemann integral or as an −π

improper integral which converges absolutely, and such that ∫ π ∫ π πan = f (t) cos nt dt, πbn = f (t) sin nt dt, −π

−π

then the trigonometrical series is called a Fourier series. Trigonometrical series that are not Fourier series first appeared in analysis in connexion with the investigations of Daniel Bernoulli (1700–1782) on vibrating strings; d’Alembert had d2 y previously solved the equation of motion yÜ = a2 2 in the form y = 21 { f (x + at) + f (x − at)}, dx where y = f (x) is the initial shape of the string starting from rest; and Bernoulli shewed that a formal solution is ∞ Õ nπx nπat bn sin cos , y= ` ` n=1 the fixed ends of the string being (0, 0) and (`, 0); and he asserted that this was the most general solution of the problem. This appeared to d’Alembert and Euler to be impossible, since such a series, having period 2`, could not possibly represent such a function as 1 cx(` − x) when t = 0. A controversy arose between these mathematicians, of which an account is given in Hobson [315]. Fourier, in his Théorie de la Chaleur [223] investigated a number of trigonometrical series and shewed that, in a large number of particular cases, a Fourier series actually converged to the sum f (x). Poisson [531] attempted a general proof of this theorem. Two proofs were given by Cauchy [122] and [123, vol. 2, p. 341–376]. These proofs, which are based on the 1

This function gives a simple form to the initial shape of the string.

163

Fourier Series and Trigonometric Series

164

theory of contour integration, are concerned with rather particular classes of functions and one is invalid. The second proof has been investigated by Harnack [283]. In 1829, Dirichlet [172] gave the first rigorous proof that, for a general class of functions, the Fourier series, defined as above, does converge to the sum f (x). A modification of this proof was given later by Bonnet [82]. He employs the second mean-value theorem directly, while Dirichlet’s original proof makes use of arguments precisely similar to those by which that theorem is proved. See §9.43. The result of Dirichlet is that 2 if f (t) is defined and bounded in the range (−π, π) and if f (t) has only a finite number of maxima and minima and a finite number of discontinuities in this range and, further, if f (t) is defined by the equation f (t + 2π) = f (t) outside the range (−π, π), then, provided that ∫ π ∫ π πan = f (t) cos nt dt, πbn = f (t) sin nt dt, −π

the series 21 a0 +

∞ Í

−π

(an cos nx + bn sin nx) converges to 12 { f (x + 0) + f (x − 0)}.

n=1

Later, Riemann and Cantor developed the theory of trigonometrical series generally, while still more recently Hurwitz, Fejér and others have investigated properties of Fourier series when the series does not necessarily converge. Thus Fejér has proved the remarkable theorem that a Fourier series (even if not convergent) is ‘summable (C1)’ at∫all points at which f (x ±0) π

exist, and its sum (C1) is 21 { f (x + 0) + f (x − 0)}, provided that

f (t) dt is an absolutely −π

convergent integral. One of the investigations of the convergence of Fourier series which we shall give later (§9.42) is based on this result. For a fuller account of investigations subsequent to Riemann, the reader is referred to Hobson [323], and to de la Vallée Poussin [639].

9.11 Nature of the region within which a trigonometrical series converges Consider the series ∞

Õ 1 (an cos nz + bn sin nz), a0 + 2 n=1 where z may be complex. If we write eis = ζ, the series becomes ∞ Õ 1 1 1 n −n a0 + (an − ibn )ζ + (an + ibn )ζ . 2 2 2 n=1 This Laurent series will converge, if it converges at all, in a region in which a ≤ |ζ | ≤ b, where a, b are positive constants. But, if z = x + iy, |ζ | = e−v , and so we get, as the region of convergence of the trigonometrical series, the strip in the z plane defined by the inequality log a ≤ −y ≤ log b. 2

The conditions postulated for f (t) are known as Dirichlet’s conditions; as will be seen in §§9.2, 9.42, they are unnecessarily stringent.

9.1 Definition of Fourier series

165

The case which is of the greatest importance in practice is that in which a = b = 1, and the strip consists of a single line, namely the real axis. Example 9.1.1 Let f (z) = sin z −

1 1 1 sin 2z + sin 3z − sin 4z + · · · , 2 3 4

where z = x + iy. Writing this in the form 1 1 1 1 1 1 f (z) = − i eis − e2is + e3is − · · · + i e−is − e−2is + e−3is − · · · 2 2 3 2 2 3 we notice that the first series converges 3 only if y ≥ 0, and the second only if y ≤ 0. Writing x in place of z (x being real), we see that by Abel’s theorem (§3.71), 1 1 f (x) = lim r sin x − r 2 sin 2x + r 3 sin 3x − · · · r→1 2 3 1 1 2 2ix 1 3 3ix ix = lim − i re − r e + r e − · · · r→1 2 2 3 1 1 2 −2ix 1 3 −3ix −ix + i re − r e + r e −··· . 2 2 3 This is the limit of one of the values of 1 1 − i log 1 + reix + i log 1 + re−ix , 2 2 and as r → 1 (if −π < x < π), this tends to 12 x + kπ, where k is some integer. ∞ (−1)n−1 sin nx Í Now converges uniformly (Example 3.3.6) and is therefore continuous n n=1 in the range −π + δ ≤ x ≤ π − δ, where δ is any positive constant. Since 12 x is continuous, k has the same value wherever x lies in the range; and putting x = 0, we see that k = 0. Therefore, when −π < x < π, f (x) = 21 x. But, when π < x < 3π, f (x) = f (x − 2π) =

x − 2π x = − π, 2 2

and generally, if (2n − 1)π < x < (2n + 1)π, f (x) = 21 x − nπ. We have thus arrived at an example in which f (x) is not represented by a single analytical expression. It must be observed that this phenomenon can only occur when the strip in which the Fourier series converges is a single line. For if the strip is not of zero breadth, the associated Laurent series converges in an annulus of non-zero breadth and represents an analytic function of ζ in that annulus; and, since ζ is an analytic function of z, the Fourier series represents an analytic function of z; such a series is given by 1 1 r sin x − r 2 sin 2x + r 3 sin 3x − · · · , 2 3 3

The series do converge if y = 0, see Example 2.3.2.

Fourier Series and Trigonometric Series r sin x where 0 < r < 1; its sum is arctan , the arctan always representing an angle 1 + r cos x between ± 12 π. 166

Example 9.1.2 When −π ≤ x ≤ π, ∞ Õ (−1)n−1 cos nx

n2

n=1

=

1 2 1 2 π − x . 12 4

The series converges only when x is real; by §3.34 the convergence is then absolute and uniform. Since 1 1 1 x = sin x − sin 2x + sin 3x − · · · 2 2 3

(−π + δ ≤ x ≤ π − δ;

δ > 0),

and this series converges uniformly, we may integrate term-by-term from 0 to x (§4.7), and consequently ∞

1 2 Õ (−1)n−1 (1 − cos nx) x = 4 n2 n=1

(−π + δ ≤ x ≤ π − δ).

That is to say, when −π + δ ≤ x ≤ π − δ, ∞

C−

1 2 Õ (−1)n−1 cos nx , x = 4 n2 n=1

where C is a constant, at present undetermined. But since the series on the right converges uniformly throughout the range −π ≤ x ≤ π, its sum is a continuous function of x in this extended range; and so, proceeding to the limit when x → ±π, we see that the last equation is still true when x = ±π. To determine C, integrate each side of the equation (§4.7) between the limits −π, π; and we get 1 2πC − π 3 = 0. 6 Consequently ∞

1 2 1 2 Õ (−1)n−1 cos nx π − x = 12 4 n2 n=1

(−π ≤ x ≤ π).

Example 9.1.3 By writing π − 2x for x in Example 9.1.2, shew that ∞ Õ sin2 nx n=1

n2

=

1 x(π − x) 2 1 {π|x| − x2 } 2

(0 ≤ x ≤ π), (−π ≤ x ≤ π).

9.2 On Dirichlet’s conditions and Fourier’s theorem

167

9.12 Values of the coefficients in terms of the sum of a trigonometrical series ∞ Í Let the trigonometrical series 12 c0 + (cn cos nx + dn sin nx) be uniformly convergent in the n=1

range (−π, π) and let its sum be f (x). Using the obvious results π

∫

cos mx cos nx dx =

0 (m , n) π (m = n , 0),

0 (m , n) π (m = n , 0),

−π

∫

π

sin mx sin nx dx =

−π

we find, on multiplying the equation 21 c0 +

∞ Í

∫

π

dx = 2π,

−π

(cn cos nx + dn sin nx) = f (x) by 4 cos nx; or

n=1

by sin nx and integrating term-by-term (§4.7), πcn =

∫

π

f (x) cos nx dx, −π

πdn =

∫

π

f (x) sin nx dx. −π

These were given by Euler [203]. Corollary 9.1.1 Fourier series.

A trigonometrical series uniformly convergent in the range (−π, π) is a

Note

Lebesgue [419, p. 124] has given a proof of a theorem communicated to him by ∞ Í sin nx/log n, which converges for all real values of Fatou that the trigonometrical series n=2

x (Example 2.3.1), is not a Fourier series.

9.2 On Dirichlet’s conditions and Fourier’s theorem A theorem, of the type described in §9.1, concerning the expansibility of a function of a real variable into a trigonometrical series is usually described as Fourier’s theorem. On account of the length and difficulty of a formal proof of the theorem (even when the function to be expanded is subjected to unnecessarily stringent conditions), we defer the proof until §9.42, §9.43. It is, however, convenient to state here certain sufficient conditions under which a function can be expanded into a trigonometrical series. Let f (t) be defined arbitrarily when −π ≤ t < π and defined 5 for all other real values of t by means of the equation f∫(t + 2π) = f (t), so that f (t) is a periodic function with period 2π. π

Let f (t) be such that

f (t) dt exists; and if this is an improper integral, let it be −π

absolutely convergent. 4 5

Multiplying by these factors does not destroy the uniformity of the convergence. This definition frequently results in f (t) not being expressible by a single analytical expression for all real values of t; cf. Example 9.1.1.

Fourier Series and Trigonometric Series

168

Let an, bn be defined by the equations 6 πan =

∫

π

f (t) cos nt dt,

πbn =

−π

∫

π

f (t) sin nt dt

(n = 0, 1, 2, . . .).

−π

Then, if x be an interior point of any interval (a, b) in which f (t) has limited total fluctuation, the series ∞

Õ 1 a0 + (an cos nx + bn sin nx) 2 n=1 is convergent, and its sum 7 is 21 { f (x + 0) + f (x − 0)}. If f (t) is continuous at t = x, this sum reduces to f (x). This theorem will be assumed in §§9.21–9.32; these sections deal with theorems concerning Fourier series which are of some importance in practical applications. It should be stated here that every function which Applied Mathematicians need to expand into Fourier series satisfies the conditions just imposed on f (t), so that the analysis given later in this chapter establishes the validity of all the expansions into Fourier series which are required in physical investigations. The reader will observe that in the theorem just stated, f (t) is subject to less stringent conditions than those contemplated by Dirichlet, and this decrease of stringency is of con∞ Í siderable practical importance. Thus, so simple a series as (−1)n−1 cosnnx is the expansion n=1 of the function 8 log 2 cos 12 x ; and this function does not satisfy Dirichlet’s condition of boundedness at ±π. ∞ Í It is convenient to describe the series 12 a0 + (an cos nx + bn sin nx) as the Fourier n=1

series associated with f (t). This description must, however, be taken as implying nothing concerning the convergence of the series in question.

9.21 The representation of a function by Fourier series for ranges other than (−π, π) Consider a function f (x) with an (absolutely) convergent integral, and with limited total fluctuation in the range a ≤ x ≤ b. Write x = 12 (a + b) − 12 (a − b)π −1 x 0, f (x) = F(x 0). Then it is known (§9.2) that ∞

Õ 1 1 {F(x 0 + 0) + F(x 0 − 0)} = a0 + (an cos nx 0 + bn sin nx 0), 2 2 n=1 6

7 8

The numbers a n , b n are called the Fourier constants of f (t), and the symbols a n , b n will be used in this sense throughout §§9.2–9.5. It may be shewn that the convergence and absolute convergence of the ∫ πintegrals defining the Fourier constants are consequences of the convergence and absolute convergence of −π f (t) dt; cf. §§2.32, 4.5. The limits f (x ± 0) exist, by Example 3.6.3. Example 9.6 at the end of the chapter.

9.2 On Dirichlet’s conditions and Fourier’s theorem

169

and so 1 { f (x + 0) + f (x − 0)} = 2 ∞ Õ πn(2x − a − b) 1 πn(2x − a − b) a0 + an cos + bn sin , 2 b−a b−a n=1 where by an obvious transformation ∫ b 1 πn(2x − a − b) (b − a)an = f (x) cos dx, 2 b−a a ∫ b 1 πn(2x − a − b) (b − a)bn = dx. f (x) sin 2 b−a a

9.22 The cosine series and the sine series Let f (x) be defined in the range (0, `) and let it have an (absolutely) convergent integral and also let it have limited total fluctuation in that range. Define f (x) in the range (−`, 0) by the equation f (−x) = f (x). Then ∞ n Õ 1 πnx πnx o 1 { f (x + 0) + f (x − 0)} = a0 + an cos + bn sin , 2 2 ` ` n=1

where, by §9.21, `an =

∫

`bn =

∫

`

−` ` −`

πnt f (t) cos dt = 2 ` πnt dt = 0, f (t) sin `

`

∫

f (t) cos 0

πnt dt, `

so what when −` ≤ x ≤ `, ∞

Õ 1 πnx 1 { f (x + 0) + f (x − 0)} = a0 + an cos ; 2 2 ` n=1 this is called the cosine series. If, however, we define f (x) in the range (−`, 0) by the equation f (−x) = − f (−x), we get, when −` ≤ x ≤ `, ∞

Õ 1 πnx { f (x + 0) + f (x − 0)} = , bn sin 2 ` n=1 where `bn = 2

∫

`

f (t) sin 0

πnt dt; this is called the sine series. `

Fourier Series and Trigonometric Series

170

Thus the series ∞

∞

Õ πnx Õ πnx 1 a0 + an cos + bn sin , 2 ` ` n=1 n=1 ∫ ` ∫ ` `an `bn πnt πnt where = dt, = dt, have the same sum when f (t) cos f (t) sin 2 ` 2 ` 0 0 0 ≤ x ≤ `; but their sums are numerically equal and opposite in sign when 0 ≥ x ≥ −`. The cosine series was given by Clairaut [147] in a memoir dated July 9, 1757; the sine series was obtained between 1762 and 1765 by Lagrange [396, vol. I, p. 553]. Example 9.2.1 Expand 12 (π − x) sin x in a cosine series in the range 0 ≤ x ≤ π. Solution. We have, by the formula just obtained, ∞

Õ 1 1 (π − x) sin x = a0 + an cos nx, 2 2 n=1 where 1 πan = 2

∫ 0

π

1 (π − x) sin x cos nx dx. 2

But, integrating by parts, if n , 1, ∫ π ∫ π 2(π − x) sin x cos nx dx = (π − x){sin(n + 1)x − sin(n − 1)x} dx 0 0 π cos(n + 1)x cos(n + 1)x − = (x − π) n+1 n−1 0 ∫ π cos(n + 1)x cos(n − 1)x − − dx n+1 n−1 0 1 2π 1 =π =− − . n+1 n−1 (n + 1)(n − 1) ∫ π Whereas if n = 1, we get 2(π − x) sin x cos x dx = 21 π. 0

Therefore the required series is 1 1 1 1 1 + cos x − cos 2x − cos 3x − cos 4x − · · · 2 4 1·3 2·4 3·5 It will be observed that it is only for values of x between 0 and π that the sum of this series is proved to be 21 (π − x) sin x; thus for instance when x has a value between 0 and −π, the sum of the series is not 21 (π − x) sin x, but − 21 (π + x) sin x; when x has a value between π and 2π, the sum of the series happens to be again 12 (π − x) sin x, but this is a mere coincidence arising from the special function considered, and does not follow from the general theorem. Example 9.2.2 Expand 81 πx(π − x) in a sine series, valid when 0 ≤ x ≤ π. sin 3x sin 5x Answer. The series is sin x + + + · · ·. 33 53

9.3 The nature of the coefficients in a Fourier series

171

Example 9.2.3 Shew that, when 0 ≤ x ≤ π, cos 3x cos 5x 1 π(π − 2x)(π 2 + 2πx − 2x 2 ) = cos x + + +··· . 96 34 54 Hint. Denoting the left-hand side by f (x), we have, on integrating by parts and observing that f 0(0) = f 0(π) = 0, ∫ π ∫ 1 1 n 0 π f (x) cos nx dx = [ f (x) sin nx]0 − f (x) sin nx dx n n 0 0 ∫ π 1 1 = 2 [ f 0(x) cos nx]0π − 2 f 00(x) cos nx dx n n 0 ∫ π 1 1 00 π f 000(x) sin nx dx = − 3 [ f (x) sin nx]0 + 3 n n 0 1 = − 4 [ f 000(x) cos nx]0π n π = 4 (1 − cos nπ). 4n Example 9.2.4 cosine series

Shew that for values of x between 0 and π, esx can be expanded in the 2s sπ cos 2x 1 cos 4x + (e − 1) + + · · · π 2s2 s2 + 4 s2 + 16 2s sπ cos x cos 3x − (e + 1) 2 + +··· , π s + 1 s2 + 9

and draw graphs of the function esx and of the sum of the series. Example 9.2.5 Shew that for values of x between 0 and π, the function 81 π(π − 2x) can be expanded in the cosine series cos x +

cos 3x cos 5x + +··· , 32 52

and draw graphs of the function 18 π(π − 2x) and of the sum of the series.

9.3 The nature of the coefficients in a Fourier series The analysis of this section and of §9.31 is contained in Stokes’ great memoir [608] (reproduced in [611, vol. I, pp. 236–313]). Suppose that (as in the numerical examples which have been discussed) the interval (−π, π) can be divided into a finite number of ranges (−π, k1 ), (k1, k2 ), . . . , (k n, π) such that throughout each range f (x) and all its differential coefficients are continuous with limited total fluctuation and that they have limits on the right and on the left (§3.2) at the end points of these ranges. Then ∫ k1 ∫ k2 ∫ π πam = f (t) cos mt dt + f (t) cos mt dt + · · · + f (t) cos mt dt. −π

k1

kn

Fourier Series and Trigonometric Series

172

Integrating by parts we get k1 k2 π πam = m−1 f (t) sin mt −π + m−1 f (t) sin mt k1 + · · · + m−1 f (t) sin mt kn ∫ k1 ∫ k2 ∫ −1 0 −1 0 −1 −m f (t) sin mt dt − m f (t) sin mt dt − · · · − m −π

k1

π

f 0(t) sin mt dt,

kn

0 n Í Am bm 0 is a Fourier − , where π Am = sin mkr [ f (kr − 0) − f (kr + 0)], and bm m m r=1 0 Bm am constant of f 0(x). Similarly bm = + , where m m

so that am =

πBm = −

n Õ

cos mkr [ f (kr − 0) − f (kr + 0)] − cos mπ [ f (π − 0) − f (−π + 0)] ,

r=1 0 and am is a Fourier constant of f 0(x). Similarly, we get 0 am =

0 Am b00 − m, m m

0 bm =

Bm0 a 00 + m, m m

00 00 where am , bm are the Fourier constants of f 00(x) and 0 = π Am

n Õ

sin mkr { f 0(kr − 0) − f 0(kr + 0)},

r=1

πBm0 = −

n Õ

cos mkr { f 0(kr − 0) − f 0(kr + 0)} − cos mπ { f 0(π − 0) − f 0(−π + 0)}.

r=1

Therefore am =

a 00 Am Bm0 − 2 − m2 , m m m

bm =

0 b00 Bm Am + 2 − m2 . m m m

0 = O(1), Bm0 = O(1), and, since the integrands involved in Now as m → ∞, we see that Am 00 00 00 and bm are bounded, it is evident that am = O(1), bm = O(1). Hence if Am = 0, Bm = 0, the Fourier series for f (x) converges absolutely and uniformly, by §3.34. The necessary and sufficient conditions that Am = Bm = 0 for all values of m are that 00 am

f (kr − 0) = f (kr + 0),

f (π − 0) = f (−π + 0),

that is to say that 9 f (x) should be continuous for all values of x.

9.31 Differentiation of Fourier series The result of differentiating ∞ Õ 1 a0 + (am cos mx + bm sin mx) 2 m=1 9

Of course f (x) is also subject to the conditions stated at the beginning of the section.

9.3 The nature of the coefficients in a Fourier series

173

∞ Í term by term is {mbm cos mx − mam sin mx}. With the notation of §9.3, this is the same m=1 as ∞ 1 0 Õ 0 0 (am cos mx + bm sin mx), a0 + 2 m=1

provided that Am = Bm = 0 and

∫

π

f 0(x)dx = 0; these conditions are satisfied if f (x) is

−π

continuous for all values of x. Consequently sufficient conditions for the legitimacy of differentiating a Fourier series term by term are that f (x) should be continuous for all values of x and f 0(x) should have only a finite number of points of discontinuity in the range (−π, π), both functions having limited total fluctuation throughout the range.

9.32 Determination of points of discontinuity The expressions for am and bm which have been found in §9.3 can frequently be applied in practical examples to determine the points at which the sum of a given Fourier series may be discontinuous. Thus, let it be required to determine the places at which the sum of the series sin x +

1 1 sin 3x + sin 5x + · · · 3 5

is discontinuous. Assuming that the series is a Fourier series and not any trigonometrical series and observing that am = 0, bm = (2m)−1 (1 − cos mπ), we get on considering the formula found in §9.3, Am = 0,

Bm =

1 2

− 12 cos mπ,

0 0 = 0. = bm am

Hence if k 1, k 2, . . . are the places at which the analytic character of the sum is broken, we have 0 = π Am = sin mk1 { f (k1 − 0) − f (k1 + 0)} + sin mk 2 { f (k 2 − 0) − f (k2 + 0)} + · · · . Since this is true for all values of m, the numbers k1, k2, . . . must be multiples of π; but there is only one even multiple of π in the range −π < x ≤ π, namely zero. So k 1 = 0, and k 2, k 3, . . . do not exist. Substituting k1 = 0 in the equation Bm = 21 − 12 cos mπ, we have π 21 − 21 cos mπ = −[cos mπ{ f (π − 0) − f (−π + 0)} + f (−0) − f (+0)]. Since this is true for all values of m, we have 1 π 2

= f (+0) − f (−0),

1 π 2

= f (π − 0) − f (−π + 0).

This shews that, if the series is a Fourier series, f (x) has discontinuities at the points nπ 0 0 (n any integer) and since am = bm = 0, we should expect to be constant in the open range (−π, 0) and to be another constant in the open range (0, π). In point of fact f (x) = −π/4 (−π < x < 0) and π/4 (0 < x < π).

Fourier Series and Trigonometric Series

174

9.4 Fejér’s theorem We now begin the discussion of the theory of Fourier series by proving the following theorem, due to Fejér [210], concerning the summability of the Fourier series associated with an arbitrary function f (t): Let f (t) be a function of the real variable t, defined arbitrarily when −π ≤ ∫ t < π, and π

defined by the equation f (t + 2π) = f (t) for all other real values of t; and let

f (t) dt −π

exist and (if it is an improper integral) let it be absolutely convergent. Then the Fourier series associated with the function f (t) is summable (C1) at all points x at which, the two limits f (x ± 0) exist. (See §8.43.) And its sum (C1) is 1 { f (x + 0) + f (x − 0)}. 2 Let an , bn , (n = 0, 1, 2, . . .) denote the Fourier constants (§9.2) of f (t) and let 1 a0 = A0, 2

m Õ

an cos nx + bn sin nx = An (x),

An (x) = Sm (x).

n=0

Then we have to prove that lim

m→∞

1 1 {A0 + S1 (x) + S2 (x) + · · · + Sm−1 (x)} = { f (x + 0) + f (x − 0)} , m 2

provided that the limits on the right exist. If we substitute for the Fourier constants their values in the form of integrals (§9.2), it is easy to verify that 10 A0 +

m−1 Õ

Sn (x) = mA0 + (m − 1)A1 (x) + (m − 2)A2 (x) + · · · + Am−1 (x)

n=1

=

1 π

∫

π

−π

1 2

m + (m − 1) cos(x − t) + (m − 2) cos 2(x − t)

+ · · · + cos(m − 1)(x − t)] f (t)dt ∫ π sin2 21 m(x − t) 1 f (t)dt = 2π −π sin2 12 (x − t) ∫ π+x sin2 12 m(x − t) 1 f (t)dt, = 2π −π+x sin2 12 (x − t) the last step following from the periodicity of the integrand. If now we bisect the path of integration and write x ∓ 2θ in place of t in the two parts of 10

It is obvious that, if we write λ for e t (x − t) in the second line, then m + (m − 1)(λ + λ−1 ) + (m − 2)(λ2 + λ−2 ) + · · · + (λ m−1 + λ1−m ) = (1 − λ)−1 {λ1−m + λ2−m + · · · + λ− 1 + 1 − λ − λ2 − · · · − λ m } 1

1

1

1

= (1 − λ)−2 {λ1−m − 2λ + λ m+1 } = (λ 2 m − λ− 2 m )2 /(λ 2 − λ− 2 )2 .

9.4 Fejér’s theorem

175

the path, we get m−1 Õ

1 A0 + Sn (x) = π n=1

∫

π/2

0

1 sin2 mθ f (x + 2θ)dθ + 2 π sin θ

∫

π/2

0

sin2 mθ f (x − 2θ)dθ. sin2 θ

Consequently it is sufficient to prove that, as m → ∞, then ∫ π/2 2 1 sin mθ π f (x + 2θ)dθ → f (x + 0), m 0 2 sin2 θ ∫ π/2 2 1 π sin mθ f (x − 2θ)dθ → f (x − 0). 2 m 0 2 sin θ Now, if we integrate the equation 1 sin2 mθ 1 = m + (m − 1) cos 2θ + · · · + cos 2(m − 1)θ, 2 sin2 θ 2 we find that ∫

π/2

0

sin2 mθ πm dθ = , 2 2 sin θ

and so we have to prove that ∫ π/2 2 sin mθ 1 φ(θ)dθ → 0 as m → ∞, m 0 sin2 θ where φ(θ) stands in turn for each of the two functions f (x + 2θ) − f (x + 0),

f (x − 2θ) − f (x − 0).

Now, given an arbitrary positive number ε, we can choose δ so that |φ(θ)| < ε whenever 0 < θ ≤ 21 δ, (on the assumption that f (x ±0) exist). This choice of δ is obviously independent of m. Then ∫ π/2 2 ∫ δ/2 2 ∫ π/2 2 1 1 1 sin mθ sin mθ sin mθ ≤ |φ(θ)| |φ(θ)| dθ φ(θ) dθ dθ + m 2 2 m 0 m δ/2 sin2 θ sin θ sin θ 0 ∫ π/2 ∫ δ/2 2 ε sin mθ 1 |φ(θ)| dθ < dθ + m 0 sin2 θ m sin2 (δ/2) δ/2 ∫ π/2 ∫ π/2 2 ε sin mθ 1 |φ(θ) | dθ ≤ dθ + m 0 sin2 θ m sin2 (δ/2) 0 ∫ π/2 πε 1 |φ(θ)| dθ. = + 2 m sin2 (δ/2) 0 ∫ π ∫ π/2 | f (t)| dt entails the convergence of |φ(θ)| dθ, and so, Now the convergence of −π

0

given ε (and therefore δ), we can make πm δ ε sin2 > 2 2

∫

π/2

|φ(θ)| dθ, 0

176

Fourier Series and Trigonometric Series

by taking m sufficiently large. Hence, by taking m sufficiently large, we can make ∫ π/2 1 sin2 mθ φ(θ) dθ < πε, m 2 sin θ 0 where ε is an arbitrary positive number; that is to say, from the definition of a limit, ∫ π/2 1 sin2 mθ lim φ(θ) dθ = 0, m→∞ m 0 sin2 θ and so Fejér’s theorem is established. Corollary 9.4.1 Let U and L be the upper and lower bounds of f (t) in any interval (a, b) whose length does not exceed 2π, and let ∫π

| f (t)|dt = π A.

−π

Then, if a + η ≤ x ≤ b − η, where η is any positive number, we have ) ( m−1 Õ 1 U− Sn (x) A0 + m n=1 ∫ x−η ∫ x+η ∫ x+η sin2 12 m(x − t) 1 {U − f (t)} dt = + + 2mπ −π+x sin2 12 (x − t) x−η x+η ∫ x−η ∫ π+x sin2 12 m(x − t) 1 {U − f (t)} dt ≥ + 2mπ −π+x sin2 21 (x − t) x+η ∫ x−η ∫ π+x |U| + | f (t)| 1 + dt, ≥− 2mπ −π+x sin2 12 η x+η so that # " m−1 Õ |U| + 12 A 1 Sn (x) ≤ U + A0 + . m m sin2 (η/2) n=1 Similarly # " m−1 Õ |L| + 12 A 1 Sn (x) ≥ L − A0 + . m m sin2 (η/2) n=1 Corollary 9.4.2 Let f (t) be continuous in the interval a ≤ t ≤ b. Since continuity implies uniformity of continuity (§3.61), the choice of δ corresponding to any value of x in (a, b) is independent of x, and the upper bound of | f (x ± 0)|, i.e. of | f (x)|, is also independent of x, so that ∫ π/2 ∫ π/2 |φ(θ)| dθ = | f (x ± 2θ) − f (x ± 0)|dθ 0 0 ∫ π 1 1 ≤ | f (t)| dt + π| f (x ± 0)|, 2 −π 2 and the upper bound of the last expression is independent of x.

9.4 Fejér’s theorem

177

Hence the choice of m, which makes ∫ π/2 2 1 sin mθ φ(θ) dθ < πε, m 2 sin θ 0 ( ) m−1 Õ 1 is independent of x, and consequently A0 + Sn (x) tends to the limit f (x), as m → ∞, m n=1 uniformly throughout the interval a ≤ x ≤ b.

9.41 The Riemann–Lebesgue lemmas In order to be able to apply Hardy’s theorem; (§8.5) to deduce the convergence of Fourier series from Fejér’s theorem, we need the two following lemmas: ∫

b

ψ(θ) dθ exist and (if it is an improper integral) let it be absolutely convergent. ∫ b Then, as λ → ∞, ψ(θ) sin(λθ) dθ is O(1).

(I) Let

a

a

(II) If, further, ψ(θ) has limited total fluctuation in the range (a, b) then, as λ → ∞, ∫b

ψ(θ) sin(λθ) dθ is

O(1/λ).

a

Of these results (I) was stated by W. R. Hamilton [270] and by Riemann [558, p. 241]. For Lebesgue’s investigation see his [419, ch. III] in the case of bounded functions. The truth of (II) seems to have been well known before its importance was realised; it is a generalisation of a result established by Dirksen [180] and Stokes [608] (§9.3) in the case of functions with a continuous differential coefficient. The reader should observe that the analysis of this section remains valid when the sines are replaced throughout by cosines. (I) It is convenient 11 to establish this lemma first in the case in which ψ(θ) is bounded in the range (a, b). In this case, let K be the upper bound of |ψ(θ)|, and let ε be an arbitrary positive number. Divide the range (a, b) into n parts by the points x1, x2, . . . , xn−1, and form the sums Sn , sn associated with the function ψ(θ) after the manner of §4.1. Take n so large that Sn − sn < ε; this is possible since ψ(θ) is integrable. In the interval (xr−1, xr ) write ψ(θ) = ψr (xr−1 ) + ωr (θ), so that |ωr (θ)| ≤ Ur − Lr , where Ur and Lr are the upper and lower bounds of ψ(θ) in the interval (xr−1, xr ). It is then clear 11

For this proof we are indebted to Mr Hardy; it seems to be neater than the proofs given by other writers, e.g. de la Vallée Poussin [639, pp. 140–141].

178

Fourier Series and Trigonometric Series

that b ∫ Õ ∫ xr n ∫ n Õ n ψ(θ) sin(λθ) dθ = ψr (xr−1 ) sin(λθ) dθ + ωr (θ) sin(λθ) dθ xr −1 r=1 r=1 xr −1 a ∫ x r Õ n n ∫ xr Õ |ψr (xr−1 )| |ωr (θ)| dθ ≤ sin(λθ) dθ + xr −1

r=1

r=1

xr −1

≤ nK · (2/λ) + (Sn − sn ) < (2nK/λ) + ε. By taking λ large (n remaining fixed after ε has been chosen), the last expression may be made less than 2ε so that ∫ b lim ψ(θ) sin(λθ) dθ = 0, λ→∞

a

and this is the result stated. When ψ(θ) is unbounded, if it has an absolutely convergent integral, by §4.5, we may enclose the points at which it is unbounded in a finite number of intervals δ1, δ2, . . . , δp (the finiteness of the number of intervals is assumed in the definition of an improper integral, §4.5) such that p ∫ Õ |ψ (θ)| dθ < ε. r=1 δ r

If K denotes the upper bound of |ψ (θ)| for values of θ outside these intervals, and if γ1, γ2, . . . , γ p+1 denote the portions of the interval (a, b) which do not belong to δ1, δ2, . . . , δp we may prove as before that ∫ b Õ ∫ p ∫ Õ p+1 = (θ) (λθ) (θ) (λθ) (θ) (λθ) ψ sin dθ ψ sin dθ + ψ sin dθ a r=1 δr r=1 γr p+1 ∫ p ∫ Õ Õ |ψ (θ) sin (λθ)| dθ ψ (θ) sin (λθ) dθ + ≤ r=1 δr r=1 γr < (2nK/λ) + 2ε. Now the choice of ε fixes n and K, so that the last expression may be made less than 3ε by taking λ sufficiently large. That is to say that, even if ψ(θ) be unbounded, ∫ b lim ψ (θ) sin (λθ) dθ = 0, λ→∞

a

provided that ψ(θ) has an (improper) integral which is absolutely convergent. The first lemma is therefore completely proved. (II) When ψ (θ) has limited total fluctuation in the range (a, b), by Example 3.6.2, we may write ψ (θ) = χ1 (θ) − χ2 (θ), where χ1 (θ), χ2 (θ) are positive increasing bounded functions. Then, by the second mean-value theorem (§4.14) a number ξ exists such that a ≤ ξ ≤ b

9.4 Fejér’s theorem

179

and ∫

a

b

∫ χ1 (θ) sin (λθ) dθ = χ1 (b)

ξ

b

sin (λθ) dθ ≤ 2 χ1 (b)/λ.

If we treat χ2 (θ) in a similar manner, it follows that ∫ ∫ b ∫ b ψ (θ) sin (λθ) dθ ≤ χ1 (θ) sin (λθ) dθ + a

a

a

b

χ2 (θ) sin (λθ) dθ

≤ 2 ( χ1 (b) + χ2 (b)) /λ = O(1/λ), and the second lemma is established. ∫ Corollary 9.4.3 If f (t) be such that

π

f (t) exists and is an absolutely convergent integral,

−π

the Fourier constants an, bn of f (t) are o(1) as n → ∞; and if, further, f (t) has limited total fluctuation in the range (−π, π), the Fourier constants are O(1/n). Note Of course these results are not sufficient to ensure the convergence of the Fourier series associated with f (t); for a series, in which the terms are of the order of magnitude of the terms in the harmonic series (§2.3), is not necessarily convergent.

9.42 The proof of Fourier’s theorem We shall now prove the theorem enunciated in §9.2, namely: Let f (t) be a function defined arbitrarily when −π∫≤ t < π, and defined by the equation π f (t + 2π) = f (t) for all other real values of t; and let f (t) dt exist and (if it is an improper −π

integral) let it be absolutely convergent. Let an, bn be defined by the equations ∫ π ∫ π πan = f (t) cos nt dt, πbn = f (t) sin nt dt. −π

−π

Then, if x be an interior point of any interval (a, b) within which f (t) has limited total fluctuation, the series ∞ Õ 1 an cos nx + bn sin nx a0 + 2 n=1 is convergent and its sum is 12 ( f (x + 0) + f (x − 0)) . It is convenient to give two proofs, one applicable to functions for which it is permissible to take the interval (a, b) to be the interval (−π + x, π + x), the other applicable to functions for which it is not permissible. (I) When the interval (a, b) may be taken to be (−π + x, π + x), it follows from §9.41(II) that an cos nx + bn sin nx is as O(1/n) as n → ∞. Now by Fejér’s theorem (§9.4) the series under consideration is summable (C1) and its sum (C1) is 12 ( f (x + 0) + f (x − 0)) . (The limits f (x ± 0) exist, by Example 3.6.3.) Therefore, by Hardy’s convergence theorem (§8.5), the series under consideration is convergent and its sum (by §8.43) is 12 ( f (x + 0) + f (x − 0)) .

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Fourier Series and Trigonometric Series

(II) Even if it is not permissible to take the interval (a, b) to be the whole interval (−π+x, π+x), it is possible, by hypothesis, to choose a positive number δ, less than π, such that f (t) has limited total fluctuation in the interval (x − δ, x + δ). We now define an auxiliary function g(t), which is equal to f (t) when x − δ ≤ t ≤ x + δ, and which is equal to zero throughout the rest of the interval (−π + x, π + x); and g(t + 2π) is to be equal to g(t) for all real values of t. Then g(t) satisfies the conditions postulated for the functions under consideration in (I), namely that it has an integral which is absolutely convergent and it has limited total fluctuation in the interval (−π + x, π + x); and so, if an(1) , and b(1) n denote the Fourier constants of g(t), the arguments used in (I) prove that the Fourier series associated with g(t), namely 1 (1) a 2 0

∞ Õ (an(1) cos nx + b(1) + n sin nx), n=1

is convergent and has the sum 21 (g(x + 0) + g(x − 0)), and this is equal to 1 2

( f (x + 0) + f (x − 0)) .

(1) Now let Sm (x) and Sm (x) denote the sums of the first m + 1 terms of the Fourier series associated with f (t) and g(t) respectively. Then it is easily seen that ∫ 1 π 1 + cos(x − t) + cos 2(x − t) + · · · + cos m(x − t) f (t) dt Sm (x) = π −π 2 ∫ π sin m + 21 (x − t) 1 f (t) dt = 2π −π sin 12 (x − t) ∫ π+x sin m + 21 (x − t) 1 = f (t) dt 2π −π+x sin 12 (x − t) ∫ ∫ 1 π/2 sin (2m + 1) θ 1 π/2 sin (2m + 1) θ = f (x + 2θ) dθ + f (x − 2θ) dθ, π 0 sin θ π 0 sin θ

by steps analogous to those given in §9.4. In like manner ∫ ∫ 1 π/2 sin(2m + 1)θ 1 π/2 sin(2m + 1)θ (1) Sm (x) = g(x + 2θ) dθ + g(x − 2θ) dθ, π 0 sin θ π 0 sin θ and so, using the definition of g(t), we have ∫ 1 π/2 sin(2m + 1)θ (1) Sm (x) − Sm (x) = f (x + 2θ) dθ π δ/2 sin θ ∫ 1 π/2 sin(2m + 1)θ + f (x − 2θ) dθ. π δ/2 sin θ Since cosec θ is a continuous function in the range 21 δ, 12 π , it follows that f (x±2θ) cosec θ are integrable functions with absolutely convergent integrals; and so, by the Riemann– Lebesgue lemma of §9.41(I), both the integrals on the right in the last equation tend to (1) (1) zero as m → ∞. That is to say lim Sm (x) − Sm (x) = 0. Hence, since lim Sm (x) = m→∞

m→∞

9.4 Fejér’s theorem 1 { f (x 2

181

+ 0) + f (x − 0)}, it follows also that lim Sm (x) =

m→∞

1 ( f (x + 0) + f (x − 0)) . 2

We have therefore proved that the Fourier series associated with f (t), namely Õ 1 a + (an cos nx + bn sin nx), 0 2 is convergent and its sum is 21 { f (x + 0) + f (x − 0)}.

9.43 The Dirichlet–Bonnet proof of Fourier’s theorem It is of some interest to prove directly the theorem of §9.42, without making use of the theory of summability; accordingly we now give a proof which is on the same general lines as the proofs due to Dirichlet and Bonnet. As usual we denote the sum of the first m + 1 terms of the Fourier series by Sm (x), and then, by the analysis of §9.42, we have ∫ ∫ 1 π/2 sin(2m + 1)θ 1 π/2 sin(2m + 1)θ Sm (x) = f (x + 2θ) dθ + f (x − 2θ) dθ. π 0 sin θ π 0 sin θ Again, on integrating the equation sin(2m + 1)θ = 1 + 2 cos 2θ + 2 cos 4θ + · · · + 2 cos 2mθ, sin θ we have ∫ 0

π/2

sin(2m + 1)θ π dθ = , sin θ 2

so that 1 Sm (x) − { f (x + 0) + f (x − 0)} 2 ∫ 1 π/2 sin(2m + 1)θ = { f (x + 2θ) − f (x + 0)} dθ π 0 sin θ ∫ 1 π/2 sin(2m + 1)θ { f (x − 2θ) − f (x − 0)} dθ. + π 0 sin θ In order to prove that lim Sm (x) =

m→∞

1 2

( f (x + 0) + f (x − 0)) ,

it is therefore sufficient to prove that ∫ π/2 sin(2m + 1)θ φ(θ) dθ = 0, lim m→∞ 0 sin θ where φ(θ) stands in turn for each of the functions f (x + 2θ) − f (x + 0),

f (x − 2θ) − f (x − 0).

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Fourier Series and Trigonometric Series

Now, by Example 3.6.4 θ φ(θ)/sin θ is a function with limited total fluctuation in an interval of which θ = 0 is an end-point 12 ; and so we may write θ φ(θ) = χ1 (θ) − χ2 (θ), sin θ where χ1 (θ), χ2 (θ) are bounded positive increasing functions of θ such that χ1 (+0) + χ2 (+0) = 0. Hence, given an arbitrary positive number ε, we can choose a positive number δ such that 0 ≤ χ1 (θ) < ε, 0 ≤ χ2 (θ) < ε whenever 0 ≤ θ ≤ δ/2. We now obtain inequalities satisfied by the three integrals on the right of the obvious equation ∫ π/2 ∫ π/2 sin(2m + 1)θ sin(2m + 1)θ φ(θ) dθ = φ (θ) dθ sin θ sin θ 0 δ/2 ∫ δ/2 ∫ δ/2 sin(2m + 1)θ sin(2m + 1)θ + χ1 (θ) dθ − χ2 (θ) dθ. θ θ 0 0 The modulus of the first integral can be made less than ε by taking m sufficiently large; this follows from §9.41(I) since φ(θ)/sin θ has an integral which converges absolutely in the interval ( 21 δ, 21 π). Next, from the second mean-value theorem, it follows that there is a number ξ between 0 and δ such that ∫ δ/2 ∫ δ/2 δ sin(2m + 1)θ sin(2m + 1)θ χ1 (θ) dθ = χ1 dθ θ 2 θ 0 ξ/2 ∫ (m+ 1 )δ 2 δ sin u = χ1 du . 2 (m+ 21 )ξ u ∫ ∞ ∫ ∞ sin t sin u Since dt is convergent, it follows that du has an upper bound 13 B which t u β is independent of β, and it is then clear that ∫ δ/2 δ sin(2m + 1)θ ≤ 2B χ < 2Bε. χ (θ) dθ 1 1 θ 2 0 On treating the third integral in a similar manner, we see that we can make ∫ π/2 sin(2m + 1)θ φ(θ) dθ < (4B + 1)ε sin θ 0 by taking m sufficiently large; and so we have proved that ∫ π/2 sin(2m + 1)θ lim φ(θ) dθ = 0. m→∞ 0 sin θ 12

13

The other end-point is θ = 21 (b − x) or θ = 21 (x − a), according as φ(θ) represents one or other of the two functions. ∫ ∞ sin u π The reader will find it interesting to prove that B = du = . u 2 0

9.4 Fejér’s theorem

183

But it has been seen that this is a sufficient condition for the limit of Sm (x) to be 1 ( f (x + 0) + f (x − 0)) ; and we have therefore established the convergence of a Fourier 2 series in the circumstances enunciated in §9.42. Note The reader should observe that in either proof of the convergence of a Fourier series the second mean-value theorem is required; but to prove the summability of the series, the first mean-value theorem is adequate. It should also be observed that, while restrictions are laid upon f (t) throughout the range (−π, π) in establishing the summability at any point x, the only additional restriction necessary to ensure convergence is a restriction on the behaviour of the function in the immediate neighbourhood of the point x. The fact that the convergence depends only on the behaviour of the function in the immediate neighbourhood of x (provided that the function has an integral which is absolutely convergent) was noticed by Riemann and has been emphasised by Lebesgue [418, p. 60]. It is obvious that the condition (due to Jordan [360]) that x should be an interior point of an interval in which f (t) has limited total fluctuation is merely a sufficient condition for the convergence of the Fourier series; and it may be replaced by any condition which makes ∫ π/2 sin(2m + 1)θ φ(θ) dθ = 0. lim m→∞ 0 sin θ Jordan’s condition is, however, a natural modification of the Dirichlet condition that the function f (t) should have only a finite number of maxima and minima, and it does not increase the difficulty of the proof. Another condition with the same effect is due to Dini [170], namely that, if Φ(θ) = ∫

1 [ f (x + 2θ) + f (x − 2θ) − f (x + 0) − f (x − 0)] , θ

a

Φ(θ)dθ should converge absolutely for some positive value of a. If the condition is ∫ δ/2 |Φ(θ)| dθ < ε, and then satisfied, given ε we can find δ so that

then

0

0 δ/2

πε sin(2m + 1)θ θΦ(θ)dθ < ; sin θ 2 0 ∫ π/2 sin(2m + 1)θ the proof that φ(θ)dθ < ε for sufficiently large values of m follows from sin θ δ/2 the Riemann–Lebesgue lemma. A more stringent condition than Dini’s is due to Lipschitz [440], namely |φ(θ)| < Cθ k , where C and k are positive and independent of θ. For other conditions due to Lebesgue and to de la Vallée Poussin, see the latter’s [639, II, pp. 149–150]. It should be noticed that Jordan’s condition differs in character from Dini’s condition; the latter is a condition that the series may converge at a point, the former that the series may converge throughout an interval. ∫

9.44 The uniformity of the convergence of Fourier series Let f (t) satisfy the conditions enunciated in §9.42, and further let it be continuous (in addition to having limited total fluctuation) in an interval (a, b). Then the Fourier series associated

184

Fourier Series and Trigonometric Series

with f (t) converges uniformly to the sum f (x) at all points x for which a + δ ≤ x ≤ b − δ, where δ is any positive number. Let h(t) be an auxiliary function defined to be equal to f (t) when a ≤ t ≤ b and equal to zero for other values of t in the range (−π, π), and let an , bn denote the Fourier constants of (2) (x) denote the sum of the first m + 1 terms of the Fourier series associated h(t). Also let Sm with h(t). ∞ Í Then, by Corollary 9.4.2, it follows that 21 a0 + (an cos nx + bn sin nx) is uniformly n=1

summable throughout the interval (a + δ, b − δ); and since |an cos nx + bn sin nx| ≤ (an2 + b2n )1/2, which is independent of x and which, by §9.41(II), is O(1/n), it follows from Corollary 8.5.1 that ∞ Õ 1 (an cos nx + bn sin nx) a0 + 2 n=1 converges uniformly to the sum h(x), which is equal to f (x). Now, as in §9.42, ∫ 1 π/2 sin(2m + 1)θ (2) Sm (x) − Sm (x) = f (x + 2θ) dθ π 12 (b−x) sin θ ∫ 1 π/2 sin(2m + 1)θ + f (x − 2θ) dθ. π 12 (x−a) sin θ As in §9.41 we choose an arbitrary positive number ε and then enclose the points at which f (t) is unbounded in a set of intervals δ1, δ2, . . . , δp such that p ∫ Õ r=1

| f (t)| dt < ε.

δr

(9.1)

If K be the upper bound of | f (t)| outside these intervals, we then have, as in §9.41, 2nK (2) |Sm (x) − Sm (x)| < + 2ε cosec δ, 2m + 1 where the choice of n depends only on a and b and the form of the function f (t). Hence, (2) by a choice of m independent of x we can make |Sm (x) − Sm (x)| arbitrarily small; so that (2) (2) Sm (x) − Sm (x) tends uniformly to zero. Since Sm (x) → f (x) uniformly, it is then obvious that Sm (x) → f (x) uniformly; and this is the result to be proved. Note It must be observed that no general statement can be made about uniformity or absoluteness of convergence of Fourier series. Thus the series of Example 9.1.1 converges uniformly except near x = (2n + 1)π but converges absolutely only when x = nπ, whereas the series of Example 9.1.2 converges uniformly and absolutely for all real values of x.

9.5 The Hurwitz–Liapounoff theorem concerning Fourier constants

185

Example 9.4.1 If φ(θ) satisfies suitable conditions in the range (0, π), shew that ∫ π/2 ∫ π sin(2m + 1)θ sin(2m + 1)θ φ(θ) dθ = lim φ(θ) dθ lim m→∞ 0 m→∞ 0 sin θ sin θ ∫ π/2 sin(2m + 1)θ + lim φ(π − θ) dθ m→∞ 0 sin θ π = (φ(+0) + φ(π − 0)) . 2 Example 9.4.2 (Math. Trip. 1894) Prove that, if a > 0, ∫ ∞ sin(2n + 1)θ −aθ π πa lim e dθ = coth . n→∞ 0 sin θ 2 2 Hint. Shew that ∫ ∞ ∫ mπ sin(2n + 1)θ −aθ sin(2n + 1)θ −aθ e dθ = lim e dθ m→∞ 0 sin θ sin θ 0 ∫ π sin(2n + 1)θ −aθ = lim e + e−a(θ+π) + · · · + e−a(θ+mπ) dθ m→∞ 0 sin θ ∫ π sin(2n + 1)θ e−aθ dθ = , sin θ 1 − e−aπ 0 and use Example 9.4.1. Example 9.4.3 Discuss the uniformity of the convergence of Fourier series by means of the Dirichlet–Bonnet integrals, without making use of the theory of summability.

9.5 The Hurwitz–Liapounoff theorem concerning Fourier constants This appears in Hurwitz [328]. Liapounoff discovered the theorem in 1896 and published it in [434]. See also Stekloff [601]. ∫ π Let f (x) be bounded in the interval (−π, π) and let f (x) dx exist, so that the Fourier constants an, bn of f (x) exist. Then the series

−π

∞

1 2 Õ 2 (a + b2n ) a + 2 0 n=1 n is convergent and its sum is 1 π

∫

π

{ f (x)}2 dx.

−π

This integral exists by Example 4.1.3. A proof of the theorem has been given by de la Vallée Poussin, in which the sole restrictions on f (x) are that the (improper) integrals of f (x) and { f (x)}2 exist in the interval (−π, π). See [639, II, pp. 165–166]. It will first be shewn that, with the notation of §9.4, )2 ∫ π( m−1 1 Õ lim f (x) − Sn (x) dx = 0. m→∞ −π m n=0

Fourier Series and Trigonometric Series

186

Divide the interval (−π, π) into 4r parts, each of length δ; let the upper and lower bounds of f (x) in the interval {(2p − 1)δ − π, (2p + 3)δ − π} be Up , L p , and let the upper bound of | f (x)| in the interval (−π, π) be K. Then, by Corollary 9.4.1 m−1 1 Õ Sn (x) < Up − L p + 2K/ m sin2 21 δ f (x) − m n=0 < 2K 1 + 1/ m sin2 12 δ , when x lies between 2pδ and (2p + 2)δ. Consequently, by the first mean-value theorem, )2 ∫ π( m−1 1 Õ Sn (x) dx f (x) − m n=0 −π ( ) ( 2r−1 ) Õ 4Kr 1 (Up − L p ) + 2δ . < 2K 1 + m sin2 12 δ m sin2 12 δ p=0 Since f (x) satisfies the Riemann condition of integrability (§4.12), it follows that both r−1 r−1 Í Í 4δ (U2p − L2p ) and 4δ (U2p+1 − L2p+1 ) can be made arbitrarily small by giving r a p=0

p=0

sufficiently large value. When also δ) has been given such a value, we may r (and therefore choose m1 , so large that r/ m1 sin2 21 δ is arbitrarily small. That is to say, we can make the expression on the right of the last inequality arbitrarily small by giving m any value greater than a determinate value m1 . Hence the expression on the left of the inequality tends to zero as m → ∞. But evidently )2 )2 ∫ π( ∫ π( m−1 m−1 Õ m−n 1 Õ Sn (x) dx = f (x) − An (x) dx f (x) − m n=0 m −π −π n=0 )2 ∫ π( m−1 m−1 Õ Õ n = An (x) + An (x) dx f (x) − m −π n=0 n=0 )2 )2 ∫ π( ∫ π (Õ m−1 m−1 Õ n = An (x) dx + f (x) − An (x) dx m −π −π n=0 n=0 ) ( m−1 ) ∫ π( m−1 Õ Õ +2 f (x) − An (x) An (x) dx −π

=

∫

π

n=0

( f (x) −

−π

m−1 Õ

n=0

)2 dx +

An (x)

n=0

m−1 π Õ 2 2 n (an + b2n ), m2 n=0

since ∫

π

f (x)Ar (x) dx =

−π

when r = 0, 1, 2, . . . , m − 1.

π

( m−1 Õ

−π

n=0

∫

) An (x)

Ar (x) dx

9.6 Riemann’s theory of trigonometrical series

187

Since the original integral tends to zero and since it has been proved equal to the sum of two positive expressions, it follows that each of these expressions tends to zero; that is to say )2 ∫ π( m−1 Õ f (x) − An (x) dx → 0. −π

n=0

Now the expression on the left is equal to ) ( m−1 ) ∫ π ∫ π( m−1 Õ Õ { f (x)}2 dx − 2 f (x) − An (x) An (x) dx −π

−π π

( m−1 Õ

−π

n=0

∫ − =

An (x)

π

∫

n=0

n=0

)2 dx π

( m−1 Õ

−π

n=0

∫ { f (x)}2 dx −

−π

)2 An (x)

dx

) m−1 1 2 Õ 2 2 (a + bn ) , a + = { f (x)} dx − π 2 0 n=1 n −π (

π

∫

2

so that, as m → ∞, ! m−1 Õ 1 { f (x)}2 dx − π a02 + (an2 + b2n ) → 0. 2 −π n=0

∫

π

This is the theorem stated. For the following corollary, Parseval assumed, of course, the permissibility of integrating the trigonometrical series term-by-term. Corollary 9.5.1 (Parseval [516]) If f (x), F(x) both satisfy the conditions laid on f (x) at the beginning of this section, and if An , Bn be the Fourier constants of F(x), it follows by subtracting the pair of equations which may be combined in the one form # " ∫ π ∞ Õ 1 {(a0 ± An )2 + (bn ± Bn )2 } { f (x) ± F(x)}2 dx = π (a0 ± A0 )2 + 2 −π n=1 that ∫

π

−π

) ∞ Õ 1 f (x)F(x) dx = π (an An + bn Bn ) . a0 A0 + 2 n=1 (

9.6 Riemann’s theory of trigonometrical series The theory of Dirichlet concerning Fourier series is devoted to series which represent given functions. Important advances in the theory were made by Riemann, who considered ∞ Í properties of functions defined by a series of the type 14 12 a0 + (an cos nx +bn sin nx), where n=1

14

Throughout §§9.6–9.632 the letters a n , b n do not necessarily denote Fourier constants.

Fourier Series and Trigonometric Series

188

it is assumed that lim (an cos nx + bn sin nx) = 0. We shall give the propositions leading up n→∞ to Riemann’s theorem that if two trigonometrical series converge and are equal at all points of the range (−π, π) with the possible exception of a finite number of points, corresponding coefficients in the two series are equal. The proof given is due to G. Cantor [113, 114].

9.61 Riemann’s associated function Let the sum of the series ∞

∞

Õ Õ 1 a0 + (an cos nx + bn sin nx) = A0 + An (x), 2 n=1 n=1 at any point x where it converges, be denoted by f (x). Let ∞

Õ An (x) 1 . F(x) = A0 x 2 − 2 n2 n=1 Then, if the series defining f (x) converges at all points of any finite interval, the series defining F(x) converges for all real values of x. To obtain this result we need the following lemma due to Cantor 15 . Lemma (Cantor) If lim An (x) = 0 for all values of x such that a ≤ x ≤ b, then an → 0, n→∞ bn → 0. For take two points x, x + δ of the interval. Then, given ε, we can find n0 (the value of n0 depends on x and on δ) such that, when n > n0 |an cos nx + bn sin nx |< ε,

| an cos n(x + δ) + bn sin n(x + δ)| < ε.

Therefore | cos nδ(an cos nx + bn sin nx) + sin nδ(− an sin nx + bn cos nx)| < ε. Since | cos nδ(an cos nx + bn sin nx)| < ε, it follows that |sin nδ(−an sin nx + bn cos nx)| < 2ε, and it is obvious that |sin nδ(an cos nx + bn sin nx)| < 2ε. Therefore, squaring and adding, √ (an2 + b2n )1/2 |sin nδ| < 2ε 2. Now suppose that an , bn have not the unique limit 0; it will be shewn that this hypothesis involves a contradiction. For, by this hypothesis, some positive number ε0 exists such that there is an unending increasing sequence n1, n2, . . . of values of n, for which (an2 + b22 )1/2 > 4ε0 . Now let the range of values of δ be called the interval I1 of length L1 on the real axis. Take n10 the smallest of the integers nr such that n10 L1 > 2π; then sin n10 y goes through all its √ phases in the interval I1 ; call I2 that sub-interval 16 of I1 in which sin n10 y > 1/ 2; its length 15

16

Riemann appears to have regarded this result as obvious. The proof here given is a modification of Cantor’s proof [114, 115]. If there is more than one such sub-interval, take that which lies on the left.

9.6 Riemann’s theory of trigonometrical series

189

is π/(2n10 ) = L2 . Next take n20 the smallest of the integers nr (> n10 ) such that n20 L2 > 2π, so that sin n20 y goes through all its phases in the interval I2 ; call I3 that sub-interval of I2 √ in which sin n20 y > 1/ 2; its length is π/(2n20 ) = L3 . We thus get a sequence of decreasing intervals I1, I2, . . . each contained in all the previous ones. It is obvious from the definition of an irrational number √ that there is a certain0point a0 which is not outside any of these intervals, and sin na ≥ 1/ 2 when n = n10 , n20 , . . . (nr+1 > nr ). √ 2 2 1/2 For these values of n, (an +bn ) sin na > 2ε0 2. But it has been shewn that corresponding √ to given numbers a and ε we can find n0 such that when n > n0 , (an2 + b2n )1/2 (sin na) < 2ε 2; since some values of nr0 are greater than n0 , the required contradiction has been obtained, because we may take ε < ε0 ; therefore an → 0, bn → 0. Assuming that the series defining f (x) converges at all points of a certain interval of the real axis, we have just seen that an → 0, bn → 0. Then, for all real values of x, |an cos nx + ∞ Í An (x)/n2 = F(x) bn sin nx| ≤ (an2 + b2n )1/2 → 0, and so, by §3.34, the series 12 A0 x 2 − n=1

converges absolutely and uniformly for all real values of x; therefore (see §3.32) F(x) is continuous for all real values of x.

9.62 Properties of Riemann’s associated function; Riemann’s first lemma It is now possible to prove Riemann’s first lemma that if F(x + 2α) + F(x − 2α) − 2F(x) 4α2 ∞ Í An (x) converges for the value of x under considthen lim G(x, α) = f (x), provided that G(x, α) =

a→0

n=0

eration. Since the series defining F(x), F(x ± 2α) converge absolutely, we may rearrange them; and, observing that cos n(x + 2α) cos n(x − 2α) − 2 cos nx = −4 sin2 nα cos nx, sin n(x + 2α) + sin n(x − 2α) − 2 sin nx = −4 sin2 nα sin nx, it is evident that G(x, α) = A0 +

2 ∞ Õ sin nα n=1

nα

An (x).

It will now be shewn that this series converges uniformly with regard to α for all values of ∞ Í α, provided that An (x) converges. The result required is then an immediate consequence n=1 2 of §3.32: for, if fn (α) = sinnαnα , (α , 0), and fn (0) = 1, then fn (α) is continuous for all values of α, and so G(x, α) is a continuous function of α, therefore, by §3.2, G(x, 0) = lim G(x, α). a→0

To prove that the series defining G(x, α) converges uniformly, we employ the test given in Example 3.3.7. The expression corresponding to ωn (x) is fn (α), and it is obvious that x Í | fn (α)| ≤ 1; it is therefore sufficient to shew that | fn+1 (α) − fn (α)| < K, where K is independent of α.

n=1

Fourier Series and Trigonometric Series

190

In fact, since x −1 sin x decreases as x increases from 0 to π, if s be the integer such that s|α| ≤ π < (s + 1)|α|, when α , 0 we have s−1 Õ

s−1 Õ sin2 α sin2 sα − 2 2 . | fn+1 (α) − fn (α)| = ( fn (α) − fn+1 (α)) = α2 s α n=1 n=1

Also ∞ Õ

∞ Õ sin2 nα 1 1 sin2 nα − sin2 (n + 1)α | fn+1 (α) − fn (α)| = − + α2 n2 (n + 1)2 (n + 1)2 α2 n=s+1 n=s+1 ∞ ∞ Õ Õ 1 1 1 | sin2 nα − sin2 (n + 1)α| ≤ − + α2 n2 (n + 1)2 (n + 1)2 α2 n=s+1 n=s+1 ∞ Õ 1 | sin α sin(2n + 1)α| ≤ + (s + 1)2 α2 n=s+1 (n2 + 1)2 α2

≤

∞ 1 1 | sin α| Õ + 2 2 2 (s + 1) α α (n + 1)2 n=s+1

1 | sin α| ≤ 2+ π α2

∫∞ s

dx (x + 1)2

1 1 ≤ 2+ . π (s + 1)|α| Therefore ∞ Õ

2 sin sα sin2 (s + 1)α 1 1 sin2 α sin2 sα − 2 2 + + + 2+ | fn+1 (α) − fn (α)| ≤ 2 2 2 2 2 α s α s α (s + 1) α π π n=1 ≤ 1+

1 2 + 2. π π

Since this expression is independent of α, the result required has been obtained (this inequality is obviously true when α = 0). ∞ Í An (x) converges, the series defining G(x, α) converges uniformly with respect Hence, if n=0

to α for all values of α, and, as stated above, lim G(x, α) = G(x, 0) = A0 +

α→0

∞ Õ

An (x) = f (x).

n=1

Example (Riemann) If H(x, α, β) =

F(x + α + β) − F(x + α − β) − F(x − α + β) + F(x − α − β) 4αβ

shew that H(x, α, β) → f (x) when f (x) converges if α, β → 0 in such a way that α/β and β/α remain finite.

9.6 Riemann’s theory of trigonometrical series

191

9.621 Riemann’s second lemma With the notation of §9.6 and §9.62, if an, bn → 0, then lim

α→0

F(x + 2α) + F(x − 2α) − 2F(x) =0 4α

for all values of x. For 1 −1 α F(x 4

+ 2α) + F(x − 2α) − 2F(x) = A0 α +

∞ Õ sin2 nα n=1

n2 α

An (x);

but by Example 9.1.3 if α > 0, ∞ Õ sin2 nα n=1

n2 α

=

1 (π − α); 2

and so, since A0 (x)α +

∞ Õ sin2 nα n=1

n2 α

1 An (x) = A0 (x)α + (π − α)A1 (x) 2 ) ( n ∞ Õ Õ 1 sin2 mα {An+1 (x) − An (x)} , (π − α) − + 2 m2 α m=1 n=1

it follows from Example 3.3.7, that this series converges uniformly with regard to α for all values of α greater than, or equal to, zero 17 . But 1 −1 α {F(x + 2α) + F(x − 2α) − 2F(x)} α→+0 4 # " ∞ Õ 1 = lim+ A0 (x)α + (π − α)A1 (x) + gn (α) {An+1 (x) − An (x)} α→0 2 n=1 lim

and this limit is the value of the function when α = 0, by §3.32; and this value is zero since lim An (x) = 0. By symmetry we see that lim+ = lim− . n→∞

α→0

α→0

9.63 Riemann’s theorem on trigonometrical series Two trigonometrical series which converge and are equal at all points of the range (−π, π), with the possible exception of a finite number of points, must have corresponding coefficients equal. The proof we give is due to G. Cantor [113]. An immediate deduction from this theorem is that a function of the type considered in §9.42 cannot be represented by any trigonometrical series in the range (−π, π) other than the Fourier series. This fact was first noticed by Du Bois Reymond. 17

If we define gn (α) by the equations gn (α) = (π − α)/2 − gn (α) is continuous when α ≥ 0, and gn+1 (α) ≤ gn (α).

n Í m=1

sin2 mα m2 α

(with α , 0), and gn (0) = π/2, then

192

Fourier Series and Trigonometric Series

We observe that it is certainly possible to have other expansions of (say) the form a0 +

∞ Õ (am cos 12 mx + βm sin 12 mx), m=1

which represent f (x) between −π and π; for write x = 2ξ, and consider a function φ(ξ), which is such that φ(ξ) = f (2ξ) when −π/2 < ξ < π/2, and φ(ξ) = g(ξ) when −π < ξ < −π/2, and when π/2 < ξ < π, where g(ξ) is any function satisfying the conditions of §9.43. Then if we expand φ(ξ) in a Fourier series of the form a0 +

∞ Õ

(am cos mξ + βm sin mξ),

m=0

this expansion represents f (x) when −π < x < π; and clearly by choosing the function g(ξ) in different ways an unlimited number of such expansions can be obtained. The question now at issue is, whether other series proceeding in sines and cosines of integral multiples of x exist, which differ from Fourier’s expansion and yet represent f (x) between −π and π. If possible, let there be two trigonometrical series satisfying the given conditions, and let their difference be the trigonometrical series A0 +

∞ Õ

An (x) = f (x).

n=1

Then f (x) = 0 at all points of the range (−π, π) with a finite number of exceptions; let ξ1 , ξ2 be a consecutive pair of these exceptional points, and let F(x) be Riemann’s associated function. We proceed to establish a lemma concerning the value of F(x) when ξ1 < x < ξ2 .

9.631 Schwartz’ lemma Quoted by G. Cantor [113]. In the range ξ1 < x < ξ2 , F(x) is a linear function of x, if f (x) = 0 in this range. For if θ = 1 or if θ = −1 1 x − ξ1 {F(ξ2 ) − F(ξ1 )} − h2 (x − ξ1 )(ξ2 − x) φ(x) = θ F(x) − F(ξ1 ) − ξ2 − ξ1 2 is a continuous function of x in the range ξ1 ≤ x ≤ ξ2 and φ(ξ1 ) = φ(ξ2 ) = 0. If the first term of φ(x) is not zero throughout the range 18 there will be some point x = c at which it is not zero. Choose the sign of θ so that the first term is positive at c, and then choose h so small that φ(c) is still positive. Since φ(x) is continuous it attains its upper bound (§3.62), and this upper bound is positive since φ(c) > 0. Let φ(x) attain its upper bound at c1 , so that c1 , ξ1, c1 , ξ2 . Then, by Riemann’s first lemma, lim

a→0

φ (c1 + a) + φ (c1 − a) − 2φ (c1 ) = h2 . a2

But φ(c1 + a) ≤ φ(c1 ), φ(c1 − a) ≤ φ(c1 ), so this limit must be negative or zero. Hence, by supposing that the first term of φ(x) is not everywhere zero in the range (ξ1, ξ2 ), we have 18

If it is zero throughout the range, F(x) is a linear function of x.

9.7 Fourier’s representation of a function by an integral

193

arrived at a contradiction. Therefore it is zero; and consequently F(x) is a linear function of x in the range ξ1 < x < ξ2 . The lemma is therefore proved.

9.632 Proof of Riemann’s theorem We see that, in the circumstances under consideration, the curve y = F(x) represents a series of segments of straight lines, the beginning and end of each line corresponding to an exceptional point; and as F(x), being uniformly convergent is a continuous function of x, these lines must be connected. But, by Riemann’s second lemma, even if ξ be an exceptional point, lim

a→0

F(ξ + α) + F(ξ − α) − 2F(ξ) = 0. α

Now the fraction involved in this limit is the difference of the slopes of the two segments which meet at that point whose abscissa is ξ; therefore the two segments are continuous in direction, so the equation y = F(x) represents a single line. If then we write F(x) = cx + c 0, it follows that c and c 0 have the same values for all values of x. Thus ∞ Õ 1 An (x) , A0 x 2 − cx − c 0 = 2 n2 n=1 the right-hand side of this equation being periodic, with period 2π. The left-hand side of this equation must therefore be periodic, with period 2π. Hence ∞ A (x) Í n A0 = 0, c = 0, and −c 0 = . Now the right-hand side of this equation converges n2 n=1 uniformly, so we can multiply by cos nx or by sin nx and integrate. This process gives ∫ π πan 0 = c cos nx dx = 0, n2 −π ∫ π πbn 0 = −c sin nx dx = 0. n2 −π Therefore all the coefficients vanish, and therefore the two trigonometrical series whose ∞ Í An (x) have corresponding coefficients equal. This is the result stated in difference is A0 + n=1

§9.63.

9.7 Fourier’s representation of a function by an integral This appears in Fourier [223]. For recent work on Fourier’s integral and the modern theory of ‘Fourier transforms’, see Titchmarsh [629, 630]. It follows from §9.43 that, if f (x) be continuous except at a finite number of discontinuities and if it have limited total fluctuation in the range (−∞, ∞), then, if x be any internal point of the range (−α, β), ∫ β sin (2m + 1)(t − x) π sin θ { f ( x + 2θ) + f ( x − 2θ)} . lim f ( t ) dt = lim m→∞ −α θ→0 2 (t − x) θ Now let λ be any real number, and choose the integer m so that λ = 2m + 1 + 2η where

Fourier Series and Trigonometric Series

194

0 ≤ η < 1. Then β

∫

f (t) dt t−x −α ∫ β f (t) = dt 2 cos[(2m + 1 + η)(t − x)] sin η(t − x) t−x −α → 0, [sin λ(t − x) − sin(2m + 1)(t − x)]

as m → ∞ by §9.41, since (t − x)−1 f (t) sin η(t − x) has limited total fluctuation. Consequently, from the proof of the Riemann–Lebesgue lemma of §9.41, it is obvious ∫ ∞ ∫ 0 | f (t)| dt and | f (t)| dt converge, then 19 that if −∞

0

∫

∞

lim

λ→∞

−∞

and so ∫

∞

∫

lim

λ→∞

−∞

sin λ (t − x) 1 f (t) dt = π { f (x + 0) + f (x − 0)} , (t − x) 2 λ

cos u (t − x) du f (t) dt =

0

π { f (x + 0) + f (x − 0)} . 2

To obtain Fourier’s result, we must reverse the order of integration in this repeated integral. For any given value of λ and any arbitrary value of ε, there exists a number β such that ∫ ∞ ε | f (t)| dt < ; 2λ β writing cos u(t − x) · f (t) = φ(t, u), we have 20 ∫ ∞ ∫ λ ∫ λ ∫ ∞ φ(t, u) du dt − φ(t, u) dt du 0 0 0 0 ∫ β ∫ λ ∫ ∞ ∫ λ = φ(t, u) du dt + φ(t, u)du dt 0

0 λ

β

β

0 λ ∫ θ

− φ(t, u) dt du − φ(t, u) dt du 0 0 0 β ∫ ∞ ∫ λ ∫ λ ∫ ∞ = φ(t, u) du dt − φ(t, u) dt du β 0 0 β ∫ ∞ ∫ λ ∫ λ∫ ∞ |φ(t, u)| du dt + |φ(t, u)| dt du < β 0 0 β ∫ ∞ < 2λ | f (t)| dt < ε. ∫

∫

∫

β

Since this is true for all values of ε, no matter how small, we infer that ∫ ∞∫ λ ∫ λ∫ ∞ ∫ −∞ ∫ λ ∫ λ ∫ = ; similarly = 0

20

0

0

0

0

0

.

0

∫ ρ ∫ σ means the double limit lim . If this limit exists, it is equal to lim . ρ→∞, σ→∞ −ρ ρ→∞ −ρ −∞ ∫β∫λ ∫λ∫β The equation 0 0 = 0 0 is easily justified by §4.3, by considering the ranges within which f (x) is continuous. ∫

19

0

−∞

∞

9.8 Miscellaneous examples

195

Hence ∫ λ∫ ∞ 1 π { f (x + 0) + f (x − 0)} = lim cos u (t − x) f (t) dt du λ→∞ 0 2 −∞ ∫ ∞∫ ∞ = cos u (t − x) f (t) dt du. 0

−∞

This result is known as Fourier’s integral theorem. For a proof of the theorem when f (x) is subject to less stringent see Hobson [316, pp. ∫ 492–493]. ∫ The reader should ob∫ restrictions, ∫ λ

∞

serve that, although lim

λ→∞ −∞

∞

∞

sin u(t − x) du f (t) dt

exists, the repeated integral −∞

0

0

does not. Example 9.7.1 (Rayleigh) Verify Fourier’s integral theorem directly (i) for the function f (x) = (a2 + x 2 )−1/2 , (ii) for the function defined by the equations f (x) = 1, (−1 < x < 1);

f (x) = 0, (|x| > 1).

9.8 Miscellaneous examples Example 9.1 Obtain the expansions 1 − r cos z = 1 + r cos z + r 2 cos 2z + · · · , 1 − 2r cos z + r 2 1 1 1 (b) log(1 − 2r cos z + r 2 ) = −r cos z − r 2 cos 2z − r 3 cos 3z − · · ·, 2 2 3 r sin z 1 2 1 3 (c) arctan = r sin z + r sin 2z + r sin 3z + · · ·, 1 − r cos z 2 3 2r sin z 1 3 1 5 1 = r sin z + r sin 3z + r sin 5z + · · ·, (d) arctan 2 1 − r2 3 5 and shew that, when |r | < 1, they are convergent for all values of z in certain strips parallel to the real axis in the z-plane. (a)

Example 9.2 (Jesus, 1902) Expand x 3 and x in Fourier sine series valid when −π < x < π; and hence find the value of the sum of the series 1 1 1 sin x − 3 sin 2x − 3 sin 3x − 3 sin 4x + · · · , 2 3 4 for all values of x. Example 9.3 (Pembroke, 1907) Shew that the function of x represented by ∞ Õ

n−1 sin nx sin2 na,

n=1

is constant (0 < x < 2a) and zero (2a < x < π), and draw a graph of the function. Example 9.4 (Peterhouse, 1906) Find the cosine series representing f (x) where ( sin x + cos x (0 < x ≤ 12 π), f (x) = sin x − cos x ( 12 π ≤ x < π).

196

Fourier Series and Trigonometric Series

Example 9.5 (Trinity, 1895) Shew that 1 sin 3πx sin 5πx sin 7πx + + + · · · = π[x], 3 5 7 4 where [x] denotes +1 or −1 according as the integer next inferior to x is even or uneven, and is zero if x is an integer. sin πx +

Example 9.6 Shew that the expansions 1 1 1 log 2 cos x = cos x − cos 2x + cos 3x · · · 2 2 3 and 1 1 1 log 2 sin x = − cos x − cos 2x − cos 3x · · · 2 2 3 are valid for all real values of x, except multiples of π. Example 9.7 (Trinity, 1898) Obtain the expansion ∞ Õ (−1)m cos mx (m + 1)(m + 2) m=0

x x = (cos x + cos 2x) log 2 cos + (sin 2x + sin x) − cos x, 2 2 and find the range of values of x for which it is applicable. Example 9.8 (Trinity, 1895) Prove that, if 0 < x < 2π, then 2 sin 2x 3 sin 3x π sinh a(π − x) sin x + 2 + 2 +··· = . 2 2 2 2 a +1 a +2 a +3 2 sinh aπ Example 9.9 Shew that between the values −π and +π of x the following expansions hold: 2 2 sin 2x 3 sin 3x sin x sin mx = sin mπ 2 − + − · · · , π 1 − m2 22 − m2 32 − m2 2 1 m cos x m cos 2x m cos 3x cos mx = sin mπ + − 2 + 2 −··· , π 2m 12 − m2 2 + m2 3 − m2 emx + e−mx 2 1 m cos x m cos 2x m cos 3x = + 2 − 2 +··· . − 2 mπ −mπ e −e π 2m 1 + m2 2 + m2 3 + m2 Example 9.10 (Berger) Let x be a real variable between 0 and 1, and let n ≥ 3 be an odd number. Shew that ∞ 1 2Õ 1 mπ (−1)s = + tan cos 2mπx, n π m=1 m n if x is not a multiple of 1/n, where s is the greatest integer contained in nx but 0=

∞ 1 2Õ 1 mπ + tan cos 2mπx n π m=1 m n

if x is an integer multiple of 1/n.

9.8 Miscellaneous examples

197

Example 9.11 (Trinity, 1901) Shew that the sum of the series ∞ 1 4 Õ −1 + m sin 3 π m=1

is 1 when 0 < x < 13 , and when

2 3

2mπ 3

cos 2mπx

< x < 1, and is −1 when

1 3

< x < 23 .

Example 9.12 (Math. Trip. 1896) If ∞

Õ a nVn (x) ae ax = , e a − 1 n=0 n! shew that, when −1 < x < 1, 22n−1 π 2n cos 4πx cos 6πx + + · · · = (−1)n−1 V2n (x), 2n 2n 2 3 (2n)! sin 4πx sin 6πx 22n π 2n+1 sin 2πx + 2n+1 + 2n+1 + · · · = (−1)n+1 V2n+1 (x). 2 3 (2n + 1)!

cos 2πx +

Example 9.13 (Trinity, 1894) If m is an integer, shew that, for all real values of x, 1 · 3 · 5 · · · (2m − 1) 1 m m(m − 1) 2m cos x = 2 + cos 2x + cos 4x 2 · 4 · 6 · · · 2m 2 m+1 (m + 1)(m + 2) m(m − 1)(m − 2) + cos 6x + · · · , (m + 1)(m + 2)(m + 3) 2m−1 4 2 · 4 · 6 · · · (2m − 2) 1 2m − 1 cos x = + cos 2x π 1 · 3 · 5 · · · (2m − 1) 2 2m + 1 (2m − 1)(2m − 3) + cos 4x + · · · . (2m + 1)(2m + 3) Example 9.14 A point moves in a straight line with a velocity which is initially u, and which receives constant increments, each equal to u, at equal intervals τ. Prove that the velocity at any time t after the beginning of the motion is ∞ u ut u Õ 1 2mπt + + sin , 2 τ π m=1 m τ

and that the distance traversed is ∞ ut uτ uτ Õ 1 2mπt (t + τ) + − cos . 2τ 12 2π 2 m=1 m2 τ

Example 9.15 (Math. Trip. 1893) If f (x) =

∞ Õ sin(6n − 3)x n=1

2n − 1

−2

∞ Õ sin(2n − 1)x n=1

2n − 1 √ 3 3 sin 5x sin 7x sin 11x + sin x − + − + · · · , π 52 72 112

Fourier Series and Trigonometric Series

198

+ 0) − f ( 2π − 0) = π2 . shew that f (+0) = f (π − 0) = − π4 , and f ( π3 + 0) − f ( π3 − 0) = − π2 , f ( 2π 3 3 Observing that the last series is (2n−1)π ∞ 6 Õ sin 3 sin(2n − 1)x , π n=1 (2n − 1)2

draw the graph of f (x). Example 9.16 (Trinity, 1908) Shew that, when 0 < x < π, √ 1 1 2 3 1 cos 11x + · · · f (x) = cos x − cos 5x + cos 7x − 3 5 7 11 1 1 1 = sin 2x + sin 4x + sin 8x + sin 10x + · · · 2 4 5 where 31 π 0 < x < 13 π, 1 f (x) = 0 π < x < 23 π, 3 1 2 − π π < x < π. 3 3 Find the sum of each series when x = 0, 13 π, 23 π, π, and for all other values of x. Example 9.17 (Math. Trip. 1895) Prove that the locus represented by ∞ Õ (−1)n−1 n=1

n2

sin nx sin ny = 0

is two systems of lines at right angles, dividing the coordinate plane into squares of area π 2 . Example 9.18 (Trinity, 1903) Shew that the equation ∞ Õ (−1)n−1 sin ny cos nx n=1

n3

=0

represents the lines y =√±mπ, (m = 0, 1, 2, . . .) together with a set of arcs of ellipses whose semi-axes are π and π/ 3, the arcs being placed in squares of area 2π 2 . Draw a diagram of the locus. Example 9.19 (Math. Trip. 1904) Shew that, if the point (x, y, z) lies inside the octahedron bounded by the planes ±x ± y ± z = π, then ∞ Õ sin nx sin ny sin nz 1 (−1)n−1 = x yz. n3 2 n=1

Example 9.20 (Pembroke, 1902) Circles of radius a are drawn having their centres at the alternate angular points of a regular hexagon of side a. Shew that the equation of the trefoil formed by the outer arcs of the circles can be put in the form πr 1 1 1 1 cos 3θ − cos 6θ + cos 9θ − · · · , √ = + 2 2 · 4 5 · 7 8 · 10 6 3a the initial line being taken to pass through the centre of one of the circles.

9.8 Miscellaneous examples

199

Example 9.21 (Jesus, 1908) Draw the graph represented by ( ) ∞ 2m π 1 Õ (−1)n cos nmθ r =1+ sin + , a π m 2 n=1 1 − (nm)2 where m is an integer. Example 9.22 (Trinity, 1905) With each vertex of a regular hexagon of side 2a as centre, the arc of a circle of radius 2a lying within the hexagon is drawn. Shew that the equation of the figure formed by the six arcs is n √ o n−1 ∞ (−1) 6 + 3 3 Õ √ πr =6−3 3+2 cos 6nθ, 4a (6n − 1)(6n + 1) n=1 the prime vector bisecting a petal. Example 9.23 (Trinity, 1894) Shew that, if c > 0, ∫ ∞ 1 1 lim e−cx cot x sin(2n + 1)x dx = π tanh cπ. n→∞ 0 2 2 Example 9.24 (King’s, 1901) Shew that ∫ ∞ sin(2n + 1)x dx 1 lim = π coth 1. n→∞ 0 sin x 1 + x2 2 Example 9.25 (Math. Trip. 1905) Shew that, when −1 < x < 1 and a is real, ∫ ∞ π sinh ax sin(2n + 1)θ sin(1 + x)θ θ lim dθ = − . 2 2 n→∞ 0 sin θ a +θ 2 sinh a Example 9.26 (Math. Trip. 1898) Assuming the possibility of expanding f (x) in a Í uniformly convergent series of the form Ak sin k x, where k is a root of the equation k

k cos ak + b sin ak = 0 and the summation is extended to all positive roots of this equation, determine the constants Ak . Example 9.27 (Beau) If f (x) = 21 a0 +

∞ Í

(an cos nx + bn sin nx) is a Fourier series, shew

n=1

that, if f (x) satisfies certain general conditions, ∫ ∞ 4 t dt an = P.V. f (t) cos nt tan , π 2 t 0

4 bn = π

∫

∞

f (t) sin nt tan 0

t dt , 2 t

where P.V. means principal value. n Í sin r x Example 9.28 If Sn (x) = 2 (−1)r−1 prove that the highest maximum of Sn (x) in r r=1 nπ the interval (0, π) is at x = and prove that, as n → ∞, n+1 ∫ π nπ sin t Sn →2 dt. n+1 t 0

Deduce that, as n → ∞, the shape of the curve y = Sn (x) in the interval (0, π) tends to

200

Fourier Series and Trigonometric Series

approximate to the shape of the curve formed by the line y = x, 0 ≤ x ≤ π, together with the line x = π, 0 ≤ y ≤ G, where ∫ π sin t dt. G=2 t 0 Note The fact that G = 3.704 · · · > π is known as Gibbs’ phenomenon; see [243]. The phenomenon is characteristic of a Fourier series in the neighbourhood of a point of ordinary discontinuity of the function which it represents. For a full discussion of the phenomenon, which was discovered by Wilbraham [680], see Carslaw [119, Chapter 9].

10 Linear Differential Equations

10.1 Linear differential equations The analysis contained in this chapter is mainly theoretical; it consists, for the most part, of existence theorems. It is assumed that the reader has some knowledge of practical methods of solving differential equations; these methods are given in works exclusively devoted to the subject, such as Forsyth [221, 222]. In some of the later chapters of this work, we shall be concerned with the investigation of extensive and important classes of functions which satisfy linear differential equations of the second order. Accordingly, it is desirable that we should now establish some general results concerning solutions of such differential equations. The standard form of the linear differential equation of the second order will be taken to be du d2u + p(z) + q(z)u = 0, (10.1) dz 2 dz and it will be assumed that there is a domain S in which both p(z), q(z) are analytic except at a finite number of poles. Any point of S at which p(z), q(z) are both analytic will be called an ordinary point of the equation; other points of S will be called singular points.

10.2 Solution of a differential equation valid in the vicinity of an ordinary point This method is applicable only to equations of the second order. For a method applicable to equations of any order, see Forsyth [221]. Let b be an ordinary point of the differential equation, and let Sb be the domain formed by a circle of radius rb , whose centre is b, and its interior, the radius of the circle being such that every point of Sb is a point of S, and is an ordinary point of the equation. Let z be a variable point of Sb . ∫ 1 z In the equation write u = 3 exp − p(ζ) dζ , and it becomes 2 b d23 + J(z)3 = 0, dz 2

(10.2)

1 dp(z) 1 − {p(z)}2 . It is easily seen (§5.22) that an ordinary point of 2 dz 4 equation (10.1) is also an ordinary point of equation (10.2).

where J(z) = q(z) −

201

Linear Differential Equations

202

Now consider the sequence of functions 3n (z), analytic in Sb , defined by the equations 30 (z) = a0 + a1 (z − b), ∫ z (ζ − z) J(ζ)3n−1 (ζ) dζ, 3n (z) =

(n = 1, 2, 3, . . .)

b

where a0 , a1 are arbitrary constants. Let M, µ be the upper bounds of |J(z)| and |30 (z)| in the domain Sb . Then at all points of this domain µM n |z − b| 2n . (10.3) |3n (z)| ≤ n! For this inequality is true when n = 0; if it is true when n = 0, 1, . . . , m − 1, we have, by taking the path of integration to be a straight line, ∫ z (ζ − z) J(ζ)3m−1 (ζ) dζ |3m (z)| = b ∫ z 1 ≤ |ζ − z| |J(ζ)| µM m−1 |ζ − b| 2m−2 |dζ | (m − 1)! b ∫ |z−b | µM m |z − b| ≤ t 2m−2 dt (m − 1)! 0 µM m |z − b| 2m . ≤ m! Therefore, by induction, the inequality holds for all values of n. ∞ µM n Í µM n 2n Also, since |3n (z)| ≤ rb when z is in Sb and rb2n converges, it follows (§3.34) n! n=0 n! ∞ Í that 3(z) = 3n (z) is a series of analytic functions uniformly convergent in Sb ; while, from n=0

the definition of 3n (z), ∫ z d 3n (z) = − J(ζ)3n−1 (ζ) dζ, (n = 1, 2, 3, . . .) dz b d2 3n (z) = −J(z)3n−1 (z); dz2 hence it follows (§5.3) that ∞ d 2 3(z) d 2 30 (z) Õ d 2 3n (z) = + dz2 dz 2 dz 2 n=1 = −J(z)3(z). Therefore 3(z) is a function of z, analytic in Sb , which satisfies the differential equation d 2 3(z) + J(z)3(z) = 0, dz 2 d 3n (z), it is evident that dz d 3(b) = a0, 3 0(b) = 3(z) = a1 , dz z=b

and, from the value obtained for

10.2 Solutions in the vicinity of an ordinary point

203

where a0 , a1 are arbitrary.

10.21 Uniqueness of the solution If there were two analytic solutions of the equation for 3, say 31 (z) and 32 (z), such that 31 (b) = 32 (b) = a0, and 31 0(b) = 32 0(b) = a1, then, writing w(z) = 31 (z) − 32 (z), we should have d 2 w(z) + J(z)w(z) = 0. dz2 Differentiating this equation n − 2 times and putting z = b, we get n−2 0 w (n) (b) + J(b)w (n−2) (b) + J (b)w (n−3) (b) + · · · + J (n−2) (b)w(b) = 0. 1 Putting n = 2, 3, 4, . . . in succession, we see that all the differential coefficients of w(z) vanish at b; and so, by Taylor’s theorem, w(z) = 0; that is to say the two solutions 31 (z), 32 (z) are identical. ∫ 1 z p(ζ) dζ , we infer without difficulty that u(z) is the only Writing u(z) = 3(z) exp − 2 b analytic solution of (10.1) such that u(b) = A0 , u 0(b) = A1 , where A0 = a0,

A1 = a1 − 21 p(b) a0 .

Now that we know that a solution of (10.1) exists which is analytic in Sb and such that u(b), u 0(b) have the arbitrary values A0 , A1, the simplest method of obtaining the solution in the form of a Taylor’s series is to assume u(z) =

∞ Õ

An (z − b)n,

n=0

substitute this series in the differential equation and equate coefficients of successive powers of z − b to zero (§3.73) to determine in order the values of A2, A3, . . . in terms of A0 , A1 . Note In practice, in carrying out this process of substitution, the reader will find it much more simple to have the equation ‘cleared of fractions’ rather than in the canonical form (10.1) of §10.1. Thus the equations in Examples 10.2.1 and 10.2.2 below should be treated in the form in which they stand; the factors 1 − z2 , (z − 2), (z − 3) should not be divided out. The same remark applies to the examples of §§10.3 and 10.32. From the general theory of analytic continuation (§5.5) it follows that the solution obtained is analytic at all points of S except at singularities of the differential equation. The solution however is not, in general, analytic throughout S (see the footnote after Corollary 5.2.2), except at these points, as it may not be one-valued; i.e., it may not return to the same value when z describes a circuit surrounding one or more singularities of the equation. The property that the solution of a linear differential equation is analytic except at singularities of the coefficients of the equation is common to linear equations of all orders. When two particular solutions of an equation of the second order are not constant multiples of each other, they are said to form a fundamental system.

Linear Differential Equations

204

Example 10.2.1 Shew that the equation (1 − z2 )u 00 − 2zu 0 + 43 u = 0 has the fundamental system of solutions 21 4 3 z −··· , u1 = 1 − z 2 − 8 128

u2 = z +

5 3 15 5 z + z +··· . 24 128

Determine the general coefficient in each series, and shew that the radius of convergence of each series is 1. Example 10.2.2 Discuss the equation (z − 2)(z − 3)u 00 − (2z − 5)u 0 + 2u = 0 in a manner similar to that of Example 10.2.1.

10.3 Points which are regular for a differential equation Suppose that a point c of S is such that, although p(z) or q(z) or both have poles at c, the poles are of such orders that (z − c)p(z), (z − c)2 q(z) are analytic at c. Such a point is called a regular point 1 for the differential equation. Any poles of p(z) or of q(z) which are not of this nature are called irregular points. The reason for making the distinction will become apparent in the course of this section. If c be a regular point, the equation may be written 2 (z − c)2

du d2u + (z − c)P(z − c) + Q(z − c)u = 0, dz 2 dz

where P(z − c), Q(z − c) are analytic at c; hence, by Taylor’s theorem, P(z − c) = p0 + p1 (z − c) + p2 (z − c)2 + · · · , Q(z − c) = q0 + q1 (z − c) + q2 (z − c)2 + · · · , where p0, p1, . . . , q0, q1, . . . are constants; and these series converge in the domain Sc formed by a circle of radius r (centre c) and its interior, where r is so small that c is the only singular point of the equation which is in Sc . Let us assume as a formal solution of the equation " # ∞ Õ α n u = (z − c) 1 + an (z − c) , n=1

where α, a1, a2, . . . are constants to be determined. Substituting in the differential equation (assuming that the term-by-term differentiations 1

2

The name ‘regular point’ is due to Thomé [624]. Fuchs had previously used the phrase ‘point of determinateness’. Frobenius calls this the normal form of the equation.

10.3 Points which are regular for a differential equation

205

and multiplications of series are legitimate) we get " # ∞ Õ (z − c)α α(α − 1) + an (α + n)(α + n − 1)(z − c)n n=1

" + (z − c)α P(z − c) α +

∞ Õ

# an (α + n)(z − c)n

n=1

" α

+ (z − c) Q(z − c) 1 +

∞ Õ

# an (z − c)

n

= 0.

n=1

Substituting the series for P(z − c), Q(z − c), multiplying out and equating to zero the coefficients of successive powers of z − c, we obtain the following sequence of equations: α2 + (p0 − 1)α + q0 = 0, a1 {(α + 1)2 + (p0 − 1)(α + 1) + q0 } + αp1 + q1 = 0, a2 {(α + 2)2 + (p0 − 1)(α + 2) + q0 } + a1 {(α + 1)p1 + q1 } + αp2 + q2 = 0, .. . an {(α + n)2 + (p0 − 1)(α + n) + q0 } +

n−1 Õ

an−m {(α + n − m)pm + qm } + αpn + qn = 0.

m=1

The first of these equations, called the indicial equation (the name is due to Cayley [139]), determines two values (which may, however, be equal) for α. The reader will easily convince himself that if c had been an irregular point, the indicial equation would have been (at most) of the first degree; and he will now appreciate the distinction made between regular and irregular singular points. Let α = ρ1 , α = ρ2 be the roots of F(α) ≡ α2 + (p0 − 1)α + q0 = 0; (these roots are called the exponents of the indicial equation) then the succeeding equations (when α has been chosen) determine a1, a2, . . ., in order, uniquely, provided that F(α + n) does not vanish when n = 1, 2, 3, . . .; that is to say, if α = ρ1 , that ρ2 is not one of the numbers ρ1 + 1, ρ1 + 2, . . .; and, if α = ρ2 , that ρ1 is not one of the numbers ρ2 + 1, ρ2 + 2, . . .. Hence, if the difference of the exponents is not zero, or an integer, it is always possible to obtain two distinct series which formally satisfy the equation. Example 10.3.1 Shew that, if m is not zero or an integer, the equation ! 1 2 − m 1 4 − u=0 u 00 + z2 4 is formally satisfied by two series whose leading terms are z2 z2 z1/2+m 1 + + · · · , z1/2−m 1 + +··· ; 16(1 + m) 16(1 − m)

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206

determine the coefficient of the general term in each series, and shew that the series converge for all values of z.

10.31 Convergence of the expansion of §10.3 If the exponents ρ1 , ρ2 are not equal, let ρ1 be that one whose real part is not inferior to the real part of the other, and let ρ1 − ρ2 = s; then F (ρ1 + n) = n (s + n). Now, by §5.23, we can find a positive number M such that |qn | < Mr −n,

|pn | < Mr −n,

| ρ1 pn + qn | < Mr −n,

where M is independent of n; it is convenient to take M ≥ 1. Taking α = ρ1 , we see that |a1 |

| ρ1 p1 + q1 | M M < < , |F(ρ1 + 1)| r |s + 1| r

since |s + 1| ≥ 1. If now we assume |an | < M n r −n when n = 1, 2, . . . , m − 1, we get m−1 Í am−t {(ρ1 + m − t)p1 + qt } + ρ1 pm + qm t=1 |am | = F(ρ1 + m) m−1 Í

≤

t=1

|am−t | · | ρ1 p1 + qt | + | ρ1 pm + qm | +

(m − t) |am−t | |pt |

t=1

m |s + m| mM m r −m +

1.

10.5 Irregular singularities and confluence Near a point which is not a regular point of linear differential equations, an equation of the second order cannot have two regular integrals, for the indicial equation is at most of the first degree; there may be one regular integral or there may be none. We shall see later (e.g. §16.3) what is the nature of the solution near such points in some simple cases. A general investigation of such solutions is beyond the scope of this book. Some elementary investigations are given in Forsyth’s [221]. Complete investigations are given in his Theory of Differential Equations [218]. It frequently happens that a differential equation may be derived from another differential equation by making two or more singularities of the latter tend to coincidence. Such a limiting process is called confluence; and the former equation is called a confluent form of the latter. It will be seen in §10.6 that the singularities of the former equation may be of a more complicated nature than those of the latter.

10.6 The differential equations of mathematical physics The most general differential equation of the second order which has every point except a1, a2, a3, a4 and ∞ as an ordinary point, these five points being regular points with exponents αr , βr at ar (r = 1, 2, 3, 4) and exponents µ1 , µ2 at ∞, may be verified 3 to be ( 4 ) 4 Õ 1 − αr − βr du Õ d2u αr βr Az2 + 2Bz + C + + + u = 0, 2 2 4 r=1 (z − ar ) dz z − ar dz Î r=1 (z − ar ) r=1 3

The coefficients of du d z and u must be rational or they would have an essential singularity at some point; the Î Î denominators of p(z), q(z) must be 4r =1 (z − ar ), 4r =1 (z − ar )2 respectively; putting p(z) and q(z) into −1 partial fractions and remembering that p(z) = O(z ), q(z) = O(z −2 ) as |z | → ∞, we obtain the required result without difficulty.

10.6 The differential equations of mathematical physics

211

where A is such that µ1 and µ2 are the roots of (

) 4 4 Õ Õ µ +µ (αr + βr ) − 3 + αr βr + A = 0, 2

r=1

r=1

and B, C are constants. (It will be observed that µ1 , µ2 are connected by the relation 4 Í µ1 + µ2 + (αr + βr ) = 3.) r=1

The remarkable theorem has been proved by Klein [376] (see also [373]) and Bôcher [78] that all the linear differential equations which occur in certain branches of Mathematical Physics are confluent forms of the special equation of this type in which the difference of the two exponents at each singularity is 12 ; a brief investigation of these forms will now be given. If we put βr = αr + 12 , (r = 1, 2, 3, 4) and write ζ in place of z, the last written equation becomes ( 4 ) 4 1 2 Õ Õ 1 − 2αr du α (α + ) Aζ + 2Bζ + C d2u r r 2 2 u = 0, + + + 2 2 4 dζ ζ − ar dζ (ζ − ar ) Î r=1 r=1 (ζ − ar ) r=1 where (on account of the condition µ2 − µ1 = 21 ) A=

4 Õ

!2 αr

r=1

−

4 Õ r=1

4

αr2 −

3Õ 3 αr + . 2 r=1 16

This differential equation is called the generalised Lamé equation. It is evident, on writing a1 = a2 throughout the equation, that the confluence of the two singularities a1 , a2 yields a singularity at which the exponents α, β are given by the equations α + β = 2(α1 + α2 ),

αβ = α1 α1 +

1 2

+ α2 α2 +

1 2

+ D,

where D=

Aa12 + 2Ba1 + C . (a1 − a3 )(a1 − a4 )

Therefore the exponent-difference at the confluent singularity is not 12 , but it may have any assigned value by suitable choice of B and C. In like manner, by the confluence of three or more singularities, we can obtain one irregular singularity. By suitable confluences of the five singularities at our disposal, we can obtain six types of equations, which may be classified according to (a) the number of their singularities with exponent-difference 12 , (b) the number of their other regular singularities, (c) the number of their irregular singularities, by means of the following scheme, which is easily seen to be exhaustive:

Linear Differential Equations

212

(I) (II) (III) (IV) (V) (VI)

(a) 3 2 1 0 1 0

(b) 1 0 2 1 0 0

(c) 0 1 0 1 1 1

Lamé Mathieu Legendre Bessel Weber, Hermite Stokes

For instance the arrangement (a) 3, (b) 0, (c) 1, is inadmissible as it would necessitate six initial singularities. The last equation of this type was considered by Stokes [609] in his researches on Diffraction; it is, however, easily transformed into a particular case of Bessel. These equations are usually known by the names of the mathematicians in the last column. Speaking generally, the later an equation comes in this scheme, the more simple are the properties of its solution. The solutions of (II)–(VI) are discussed in Chapters 15–19 of this work, and of (I) in Chapter 23. For properties of equations of type (I), see the works of Klein [376] and Forsyth [218]; also Todhunter [631]. The derivation of the standard forms of the equations from the generalised Lamé equation is indicated by the following examples: Example 10.6.1 Obtain Lamé’s equation ( 3 ) 1 Õ du {n(n + 1) ζ + h} u d2u 2 + − = 0, 2 3 dζ (ζ − ar ) dζ Î r=1 4 (ζ − ar ) r=1

(where h and n are constants) by taking a1 = a2 = a3 = a4 = 0,

8B = n(n + 1)a4,

4C = ha4,

and making a4 → ∞. Example 10.6.2 Obtain the equation ! 1 1 d2u du (a − 16q + 32qζ)u + 2 + 2 − = 0, 2 dζ ζ ζ − 1 dζ 4ζ(ζ − 1) (where a and q are constants) by taking a1 = 0, a2 = 1, and making a3 = a4 → ∞. Derive Mathieu’s equation (§19.1) d2u + (a + 16q cos 2z) u = 0 dz2 by the substitution ζ = cos2 z. Example 10.6.3 Obtain the equation ( ) 1 1 du 1 n(n + 1) m2 u d2u 2 + + + − = 0, 2 dζ ζ ζ − 1 dζ 4 ζ ζ − 1 ζ(ζ − 1) by taking a1 = a2 = 1,

a3 = a4 = 0,

a1 = a2 = a3 = 0,

a4 =

1 . 4

10.6 The differential equations of mathematical physics

213

Derive Legendre’s equation (§15.13 and §15.5) 2 du m2 2 d u + n(n + 1) − u=0 (1 − z ) 2 − 2z dz dz 1 − z2 by the substitution ζ = z−2 . Example 10.6.4 By taking a1 = a2 = 0, a1 = a2 = a3 = a4 = 0, and making a3 = a4 → ∞, obtain the equation ζ2

d2u du 1 +ζ + (ζ − n2 ) u = 0. 2 dζ dζ 4

Derive Bessel’s equation (§17.11) z2

d2u du +z + (z 2 − n2 )u = 0 2 dz dz

by the substitution ζ = z2 . Example 10.6.5 By taking a1 = 0, a1 = a2 = a3 = a4 = 0, and making a2 = a3 = a4 → ∞, obtain the equation 1 1 d 2 u 1 du 1 n + − ζ u = 0. + ζ 2+ dζ 2 dζ 4 2 4 Derive Weber’s equation (§16.5) d2u 1 1 2 + n+ − z u=0 dz 2 2 4 by the substitution ζ = z2 . Example 10.6.6 By taking ar = 0, and making ar → ∞ (r = 1, 2, 3, 4), obtain the equation d2u + (B1 ζ + C1 )u = 0. dζ 2 By taking u = (B1 ζ + C1 ) 2 3, 1

B1 ζ + C1 =

3 Bz 2 1

23

,

shew that 3 d3 1 2 3 = 0. z +z + z − dz 2 dz 9 2d

2

Example 10.6.7 Shew that the general form of the generalised Lamé equation is unaltered (i) by any homographic change of independent variable such that ∞ is a singular point of the transformed equation, (ii) by any change of dependent variable of the type u = (z − ar )λ 3. Example 10.6.8 Deduce from Example 10.6.7 that the various confluent forms of the generalised Lamé equation may always be reduced to the forms given in Examples 10.6.1– 10.6.6. (Note that a suitable homographic change of variable will transform any three distinct points into the points 0, 1, ∞.)

Linear Differential Equations

214

10.7 Linear differential equations with three singularities Let d2u du + p(z) + q(z)u = 0 (10.4) 2 dz dz have three, and only three singularities, a, b, c; let these points be regular points, the exponents thereat being α, α 0; β, β 0; γ, γ 0. The point at infinity is to be an ordinary point. Then p(z) is a rational function with simple poles at a, b, c, its residues at these poles being 1 − α − α 0, 1 − β − β 0, 1 − γ − γ 0; and as z → ∞, p(z) − 2z −1 is O(z −2 ). Therefore 1 − α − α0 1 − β − β0 1 − γ − γ 0 + + z−a z−b z−c 0 0 0 and α + α + β + β + γ + γ = 1. This relation must be satisfied by the exponents. In a similar manner 0 αα (a − b)(a − c) ββ 0(b − c)(b − a) γγ 0(c − a)(c − b) q(z) = + + z−a z−b z−c 1 × , (z − a)(z − b)(z − c) p(z) =

and hence the differential equation is 1 − α − α 0 1 − β − β 0 1 − γ − γ 0 du d2u + + + dz 2 z−a z−b z−c dz 0 0 αα (a − b)(a − c) ββ (b − c) (b − a) γγ 0(c − a)(c − b) + + + z−a z−b z−c u × = 0. (z − a)(z − b)(z − c) This equation was first given by Papperitz [515]. To express the fact that u satisfies an equation of this type (which will be called Riemann’s P-equation [556]; it will be seen from this memoir that, although Riemann did not apparently construct the equation, he must have inferred its existence from the hypergeometric equation) Riemann wrote a b c u=P α β γ z . α0 β0 γ 0 The singular points of the equation are placed in the first row with the corresponding exponents directly beneath them, and the independent variable is placed in the fourth column. Example 10.7.1 Shew that the hypergeometric equation z(1 − z)

d2u du + {c − (a + b + 1)z} − abu = 0 2 dz dz

is defined by the scheme ∞ 1 0 0 a 0 z . P 1 − c b c − a − b

10.7 Linear differential equations with three singularities

215

10.71 Transformations of Riemann’s P-equation The two transformations which are typified by the equations (I)

z−a z−b

k

z−c z−b

l

a P α α0

b c β γ β0 γ 0

a z =P α+k α0 + k

b β−k−l β0 − k − l

c γ+l γ0 + l

z ,

(II) a P α α0

b c β γ β0 γ 0

a1 z =P α α0

b1 β β0

c1 γ γ0

z1

(where z1 , a1 , b1 , c1 are derived from z, a, b, c by the same homographic transformation) are of great importance. They may be derived by direct transformation of the differential equation of Papperitz and Riemann by suitable changes in the dependent and independent variables respectively; but the truth of the results of the transformations may be seen intuitively when we consider that Riemann’s P-equation is determined uniquely by a knowledge of the three singularities and their exponents, and (I) that if a u=P α α0

b c β γ β0 γ 0

z ,

k l z−c z−a u satisfies a differential equation of the second order with the then u1 = z−b z−b same three singular points and exponents α + k, α 0 + k; β − k − l, β 0 − k − l; γ + l, γ 0 + l; and that the sum of the exponents is 1. Az1 + B Also (II) if we write z = , the equation in z1 is a linear equation of the second Cz1 + D order with singularities at the points derived from a, b, c by this homographic transformation, and exponents α, α 0; β, β 0; γ, γ 0 thereat.

10.72 The connexion of Riemann’s P-equation with the hypergeometric equation By means of the results of §10.71 it follows that a P α α0

b c β γ β0 γ 0

α z−a z = z−b α z−a = z−b

a 0 P α0 − α γ 0 z−c 0 P z−b α0 − α

z−c z−b

γ

where x=

(z − a) (c − b) . (z − b) (c − a)

b β+α+γ β0 + α + γ

c 0 γ0 − γ

∞ β+α+γ β0 + α + γ

1 0 γ0 − γ

z

x ,

Linear Differential Equations

216

Hence, by Example 10.7.1, the solution of Riemann’s P-equation can always be obtained in terms of the solution of the hypergeometric equation whose elements a, b, c, x are α+ β+γ, α + β 0 + γ, 1 + α − α 0, (z − a)(c − b)/(z − b)(c − a) respectively.

10.8 Linear differential equations with two singularities If, in §10.7, we make the point c an ordinary point, we must have 1 − γ − γ 0 = 0, γγ 0 = 0 αα 0(a − b)(a − c) ββ 0(b − c)(b − a) + must be divisible by z − c, in order that p(z) and and z−a z−b q(z) may be analytic at c. Hence α + α 0 + β + β 0 = 0, αα 0 = ββ 0, and the equation is d2u 1 − α − α 0 1 + α + α 0 du αα 0(a − b)2 u + + + = 0, dz 2 z−a z−b dz (z − a)2 (z − b)2 of which the solution is

z−a u=A z−b

α

z−a +B z−b

α0 ;

that is to say, the solution involves elementary functions only. When α = α 0, the solution is α α z−a z−a z−a + B1 log . u=A z−b z−b z−b

10.9 Miscellaneous examples Example 10.1 Shew that two solutions of the equation d2u + zu = 0 dz 2 are z − 121 z4 + · · · , and 1 − 61 z3 + · · · , and investigate the region of convergence of these series. Example 10.2 Obtain integrals of the equation d 2 u 1 − z2 + u = 0, dz 2 4z2 regular near z = 0, in the form z2 z4 u1 = z1/2 1 + + +··· , 16 1024 z 3/2 u2 = u1 log z − +··· . 16 Example 10.3 Shew that the equation d2u 1 1 2 + n + − z u=0 dz 2 2 4

10.9 Miscellaneous examples

217

has the solutions 2n + 1 2 4n2 + 4n + 3 4 z + z −··· , 4 96 2n + 1 3 4n2 + 4n + 7 5 z + z −··· , z− 12 480

1−

and that these series converge for all values of z. Example 10.4 (Klein) Shew that the equation ( n ) ( n ) n Õ 1 − αr − βr du Õ αr βr Õ d2u Dr + + + u = 0, dz2 z − ar dz (z − ar )2 r=1 z − ar r=1 r=1 where n Õ

(αr + βr ) = n − 2,

n Õ

Dr = 0,

(ar Dr + αr βr ) = 0,

r=1

r=1

r=1

n Õ

n Õ

(ar 2 Dr + 2ar αr βr ) = 0,

r=1

is the most general equation for which all points (including ∞), except a1, a2, . . . , an, are ordinary points, and the points ar are regular points with exponents αr , βr respectively. Example 10.5 (Riemann) Shew that, if β + γ + β 0 + γ 0 = 0 ∞ 1 P 0 β γ 1 β0 γ 0 2

1 2

then

−1 ∞ 1 z = P γ 2β γ γ 0 2β 0 γ 0 2

z .

The differential equation in each case is d 2 u 2z(1 − γ − γ 0) du γγ 0 4u 0 + + ββ + = 0. 2 2 2 2 dz z −1 dz z −1 z −1 Example 10.6 (Riemann) Shew that, if γ + γ 0 = 13 and if ω, ω2 are the complex cube roots of unity, then 2 1 ω ω 0 ∞ 1 3 P 0 0 γ z =P γ γ γ z . γ 0 γ 0 γ 0 1 1 γ0 3 3 The differential equation in each case is d2u 2z2 du 9γγ 0 zu + + = 0. dz 2 z3 − 1 dz (z3 − 1)2 Example 10.7 (Halm) Shew that the equation (1 − z2 )

d2u du − (2a + 1) z + n (n + 2a) u = 0 2 dz dz

Linear Differential Equations

218

is defined by the scheme ∞ 1 0 −n p 1 − a n + 2a 2

−1 0 1 − a 2

z ,

and that the equation (1 + ζ 2 )2

d2u + n(n + 2)u = 0 dζ 2

may be obtained from it by taking a = 1 and changing the independent variable. Example 10.8 (Cunningham) Discuss the solutions of the equation du 1 d2u + n+1+ m u =0 z 2 + (z + 1 + m) dz dz 2 valid near z = 0 and those valid near z = ∞. Example 10.9 (Curzon) Discuss the solutions of the equation d 2 u 2µ du du + − 2z + 2(ν − µ)u = 0 dz 2 z dz dz valid near z = 0 and those valid near z = ∞. Consider the following special cases: 3 1 µ=− ; (ii) µ = ; (iii) µ + ν = 3. 2 2 Example 10.10 (Lindemann; see §19.5) Prove that the equation (i)

d2u 1 du + (1 − 2z) + (az + b)u = 0 2 dz 2 dz has two particular integrals the product of which is a single-valued transcendental function. Under what circumstances are these two particular integrals coincident? If their product be F(z), prove that the particular integrals are ( ) ∫ z p dz , u1, u2 = F(z) exp ±C p F(z) z(1 − z) z(1 − z)

where C is a determinate constant. Example 10.11 (Math. Trip. 1912) Prove that the general linear differential equation of the third order, whose singularities are 0, 1, ∞, which has all its integrals regular near each singularity (the exponents at each singularity being 1, 1, −1), is 2 2 2 1 3 1 du d u d3u + + − 2− + dz3 z z − 1 dz 2 z z(z − 1) (z − 1)2 dz 1 3 cos2 a 3 sin2 a 1 + 3− 2 − + u = 0, z z (z − 1) z(z − 1)2 (z − 1)3 where a may have any constant value.

11 Integral Equations

11.1 Definition of an integral equation An integral equation is one which involves an unknown function under the sign of integration; and the process of determining the unknown function is called solving the equation. Except in the case of Fourier’s integral (§9.7) we practically always need continuous solutions of integral equations. The introduction of integral equations into analysis is due to Laplace (1782) who considered the equations ∫ ∫ xt e φ(t) dt, g(x) = t x−1 φ(t) dt f (x) = (where in each case φ represents the unknown function), in connexion with the solution of differential equations. The first integral equation of which a solution was obtained, was Fourier’s equation ∫ ∞ f (x) = cos(xt)φ(t) dt, −∞

of which, in certain circumstances, a solution is 1 ∫ 2 ∞ φ(x) = cos(ux) f (u) du, π 0 f (x) being an even function of x, since cos(xt) is an even function. Later, Abel [6] was led to an integral equation in connexion with a mechanical problem and obtained two solutions of it; after this, Liouville investigated an integral equation which arose in the course of his researches on differential equations and discovered an important method for solving integral equations 2 , which will be discussed in §11.4. In recent years, the subject of integral equations has become of some importance in various branches of Mathematics; such equations (in physical problems) frequently involve repeated integrals and the investigation of them naturally presents greater difficulties than do those elementary equations which will be treated in this chapter. To render the analysis as easy as possible, we shall suppose throughout that the constants a, b and the variables x, y, ξ are real and further that a ≤ x, y, ξ ≤ b; also that the given function, K(x, y), which occurs under the integral sign in the majority of equations considered, is a real function of x and y and either (i) it is a continuous function of both variables in the range 1 2

If this value of φ be substituted in the equation we obtain a result which is, effectively, that of §9.7. The numerical computation of solutions of integral equations has been investigated by Whittaker [677].

219

Integral Equations

220

(a ≤ x ≤ b, a ≤ y ≤ b), or (ii) it is a continuous function of both variables in the range a ≤ y ≤ x ≤ b and K(x, y) = 0 when y > x; in the latter case K(x, y) has its discontinuities regularly distributed, and in either case it is easily proved that, if f (y) is continuous when ∫ b a ≤ y ≤ b, f (y)K(x, y) dy is a continuous function of x when a ≤ x ≤ b. a

Bôcher [80] in his important work on integral equations, always considers the more general case in which K(x, y) has discontinuities regularly distributed, i.e. the discontinuities are of the nature described in Example 4.11. The reader will see from that example that the results of this chapter can almost all be generalised in this way. To make this chapter more simple we shall not consider such generalisations.

11.11 An algebraical lemma The algebraical result which will now be obtained is of great importance in Fredholm’s theory of integral equations. Let (x1, y1, z1 ), (x2, y2, z2 ), (x3, y3, z3 ) be three points at unit distance from the origin. The greatest (numerical) value of the volume of the parallelepiped, of which the lines joining the origin to these points are conterminous edges, is +1, the edges then being perpendicular. Therefore, if xr2 + yr2 + zr2 = 1 (r = 1, 2, 3), the upper and lower bounds of the determinant x1 y1 z 1 x2 y2 z 2 x3 y3 z 3 are ±1. A lemma due to Hadamard [265] generalises this result. Let a11 a12 · · · a1n a21 a22 · · · a2n D= . .. .. .. . . . . . an1 an2 · · · ann where amr is real and

n Í r=1

2 amr = 1 (m = 1, 2, . . . , n); let Amr be the cofactor of amr in D and

let ∆ be the determinant whose elements are Amr , so that, by a well-known theorem (see Burnside and Panton [111, vol. 2, p. 40]), ∆ = D n−1 . Since D is a continuous function of its elements, and is obviously bounded, the ordinary theory of maxima and minima is applicable, and if we consider variations in a1r (r = n ∂D Í 1, 2, . . . , n) only, D is stationary for such variations if δa1r = 0, where δa1r , . . . are r=1 ∂a1r n Í variations subject to the sole condition a1r δa1r = 0; therefore 3 r=1

A1r = 3

∂D = λa1r , ∂a1r

By the ordinary theory of undetermined multipliers.

11.2 Fredholm’s equation and its tentative solution

but

n Í r=1

a1r A1r = D, and so λ

Í

221

2 a1r = D; therefore A1r = Da1r .

Considering variations in the other elements of D, we see that D is stationary for variations in all elements when Amr = Damr (m = 1, 2, . . . , n; r = 1, 2, . . . , n). Consequently ∆ = D n · D, and so D n+1 = D n−1 . Hence the maximum and minimum values of D are ±1. Corollary 11.1.1 If amr be real and subject only to the condition |amr | < M, since n Õ amr 2 ≤ 1, n1/2 M r=1

we easily see that the maximum value of |D| is (n1/2 M)n = nn/2 M n .

11.2 Fredholm’s equation and its tentative solution Fredholm’s first paper on the subject appeared in [224]. His researches are also given in [225]. An important integral equation of a general type is ∫ b φ(x) = f (x) + λ K(x, ξ) φ(ξ) dξ, a

where f (x) is a given continuous function, λ is a parameter (in general complex) and K(x, ξ) is subject to the conditions laid down in §11.1. K(x, ξ) is called the nucleus, or the kernel of the equation. The reader will observe that if K(x, ξ) = 0 (ξ > x), the equation may be written ∫ x φ(x) = f (x) + λ K(x, ξ) φ(ξ) d ξ. a

This is called an equation with variable upper limit. This integral equation is known as Fredholm’s equation or the integral equation of the second kind (see §11.3). It was observed by Volterra that an equation of this type could be regarded as a limiting form of a system of linear equations. Fredholm’s investigation involved the tentative carrying out of a similar limiting process, and justifying it by the reasoning given below in §11.21. Hilbert [305] justified the limiting process directly. We now proceed to write down the system of linear equations in question, and shall then investigate Fredholm’s method of justifying the passage to the limit. The integral equation is the limiting form (when δ → 0) of the equation φ(x) = f (x) + λ

n Õ

K(x, xq ) φ(xq ) δ,

q=1

where xq − xq−1 = δ, x0 = a, xn = b. Since this equation is to be true when a ≤ x ≤ b, it is true when x takes the values x1, x2, . . . , xn ; and so −λδ

n Õ q=1

K(x p , xq ) φ(xq ) + φ(x p ) = f (x p )

(p = 1, 2, . . . , n).

Integral Equations

222

This system of equations for φ(x p ), (p = 1, 2, . . . , n) has a unique solution if the determinant formed by the coefficients of φ(x p ) does not vanish. This determinant is 1 − λδK(x1, x1 ) −λδK(x1, x2 ) · · · −λδK(x1, xn ) −λδK(x2, x1 ) 1 − λδK(x2, x2 ) · · · −λδK(x2, xn ) Dn (λ) = .. .. .. .. . . . . −λδK(xn, x1 ) −λδK(xn, x2 ) · · · 1 − λδK(xn, xn ) n n Õ λ2 Õ 2 K(x p , x p ) K(x p , xq ) =1−λ δK(x p , x p ) + δ 2! p,q=1 K(xq , x p ) K(xq , xq ) p=1 K(x p , x p ) K(x p , xq ) K(x p , xr ) n 3 Õ λ − δ3 K(xq , x p ) K(xq , xq ) K(xq , xr ) + · · · 3! p,q,r=1 K(xr , x p ) K(xr , xq ) K(xr , xr ) on expanding 4 in powers of λ. Making δ → 0, n → ∞, and writing the summations as integrations, we are thus led to consider the series ∫ b ∫ ∫ λ2 b b K(ξ1, ξ1 ) K(ξ1, ξ2 ) D(λ) = 1 − λ K(ξ1, ξ1 ) dξ1 + dξ1 dξ2 − · · · . 2! a a K(ξ2, ξ1 ) K(ξ2, ξ2 ) a Further, if Dn (xµ, xν ) is the cofactor of the term in Dn (λ) which involves K(xν , xµ ), the solution of the system of linear equations is φ(xµ ) =

f (x1 )Dn (xµ, x1 ) + f (x2 )Dn (xµ, x2 ) + · · · + f (xn )Dn (xµ, xn ) . Dn (λ)

Now it is easily seen that the appropriate limiting form to be considered in association with Dn (xµ, xµ ) is D(λ); also that, if µ , ν, ( n Õ K(xµ, xν ) K(xµ, x p ) Dn (xµ, xν ) = λδ K(xµ, xν ) − λδ K(x p , xν ) K(x p , x p ) p=1 K(xµ, xν ) K(xµ, x p ) K(xµ, xq ) n 1 2 2 Õ K(x , x ) K(x , x ) K(x , x ) + λ δ . − · · · p ν p p p q 2! p,q=1 K(xq , xν ) K(xq , x p ) K(xq , xq ) So that the limiting form for δ−1 D(xµ, xν ) to be considered is ∫ b K(xµ, xν ) K(xµ, ξ1 ) 2 D(xµ, xν ; λ) =λK(xµ, xν ) − λ K(ξ1, xν ) K(ξ1, ξ1 ) dξ1 a ∫ b ∫ b K(xµ, xν ) K(xµ, ξ1 ) K(xµ, ξ2 ) 1 K(ξ1, xν ) K(ξ1, ξ1 ) K(ξ1, ξ2 ) dξ1 dξ2 − · · · . + λ3 2! a a K(ξ2, xν ) K(ξ2, ξ1 ) K(ξ2, ξ2 ) (The law of formation of successive terms is obvious from those written down.) 4

The factorials appear because each determinant of s rows and columns occurs s! times as p, q, . . . take all the values 1, 2, . . . , n, whereas it appears only once in the original determinant for D n (λ).

11.2 Fredholm’s equation and its tentative solution

223

Consequently we are led to consider the possibility of the equation ∫ b 1 φ(x) = f (x) + D(x, ξ; λ) f (ξ) dξ D(λ) a giving the solution of the integral equation. Example 11.2.1 Shew that, in the case of the equation ∫ 1 φ(x) = x + λ x yφ(y) dy, 0

we have D(λ) = 1 − 31 λ, D(x, y; λ) = λx y and a solution is 3x . 3−λ Example 11.2.2 Shew that, in the case of the equation ∫ 1 φ(x) = x + λ (x y + y 2 )φ(y) dy, φ(x) =

0

we have D(λ) = 1 − 32 λ −

1 2 λ, 72 2 2

D(x, y; λ) = λ(xy + y ) + λ

1 xy 2 2

− 13 x y − 13 y 2 + 41 y ,

and obtain a solution of the equation.

11.21 Investigation of Fredholm’s solution So far the construction of the solution has been purely tentative; we now start ab initio and verify that we actually do get a solution of the equation; to do this we consider the two functions D(λ), D(x; yλ) arrived at in §11.2. ∞ Í an λn so that We write the series, by which D(λ) was defined in §11.2, in the form 1 + n! n=1

K (ξ1, ξ1 ) K (ξ1, ξ2 ) · · · ∫ b∫ b ∫ b K (ξ , ξ ) K (ξ , ξ ) · · · 2 1 2 2 an = (−1)n ··· .. .. .. a a a . . . K (ξn, ξ1 ) K (ξn, ξ2 ) · · ·

K (ξ1, ξn ) K (ξ2, ξn ) dξ1 dξ2 · · · dξn ; .. . K (ξn, ξn )

since K(x, y) is continuous and therefore bounded, we have |K(x, y)| < M, where M is independent of x and y; since K(x, y) is real, we may employ Hadamard’s lemma (§11.11) and we see at once that |an | < nn/2 M n (b − a)n . Write nn/2 M n (b − a)n = n!bn ; then lim

n→∞

n/2 bn+1 (b − a)M 1 = lim 1 + = 0, n→∞ (n + 1)1/2 bn n

Integral Equations

224

since (1 + 1/n)n → e. ∞ Í The series bn λ n is therefore absolutely convergent for all values of λ; and so (§2.34) the series 1 +

∞ Í

n=1

n=1

an λn n!

converges for all values of λ and therefore (§5.64) represents an integral

function of λ. ∞ Í Now write the series for D(x, y; λ) in the form

n=0

vn (x,y)λ n+1 . n!

Then, by Hadamard’s lemma

(§11.11), |vn−1 (x, y)| < nn/2 M n (b − a)n−1, ∞ vn (x, y) < cn where cn is independent of x and y and Í cn λ n+1 is absolutely and hence n! n=0 convergent. Therefore D(x, y; λ) is an integral function of λ and the series for D(x, y; λ) − λK(x, y) is a uniformly convergent (§3.34) series of continuous 5 functions of x and y when a ≤ x ≤ b, a ≤ y ≤ b. Now pick out the coefficient of K(x, y) in D(x, y; λ); and we get D(x, y; λ) = λD(λ)K(x, y) +

∞ Õ

(−1)n λ n+1

n=1

Q n (x, y) , n!

where 0 K(x, ξ1 ) K(x, ξ2 ) · · · ∫ b∫ b ∫ b K(ξ , y) K(ξ , ξ ) K(ξ , ξ ) · · · 1 1 1 1 2 Q n (x, y) = ··· .. .. .. .. a a a . . . . K(ξn, y) K(ξn, ξ1 ) K(ξn, ξ2 ) · · ·

K(x, ξn ) K(ξ1, ξn ) dξ1 · · · dξn . .. . K(ξn, ξn )

Expanding in minors of the first column, we get Q n (x, y) equal to the integral of the sum of n determinants; writing ξ1, ξ2, . . . , ξm−1, ξ, ξm, . . . , ξn−1 in place of ξ1, ξ2, . . . , ξn in the mth of them, we see that the integrals of all the determinants 6 are equal and so ∫ b∫ b ∫ b Q n (x, y) = −n ··· K(ξ, y)Pn dξ dξ1 · · · dξn−1, a

a

a

where K(x, ξ) K(x, ξ1 ) ··· K(ξ1, ξ) K(ξ , ξ ) · ·· 1 1 Pn = .. .. .. . . . K(ξn−1, ξ) K(ξn−1, ξ1 ) · · ·

K(x, ξn−1 ) K(ξ1, ξn−1 ) . .. . K(ξn−1, ξ)

It follows at once that D(x; y; λ) = λD(λ)K(x, y) + λ

∫

b

D(x, ξ; λ)K(ξ, y) dξ.

a 5

6

It is easy to verify that every term (except possibly the first) of the series for D(x, y; λ) is a continuous function under either hypothesis (i) or hypothesis (ii) of §11.1. The order of integration is immaterial (§4.3).

11.2 Fredholm’s equation and its tentative solution

225

Now take the equation φ(ξ) = f (ξ) + λ

∫

b

K(ξ, y)φ(y) dy, a

multiply by D(x, ξ; λ) and integrate, and we get ∫ b f (ξ)D(x, ξ; λ) dξ a ∫ b ∫ b∫ = φ(ξ)D(x, ξ; λ) dξ − λ a

a

b

D(x; ξ; λ)K(ξ, y)φ(y) dy dξ,

a

the integrations in the repeated integral being in either order. That is to say ∫ b f (ξ)D(x, ξ; λ) dξ a ∫ b ∫ b [D(x, y; λ) − λD(λ)K(x, y)] φ(y) dy = φ(ξ)D(x, ξ; λ) dξ − a a ∫ b = λD(λ) K(x, y)φ(y) dy a

= D(λ) [φ(x) − f (x)] , in virtue of the given equation. Therefore if D(λ) , 0 and if Fredholm’s equation has a solution it can be none other than ∫ b D(x, ξ; λ) dξ; φ(x) = f (x) + f (ξ) D(λ) a and, by actual substitution of this value of φ(x) in the integral equation, we see that it actually is a solution. This is, therefore, the unique continuous solution of the equation if D(λ) , 0. Corollary 11.2.1 If we put f (x) ≡ 0, the ‘homogeneous’ equation ∫ b φ(x) = λ K(x, ξ)φ(ξ) dξ a

has no continuous solution except φ(x) = 0, unless D(λ) = 0. By expanding the determinant involved in Q n (x, y) in minors of its first

Example 11.2.3 row, shew that

D(x, y; λ) = λD(λ)K(x, y) + λ

∫

b

K(x, ξ)D(ξ, y; λ) dξ.

a

Example 11.2.4 By using the formulae D(λ) = 1 +

∞ Õ an λ n n=1

n!

,

D(x, y; λ) = λD(λ)K(x, y) +

∞ Õ n=1

shew that ∫ a

b

D(ξ, ξ; λ) dξ = −λ

dD(λ) . dλ

(−1)n

λ n+1 Q n (x, y) , n!

Integral Equations

226

Example 11.2.5 If ( K(x, y) =

1 if 0 if

y ≤ x, y > x;

shew that D(λ) = exp{−(b − a)λ}. Example 11.2.6 Shew that, if K(x, y) = f1 (x) f2 (y), and if ∫ b f1 (x) f2 (x) dx = A, a

then D(λ) = 1 − Aλ,

D(x, y; λ) = λ f1 (x) f2 (y),

and the solution of the corresponding integral equation is ∫ b λ f1 (x) φ(x) = f (x) + f (ξ) f2 (ξ) dξ. 1 − Aλ a Example 11.2.7 Shew that, if K(x, y) = f1 (x)g1 (y) + f2 (x)g2 (y), then D(λ) and D(x, y; λ) are quadratic in λ, and, more generally, if K(x, y) =

n Õ

fm (x)gm (y),

m=1

then D(λ) and D(x, y, λ) are polynomials of degree n in λ.

11.22 Volterra’s reciprocal functions Two functions K(x, y), k(x, y; λ) are said to be reciprocal if they are bounded in the ranges a ≤ x, y ≤ b, if any discontinuities they may have are regularly distributed (§11.1, footnote on Bôcher’s work), and if ∫ b K(x, y) + k(x, y; λ) = λ k(x, ξ; λ)K(ξ, y) dξ. a

We observe that, since the right-hand side is continuous (by Example 4.11), the sum of two reciprocal functions is continuous. Also, a function K(x, y) can only have one reciprocal if D(λ) , 0; for if there were two, their difference k1 (x, y) would be a continuous solution of the homogeneous equation ∫ b k 1 (x, y; λ) = λ k1 (x, ξ; λ)K(ξ, y) dξ, a

(where x is to be regarded as a parameter), and by Corollary 11.2.1, the only continuous solution of this equation is zero. By the use of reciprocal functions, Volterra has obtained an elegant reciprocal relation between pairs of equations of Fredholm’s type.

11.2 Fredholm’s equation and its tentative solution

227

We first observe, from the relation b

∫

D(x, y; λ) = λD(λ)K(x, y) + λ

D(x, ξ, λ)K(ξ, y) dξ,

a

proved in §11.21, that the value of k(x, y; λ) is −

D(x, y; λ) , λD(λ)

and from Example 11.2.3, the equation k(x, y; λ) + K(x, y) = λ

b

∫

K(x, ξ)k(ξ, y; λ) dξ

a

is evidently true. Then, if we take the integral equation φ(x) = f (x) + λ

∫

b

K(x, ξ)φ(ξ) dξ,

a

when a ≤ x ≤ b, we have, on multiplying the equation ∫ b φ(ξ) = f (ξ) + λ K(ξ, ξ1 )φ(ξ1 ) dξ1 a

by k(x, ξ; λ) and integrating, ∫ b ∫ k(x, ξ; λ)φ(ξ) dξ = a

a

b

k(x, ξ; λ) f (ξ) dξ ∫ b∫ b +λ k(x, ξ, λ)K(ξ, ξ1 )φ(ξ1 ) dξ1 dξ. a

a

7

Reversing the order of integration in the repeated integral and making use of the relation defining reciprocal functions, we get ∫ b ∫ b k(x, ξ; λ)φ(ξ) dξ = k(x, ξ; λ) f (ξ) dξ a a ∫ b + {K(x, ξ) + k(x, ξ1 ; λ)}φ(ξ1 ) dξ1 a

and so λ

∫

b

k(x, ξ; λ) f (ξ) dξ = −λ

a

∫

b

K(x, ξ1 )φ(ξ1 ) dξ1

a

= −φ(x) + f (x). Hence f (x) = φ(x) + λ

∫

b

k(x, ξ; λ) f (ξ) dξ; similarly, from this equation we can derive

a

the equation φ(x) = f (x) + λ

∫

b

K(x, ξ)φ(ξ) dξ,

a 7

The reader will have no difficulty in extending the result of §4.3 to the integral under consideration.

Integral Equations

228

so that either of these equations with reciprocal nuclei may be regarded as the solution of the other.

11.23 Homogeneous integral equations The equation φ(x) = λ

b

∫

K(x, ξ)φ(ξ) dξ

(11.1)

a

is called a homogeneous integral equation. We have seen (Corollary 11.2.1) that the only continuous solution of the homogeneous equation, when D(λ) , 0, is φ(x) = 0. The roots of the equation D(λ) = 0 are therefore of considerable importance in the theory of the integral equation. They are called the characteristic numbers of the nucleus. It will now be shewn that, when D(λ) = 0, a solution which is not identically zero can be obtained. It will be proved in §11.51 that, if K(x, y) ≡ K(y, x), the equation D(λ) = 0 has at least one root. Let λ = λ0 be a root m times repeated of the equation D(λ) = 0. Since D(λ) is an integral function, we may expand it into the convergent series D(λ) = cm (λ − λ0 )m + cm+1 (λ − λ0 )m+1 + · · ·

(m > 0; cm , 0).

Similarly, since D(x, y; λ) is an integral function of λ, there exists a Taylor series of the form D(x, y; λ) =

g`+1 (x, y) g` (x, y) (λ − λ0 )` + (λ − λ0 )`+1 + · · · `! (` + 1)!

(` ≥ 0; g` . 0);

by §3.34 it is easily verified that the series defining gn (x, y), (n = `, ` + 1, . . .) converges absolutely and uniformly when a ≤ x ≤ b, a ≤ y ≤ b, and thence that the series for D(x, y; λ) converges absolutely and uniformly in the same domain of values of x and y. But, by Example 11.2.4, ∫ b dD(λ) D(ξ, ξ; λ) dξ = −λ ; dλ a now the right-hand side has a zero of order m − 1 at λ0, while the left-hand side has a zero of order at least `, and so we have m − 1 ≥ `. Substituting the series just given for D(λ) and D(x, y; λ) in the result of Example 11.2.3, viz. ∫ b D(x, y; λ) = λD(λ)K(x, y) + λ K(x, ξ)D(ξ, y; λ) dξ, a `

dividing by (λ − λ0 ) and making λ → λ0, we get ∫ b g` (x, y) = λ0 K(x, ξ)g` (ξ, y) dξ. a

Hence if y have any constant value, g` (x, y) satisfies the homogeneous integral equation, and any linear combination of such solutions, obtained by giving y various values, is a solution.

11.3 Integral equations of the first and second kinds

229

Corollary 11.2.2 The equation φ(x) = f (x) + λ0

∫

b

K(x, ξ)φ(ξ) dξ

a

has no solution or an infinite number. For, if φ(x) is a solution, so is φ(x) +

Í

cy g` (x, y),

y

where cy may be any function of y. Example 11.2.8 Shew that solutions of ∫ π φ(x) = λ cosn (x − ξ)φ(ξ) dξ −π

are φ(x) = cos(n − 2r)x, and φ(x) = sin(n − 2r)x; where r assumes all positive integral values (zero included) not exceeding 21 n. Example 11.2.9 Shew that φ(x) = λ

∫

π

cosn (x + ξ)φ(ξ) dξ

−π

has the same solutions as those given in Example 11.2.8, and shew that the corresponding values of λ give all the roots of D(λ) = 0.

11.3 Integral equations of the first and second kinds Fredholm’s equation is sometimes called an integral equation of the second kind; while the equation ∫ b f (x) = λ K(x, ξ)φ(ξ) dξ a

is called the integral equation of the first kind. In the case when K(x, ξ) = 0 if ξ > x, we may write the equations of the first and second kinds in the respective forms ∫ x f (x) = λ K(x, ξ)φ(ξ) dξ, a ∫ x φ(x) = f (x) + λ K(x, ξ)φ(ξ) dξ. a

These are described as equations with variable upper limits.

11.31 Volterra’s equation The equation of the first kind with variable upper limit is frequently known as Volterra’s equation. The problem of solving it has been reduced by that writer to the solution of Fredholm’s equation. Assuming that K(x, ξ) is a continuous function of both variables when ξ ≤ x, we have ∫ x f (x) = λ K(x, ξ)φ(ξ) dξ. a

Integral Equations

230

∂K exists and is The right-hand side has a differential coefficient (see Example 4.2.1) if ∂x continuous, and so ∫ x ∂K φ(ξ) dξ. f 0(x) = λK(x, x)φ(x) + λ a ∂x This is an equation of Fredholm’s type. If we denote its solution by φ(x), we get on integrating from a to x, ∫ x f (x) − f (a) = λ K(x, ξ)φ(ξ) dξ, a

and so the solution of the Fredholm’s equation gives a solution of Volterra’s equation if f (a) = 0. The solution of the equation of the first kind with constant upper limit can frequently be obtained in the form of a series. See Example 11.6. A solution valid under fewer restrictions is given by Bôcher.

11.4 The Liouville–Neumann method of successive substitutions This appears in Liouville [438]. K. Neumann’s investigations were later (1870); see [488]. A method of solving the equation φ(x) = f (x) + λ

∫

b

K(x, ξ)φ(ξ) dξ,

a

which is of historical importance, is due to Liouville. It consists in continually substituting the value of φ(x) given by the right-hand side in the expression φ(ξ) which occurs on the right-hand side. This procedure gives the series ∫ b S(x) = f (x) + λ K(x, ξ) f (ξ) dξ a

+

∞ Õ m=2

λ

m

∫ a

b

K(x, ξ1 )

∫ a

b

K(ξ1, ξ2 ) · · ·

∫

b

K(ξm−1, ξm ) f (ξm ) dξm · · · dξ1 .

a

Since |K(x, y)| and | f (x)| are bounded, let their upper bounds be M, M 0. Then the modulus of the general term of the series does not exceed |λ| m M m M 0(b − a)m . The series for S(x) therefore converges uniformly when |λ| < M −1 (b − a)−1 ; and, by actual substitution, it satisfies the integral equation. If K(x, y) = 0 when y > x, we find by induction that the modulus of the general term in the series for S(x) does not exceed |λ| m M m M 0(x − a)m /m! ≤ |λ| m M m M 0(b − a)m /m!, and so the series converges uniformly for all values of λ; and we infer that in this case Fredholm’s solution is an integral function of λ.

11.5 Symmetric nuclei

231

It is obvious from the form of the solution that when |λ| < M −1 (b − a)−1, the reciprocal function k(x, ξ; λ) may be written in the form k(x, ξ; λ) = − K(x, ξ) ∫ ∞ Õ m−1 − λ

b

K(x, ξ1 )

a

m=2

∫

b

∫

K(ξ1, ξ2 ) · · ·

a

b

K(ξm−1, ξ) dξm−1 dξm−2 · · · dξ1,

a

for with this definition of k(x, ξ; λ), we see that ∫ b S(x) = f (x) − λ k(x, ξ; λ) f (ξ) dξ, a

so that k(x, ξ; λ) is a reciprocal function, and by §11.22 there is only one reciprocal function if D(λ) , 0. Write ∫ b K(x, ξ) = K1 (x, ξ), K(x, ξ 0)Kn (ξ 0, ξ) dξ 0 = Kn+1 (x, ξ), a

and then we have −K(x, ξ; λ) =

∞ Õ

λ m Km+1 (x, ξ),

m=0

while b

∫

Km (x, ξ 0)Kn (ξ 0, ξ) dξ 0 = Km+n (x, ξ),

a

as may be seen at once on writing each side as an (m + n − 1)-tuple integral. The functions Km (x, ξ) are called iterated functions.

11.5 Symmetric nuclei Let K1 (x, y) ≡ K1 (y, x); then the nucleus K(x, y) is said to be symmetric. The iterated functions of such a nucleus are also symmetric, i.e. Kn (x, y) = Kn (y, x) for all values of n; for, if Kn (x, y) is symmetric, then Kn+1 (x, y) =

∫

b

K1 (x, ξ)Kn (ξ, y) dξ =

a

∫

b

K1 (ξ, x)Kn (y, ξ) dξ

a

=

∫

b

Kn (y, ξ)K1 (ξ, x) dξ

a

= Kn+1 (y, x), and the required result follows by induction. Also, none of the iterated functions are identically zero; for, if possible, let K p (x, y) ≡ 0; let n be chosen so that 2n−1 < p ≤ 2n , and, since K p (x, y) ≡ 0, it follows that K2n (x, y) ≡ 0,

Integral Equations

232

from the recurrence formula. But then 0 = K2n (x, x) =

b

∫

K2n−1 (x, ξ)K2n−1 (ξ, x) dξ

a

=

b

∫

{K2n−1 (x, ξ)}2 dξ,

a

and so K2n−1 (x, ξ) ≡ 0; continuing this argument, we find ultimately that K1 (x, y) ≡ 0, and the integral equation is trivial.

11.51 Schmidt’s theorem that, if the nucleus is symmetric, the equation D(λ) = 0 has at least one root The proof given is due to Kneser [380]. To prove this theorem, let Un =

∫

b

Kn (x, x) dx, a

so that, when |λ| < M −1 (b − a)−1 , we have, by Example 11.2.4 and §11.4, ∞

−

1 dD(λ) Õ Un λ n−1 . = D(λ) dλ n=1

Now since ∫ a

b

∫

b

(µKn+1 (x, ξ) + Kn−1 (x, ξ))2 dξ dx ≥ 0

a

for all real values of µ, we have µ2U2n+2 + 2µU2n + U2n−2 ≥ 0, and so U2n+2U2n−2 ≥ U2n2 , U2n−2 > 0. Therefore U2, U4, . . . are all positive, and if U4 /U2 = ν, it follows, by induction from the inequality U2n+2U2n−2 ≥ U2n2 , that U2n+2 /U2n ≥ ν n . Therefore when |λ2 | ≥ ν −1 , ∞ Í 1 dD(λ) the terms of Un λ n−1 do not tend to zero; and so, by §5.4, the function has a D(λ) dλ n=1 1 singularity inside or on the circle |λ| = ν − 2 ; but since D(λ) is an integral function, the only 1 dD(λ) possible singularities of are at zeros of D(λ); therefore D(λ) has a zero inside D(λ) dλ 1 or on the circle |λ| = ν − 2 . Note By §11.21, D(λ) is either an integral function or else a mere polynomial; in the latter case, it has a zero by Example 6.3.1; the point of the theorem is that in the former case D(λ) 2 cannot be such a function as eλ , which has no zeros.

11.6 Orthogonal functions

233

11.6 Orthogonal functions The real continuous functions φ1 (x), φ2 (x), . . . are said to be orthogonal and normal 8 for the range (a, b) if ( ∫ b 0 (m , n), φm (x)φn (x) dx = 1 (m = n). a If we are given n real continuous linearly independent functions u1 (x), . . . , un (x), we can form n linear combinations of them which are orthogonal. For suppose we can construct m − 1 orthogonal functions φ1, . . . , φm−1 such that φ p is a linear combination of u1, u2, . . . , u p (where p = 1, 2, . . . , m − 1); we shall now shew how to construct the function φm such that φ1, φ2, . . . , φm are all normal and orthogonal. Let 1 φm (x) = c1,m φ1 (x) + c2,m φ2 (x) + · · · + cm−1 φm−1 (x) + um (x), so that 1 φm is a function of u1, u2, . . . , um . Then, multiplying by φ p and integrating, ∫ b ∫ b φ (x)φ (x) dx = c + um (x)φ p (x) dx (p < m). 1 m p p,m a

∫ Hence

a

b 1 φ m (x)φ p (x) dx = 0 if c p,m = −

a

b

∫

um (x)φ p (x) dx; a function 1 φm (x), or-

a

thogonal to φ1 (x), φ2 (x), . . . , φm−1 (x), is therefore constructed. ∫ b 2 {1 φm (x)}2 dx = 1; and take φm (x) = α (1 φm (x)). Then Now choose α so that α a

∫

(

b

φm (x)φ p (x) dx =

a

0 (p < m), 1 (p = m).

We can thus obtain the functions φ1, φ2, . . . in order. The members of a finite set of orthogonal functions are linearly independent. For, if α1 φ1 (x) + α2 φ2 (x) + · · · + αn φn (x) ≡ 0, we should get, on multiplying by φ p (x) and integrating, αp = 0; therefore all the coefficients αp vanish and the relation is nugatory. It is obvious that π −1/2 cos mx, π −1/2 sin mx form a set of normal orthogonal functions for the range (−π, π). Example 11.6.1 From the functions 1, x, x 2, . . . construct the following set of functions which are orthogonal (but not normal) for the range (−1, 1); 1 3 3 1, x, x 2, − x 3 − x, x 4 − 2x 2 + , . . . . 3 5 35 Example 11.6.2 From the functions 1, x, x 2, . . . construct a set of functions f0 (x), f1 (x), f2 (x), . . . which are orthogonal (but not normal) for the range (a, b); where dn {(x − a)n (x − b)n } . dx n A similar investigation is given in §15.14. fn (x) =

8

They are said to be orthogonal if the first equation only is satisfied; the systematic study of such functions is due to Murphy [481, 483].

Integral Equations

234

11.61 The connexion of orthogonal functions with homogeneous integral equations Consider the homogeneous equation φ(x) = λ0

b

∫

φ(ξ)K(x, ξ) dξ,

a

where λ0 is a real characteristic number for K(x, ξ). It will be seen immediately that the characteristic numbers of a symmetric nucleus are all real. We have already seen how solutions of it may be constructed; let n linearly independent solutions be taken and construct from them n orthogonal and normal functions φ1, φ2, . . . , φn . Then, since the functions φm are orthogonal and normal, #2 ∫ b "Õ ∫ b n φm (y) K(x, ξ)φm (ξ) dξ dy a

a

m=1 n ∫ Õ

=

b

φm (y)

a

m=1

b

∫

K(x, ξ)φm (ξ) dξ

2 dy,

a

and it is easily seen that the expression on the right may be written in the form 2 n ∫ b Õ K(x, ξ)φm (ξ) dξ a

m=1

on performing the integration with regard to y; and this is the same as ∫ b n ∫ b Õ K(x, y)φm (y) dy K(x, ξ)φm (ξ) dξ. a

m=1

a

Therefore, if we write K for K(x, y) and Λ for ∫ b n Õ φm (y) K(x, ξ)φm (ξ) dξ, a

m=1

∫ we have

b

Λ2 dy =

∫

b

KΛ dy, and so

a

a b

∫

Λ dy =

b

∫

2

∫

b

2

(K − Λ)2 dy.

K dy −

a

a

a

Therefore ∫ a

b

"

n Õ φm (y)φm (x) λ0 m=1

and so λ0−2

n Õ

#2

∫ dy ≤

b

[K(x, y)]2 dy,

a

[φm (x)]2 ≤

∫

b

[K(x, y)]2 dy.

a

m=1

Integrating, we get n ≤ λ0

∫

b

∫

2 a

a

b

[K(x, y)]2 dy dx.

(11.2)

11.6 Orthogonal functions

235

This formula gives an upper limit to the number, n, of orthogonal functions corresponding to any characteristic number λ0 . These n orthogonal functions are called characteristic functions (or auto-functions) corresponding to λ0 . Now let φ(0) (x), φ(1) (x) be characteristic functions corresponding to different characteristic numbers λ0 , λ1 . Then ∫ b (0) (1) φ (x)φ (x) = λ1 K(x, ξ)φ(0) (x)φ(1) (ξ) dξ, a

and so b

∫

φ (x)φ (x) dx = λ1 (0)

(1)

b

∫

a

b

∫

a

K(x, ξ)φ(0) (x)φ(1) (ξ) dξ dx

(11.3)

a

and similarly ∫

b

φ (x)φ (x) dx =λ0 (0)

(1)

a

∫

b

a

=λ0

∫ a

∫

b

K(x, ξ)φ(0) (ξ)φ(1) (x) dξ dx

a b

∫

b

K(ξ, x)φ(0) (x)φ(1) (ξ) dx dξ,

(11.4)

a

on interchanging x and ξ. We infer from (11.3) and (11.4) that if λ1 , λ0 , and if K(x, ξ) = K(ξ, x), ∫ b φ(0) (x)φ(1) (x) dx = 0, a

and so the functions φ(0) (x), φ(1) (x) are mutually orthogonal. If therefore the nucleus be symmetric and if, corresponding to each characteristic number, we construct the complete system of orthogonal functions, all the functions so obtained will be orthogonal. Further, if the nucleus be symmetric all the characteristic numbers are real; for if λ0 , λ1 be conjugate complex roots and if 9 u0 (x) = v(x) + iw(x) be a solution for the characteristic number λ0 , then u1 (x) = v(x) − iw(x) is a solution for the characteristic number λ1 ; replacing φ(0) (x), φ(1) (x) in the equation ∫ b φ(0) (x)φ(1) (x) dx = 0 a

by v(x) + iw(x), v(x) − iw(x) (which is obviously permissible), we get ∫ b (v(x))2 + (w(x))2 dx = 0, a

which implies v(x) ≡ w(x) ≡ 0, so that the integral equation has no solution except zero corresponding to the characteristic numbers λ0 , λ1 ; this is contrary to §11.23; hence, if the nucleus be symmetric, the characteristic numbers are real. 9

v(x) and w(x) being real.

236

Integral Equations

11.7 The development of a symmetric nucleus This investigation is due to Schmidt, the result to Hilbert. Let φ1 (x), φ2 (x), φ3 (x), . . . be a complete set of orthogonal functions satisfying the homogeneous integral equation with symmetric nucleus ∫ b φ(x) = λ K(x, ξ)φ(ξ) dξ, a

the corresponding characteristic numbers being 10 λ1, λ2, λ3, . . .. ∞ φ (x) φ (y) Í n n is uniformly convergent when a ≤ x ≤ b, Now suppose 11 that the series λ n=1 n a ≤ y ≤ b. Then it will be shewn that ∞ Õ φn (x)φn (y) . λn n=1

K(x, y) = For consider the symmetric nucleus

H(x, y) = K(x, y) −

∞ Õ φn (x)φn (y) . λn n=1

If this nucleus is not identically zero, it will possess (§11.51) at least one characteristic number µ. Let ψ(x) be any solution of the equation ∫ b ψ(x) = µ H(x, ξ)ψ(ξ) dξ, a

which does not vanish identically. Multiply by φn (x) and integrate and we get ) ∫ b∫ b( ∫ b ∞ Õ φm (x)φm (ξ) ψ(ξ)φn (x) dx dξ; ψ(x)φn (x) dx = µ K(x, ξ) − λm a a a m=1 since the series converges uniformly, we may integrate term by term and get ∫ b ∫ b ∫ b µ µ ψ(x)φn (x) dx = ψ(ξ)φn (ξ) dξ − φn (ξ)ψ(ξ) dξ = 0. λn a λn a a Therefore ψ(x) is orthogonal to φ1 (x), φ2 (x), . . . ; and so taking the equation ) ∫ b( ∞ Õ φn (x)φn (ξ) ψ(ξ) dξ, ψ(x) = µ K(x, ξ) − λn a n=1 we have ψ(x) = µ

∫

b

K(x, ξ)ψ(ξ) dξ.

a 10 11

These numbers are not all different if there is more than one orthogonal function to each characteristic number. The supposition is, of course, a matter for verification with any particular equation.

11.7 The development of a symmetric nucleus

237

Therefore µ is a characteristic number of K(x, y), and so ψ(x) must be a linear combination of the (finite number of) functions φn (x) corresponding to this number; let Õ ψ(x) = am φm (x). m

Multiply by φm (x) and integrate; then since ψ(x) is orthogonal to all the functions φn (x), we see that am = 0, so, contrary to hypothesis, ψ(x) ≡ 0. The contradiction implies that the nucleus H(x, y) must be identically zero; that is to say, K(x, y) can be expanded in the given series, if it is uniformly convergent. Example 11.7.1 Shew that, if λ0 be a characteristic number, the equation ∫ b φ(x) = f (x) + λ0 K(x, ξ)φ(ξ) dξ a

certainly has no solution when the nucleus is symmetric, unless f (x) is orthogonal to all the characteristic functions corresponding to λ0 .

11.71 The solution of Fredholm’s equation by a series Retaining the notation of §11.7, consider the integral equation ∫ b Φ(x) = f (x) + λ K(x, ξ)Φ(ξ) dξ, a

where K(x, ξ) is symmetric. If we assume that Φ(ξ) can be expanded into a uniformly convergent series

∞ Í

an φn (ξ),

n=1

we have ∞ Õ

an φn (x) = f (x) +

n=1

∞ Õ λ an φn (x), λ n=1 n

so that f (x) can be expanded in the series ∞ Õ

an

n=1

λn − λ φn (x). λn

Hence if the function f (x) can be expanded into the convergent series

∞ Í

bn φn (x), then

n=1

the series ∞ Õ bn λn φn (x), λ −λ n=1 n

if it converges uniformly in the range (a, b), is the solution of Fredholm’s equation. ∞ Í To determine the coefficients bn we observe that bn φn (x) converges uniformly by §3.35; n=1

then, multiplying by φn (x) and integrating, we get ∫ b bn = φn (x) f (x) dx. a

Integral Equations

238

Since the numbers λn are all real we may arrange them in two sets, one negative the other positive, the members in each set being in order of magnitude; then, when |λn | > λ, it is evident that λn /(λn − λ) is a monotonic sequence in the case of either set.

11.8 Solution of Abel’s integral equation This equation is of the form ∫ f (x) =

x

a

u(ξ) dξ (x − ξ)µ

(0 < µ < 1,

a ≤ x ≤ b),

where f 0(x) is∫continuous and f (a) = 0; we proceed to find a continuous solution u(x). x

Let φ(x) =

u(ξ) dξ, and take the formula (this follows from Example 6.2.14, by writing a

(z − x)/(x − ξ) in place of x) π = sin µπ

∫ ξ

z

(z −

dx , (x − ξ)µ

x)1−µ

multiply by u(ξ) and integrate, and we get, on using Dirichlet’s formula (Corollary 4.5.1), ∫ z ∫ z π u(ξ) dx {φ(z) − φ(a)} = dξ 1−µ (x − ξ)µ sin µπ a ξ (z − x) ∫ z ∫ z u(ξ) dξ = dx 1−µ (x − ξ)µ a (z − x) ∫a z f (x) dx = . 1−µ a (z − x) Since the original expression has a continuous derivate, so has the final one; therefore the continuous solution, if it exists, can be none other than ∫ z sin µπ d f (x) dx u(z) = ; π dz a (z − x)1−µ and it can be verified by substitution 12 that this function actually is a solution.

11.81 Schlömilch’s integral equation This comes from [580]. The reader will easily see that this is reducible to a case of Volterra’s equation with a discontinuous nucleus. Let f (x) have a continuous differential coefficient when −π ≤ x ≤ π. Then the equation ∫ 2 π/2 f (x) = φ(x sin θ) dθ π 0 has one solution with a continuous differential coefficient when −π ≤ x ≤ π, namely ∫ π/2 φ(x) = f (0) + x f 0(x sin θ) dθ. 0 12

For the details we refer to Bôcher’s tract [80].

11.9 Miscellaneous examples

239

From §4.2 it follows that 2 f (x) = π 0

∫

π/2

sin θ φ 0(x sin θ) dθ

0

(so that we have φ(0) = f (0), φ 0(0) = π2 f 0(0)). Write x sin ψ for x, and we have on multiplying by x and integrating ∫ π/2 ∫ ∫ π/2 2x π/2 0 0 x sin θ φ (x sin θ sin ψ) dθ dψ. f (x sin ψ) dψ = π 0 0 0 Change the order of integration in the repeated integral (§4.3) and take a new variable χ in place of ψ, defined by the equation sin χ = sin θ sin ψ. Then ∫ θ 0 ∫ π/2 ∫ 2x π/2 φ (x sin χ) cos χdχ 0 x dθ. f (x sin ψ) dψ = π 0 cos ψ 0 0 Changing the order of integration again (§4.51), ) ∫ π/2 ∫ π/2 (∫ π/2 0 φ (x sin χ) cos χ sin θ 2x dθ dχ. x f 0(x sin ψ) dψ = p π 0 0 χ sin2 θ − sin2 χ But π/2

∫ χ

sin θ dθ

cos θ = − arcsin p cos χ cos2 χ − cos2 θ

π/2

=

χ

π , 2

and so ∫

π/2

x

f (x sin ψ) dψ = x 0

0

π/2

∫

φ 0(x sin χ) cos χ dχ

0

= φ(x) − φ(0).

Since φ(0) = f (0), we must have φ(x) = f (0) + x

π/2

∫

f 0(x sin ψ) dψ;

0

and it can be verified by substitution that this function actually is a solution.

11.9 Miscellaneous examples Example 11.1 (Abel) Shew that if the time of descent of a particle down a smooth curve to its lowest point is independent of the starting-point (the particle starting from rest) the curve is a cycloid. Example 11.2 Shew that, if f (x) is continuous, the solution of ∫ ∞ φ(x) = f (x) + λ cos(2xs)φ(s) ds 0

is φ(x) =

1 1 − 41 λ2 π

f (x) + λ

∫ 0

∞

f (s) cos(2xs) ds ,

Integral Equations

240

assuming the legitimacy of a certain change of order of integration. Example 11.3 (A. Milne) Shew that the Weber–Hermite functions n 1 2 d − 21 x 2 e Dn (x) = (−1)n e 4 x dx n satisfy ∫ ∞ 1 φ(x) = λ e 2 isx φ(s) ds −∞

for the characteristic values of λ. Example 11.4 (Whittaker; see §19.21, [673]) period 2π) of the differential equation

Shew that even periodic solutions (with

d 2 φ(x) + (a2 + k 2 cos2 x)φ(x) = 0 dx 2 satisfy the integral equation φ(x) = λ

∫

π

ek cos x cos s φ(s) ds.

−π

Example 11.5 Shew that the characteristic functions of the equation ∫ π 1 1 φ(x) = λ (x − y)2 − |x − y| φ(y) dy 2 −π 4π are φ(x) = cos mx, sin mx, where λ = m2 and m is any integer. Example 11.6 (Bôcher) Shew that φ(x) =

∫

x

ξ x−ξ φ(ξ) dξ

0

has the discontinuous solution φ(x) = k x x−1 . Example 11.7 Shew that a solution of the integral equation with a symmetric nucleus ∫ b ∞ Õ f (x) = an λn φn (x), K(x, ξ)φ(ξ) dξ is φ(x) = a

n=1

provided that this series converges uniformly, where λn , φn (x) are the characteristic numbers ∞ Í and functions of K(x, ξ) and an φn (x) is the expansion of f (x). n=1

Example 11.8 Shew that, if |h| < 1, the characteristic functions of the equation ∫ π 1 − h2 λ φ(ξ) dξ φ(x) = 2π −π 1 − 2h cos(ξ − x) + h2 are 1, cos mx, sin mx, the corresponding characteristic numbers being 1, 1/h m , 1/h m , where m takes all positive integral values.

Part II The Transcendental Functions

241

12 The Gamma-Function

12.1 Definitions of the Gamma-function. The Weierstrassian product Historically, the Gamma-function Γ(z) was first defined by as the limit of a product ∫ Euler ∞ (§12.11) from which can be derived the infinite integral t z−1 e−t dt. The notation Γ(z) 0

was introduced by Legendre in 1814. But in developing the theory of the function, it is more convenient to define it by means of an infinite product of Weierstrass’ canonical form. Consider the product ∞ n Ö z − nz o , e zeγz 1+ n n=1 where γ = lim

m→∞

1 1

+

1 2

+···+

1 m

− log m ∼ 0.5772157 · · · .

(12.1)

(The constant γ is known as Euler’s constant, or the Euler–Mascheroni constant.) To prove that it exists we observe that, if ∫ 1 t dt 1 n+1 un = = − log , n (n + t) n n 0 ∫ 1 ∞ Í dt 1 un converges, and un is positive and less than = ; therefore 2 n2 n=1 0 n ( m ) ∞ Õ Õ 1 1 m+1 1 lim un + log un . + + · · · + − log m = lim = m→∞ 1 m→∞ 2 m m n=1 n=1 The value of γ has been calculated by J. C. Adams to 260 places of decimals [10]. The product under consideration represents an analytic function of z, for all values of z; for, if N be an integer such that |z| ≤ 12 N, we have 1 if n > N, z z 1 z2 1 z3 − = − + − · · · log 1 + n n 2 n2 3 n3 2 2 |z| z z ≤ 2 1 + + 2 + ··· n n n 2 1 N 1 1 1 N2 ≤ 1 + + + · · · ≤ . 2 2 4 n 2 2 2 n2 1

Taking the principal value of log(1 + z/n).

243

The Gamma-Function

244

Since the series

∞ Í

N2 converges, it follows that, when |z| ≤ 2 n=N +1 2n ∞ Õ

1 2

N,

{log (1 + z/n) − z/n}

n=N +1

is an absolutely and uniformly convergent series of analytic functions, and so it is an analytic function (§5.3); consequently its exponential ∞ Ö

(1 + z/n) e−z/n

n=N +1

is an analytic function and zeγz

∞ Î n=1

(1 + z/n) e−z/n is an analytic function when |z| ≤

1 2

N,

where N is any integer; that is to say, the product is analytic for all finite values of z. The Gamma-function was defined by Weierstrass [659]. This formula for Γ(z) had been obtained from Euler’s formula (§12.11) in 1848 by F. W. Newman [493] by the equation ∞ n Ö 1 z −z/n o = zeγz 1+ e ; Γ(z) n n=1

from this equation it is apparent that Γ(z) is analytic except at the points z = 0, −1, −2, where it has simple poles. Note Proofs have been published by Hölder [325], Moore [472] and Barnes [44] of a theorem known to Weierstrass that the Gamma-function does not satisfy any differential equation with rational coefficients. Example 12.1.1 Prove that Γ(1) = 1,

Γ 0(1) = −γ,

where γ is Euler’s constant. Hint. Justify differentiating logarithmically the equation ∞ n Ö 1 z −z/n o 1+ = zeγz e Γ(z) n 1

by §4.7, and put z = 1 after the differentiations have been performed. Example 12.1.2 Shew that 1 1 1 1+ + +···+ = 2 3 n

∫ 0

1

1 − (1 − t)n dt, t

and hence that Euler’s constant γ is given by ∫ 1 ∫ n t n dt t n dt 1− 1− lim − 1− . n→∞ n t n t 0 1 The reader will see later (Example 11.6.2) that this limit may be written ∫ 1 ∫ ∞ −t e dt −t dt (1 − e ) − . t t 0 1

12.1 Definitions of the Gamma-function

245

Example 12.1.3 Shew that ∞ Ö

1−

n=1

x eγx Γ(z + 1) . e x/n = z+n Γ(z − x + 1)

12.11 Euler’s formula for the Gamma-function By the definition of an infinite product we have " # m n h i Ö z − nz o 1 1 1+ 21 +···+ m −log m)z ( = z lim e e lim 1+ m→∞ m→∞ Γ(z) n n=1 " # m n zo Ö z 1 1 = z lim e(1+ 2 +···+ m −log m)z 1+ e− n m→∞ n n=1 # " m Ö z 1+ = z lim m−z m→∞ n n=1 " m−1 # m −z Ö Ö 1 z 1+ 1+ = z lim m→∞ n n n=1 n=1 " m −z z# Ö z 1 1 = z lim 1+ 1+ 1+ . m→∞ n n m n=1 Hence ∞

1Ö Γ(z) = z n=1

1 1+ n

z

z −1 . 1+ n

(12.2)

This formula is due to Euler. It was given in 1729 in a letter to Goldbach, printed in Fuss [231]. It is valid except when z = 0, −1, −2, . . . . Example 12.1.4 (Euler) Prove that Γ(z) = lim

n→∞

1 · 2 · · · · · (n − 1) z n . z (z + 1) · · · (z + n − 1)

12.12 The difference equation satisfied by the Gamma-function We shall now shew that the function Γ(z) satisfies the difference equation Γ(z + 1) = zΓ(z).

The Gamma-Function

246

For, by Euler’s formula, if z is not a negative integer " z+1 # " z # −1 m m Ö Ö 1 + n1 1 + n1 1 1 Γ(z + 1) = lim lim Γ(z) z + 1 m→∞ n=1 1 + z+1 z m→∞ n=1 1 + nz n ( ) m Ö 1 + n1 (z + n) z = lim z + 1 m→∞ n=1 z+n+1 m+1 = z. z+m+1 This is one of the most important properties of the Gamma-function. Since Γ(1) = 1, it follows that, if z is a positive integer, Γ(z) = (z − 1)!. = z lim

m→∞

Example 12.1.5 Prove that 1 1 1 + + +··· Γ (z + 1) Γ (z + 2) Γ (z + 3) e 1 1 1 1 1 = − + −··· . Γ (z) z 1 ! z + 1 2 ! z + 2 Hint. Consider the expression 1 1 1 1 + + +···+ . z z (z + 1) z (z + 1)(z + 2) z(z + 1) · · · (z + m) m Í an It can be expressed in partial fractions in the form , where n=0 z + n ( ) ∞ Õ 1 (−1)n 1 (−1)n 1 1 1+ = an = + +···+ e− . n! 1! 2! (m − n) ! n! r! r=m−n+1 Noting that

∞ Í r=m−n+1

displayst yle r!1

0, the integral is equal to Γ(z). Denoting the real part of z by x, we have x > 0. Now, if 2 ∫ n t n z−1 t dt, Π(z, n) = 1− n 0 ∫ 1 we have Π(z, n) = nz (1 − τ)n τ z−1 dτ, if we write t = nτ; it is easily shewn by repeated 0

integrations by parts that, when x > 0 and n is a positive integer, 1 ∫ 1 ∫ n 1 1 2 n z−1 n (1 − τ) τ dτ = τ (1 − τ) + (1 − τ)n−1 τ 2 dτ z z 0 0 0 .. . ∫ 1 n(n − 1) · · · 1 τ z+n−1 dτ, = z(z + 1) · · · (z + n − 1) 0 and so Π(z, n) =

1 · 2···n nz . Hence, by Example 12.1.4, Π(z, n) → Γ(z) as n → ∞. z(z + 1) · · · (z + n)

Consequently Γ(z) = lim

∫

n→∞

And so, if Γ1 (z) =

∫

0

n

1−

t n z−1 t dt. n

∞

e−t t z−1 dt, we have

0

Γ1 (z) − Γ(z) = lim

n→∞

∫

e t

Now lim

n→∞

∞ −t z−1

n

n

∫

dt = 0, since

0

∫

∫ ∞ t n z−1 t dt + e−t − 1 − e−t t z−1 dt . n n

∞

e−t t z−1 dt converges. To shew that zero is the limit of

0

the first of the two integrals in the formula for Γ1 (z) − Γ(z) we observe that t n ≤ n−1 t 2 e−t . 0 ≤ e−t − 1 − n Hint. To establish these inequalities, we proceed as follows: when 0 ≤ y < 1, 1 + y ≤ ey ≤ (1 − y)−1, from the series for ey and (1 − y)−1 . Writing t/n for y, we have t n t −n ≥ e−t ≥ 1 − , 1+ n n 2

The many-valued function t z−1 is made precise by the equation t z−1 = e(z−1) log t , log t being purely real.

12.2 Euler’s expression of Γ(z) as an infinite integral

251

and so n t n t n t2 −t −t −t = e 1−e 1− 0 ≤ e − 1− ≤ e 1− 1− 2 . n n n −t

Now, if 0 ≤ a ≤ 1, (1 − a)n ≥ 1 − na by induction when na < 1 and obviously when na ≥ 1; and, writing t 2 /n2 for a, we get n t2 t2 1− 1− 2 ≤ n n and so

t n 0 ≤ e−t − 1 − ≤ e−t t 2 /n, n which is the required result. This analysis is a modification of that given by Schlömilch [583, Vol. 2, p. 243]. A simple method of obtaining a less precise inequality (which is sufficient for the object required) is given by Bromwich [102, p. 459]. From the inequalities, it follows at once that ∫ n ∫ n t n z−1 −t e − 1− t dt ≤ n−1 e−t t x+1 dt n 0 0 ∫ ∞ −1 0 and Re(s) > 0, ∫ ∞ Γ(s) e−zx x s−1 dx = s . z 0 Example 12.2.3 Prove that, if Re(z) > 0 and Re(s) > 1, 1 1 1 1 + + +··· = (z + 1)s (z + 2)s (z + 3)s Γ(s)

∫

∞

0

Example 12.2.4 From Example 12.1.2 by using the inequality t n t 2 e−t ≤ 0 ≤ e−t − 1 − n n

e−xz x s−1 dx ex − 1

252

The Gamma-Function

deduce that γ=

∫ 0

1

1 − e−t − e−1/t dt. t

12.21 Extension of the infinite integral to the case in which the argument of the Gamma-function is negative The formula of the last article is no longer applicable when the real part of z is negative. Cauchy [123, volume 2, pp. 91–92] and Saalschütz [568, 569] have shewn, however, that, for negative arguments, an analogous theorem exists. This can be obtained in the following way. Consider the function ∫ ∞ k t2 −t k+1 t z−1 Γ2 (z) = e −1+t− + · · · + (−1) dt, t 2! k! 0 where k is the integer so chosen that −k > x > −k − 1, x being the real part of z. By partial integration we have, when z < −1, ∞ z k t2 t −t k+1 t e −1+t− Γ2 (z) = + · · · + (−1) z 2! k! 0 ∫ ∞ 1 t k−1 dt. + t z e−t − 1 + t − · · · + (−1)k z 0 (k − 1)! The integrated part tends to zero at each limit, since x + k is negative and x + k + 1 is positive: so we have Γ2 (z + 1) Γ2 (z) = . z The same proof applies when x lies between 0 and −1, and leads to the result Γ(z+1) = zΓ2 (z) (0 > x > −1). The last equation shews that, between the values 0 and −1 of x, Γ2 (z) = Γ(z). The preceding equation then shews that Γ2 (z) is the same as Γ(z) for all negative values of Re(z) less than −1. Thus, for all negative values of Re(z), we have the result of Cauchy and Saalschütz ∫ ∞ k t2 k+1 t s−1 −t + · · · + (−1) dt, Γ(z) = t e −1+t− 2! k! 0 where k is the integer next less than − Re(z). Example 12.2.5 (Saalschütz) have

If a function P(µ) be such that for positive values of µ we P(µ) =

∫

1

x µ−1 e−x dx,

0

and if for negative values of µ we define P1 (µ) by the equation ∫ 1 xk P1 (µ) = x µ−1 e−x − 1 + x − · · · + (−1)k+1 dx, k! 0

12.2 Euler’s expression of Γ(z) as an infinite integral

253

where k is the integer next less than −µ, shew that P1 (µ) = P(µ) −

1 1 1 + − · · · + (−1)k−1 . µ 1!(µ + 1) k!(µ + k)

12.22 Hankel’s expression of Γ(z) as a contour integral The integrals obtained for Γ(z) in §§12.2, 12.21 are members of a large class of definite integrals by which the Gamma-function can be defined. The most general integral of the class in question is due to Hankel [272]; this integral will now be investigated. Let D be a contour which starts from a point∫ ρ on the real axis, encircles the origin once counter-clockwise and returns to ρ. Consider

(−t)z−1 e−t dt, when Re z > 0 and z is not

D

an integer. The many-valued function (−t)z−1 is to be made definite by the convention that (−t)z−1 = e(z−1) log(−t) and log(−t) is purely real when t is on the negative part of the real axis, so that −π ≤ arg(−t) ≤ π on D. The integrand is not analytic inside D, but, by Corollary 5.2.1, the path of integration may be deformed (without affecting the value of the integral) into the path of integration which starts from ρ, proceeds along the real axis to δ, describes a circle of radius δ counter-clockwise round the origin and returns to ρ along the real axis. On the real axis in the first part of this new path we have arg(−t) = −π, so that (−t)z−1 = e−iπ(z−1) t z−1 (where log t is purely real); and on the last part of the new path (−t)z−1 = eiπ(z−1) t z−1 . On the circle we write −t = δeiθ ; then we get ∫ ∫ δ ∫ π z−1 −t −iπ(z−1) z−1 −t (−t) e dt = e t e dt + (δeiθ )z−1 e δ(cos θ+i sin θ) δeiθ i dθ D ρ −π ∫ ρ + eiπ(z−1) t s−1 e−t dt δ ∫ ρ ∫ π z−1 −t z = 2i sin(πz) t e dt + iδ eizθ+δ(cos θ+i sin θ) dθ. δ

−π

This is true for all positive values of δ ≤ ρ; now make δ → 0; then δ z → 0 and ∫ π ∫ π eizθ+δ(cos θ+i sin θ) dθ → eizθ dθ −π

−π

since the integrand tends to its limit uniformly. We consequently infer that ∫ ∫ ρ z−1 −t (−t) e dt = −2i sin(πz) t z−1 e−t dt. D

0

This is true for all positive values of ρ; make ρ → ∞, and let C be the limit of the contour D. Then ∫ ∫ ∞ (−t)z−1 e−t dt = −2i sin(πz) t z−1 e−t dt. C

0

Therefore 1 Γ(z) = − 2i sin πz

∫ C

(−t)z−1 e−t dt.

(12.5)

The Gamma-Function

254

Now, since the contour C does not pass through∫the point t = 0, there is no need longer

(−t)z−1 e−t dt is a one-valued analytic

to stipulate that the real part of z is positive; and C

function of z for all values of z. Hence, by §5.5, the equation, just proved when the real part of z is positive, persists for all values of z with the exception of the values 0, ±1, ±2, . . .. Consequently, for all except integer values of z, ∫ 1 (−t)z−1 e−t dt. Γ(z) = − 2i sin πz C This is Hankel’s formula; if we write 1 − z for z and make use of §12.14, we get the further result that ∫ 1 i = (−t)−z e−t dt. Γ(z) 2π C ∫ (0+) ∫ We shall write for , meaning thereby that the path of integration starts at ‘infinity’ ∞

C

on the real axis, encircles the origin in the positive direction and returns to the starting point. ∫ Example 12.2.6 Shew that, if Re z > 0 and if a be any positive constant, (−t)−z e−t dt tends to zero as ρ → ∞, when the path of integration is either of the quadrants of circles of radius ρ + a with centres at −a, the end points of one quadrant being ρ and −a + i(ρ + a), and of the other ρ and −a − i(ρ + a). Deduce that ∫ −a−iρ ∫ −z −t lim (−t) e dt = lim (−t)−z e−t dt, ρ→∞

−a+iρ

ρ→∞

C

and hence, by writing t = −a − iu, shew that ∫ ∞ 1 1 = e a+iu (a + iu)−z du. Γ(z) 2π −∞ This formula was given by Laplace [410, p. 134], and it is substantially equivalent to Hankel’s formula involving a contour integral. Example 12.2.7 (Bourguet, L. [96, 97]) By taking a = 1, and putting t = −1 + i tan θ in Example 12.2.6, shew that ∫ 12 π 1 e = cos(tan θ − zθ) cosz−2 θ dθ. Γ(z) π 0 Example 12.2.8 By taking as contour of integration a parabola whose focus is the origin, shew that, if a > 0, then ∫ 2az e a ∞ −at 2 1 Γ(z) = e (1 + t 2 )z− 2 cos {2at + (2z − 1) arctan t} dt. sin πz 0

12.3 Gauss’ infinite integral for Γ 0(z)/Γ(z)

255

Example 12.2.9 (St John’s, 1902) Investigate the values of x for which the integral ∫ 2 ∞ x−1 t sin t dt π 0 converges; for such values of x express it in terms of Gamma-functions, and thence shew that it is equal to e−γx

∞ n Ö

∞

1−

n=1

x x/(2n) o . Ö n x −x/(2n−1) o 1+ . e e 2n 2n − 1 n=1 ∫

∞

sin t dt converges when m > 0, t 0 and, by means of Example 12.2.9, evaluate it when m = 1 and when m = 2.

Example 12.2.10 (St John’s, 1902) Prove that

(log t)m

12.3 Gauss’ expression for the logarithmic derivate of the Gamma-function as an infinite integral d Γ 0(z) We shall now express the function log Γ(z) = as an infinite integral when Re z > 0; dz Γ(z) the function in question is frequently written ψ(z). (The results appear in [236, p. 159].) We first need a new formula for γ. Take the formula in Example 11.6.2 ∫ 1 ∫ 1 ∫ ∞ −t ∫ ∞ −t 1 − e−t e dt e γ= dt − dt = lim − dt δ→0 t t t t 0 0 δ δ ∫ 1 ∫ ∞ −t dt e = lim − dt , δ→0 t t ∆ δ where ∆ = 1 − e δ , since ∫ ∆

δ

dt δ →0 = log t 1 − e−δ

as

δ → 0.

Writing t = 1 − e−u in the first of these integrals and then replacing u by t we have ∫ ∞ ∫ ∞ −t ∫ ∞ 1 1 −t e−t e γ = lim dt − dt = − e dt. ∞→0 1 − e−t t 1 − e−t t δ δ 0 This is the formula for γ which was required. To get Gauss’ formula, take the equation (§12.16) n Õ Γ 0(z) 1 1 1 = −γ − + lim − , Γ(z) z n→∞ m=1 m z + m 1 and write = z+m

∫ 0

∞

e−t( z+m) dt; this is permissible when m = 0, 1, 2, . . . if Re z > 0. It

The Gamma-Function

256

follows that Γ 0(z) = −γ − Γ(z)

∞

∫

e

−zt

∫

dt + lim

n→∞

0

∞

0

n Õ

(e−mt − e−(m+z)t ) dt

m=1

e−t − e−zt − e−(n+1)t + e−(z+n+1) t dt = −γ + lim n→∞ 0 1 − e−t ∫ ∞ −t ∫ ∞ e−zt e 1 − e−zt −(n+1)t = − dt − lim e dt. −t n→∞ 0 t 1−e 1 − e−t 0 1 − e−zt is a bounded function of t whose limit as t → 0 is finite; Now, when 0 < t ≤ 1, 1 − e−t and when t ≥ 1, 1 − e−zt 1 + |e−zt | 2 1 − e−t < 1 − e−t < 1 − e−t . ∫

∞

Therefore we can find a number K independent of t such that, on the path of integration, 1 − e−zt 1 − e−t < K; and so ∫

0

∞

∫ ∞ 1 − e−zt (n+1)t K e dt < K e−(n+1)t dt = → 0 as 1 − e−t n+1 0

n → ∞.

We have thus proved the formula ∞

∫

d ψ(z) = log Γ(z) = dz

0

−t e e−zt t − 1 − e−t dt,

which is Gauss’ expression of ψ(z) as an infinite integral. It may be remarked that this is the first integral which we have encountered connected with the Gamma-function in which the integrand is a single-valued function. Writing t = log(1 + x) in Gauss’ result, we get, if ∆ = e δ − 1, ∫ ∞ −t Γ 0(z) e−zt e dt = lim − Γ(z) δ→0 δ t 1 − e−t ∫ ∞ −t ∫ ∞ e dx = lim dt − δ→0 t x(1 + x)z δ ∆ ∫ ∞ −t ∫ ∞ e dx = lim dt − , δ→0 t x(1 + x)z ∆ ∆ since 0

0, ∫ 1 1 t z−1 ψ(z) = − dt. − log t 1 − t 0 Example 12.3.2 (Dirichlet). Shew that ∫ ∞ γ= (1 + t)−1 − e−t t −1 dt. 0

12.31 Binet’s first expression for log Γ(z) in terms of an infinite integral Binet [73] has given two expressions for log Γ(z) which are of great importance as shewing the way in which log Γ(z) behaves as |z| → ∞. To obtain the first of these expressions, we observe that, when the real part of z is positive, ∫ ∞ −t e e−tz Γ 0(z + 1) dt, = − t Γ(z + 1) t e −1 0 writing z + 1 for z in §12.3. Now, by Example 6.2.13, we have ∫ ∞ −t e − e−tz log z = dt, t 0 ∫ ∞ 1 and so, since = e−tz dt, we have z 0 ∫ ∞ 1 1 1 1 d e−tz dt. log Γ(z + 1) = + log z − − + dz 2z 2 t et − 1 0 1 1 1 The integrand in the last integral is continuous as t → 0; and since − + t is 2 t e −1 bounded as t → ∞, it follows without difficulty that the integral converges uniformly when the real part of z is positive; we may consequently integrate from 1 to z under the sign of integration (§4.44) and we get 3 −tz ∫ ∞ 1 1 1 e − e−t 1 log Γ(z + 1) = (z + 2 ) log z − z + 1 + − + t dt. 2 t e −1 t 0 1 1 1 Since − + is continuous as t → 0 by §7.2, and since 2 t et − 1 log Γ(z + 1) = log z + log Γ(z), 3

log Γ(z + 1) means the sum of the principal values of the logarithms in the factors of the Weierstrassian product.

The Gamma-Function

258

we have log Γ(z) = z −

1 2

∞

−tz 1 e 1 1 − + t dt log z − z + 1 + 2 t e −1 t 0 −t ∫ ∞ 1 1 1 e − − + t dt. 2 t e − 1 t 0 ∫

To evaluate the second of these integrals, let 4 −t −1t ∫ ∞ ∫ ∞ 1 1 1 1 e 1 1 e 2 − + t dt = I, − + t dt = J; 2 t e −1 t 2 t e −1 t 0 0 so that, taking z =

1 2

in the last expression for log Γ(z), we get 1 1 log π = + J − I. 2 2

Also, since I=

∫

∞

0

−t/2 1 e 1 2 − + dt, 2 t et/2 − 1 t

we have ∞

1 et/2 e−t/2 dt − t t e −1 t 0 ∫ ∞ −t/2 dt e 1 = − t . t e − 1 t 0

J−I =

∫

And so ∞

e−t/2 dt 1 1 e−t e−t − t + e−t − + t t e − 1 2 t e − 1 t 0 ∫ ∞ −t/2 −t e −e 1 dt = − e−t t 2 t 0 ) −t/2 1 −t/2 ∫ ∞( −t −t −t e − e d e −e e = − − 2 − dt dt t t 2t 0 ∞ −t/2 ∫ 1 ∞ e−t − e−t/2 e − e−t + dt = − t 2 0 t 0

J=

=

∫

1 2

+ 12 log 12 .

Consequently I = 1 − 12 log(2π). We therefore have Binet’s result that, when Re z > 0, −tz ∫ ∞ 1 1 1 e 1 1 log Γ(z) = z − 2 log z − z + 2 log(2π) + − + t dt. 2 t e − 1 t 0 If z = x + iy, we see that, if the upper bound of 1 1 − + 1 2 t et − 1 4

This artifice is due to Pringsheim [538].

1 t

12.3 Gauss’ infinite integral for Γ 0(z)/Γ(z)

259

for real values of t is K, then ∫ ∞ log Γ(z) − z − 1 log z + z − 1 log(2π) < K e−tz dt 2 2 0 = K x −1, so that, when x is large, the terms z − expression for log Γ(z).

1 2

log z − z +

1 2

log(2π) furnish an approximate

Example 12.3.3 (Malmstén) Prove that, when Re(z) > 0, ∫ ∞ −zt dt e − e−t −t + (z − 1)e . log Γ(z) = −t 1−e t 0 Example 12.3.4 (Féaux) Prove that, when Re(z) > 0, ∫ ∞ (1 + t)−z − (1 + t)−1 dt −t . log Γ(z) = (z − 1)e + log(1 + t) t 0 Example 12.3.5 (Kummer) From the formula of §12.14 shew that, if 0 < x < 1, ) ∫ ∞( sinh( 12 − x)t dt 2 log Γ(x) − log π + log sin πx = − (1 − 2x)e−t . 1 t sinh 2 t 0 Example 12.3.6 (Kummer [390]) By expanding sinh( 21 − x)t and 1 − 2x in Fourier sine series, shew from Example 12.3.5 that, if 0 < x < 1, ∞ Õ 1 1 log Γ(x) = log π − log sin πx + 2 an sin 2nπx, 2 2 n=1

where an =

∫ 0

∞

2nπ e−t − t 2 + 4n2 π 2 2nπ

dt . t

Deduce from Example 12.3.2, that an =

1 (γ + log 2π + log n). 2nπ

12.32 Binet’s second expression for log Γ(z) in terms of an infinite integral Consider the application of Example 7.7 to the equation (12.4), ∞

Õ d2 1 log Γ(z) = . dz2 (z + n)2 n=0 The conditions there stated as sufficient for the transformation of a series into integrals are obviously satisfied by the function φ(ζ) = 1/(z + ζ)2, if Re z > 0; and we have ∫ ∞ ∫ ∞ ∫ ∞ d2 1 dξ q(t, z) q(t, z + n) log Γ(z) = 2 + −2 dt + 2 lim dt, 2 2π − 1 n→∞ dz 2 2z (z + ξ) e e2πt − 1 0 0 0

260

The Gamma-Function

1 1 − . Since |q(t, z + n)| is easily seen to be less than K1 t/n, (z + it)2 (z − it)2 where K1 is independent of t and n, it follows that the limit of the last integral is zero. Hence ∫ ∞ d2 1 1 dt 4tz log Γ(z) = 2 + + . (12.6) 2 2 2 2 2π dz 2z z (z + t ) e t − 1 0 Since z 22z+t 2 does not exceed K (where K depends only on δ) when the real part of z exceeds δ, the integral converges uniformly and we may integrate under the integral sign §4.44 from 1 to z. We get ∫ ∞ d 1 t dt log Γ(z) = + log z + C − 2 , 2 2 dz 2z (z + t )(e2πt − 1) 0 where 2iq(t) =

where C is a constant. Integrating again, ∫ ∞ 1 arctan(t/z) dt, log Γ(z) = z − log z + (C − 1)z + C 0 + 2 2 e2πt − 1 0 where C 0 is a constant. Now, if z is real, 0 ≤ arctan t/z ≤ t/z, and so ∫ ∞ t log Γ(z) − z − 1 log z − (C − 1) z − C 0 < 2 dt. 2πt 2 z 0 e −1 But it has been shewn in §12.31 that 1 1 log Γ(z) − z − log z + z − log(2π) → 0, 2 2 as z → ∞ through real values. Comparing these results we see that C = 0, C0 =

1 log(2π). 2

Hence for all values of z whose real part is positive, ∫ ∞ 1 1 arctan(t/z) log Γ(z) = z − log z − z + log(2π) + 2 dt, 2 2 e2πt − 1 0 where arctan u is defined by the equation arctan u =

∫ 0

u

dt , 1 + t2

in which the path of integration is a straight line. This is Binet’s second expression for log Γ(z). Example 12.3.7 Justify differentiating with regard to z under the sign of integration, so as to get the equation ∫ ∞ Γ 0(z) 1 tdt = log z − −2 . 2 + z 2 ) (e2πt − 1) Γ(z) 2z (t 0

12.3 Gauss’ infinite integral for Γ 0(z)/Γ(z)

261

12.33 The asymptotic expansion of the logarithms of the Gamma-function We can now obtain an expansion which represents the function log Γ(z) asymptotically (§8.2) for large values of |z|, and which is used in the calculation of the Gamma-function. Let us assume that, if z = x + iy, then x ≥ δ > 0; and we have, by Binet’s second formula, log Γ (z) = z − 12 log z − z + 12 log (2π) + φ(z), where φ(z) = 2

∫ 0

∞

arctan(t/z) dt. e2πt − 1

(12.7)

Now t 1 t3 1 t5 (−1)n−1 t 2n−1 (−1)n arctan(t/z) = − + − · · · + + n−1 z 3 z3 5 z5 2n − 1 z2n−1 z

∫ 0

t

u2n du . u2 + z 2

Substituting and remembering (§7.2) that ∫ ∞ 2n−1 Bn t dt = , 2πt e − 1 4n 0 where B1, B2, . . . are Bernoulli’s numbers, we have ∫ ∫ t 2n n Õ (−1)r−1 Br 2(−1)n ∞ u du dt + 2n−1 φ(z) = . 2r−1 2 2 2πt 2r(2r − 1)z z e −1 0 0 u +z r=1 z2 for positive values of u be Kz . Then Let the upper bound 5 of 2 u + z2 ∫ ∞ ∫ t 2n ∫ ∞ ∫ t dt u du dt −2 2n ≤ Kz |z| u du 2πt 2 2 2πt e −1 e −1 0 0 u +z 0 0 Kz Bn+1 ≤ . 4 (n + 1) (2n + 1) |z| 2 Hence ∫ ∫ t 2n 2(−1)n ∞ dt Kz Bn+1 u du < , z2n−1 2 2 2πt−1 e 2(n + 1)(2n + 1)|z| 2n+1 0 0 u +z and it is obvious that this tends to zero uniformly as |z| → ∞ if | arg z| ≤ 12 π − ∆, where 1 π > ∆ > 0, so that Kz ≤ cosec 2∆. 4 Also it is clear that if | arg z| ≤ 41 π (so that Kz = 1) the error in taking the first n terms of the series ∞ Õ (−1)r−1 Br 1 r=1

2r (2r − 1) z2r−1

as an approximation to φ(z) is numerically less than the (n + 1)th term. 5

K s−2 is the lower bound of x 2 < y 2 or x 2 ≥ y 2 .

{u 2 + (x 2 − y 2 )}2 + 4x 2 y 2 4x 2 y 2 and is consequently equal to 2 or 1 as (x 2 + y 2 )2 (x + y 2 )2

The Gamma-Function

262

Since, if | arg z| ≤ 21 π − ∆, ( ) n Õ (−1)r−1 Br Bn+1 2n−1 φ(z) − |z| −2 z < 2 2r (2r − 1) 2(n + 1)(2n + 1) sin (2∆) r=1 → 0, as z → ∞, it is clear that B1 B2 B3 − + −··· 3 1·2·z 3·4·z 5 · 6 · z5 is the asymptotic expansion of φ(z) (§8.2). (The development is asymptotic; for if it converged when |z| ≥ ρ, by §2.6 we could find K, such that Bn < (2n − 1)2nK ρ2n ; and then the series ∞ (−1)n−1 B t 2n Í n would define an integral function; this is contrary to §7.2.) (2n)! n=1 We see therefore that the series ∞ Õ (−1)r−1 Br z − 12 log z − z + 12 log(2π) + 2r (2r − 1)z2r−1 r=1 is the asymptotic expansion of log Γ(z) when | arg z| ≤ π/2 − ∆. This is generally known as Stirling’s series. In §13.6 it will be established over the extended range | arg z| ≤ π − ∆. In particular when z is positive (= x), we have ∫ ∞ ∫ t 2n dt Bn+1 u du . 0 0, the value of φ(x) always lies between the sum of n terms and the sum of n + 1 terms of the series for all values of n. In particular 0 < φ(x)

0, loge Γ(z) = z − 12 loge z − z + 21 loge (2π) + J(z),

12.4 The Eulerian integral of the first kind

263

where J(z) =

c2 c3 c1 1 + + +··· , 2 z + 1 2(z + 1)(z + 2) 3(z + 1)(z + 2)(z + 3)

in which 1 c1 = , 6

1 c2 = , 3

c3 =

59 , 90

c4 =

227 , 60

and generally cn =

∫

1

(x + 1)(x + 2) · · · (x + n − 1)(2x − 1)x dx.

0

12.4 The Eulerian integral of the first kind The name Eulerian Integral of the First Kind was given by Legendre to the integral ∫ 1 B(p, q) = x p−1 (1 − x)q−1 dx, 0

which was first studied by Euler [200] and Legendre [421, vol. 1, p. 221]. In this integral, the real parts of p and q are supposed to be positive; and x p−1 , (1 − x)q−1 are to be understood to mean those values of e(p−1) log x and e(q−1) log(1−x) which correspond to the real determinations of the logarithms. With these stipulations, it is easily seen that B(p, q) exists, as a (possibly improper) integral (see formula (2) in Example 4.5.1). We have, on writing (1 − x) for x, B(p, q) = B(q, p). Also, integrating by parts, 1 p ∫ ∫ 1 q 1 p x (1 − x)q p−1 q + x (1 − x)q−1 dx, x (1 − x) dx = p p 0 0 0 q so that B(p, q + 1) = B(p + 1, q). p Example 12.4.1 Shew that B(p, q) = B(p + 1, q) + B(p, q + 1). Example 12.4.2 Deduce from Example 12.4.1 that q B(p, q). B(p, q + 1) = p+q Example 12.4.3 Prove that if n is a positive integer, 1 · 2···n . p(p + 1) · · · (p + n)

B(p, n + 1) = Example 12.4.4 Prove that B(x, y) =

∫ 0

∞

a x−1 da. (1 + a)x+y

The Gamma-Function

264

Example 12.4.5 Prove that Γ(z) = lim nz B(z, n). n→∞

12.41 Expression of the Eulerian integral of the first kind in terms of the Gamma-function We shall now establish the important theorem that B(m, n) =

Γ(m)Γ(n) . Γ(m + n)

First let the real parts of m and n exceed 12 ; then Γ(m) Γ(n) =

∞

∫

e−x x m−1 dx ×

∞

∫

0

e−y y n−1 dy.

0

On writing x 2 for x and y 2 for y, this gives ∫ R 2 2 e−y y 2n−1 dy e−x x 2m−1 dx × R→∞ 0 0 ∫ R∫ R 2 2 = 4 lim e−(x +y ) x 2m−1 y 2n−1 dx dy.

Γ(m) Γ(n) = 4 lim

∫

R→∞

R

0

0

Now, for the values of m and n under consideration, the integrand is continuous over the range of integration, and so the integral may be considered as a double integral taken over a square SR . Calling the integrand f (x, y), and calling Q R the quadrant with centre at the origin and radius R, we have, if TR be the part of SR outside Q R, ∫ ∫ ∫ ∫ ∫∫ f (x, y) dx dy f (x, y) dx dy = f (x, y) dx dy − T QR SR ∫∫ R ≤ | f (x, y)| dx dy TR ∫∫ ∫∫ ≤ | f (x, y) | dx dy − | f (x, y)| dx dy SR

S R/2

→ 0 as R → ∞, since

∫∫ SR

| f (x, y)| dx dy converges to a limit, namely ∫

∞

e

2 0

−x 2 2m−1

x

∫ dx × 2

∞

e−y y 2n−1 dy.

0

Therefore ∫∫ lim

R→∞

SR

f (x, y) dx dy = lim

R→∞

∫∫ f (x, y) dx dy. QR

12.4 The Eulerian integral of the first kind

265

Changing to polar 6 coordinates (x = r cos θ, y = r sin θ), we have ∫ ∫ ∫ R ∫ π/2 2 f (x, y) dx dy = e−r (r cos θ)2m−1 (r sin θ)2n−1 r dr dθ. QR

0

0

Hence Γ(m)Γ(n) = 4

∫

∞

e

−r 2 2(m+n)−1

0

= 2Γ(m + n)

r ∫

∫ dr

π/2

cos2m−1 θ sin2n−1 θ dθ

0

π/2

cos2m−1 θ sin2n−1 θ dθ.

0

Writing cos2 θ = u we at once get Γ(m)Γ(n) = Γ(m + n) · B(m, n). This has only been provided when the real parts of m and n exceed 1/2; but it can obviously be deduced when these are less than 1/2 by Example 12.4.2. This result, discovered by Euler, connects the Eulerian Integral of the First Kind with the Gamma-function. Example 12.4.6 Shew that ∫ 1 Γ(p)Γ(q) . (1 + x) p−1 (1 − x)q−1 dx = 2 p+q−1 Γ(p + q) −1 Example 12.4.7 (Jesus, 1901) Shew that, if f (x, y) =

1 y(y − 1) 1 y(y − 1)(y − 2) 1 1 −y + − +··· , x x+1 2! x+2 3! x+3

then f (x, y) = f (y + 1, x − 1), where x and y have such values that the series are convergent. Example 12.4.8 (Math. Trip. 1894) Prove that ∫ ∫ 1∫ 1 Γ(µ)Γ(ν) 1 µ−1 µ ν−1 f (z)(1 − z)µ+ν−1 dz. f (xy)(1 − x) y (1 − y) dx dy = Γ(µ + ν) 0 0 0

12.42 Evaluation of trigonometrical integrals in terms of the Gamma-function ∫ π/2 We can now evaluate the integral cosm−1 x sinn−1 x dx, where m and n are not restricted 0

to be integers, but have their real parts positive. For, writing cos2 x = t, we have, as in §12.41, ∫ π/2 Γ( m2 )Γ( n2 ) m−1 n−1 cos x sin x dx = . 2Γ( m+n ) 0 2 6

It is easily provided by the methods of §4.11 that the areas A m, µ of §4.3 need not be rectangles provided only that their greatest diameters can be made arbitrarily small by taking the number of areas sufficiency large; so the areas may be taken to be the regions bounded by radii vectors and circular arcs.

The Gamma-Function

266

The well-known elementary formulae for the cases in which m and n are integers can be at once derived from this result. Example 12.4.9 (Trinity, 1898) Prove that, when |k | < 1, ∫ π/2 n+1 ∫ m+1 cosm+n θ dθ cosm θ sinn θ dθ Γ( 2 )Γ( 2 ) π/2 = . √ (1 − k sin2 θ)1/2 (1 − k sin2 θ)(n+1)/2 Γ( m+n+1 ) π 0 0 2

12.43 Pochhammer’s extension of the Eulerian integral of the first kind This appears in [527]. The use of the double circuit integrals of this section seems to be due to Jordan [363]. We have seen in §12.22 that it is possible to replace the second Eulerian integral for Γ(z) by a contour integral which converges for all values of z. A similar process has been carried out by Pochhammer for Eulerian integrals of the first kind. Let P be any point on the real axis between 0 and 1; consider the integral ∫ (1+,0+,1−,0−) e−πi(α+β) t α−1 (1 − t)β−1 dt = ε(α, β). p

The notation employed is that introduced at the end of §12.22 and means that the path of integration starts from P, encircles the point 1 in the positive (counter-clockwise) direction and returns to P, then encircles the origin in the positive direction and returns to P, and so on. At the starting-point the arguments of t and 1 − t are both zero; after the circuit (1+) they are 0 and 2π; after the circuit (0+) they are 2π and 2π; after the circuit (1−) they are 2π and 0 and after the circuit (0−) they are both zero, so that the final value of the integrand is the same as the initial value. It is easily seen that, since the path of integration may be deformed in any way so long as it does not pass over the branch points 0, 1 of the integrand, the path may be taken to be that shewn in the figure, wherein the four parallel lines are supposed to coincide with the real axis.

If the real parts of α and β are positive the integrals round the circles tend to zero as the radii of the circles tend to zero (the reader ought to have no difficulty in proving this); the integrands on the paths marked a, b, c, d are t a−1 (1 − t)β−1, t a−1 (1 − t)β−1 e2πi(β−1), t a−1 e2πi(α−1) (1 − t)β−1 e2πi(β−1), t a−1 e2πi(α−1) (1 − t)β−1 respectively, the arguments of t and 1 − t now being zero in each case.

12.5 Dirichlet’s integral

267

Hence we may write ε(α, β) as the sum of four (possibly improper) integrals, thus: ε(α, β) = e

∫1 ∫ 0 t α−1 (1 − t)β−1 dt + t α−1 (1 − t)β−1 e2πiβ dt 1 0 ∫ 1 ∫ 0 + t α−1 (1 − t)β−1 e2πi(α+β) dt + t α−1 (1 − t)β−1 e2πia dt .

πi(α+β)

0

1

Hence ε(α, β) = e

πi(α+β)

∫ (1 − e

2πiα

)(1 − e

2πiβ

)

1

t α−1 (1 − t)β−1 dt

0

= −4 sin(απ) sin(βπ) =

Γ(α)Γ(β) Γ(α + β)

−4π 2 . Γ(1 − α)Γ(1 − β)Γ(α + β)

Now ε(α, β) and this last expression are analytic functions of α and of β for all values of α and β. So, by the theory of analytic continuation, this equality, proved when the real parts of α and β are positive, holds for all values of α and β. Hence for all values of α and β we have proved that ε(α, β) =

−4π 2 . Γ(1 − α)Γ(1 − β)Γ(α + β)

12.5 Dirichlet’s integral This material appears in [178, pp. 375, 391]. We shall now shew how the repeated integral ∫ ∫ ∫ ··· f (t1 + t2 + · · · + tn )t1α1 −1 t2α2 −1 · · · tnαn −1 dt1 dt2 · · · dtn I= may be reduced to a simple integral, where f is continuous, αr > 0 (r = 1, 2, . . . , n) and the integration is extended over all positive values of the variables such that t1 + t2 + · · · + tn ≤ 1. To simplify ∫ 1−λ ∫ 1−λ−T f (t + T + λ)t α−1T β−1 dt dT 0

0

(where we have written t,T, α, β for t1, t2, α1, α2 and λ for t3 + t4 + · · · + tn ), put t = T(1 − v)/v; the integral becomes (if λ , 0) ∫ 1−λ ∫ 1 f (λ + T/v)(1 − v)α−1 v −α−1T α+β−1 dv dT . T /(1−λ)

0

Changing the order of integration (§4.51), the integral becomes ∫ 1 ∫ (1−λ)v f (λ + T/v)(1 − v)α−1 v −α−1T α+β−1 dT dv. 0

0

The Gamma-Function

268

Putting T = vτ2 , the integral becomes ∫ 1 ∫ 1−λ ∫ Γ(α)Γ(β) 1−λ α−1 β−1 α+β−1 f (λ + τ2 )(1 − v) v τ2 dτ2 dv = f (λ + τ2 )τ2α+β−1 dτ2 . Γ(α + β) 0 0 0 Hence I=

Γ(α1 )Γ(α2 ) Γ(α1 + α2 )

∫ ∫

∫ ···

f (τ2 + t3 + · · · + tn )τ2α1 +α2 −1 t3a3 −1 · · · tnan −1 dτ2 dt3 · · · dtn,

the integration being extended over all positive values of the variables such that τ2 + t3 + · · · + tn ≤ 1. Continually reducing in this way we get ∫ 1 Í Γ(α1 )Γ(α2 ) · · · Γ(αn ) I= f (τ)τ α−1 dτ, Γ(α1 + α2 + · · · + αn ) 0 which is Dirichlet’s result. Example 12.5.1 (Dirichlet) Reduce ∫ ∫ ∫ x α y β z γ p−1 q−1 r−1 + + x y z dx dy dz f a b c to a simple integral; the range of integration being extended over all positive values of the variables such that x α y β z γ + + ≤ 1, a b c it being assumed that a, b, c, α, β, γ, p, q, r are positive. ∫ ∫ Example 12.5.2 (Pembroke, 1907) Evaluate x p y q dx dy, where m and n are positive and x ≥ 0, y ≥ 0, x m + y n ≤ 1. Example 12.5.3 Shew that the moment of inertia of a homogeneous ellipsoid of unit density, taken about the axis of z, is 4π 2 (a + b2 )abc, 15 where a, b, c are the semi-axes. Example 12.5.4 Shew that the area of the hypocycloid x 2/3 + y 2/3 = ` 2/3 is 3π` 2 /8.

12.6 Miscellaneous examples Example 12.1 (Trinity, 1897) Shew that √ z z z π (1 − z) 1 + 1− 1+ ··· = . 2 3 4 Γ(1 + 12 z)Γ( 12 − 21 z) Example 12.2 (Trinity, 1885) Shew that lim

n→∞

1 1 1 1 ··· n x = Γ(x + 1). 1 1 1 + x 1 + 2x 1 + 3x 1 + n1 x

12.6 Miscellaneous examples

Example 12.3 (Jesus, 1903) Prove that 1 Γ 0(1) Γ 0( 2 ) − = 2 log 2. Γ(1) Γ( 12 )

Example 12.4 (Trinity, 1891) Shew that 1 4 Γ( 4 ) 32 52 − 1 72 92 − 1 112 = · · · · ··· 16π 2 32 − 1 52 72 − 1 92 112 − 1 Example 12.5 (Trinity, 1905) Shew that ∞ Ö α 1 (n − α) (n + β + γ) 1+ = − sin(πα)B(β, γ). (n + β) (n + γ) n+1 π n=0 Example 12.6 (Peterhouse, 1906) Shew that 3 ∞ Ö k 640 π Γ = 6 √ . 3 3 3 k=1 Example 12.7 (Trinity, 1904) Shew that, if z = iζ where ζ is real, then r π . |Γ(z)| = ζ sinh(πζ) Example 12.8 (Math. Trip. 1897) When x is positive, shew that Γ(x)Γ( 21 ) Γ(x + 12 )

=

∞ Õ (2n)! 1 . 2n n!2 x + n 2 n=0

(This and some other examples are most easily proved by the result of §14.11.) Example 12.9 If a is positive, shew that ∞

Γ(z)Γ(a + 1) Õ (−1)n (a)(a − 1)(a − 2) · · · (a − n) 1 = . Γ(z + a) n! z+n n=0 Example 12.10 If x > 0 and P(x) =

∫

1

e−t t x−1 dt,

0

shew that P(x) =

1 1 1 1 1 1 1 − + − +··· , x 1! x + 1 2! x + 2 3! x + 3

and P(x + 1) = xP(x) − 1/e. Example 12.11 (Euler) Shew that if λ > 0, x > 0, −π/2 < a < π/2, then ∫ ∞ t x−1 e−λt cos a cos(λt sin a) dt = λ−x Γ(x) cos ax, 0 ∫ ∞ t x−1 e−λt cos a sin(λt sin a) dt = λ−x Γ(x) sin ax. 0

269

The Gamma-Function

270

Example 12.12 (Euler) Prove that, if b > 0, then, when 0 < z < 2, ∫ ∞ πz π bz−1 sin bx dx = cosec , xz 2 Γ(z) 2 0 and, when 0 < z < 1, ∞

∫ 0

πz π bz−1 cos bx dx = sec . xz 2 Γ(z) 2

Example 12.13 (Peterhouse, 1895) If 0 < n < 1, prove that ∫ ∞ nπ 1 1 (1 + x)n−1 cos x dx = Γ(n) cos −1 − + −··· . 2 Γ(n + 1) Γ(n + 3) 0 Example 12.14 (Bourguet [97]) By taking as contour of integration a parabola with its vertex at the origin, derive from the formula ∫ (0+) 1 Γ(a) = − (−z)a−1 e−z dz 2i sin aπ ∞ the result 1 Γ(a) = 2 sin aπ

∫

∞

e−x x a−1 (1 + x 2 )a/2 [3 sin {x + a arccot(−x)} 2

0

+ sin {x + (a − 2) arccot(−x)}] dx, the arccot denoting an obtuse angle. Example 12.15 (Math. Trip. 1907)

Shew that, if Re an > 0 and

∞ Í n=1

then ∞ Ö n=1

"

( m )# Õ 2s Γ(an ) (s) exp ψ (an ) Γ(z + an ) s! n=1

is convergent when m > 2, where ψ (s) (z) =

ds log Γ(z). dz s

Example 12.16 (Legendre) Prove that ∫ ∞ −a d log Γ(z) e − e−za = da − γ dz 1 − e−a 0 ∫ ∞ da = (1 + a)−1 − (1 + a)−s −γ a 0 ∫ 1 s−1 x −1 = dx − γ. x−1 0 Example 12.17 (Binet) Prove that, when Re(z) > 0, ∫ 1 s x −x dx log Γ(z) = − x(z − 1) . x − 1 x log x 0

1/an2 is convergent,

12.6 Miscellaneous examples

271

Example 12.18 Prove that, for all values of z except negative real values, ( ∞ 1 1 1 1 Õ 1 log Γ(z) = z − log z − z + log(2π) + 2 2 2 2 · 3 r=1 (z + r)2 ) ∞ ∞ 3 Õ 2 Õ 1 1 + +··· . + 3 · 4 r=1 (z + r)3 4 · 5 r=1 (z + r)4 Example 12.19 Prove that, when Re(z) > 0, ∫ 1 d x s−1 [1 − x + log x] dx. log Γ(z) = log z − dz 0 (1 − x) log x Example 12.20 Prove that, when Re(z) > 0, ∫ ∞ −xs d2 xe dx log Γ(z) = . dz2 1 − e−x 0 ∫ s+1 Example 12.21 (Raabe, [546]) If log Γ(t) dt = u, shew that s

du = log z, dz and deduce from §12.33 that, for all values of z except negative real values, u = z log z − z + 21 log(2π). Example 12.22 (Bourguet) Prove that, for all values of z except negative real values, ∞ ∫ ∞ Õ dx sin 2nπx 1 1 log Γ(z) = z − 2 log z − z + 2 log(2π) + . x + z nπ n=1 0 This result is attributed to Bourguet by Stieltjes [606]. Example 12.23 (Binet) Prove that B(p, p)B p + 12 , p +

1 2

=

π 24p−1 p

.

Example 12.24 Prove that, when −t < r < t, B(t + r, t − r) =

1 4t−1

∫ 0

∞

cosh(2ru) du . cosh2t u

Example 12.25 Prove that, when q > 1, B(p, q) + B(p + 1, q) + B(p + 2, q) + · · · = B(p, q − 1). Example 12.26 Prove that, when p − a > 0, B(p − a, q) aq a(a + 1)q(q + 1) =1+ + +··· . B(p, q) p + q 1 · 2 · (p + q)(p + q + 1) Example 12.27 (Euler) Prove that B(p, q)B(p + q, r) = B(q, r)B(q + r, p).

The Gamma-Function

272

Example 12.28 (Trinity, 1908) Shew that ∫ 1 1 dx Γ(a)Γ(b) x a−1 (1 − x)b−1 = , a+b Γ(a + b) (x + p) (1 + p)a pb 0 if a > 0, b > 0, p > 0. Example 12.29 (St John’s, 1904) Shew that, if m > 0, n > 0, then ∫ 1 Γ(m)Γ(n) (1 + x)2m−1 (1 − x)2n−1 dx = 2m+n−2 ; 2 m+n (1 + x ) Γ(m + n) −1 and deduce that, when a is real and not an integer multiple of π/2, cos 2a ∫ π/2 π cos θ + sin θ . dθ = 2 sin(π cos2 a) −π/2 cos θ − sin θ Example 12.30 (Kummer) Shew that, if α > 0, β > 0, ∫ 1 α−1 t 1 α 1 α+1 − ψ dt = ψ , 1+t 2 2 2 2 0 and ∫ 0

1

Γ( α+1 )Γ( β2 ) t α−1 − t β−1 2 dt = log α . (1 + t) log t Γ( 2 ) Γ( β+1 ) 2

Example 12.31 Shew that, if a > 0, a + b > 0, ∫ 1 a−1 x (1 − x b ) Γ(a) Γ(δ) Γ(a + b) Γ(δ) dx = lim − δ→0 Γ(a + δ) 1−x Γ(a + b + δ) 0 = ψ(a + b) − ψ(a). Deduce that, if in addition a + c > 0, a + b + c > 0, ∫ 1 a−1 x (1 − x b )(1 − x c ) Γ(a) Γ(a + b + c) dx = log . (1 − x)(− log x) Γ(a + b) Γ(a + c) 0 Example 12.32 Shew that, if a, b, c be such that the integral converges, ∫ 1 (1 − x a ) (1 − x b ) (1 − x c ) dx (1 − x) (− log x) 0 Γ(b + c + 1) Γ(c + a + 1) Γ(a + b + 1) = log . Γ(a + 1) Γ(b + 1) Γ(c + 1) Γ(a + b + c + 1) Example 12.33 (St John’s, 1896) By the substitution cos θ = 1 − 2 tan 12 φ, shew that ∫ 0

Example 12.34 (Clare, 1898)

π

1 2 Γ( 4 ) dθ = √ . 1/2 (3 − cos θ) 4 π Evaluate in terms of Gamma-functions the integral

12.6 Miscellaneous examples

273

∞

sin p x dx, when p is a fraction greater than unity whose numerator and denominax 0 tor are both odd integers. Shew that the integral is " # ∫ π ∞ 1 1 1 1 Õ n p + (−1) + sin x dx. 2 0 x n=1 x + nπ x − nπ

∫

Example 12.35 Shew that ∫ π/2 n−1/2 dx = 1 + 12 sin2 x 0

2 n n! Õ 23r 2r + 1 Γ . √ 4 2n+2 π r=0 (2r)!(n − r)!

Example 12.36 (Euler) Prove that log B(p, q) = log

∫ 1 p+q (1 − v p )(1 − v q ) dv. + pq (1 − v) log v 0

Example 12.37 (Binet) Prove that, if p > 0, p + s > 0, then s(s − 1) s(s − 1)(s − 2)(s − 3) B(p, p) 1 + + + · · · . B(p, p + s) = 2s 2(2p + 1) 2 · 4(2p + 1)(2p + 3) Example 12.38 The curve r m = 2m−1 a m cos mθ is composed of m equal closed loops. Shew that the length of the arc of half of one of the loops is ∫ π/2 1/m−1 a 1 cos x dx, 2 m 0 and hence that the total perimeter of the curve is 1 2 Γ 2m a . Γ m1 Example 12.39 (Cauchy) Draw the straight line joining the points ±i, and the semicircle of |z| = 1 which lies on the right of this line. Let C be the contour formed by indenting this figure at −i, 0, i. By considering ∫ z p−q−1 (z + z−1 ) p+q−2 dz, C

shew that, if p + q > 1, q < 1, ∫ π/2 cos p+q−2 θ cos(p − q)θ dθ = 0

π . (p + q − 1)2 p+q−1 B(p, q)

Prove that the result is true for all values of p and q such that p + q > 1. Example 12.40 If s is positive (not necessarily integral), and − 12 π ≤ x ≤ 12 π, shew that 1 Γ(s + 1) 1 s s(s − 2) s (cos x) = s−1 2 2 + s + 2 cos 2x + (s + 2)(s + 4) cos 4x + · · · , 2 Γ( 12 s + 1) and draw graphs of the series and of the function coss x.

The Gamma-Function

274

Example 12.41 (Cauchy) Obtain the expansion (cos x)s = a 2s−1

" Γ(s + 1)

# cos ax cos 3ax + +··· , Γ( 12 s + 12 a + 1)Γ( 21 s − 12 a + 1) Γ( 12 s + 32 a + 1)Γ( 12 s − 32 a + 1)

and find the values of x for which it is applicable. Example 12.42 (Binet) Prove that, if p > 12 , 1/2 22p−1 12 · 32 2p2 12 2 Γ(2p) = √ {Γ(p)} + +··· . 1+ 2p + 1 2(2p + 3) 2 · 4 · (2p + 3)(2p + 5) π Example 12.43 Shew that, if x < 0, x + z > 0, then Γ(−x) −x 1 (−x)(1 − x) 1 (−x)(1 − x)(2 − x) + + +··· Γ(z) z 2 z(1 + z) 3 z(1 + z)(2 + z) ∫ 1 1 = t −x−1 {− log(1 − t)} (1 − t)x+z−1 dt, Γ(x + z) 0 and deduce that, when x + z > 0, d Γ(z + x) x 1 x(x − 1) 1 x(x − 1)(x − 2) log = − + −··· . dz Γ(z) z 2 z(z + 1) 3 z(z + 1)(z + 2) Example 12.44 (Binet [73]) Using the result of Example 12.43 above, prove that a − a2 log Γ(z + a) = log Γ(z) + a log z − 2z ∫1 ∫a ∞ Õ a 0 t(l − t)(2 − t) · · · (n − t) dt − 0 t(l − t)(2 − t) · · · (n − t) dt − , (n + 1)z(z + 1)(z + 2) · · · (z + n) n=1 investigating the region of convergence of the series. Example 12.45 Prove that, if p > 0, q > 0, then 1

B(p, q) =

1

p p− 2 q q− 2 (p + q)

p+q− 12

(2π) 2 e M(p,q), 1

where M(p, q) = 2ρ

∫ 0

∞

(t 3 + t)ρ3 arctan dt, e2πtρ − 1 pq(p + q) 1

and ρ2 = p2 + q2 + pq. Example 12.46 If U = 2x/2 /Γ(1 − 12 x), V = 2x/2 /Γ( 12 − 12 x), and if the function F(x) be defined by the equation √ dU dV F(x) = π V −U , dx dx shew:

12.6 Miscellaneous examples

275

(a) that F(x) satisfies the equation F(x + 1) = xF(x) +

1 ; Γ(1 − x)

(b) that, for all positive integral values of x, F(x) = Γ(x); (c) that F(x) is analytic for all finite values of x; (d) that Γ 1−x 1 d 2 . F(x) = log Γ(1 − x) dx Γ 1 − x2 Example 12.47 Expand 1/Γ(a) as a series of ascending powers of a. Note Various evaluations of the coefficients in this expansion have been given by Bourguet [94]; Schlömilch [585, 586]. Example 12.48 (Alexeiwksky) Prove that the G-function, defined by the equation ∞ n Ö z n −z+z 2 /(2n) o 1 1 1 2 1+ e G(z + 1) = (2π) 2 z e− 2 z(z+1)− 2 γz n n=1 is an integral function which satisfies the relations G(z + 1) = Γ(z)G(z), (n!)n = 11 · 22 · 33 · · · nn . G(n + 1)

G(1) = 1,

The most important properties of the G-function are discussed in Barnes [42]. Example 12.49 Shew that G 0(z + 1) 1 1 Γ 0(z) = log(2π) + − z + z , G(z + 1) 2 2 Γ(z) and deduce that G(1 − z) log = G(1 + z)

∫

s

πz cot πz dz − z log(2π).

0

Example 12.50 Shew that ∫ s 1 1 log Γ(t + 1) dt = z log(2π) − z(z + 1) + z log Γ(z + 1) − log G(z + 1). 2 2 0

13 The Zeta-Function of Riemann

13.1 Definition of the zeta-function Let s = σ + it where σ and t are real 1 ; then, if δ > 0, the series ∞ Õ 1 ns n=1

ζ(s) =

is a uniformly convergent series of analytic functions (§§2.33, 3.34) in any domain in which σ ≥ 1 + δ; and consequently the series is an analytic function of s in such a domain. The function is called the zeta-function; although it was known to Euler [197], its most remarkable properties were not discovered before Riemann [557] who discussed it in his memoir on prime numbers; it has since proved to be of fundamental importance, not only in the Theory of Prime Numbers, but also in the higher theory of the Gamma-function and allied functions.

13.11 The generalised zeta-function The definition of this function appears to be due to Hurwitz [327]. Many of the properties possessed by the zeta-function are particular cases of properties possessed by a more general function defined, when σ ≥ 1 + δ, by the equation ζ(s, a) =

∞ Õ n=0

1 , (a + n)s

(13.1)

where a is a constant. For simplicity, we shall suppose that 0 < a ≤ 1, and then we take arg(a + n) = 0. It is evident that ζ(s, 1) = ζ(s). (When a has this range of values, the properties of the function are, in general, much simpler than the corresponding properties for other values of a. The results of §13.14 are true for all values of a (negative integer values excepted); and the results of §§13.12, 13.13, 13.2 are true when Re(a) > 0.)

13.12 The expression of ζ(s, a) as an infinite integral Since (a + n)−s Γ(s) =

∫

∞

x s−1 e−(n+a)x dx,

0 1

The letters σ, t will be used in this sense throughout the chapter.

276

13.1 Definition of the zeta-function

277

when arg x = 0 and σ > 0 (and a fortiori when σ ≥ (1 + δ)), we have, when σ ≥ 1 + δ, N ∫ ∞ Õ Γ(s) ζ(s, a) = lim x s−1 e−(n+a)x dx N →∞

= lim

N →∞

n=0

∫

0 ∞

x s−1 e−ax dx − 1 − e−x

0

∫

∞

0

x s−1 −(N +1+a)x e dx . 1 − e−x

Now, when x ≥ 0, e ≥ 1 + x, and so the modulus of the second of these integrals does not exceed ∫ ∞ x σ−2 e−(N +a)x dx = (N + a)1−σ Γ(σ − 1), x

0

which (when σ ≥ 1 + δ) tends to 0 as N → ∞. Hence, when σ ≥ 1 + δ and arg x = 0, ∫ ∞ s−1 −ax 1 x e dx; ζ(s, a) = Γ(s) 0 1 − e−x this formula corresponds in some respects to Euler’s integral for the Gamma-function.

13.13 The expression of ζ(s, a) as a contour integral . This was given by Riemann for the ordinary zeta-function. When σ ≥ 1 + δ, consider ∫

(0+)

∞

(−z)s−1 e−az dz, 1 − e−z

the contour of integration being of Hankel’s type (§12.22) and not containing the points ±2nπi (n = 1, 2, 3, . . .) which are poles of the integrand; it is supposed (as in §12.22) that | arg(−z)| ≤ π. It is legitimate to modify the contour, precisely as in §12.22, when 2 σ ≥ 1 + δ; and we get ∫ (0+) ∫ πi(s−1) ∞ x s−1 e−ax (−z)s−1 e−az −πi(s−1) dz = e −e dx. 1 − e−z 1 − e−x ∞ 0 Therefore Γ (1 − s) ζ(s, a) = − 2πi

∫

(0+)

∞

(−z)s−1 e−az dz. 1 − e−z

Now this last integral is a one-valued analytic function of s for all values of s. Hence the only possible singularities of ζ(s, a) are at the singularities of Γ(1 − s), i.e. at the points 1, 2, 3, . . . , and, with the exception of these points, the integral affords a representation of ζ(s, a) valid over the whole plane. The result obtained corresponds to Hankel’s integral for the Gamma-function. Also, we have seen that ζ(s, a) is analytic when σ ≥ 1 + δ, and so the only singularity of ζ(s, a) is at the point s = 1. Writing s = 1 in the integral, we get ∫ (0+) −az 1 e dz , 2πi ∞ 1 − e−z 2

If σ ≤ 1, the integral taken along any straight line up to the origin does not converge.

The Zeta-Function of Riemann

278

which is the residue at z = 0 of the integrand, and this residue is 1. Hence lim

s→1

ζ(s, a) = −1. Γ(1 − s)

Since Γ(1 − s) has a single pole at s = 1 with residue −1, it follows that the only singularity of ζ(s, a) is a simple pole with residue +1 at s = 1. Example 13.1.1 Shew that, when Re(s) > 0, 1 1 1 1 − + − +··· 1s 2s 3s 4s ∫ ∞ s−1 1 x dx. = Γ(s) 0 e x + 1

(1 − 21−s ) ζ(s) =

Example 13.1.2 Shew that, when Re(s) > 1, (2s − 1)ζ(s) = ζ s, 12 ∫ ∞ s−1 x 2s x e dx. = Γ(s) 0 e2x − 1 Example 13.1.3 Shew that ζ(s) = −

21−s Γ(1 − s) 2πi(21−s − 1)

∫

(0+)

∞

(−z)s−1 dz, es + 1

where the contour does not include any of the points ±πi, ±3πi, ±5πi, . . . .

13.14 Values of ζ(s, a) for special values of s In the special case when s is an integer (positive or negative), (−z)s−1 e−az 1 − e−z is a one-valued function of z. We may consequently apply Cauchy’s theorem, so that ∫ (0+) 1 (−z)s−1 e−az dz 2πi ∞ 1 − e−z is the residue of the integrand at z = 0, that is to say, it is the coefficient of z−s in To obtain this coefficient we differentiate the expansion (§7.2)

(−1)s−1 e−az . 1 − e−z

∞

e−az − 1 Õ (−1)n φn (a) z n −z −z = e −1 n! n=1 term-by-term with regard to a, where φn (a) denotes the Bernoullian polynomial. (This is z2 e−az obviously legitimate, by §4.7, when |z| < 2π, since −z can be expanded into a power e −1 series in z uniformly convergent with respect to a.) Then ∞

Õ (−1)n φ 0 (a)z n z2 e−az n = . e−z − 1 n=1 n!

(13.2)

13.1 Definition of the zeta-function

279

Therefore if s is zero or a negative integer (= −m), we have ζ(−m, a) = −

0 φm+2 (a) . (m + 1)(m + 2)

In the special case when a = 1, if s = −m, then ζ(s) is the coefficient of z 1−s in the expansion (−1)s m!z . Hence, by §7.2 of z e −1 ζ(−2m) = 0;

ζ(1 − 2m) =

(−1)m Bm 2m

(m = 1, 2, 3, . . .);

1 ζ(0) = − . 2

These equations give the value of ζ(s) when s is a negative integer or zero.

13.15 The formula of Hurwitz for ζ(s, a) when σ < 0 ∫ 1 (−z)s−1 e−az This appears in [327, p. 95]. Consider − dz taken round a contour C 2πi c 1 − e−z consisting of a (large) circle of radius (2N +1)π, (N an integer), starting at the point (2N +1)π and encircling the origin in the positive direction, arg(−z) being zero at z = −(2N + 1)π. In the region between C and the contour (2N π + π, 0+), of which the contour of §13.13 is the limiting form, (−z)s−1 e−az (1 − e−z )−1 is analytic and one-valued except at the simple poles ±2πi, ±4πi, . . . , ±2N πi. Hence ∫ ∫ (0+) N Õ (−z)s−1 e−az 1 (−z)s−1 e−az 1 (Rn + Rn0 ), dz − dz = 2πi C 1 − e−z 2πi (2N +1)π 1 − e−z n=1 where Rn , Rn0 are the residues of the integrand at 2nπi, −2nπi respectively. At the point at which −z = 2nπe−πi/2, the residue is (2nπ)s−1 e− 2 πi(s−1) e−2anπi , 1

and hence Rn + Rn0 = 2(2nπ)s−1 sin (sπ/2 + 2πan) . Hence ∫ (0+) 1 (−z)s−1 e−az − dz 2πi (2N +1)π 1 − e−z N 2 sin(sπ/2) Õ cos(2πan) = (2π)1−s n=1 n1−s ∫ N 2 cos(sπ/2) Õ sin(2πan) 1 (−z)s−1 e−az + − dz. (2π)1−s n=1 n1−s 2πi C 1 − e−z Now, since 0 < a ≤ 1, it is easy to see that we can find a number K independent of N such that e−az (1 − e−z )−1 < K when z is on C. Hence ∫ ∫ π s−1 −az 1 (−z) e [(2N + 1)π]s esiθ dθ < K dz 2πi −z 1−e 2π −π C < K [(2N + 1)π]s e π |s | → 0 as N → ∞

if σ < 0.

The Zeta-Function of Riemann

280

Making N → ∞, we obtain the result of Hurwitz that, if σ < 0, " # ∞ ∞ sπ Õ sπ Õ 2Γ(1 − s) cos(2πan) sin(2πan) + cos ζ(s, a) = sin , (2π)1−s 2 n=1 n1−s 2 n=1 n1−s each of these series being convergent.

13.151 Riemann’s relation between ζ(s) and ζ(1 − s) If we write a = 1 in the formula of Hurwitz given in §13.15, and employ §12.14, we get the remarkable result, due to Riemann, that sπ = π s ζ(1 − s). (13.3) 21−s Γ(s)ζ(s) cos 2 Since both sides of this equation are analytic functions of s, save for isolated values of s at which they have poles, this equation, proved when σ < 0, persists (by §5.5) for all values of s save those isolated values. Example 13.1.4 If m be a positive integer, shew that ζ(2m) =

22m−1 π 2m Bm . (2m)!

Example 13.1.5 (Riemann) Shew that Γ(s/2)π −s/2 ζ(s) is unaltered by replacing s by 1 − s. Example 13.1.6 Deduce from Riemann’s relation that the zeros of Hermite’s integral for ζ(s) at −2, −4, −6, . . . are zeros of the first order.

13.2 Hermite’s formula for ζ(s, a) This appears in [296]. Let us apply Plana’s theorem (Example 7.7 in Chapter 7) to the function φ(z) = (a + z)−s , where arg(a + z) has its principal value. Define the function q(x, y) by the equation 1 [(a + x + iy)−s − (a + x − iy)−s ] 2i −s/2 y sin s arctan = − (a + x)2 + y 2 . x+a |y| π y , we have Since 3 arctan does not exceed the smaller of and x+a 2 x+a πs (1−σ)/2 −1 |q(x, y)| ≤ (a + x)2 + y 2 y sinh , 2 −σ/2 sinh y|s| . ≤ (a + x)2 + y 2 x + a q(x, y) =

Using∫ the first result when |y| > a and the second when |y| < a, it is evident that, if ∞ σ > 0, q(x, y) (e2πy − 1)−1 dy is convergent when x ≥ 0 and tends to 0 as x → ∞; also 0 3

If ξ > 0, arctan ξ =

∫ 0

ξ

dt < 1 + t2

∫ 0

∞

dt ; and arctan ξ < 1 + t2

∫

ξ

dt. 0

13.2 Hermite’s formula for ζ(s, a)

281

∞

∫

(a + x)−s dx converges if σ > 1. Hence, if σ > 1, it is legitimate to make x2 → ∞ in

0

the result contained in the example cited; and we have ∫ ∞ ∫ ∞ h dy y i 1 −s −s . (a + x) dx + 2 ζ(s, a) = a + (a2 + y 2 )−s/2 sin s arctan 2πy 2 a e −1 0 0 So ∞

∫

1 a1−s ζ(s, a) = a−s + +2 2 s−1

0

h y i dy . (a2 + y 2 )−s/2 sin s arctan 2πy a e −1

This is Hermite’s formula 4 ; using the results that, if y ≥ 0, arctan

y y ≤ , a a

y

πa 2

,

we see that the integral involved in the formula converges for all values of s. Further, the integral defines an analytic function of s for all values of s. To prove this, it is sufficient (§5.31) to shew that the integral obtained by differentiating under the sign of integration converges uniformly; that is to say we have to prove that ∫ ∞ 1 dy y 2 2 2 2 −s/2 − log(a + y )(a + y ) sin s arctan 2πy − 1 2 a e 0 ∫ ∞h y i dy y + (a2 + y 2 )−s/2 arctan cos s arctan a a e2πy − 1 0 converges uniformly with respect to s in any domain of values of s. Now when |s| ≤ ∆, where ∆ is any positive number, we have y y y 2 (a + y 2 )−s/2 arctan cos s arctan < (a2 + y 2 )∆/2 cosh 12 π∆ ; a a a since ∆ a

∫ 0

∞

(a2 + y 2 )∆/2

y dy −1

e2πy

converges, uniform convergence of the second integral is justified using de la Vallée Poussin’s test in (I) of §4.431. By dividing the path of integration of the first integral into two parts (0, 12 πa), ( 12 πa, ∞) and using the results y ∆y y 1 , sin s arctan sin s arctan < sinh < sinh π∆ a a a 2 in the respective parts, we can similarly shew that the first integral converges uniformly. Consequently Hermite’s formula is valid (§5.5) for all values of s, and it is legitimate to differentiate under the sign of integration, and the differentiated integral is a continuous function of s. 4

The corresponding formula when a = 1 had been previously given by Jensen.

The Zeta-Function of Riemann

282

13.21 Deductions from Hermite’s formula Writing s = 0 in Hermite’s formula, we see that ζ (0, a) = 21 − a. Making s → 1, from the uniformity of convergence of the integral involved in Hermite’s formula we see that ∫ ∞ 1 1 a1−s − 1 y dy lim ζ(s, a) − + +2 . = lim 2 2 s→1 s→1 s−1 s−1 2a (a + y )(e2πy − 1) 0 Hence, by the example of §12.32, we have lim ζ(s, a) − s→1

1 Γ 0(a) . =− s−1 Γ(a)

Further, differentiating the formula for ζ(s, a) and then making s → 0 (this was justified in §13.2), we get 1 a1−s log a a1−s d = lim − a−s log a − ζ (s, a) − s→0 ds 2 s−1 (s − 1)2 s=0 ∫ ∞ 1 y 1 +2 − log(a2 + y 2 ) · (a2 + y 2 )− 2 s sin s arctan 2 a 0 o y dy y +(a2 + y 2 )−s/2 arctan cos s arctan a a e2πy − 1 ∫ ∞ 1 arctan(y/a) = a− log a − a + 2 dy. 2 e2πy − 1 0 Hence, by §12.32,

d ζ(s, a) ds

s=0

= log Γ(a) − 21 log(2π).

These results had previously been obtained in a different manner by Lerch [425]. The formula for ζ(s, a) from which Lerch derived these results is given in a memoir published by the Academy of Sciences of Prague. A summary of his memoir is contained in [429]. 1 = γ, ζ 0 (0) = − 21 log(2π). Corollary 13.2.1 lim ζ (s) − s−1 s→1

13.3 Euler’s product for ζ(s) Let σ ≥ 1 + δ; and let 2, 3, 5, . . . , p, . . . be the prime numbers in order. Then, subtracting the series for 2−s ζ(s) from the series for ζ(s), we get 1 1 1 1 + + + +··· , 1s 3s 5s 7s for which n is a multiple of 2 being omitted; then in like manner ζ(s) · (1 − 2−s ) =

all the terms of

Í

n−s

1 1 1 + + +··· , 1s 5s 7s all the terms for which n is a multiple of 2 or 3 being omitted, and so on; so that Õ0 ζ(s) · (1 − 2−s )(1 − 3−s ) · · · (1 − p−s ) = 1 + n−s , ζ(s) · (1 − 2−s )(1 − 3−s ) =

13.4 Riemann’s integral for ζ(s)

283

the prime denoting that only those values of n (greater than p) which are prime to 2, 3, . . . , p Í occur in the summation. Now, since the first term of 0 starts with the prime next greater than p, ∞ Õ0 Õ 0 Õ n−s ≤ n−1−s ≤ n−1−s → 0 as p → ∞. n=p+1

Therefore if σ ≥ 1 + δ, the product ζ(s)

(1 − p−s ) converges to 1, where the number Î p assumes the prime values 2, 3, 5, . . . only. But the product (1 − p−s ) converges when Î p

p

σ ≥ 1 + δ, for it consists of some of the factors of the absolutely convergent product ∞ Î (1 − n−s ). Consequently we infer that ζ(s) has no zeros at which σ ≥ 1 + δ; for if it had n=2 Î any such zeros, (1 − p−s ) would not converge at them. Therefore, if σ ≥ 1 + δ, p

Ö

1 1 = . ps ζ(s)

1−

p

This is Euler’s result.

13.31 Riemann’s hypothesis concerning the zeros of ζ(s) It has just been proved that ζ(s) has no zeros at which σ > 1. From the formula (13.3) sπ 2s−1 π s ζ(s) = sec ζ(1 − s) Γ(s) 2 it is now apparent that the only zeros of ζ(s) for which σ < 0 are the zeros of sec (sπ/2) /Γ(s), i.e. the points s = −2, −4, . . .. Hence all the zeros of ζ(s) except those at −2, −4, . . . lie in that strip of the domain of the complex variable s which is defined by 0 ≤ σ ≤ 1. It was conjectured by Riemann, but it has not yet been proved, that all the zeros of ζ(s) in this strip lie on the line σ = 12 ; while it has quite recently been proved by Hardy [279] that an infinity of zeros of ζ(s) actually lie on σ = 12 . It is highly probable that Riemann’s conjecture is correct, and the proof of it would have far-reaching consequences in the theory of Prime Numbers.

13.4 Riemann’s integral for ζ(s) It is easy to see that, if σ > 0, n−s Γ

s 2

π −s/2 =

∞

∫

e−n

2

πx

x s/2−1 dx.

0

Hence, when σ > 0, ζ(s)Γ

s 2

π

−s/2

= lim

N →∞

∫ 0

N ∞Õ n=1

e−n

2

πx

x s/2−1 dx.

The Zeta-Function of Riemann

284

Now, if ϑ(x) =

∞ Í

e−n

2

πx

n=1

, since, by Example 6.17 of Chapter 6,

1 + 2ϑ(x) = x −1/2 (1 + 2ϑ(1/x)) , (13.4) ∫ ∞ we have lim x 1/2 ϑ (x) = 1/2; and hence ϑ (x)x s/2−1 dx converges when σ > 1. x→0

0

Consequently, if σ > 2, ζ(s)Γ

s

π

−s/2

2

= lim

"∫

N →∞

∞

ϑ(x)x

s/2−1

∞

∫ dx −

0

0

∞ Õ

# e

−n2 πx s/2−1

x

dx .

n=N +1

Now, as in §13.12, the modulus of the last integral does not exceed ) ∫ ∞( Õ ∫ ∞ −(N +1)2 πx σ/2−1 ∞ e x −n(N +1)πx σ/2−1 dx e x dx = −(N +1)πx 1−e 0 0 n=N +1 ∫ ∞ 2 −1 < {π(N + 1)} e−(N +2N )πx x σ/2−2 dx 0

= {π(N + 1)}

−1

(N 2 + 2N)π

→ 0 as N → ∞,

since

1−σ/2

Γ (σ/2 − 1)

σ > 2.

Hence, when σ > 2, ∫ ∞ s π −s/2 = ϑ(x)x s/2−1 dx ζ(s)Γ 2 0 ∫ 1n ∫ ∞ o − 12 s/2−1 1 1 − 12 ϑ(x) x s/2−1 dx dx + = − 2 + 2 x + x ϑ(1/x) x 1 0 ∫ 1 ∫ ∞ 1 1 1 1/2 −s/2+1 =− + + x ϑ(x)x ϑ(x) x s/2−1 dx. − 2 dx + s s−1 x ∞ 1 Consequently ζ(s)Γ

s 2

π

−s/2

1 − = s(s − 1)

∫

∞

(x (1−s)/2 + x s/2 )

1

ϑ(x) dx. x

Now the integral on the right represents an analytic function of s for all values of s, by §5.32, since on the path of integration ϑ(x) < e

−πx

x Õ

e−mπ x ≤ e−πx (1 − e−π )−1 .

n=0

Consequently, by §5.5, the above equation, proved when σ > 2, persists for all values of s. If now we put s 1 1 s = + it, s(s − 1)ζ(s)Γ π −s/2 = ξ(t), 2 2 2 we have ξ(t) =

1 − t 2 + 14 2

∫ 1

∞

x −3/4 ϑ(x) cos

1 t 2

log x dx.

13.5 Inequalities satisfied by ζ(s, a) when σ > 0

285

Since ∫

∞

x −3/4 ϑ(x)

1 2

log x

n

cos

1 t 2

log x + 12 nπ dx

1

satisfies the test of Corollary 4.4.1, we may differentiate any number of times under the sign of integration, and then put t = 0. Hence, by Taylor’s theorem, we have for all values 5 of t ξ(t) =

∞ Õ

a2n t 2n ;

(13.5)

n=0

by considering the last integral a2n is obviously real. This result is fundamental in Riemann’s researches.

13.5 Inequalities satisfied by ζ(s, a) when σ > 0 We shall now investigate the behaviour of ζ(s, a) as t → ±∞, for given values of σ. When σ > 1, it is easy to see that, if N be any integer, ζ(s, a) =

N Õ n=0

∞

Õ 1 1 fn (s), − − (a + n)s (1 − s)(N + a)s−1 n=N

where 1 1 1 1 − − 1 − s (n + 1 + a)s−1 (n + a)s−1 (n + 1 + a)s ∫ n+1 u−n du. =s (u + a)s+1 n

fn (s) =

Now, when σ > 0, ∫

n+1

u−n du (u + a)σ+1 n ∫ n+1 du < |s| (n + a)σ+1 n = |s| (n + a)−σ−1 .

| fn (s)| ≤ |s|

Therefore the series σ > 0; so that

∞ Í n=N

∞ Í n=N

fn (s) is a uniformly convergent series of analytic functions when

fn (s) is an analytic function when σ > 0; and consequently, by §5.5, the

function ζ(s, a) may be defined when σ > 0 by the series ζ(s, a) =

N Õ n=0

5

∞

Õ 1 1 − − fn (s). (a + n)s (1 − s)(N + a)s−1 n=N

In this particular piece of analysis it is convenient to regard t as a complex variable, defined by the equation s = 12 + it; and then ξ(t) is an integral function of t.

The Zeta-Function of Riemann

286

Now let btc be the greatest integer in |t|; and take N = btc. Then |ζ(s, a)| ≤

bt c Õ

∞ Õ |(a + n)−s | + (1 − s)−1 (btc + a)1−s + |s|(n + a)−σ−1

n=0

0, we have |ζ(s, a)| < a−σ + (1 − σ)−1 (a + btc)1−σ − a1−σ + |t| −1 (btc + a)1−σ + |s|σ −1 (btc − 1 + a)−σ . Hence ζ(s, a) = O(|t| 1−σ ), the constant implied in the symbol O being independent of s. But, when 1 − δ ≤ σ ≤ 1 + δ, we have ∫ bt c 1−σ |ζ(s, a)| = O(|t| ) + (a + x)−σ dx 0 ∫ bt c 1−σ 1−σ 1−σ < O(|t| ) + a + (a + t) (a + x)−1 dx, 0

since (a + x)−σ ≤ a1−σ (a + x)−1 when σ ≥ 1, and (a + x)−σ ≤ (a + btc)1−σ (a + x)−1 when σ ≤ 1, and so ζ(s, a) = O |t| 1−σ log |t| . When σ ≥ 1 + δ, |ζ(s, a)| ≤ a−σ +

∞ Õ

(a + n)−1−δ = O(1).

n=1

13.51 Inequalities satisfied by ζ(s, a) when σ ≤ 0 We next obtain inequalities of a similar nature when σ ≤ δ. In the case of the function ζ(s) we use Riemann’s relation sπ . ζ(s) = 2s π s−1 Γ(1 − s) ζ(1 − s) sin 2 Now, when σ < 1 − δ, we have, by §12.33, 1 Γ(1 − s) = O e( 2 −s) log(1−s)−(1−s) and so π ζ (s) = O exp |t| + 2

1 2

− σ − it log |1 − s| + i arctan

t (1 − σ)

ζ (1 − s).

13.5 Inequalities satisfied by ζ(s, a) when σ > 0

287

Since arctan t/(1 − σ) = ± 12 π + O(t −1 ), according as t is positive or negative, we see, from the results already obtained for ζ(s, a), that ζ (s) = O |t| 1/2−σ ζ(1 − s). In the case of the function ζ(s, a), we have to use the formula of Hurwitz (§13.15) to obtain the generalisation of this result; we have, when σ < 0, ζ(s, a) = −i(2π)s−1 Γ(1 − s) esπi/2 ζa (1 − s) − e−sπi/2 ζ−a (1 − s) , where ζa (1 − s) =

∞ Õ e2nπia

n1−s

n=1

.

Hence (1 − e2πia )ζa (1 − s) = e2πia +

N Õ

e2nπia ns−1 − (n − 1)s−1

n=2 ∞ Õ

+(s − 1)

∫

n

us−2 du;

e2nπia

n=N +1

n−1

since the series on the right is a uniformly convergent series of analytic functions whenever σ ≤ 1 − δ, this equation gives the continuation of ζa (1 − s) over the range 0 ≤ σ ≤ 1 − δ; so that, whenever σ ≤ 1 − δ, we have | sin πa ζa (1 − s)| ≤ 1 +

N Õ

∞ ∫ Õ nσ−1 + (n − 1)σ−1 + |s − 1| n=N +1

n=2

n−1

Taking N = btc, we obtain, as in §13.5, ζa (1 − s) = O(|t| σ )

(δ ≤ σ ≤ 1 − δ)

σ

= O( | t | log | t | )

(−δ ≤ σ < δ),

and obviously ζa (1 − s) = O(1)

(σ < −δ).

Consequently, whether a is unity or not, we have the results ζ(s, a) = O(|t| 1/2−σ )

(σ ≤ δ)

= O(|t|

1/2

)

= O(|t|

1/2

log |t|)

(δ ≤ σ ≤ 1 − δ) (−δ ≤ σ ≤ δ).

We may combine these results and those of §13.5, into the single formula ζ(s, a) = O(|t|τ(σ) log |t|),

n

uσ−2 du.

288

The Zeta-Function of Riemann

where 6 τ(σ) =

1 −σ 2 1 2

σ ≤ 0, 0 ≤ σ ≤ 21 , 1 ≤ σ ≤ 1, 2 σ ≥ 1,

1−σ 0 and the log |t| may be suppressed except when −δ ≤ σ ≤ δ or when 1 − δ ≤ σ ≤ 1 + δ.

13.6 The asymptotic expansion of log Γ(z + a) From Example 12.1.3 it follows that ∞ z Ö n z −z/n o e−γz Γ(a) = 1+ 1+ e . a n=1 a+n Γ(z + a) Now, the principal values of the logarithms being taken, ∞ n Ö z z −z/n o log 1 + + log 1+ e a a+n n=1 # " Õ ∞ ∞ ∞ Õ Õ (−1)m−1 z m (−1)m−1 z m −az + . + = n(a + n) m (a + n)m m am m=2 m=1 n=1 If |z| < a, the double series is absolutely convergent since ∞ Õ a|z| |z| |z| + − log 1 − n(a + n) a+n a+n n=1 converges. Consequently ∞ ∞ Õ e−γz Γ(a) z Õ az (−1)m−1 m log = − + z ζ(m, a). Γ(z + a) a n=1 n(a + n) m=2 m ∫ πz s 1 Now consider − ζ(s, a) ds, the contour of integration being similar to that 2πi C s sin πs of §12.22 enclosing the points s = 2, 3, 4, . . . but not the points 1, 0, −1, −2, . . .; the residue of (−1)m m the integrand at s = m (m ≥ 2) is z ζ(m, a); and since, as σ → ∞ (where s = σ + it), m ζ(s, a) = O(1), the integral converges if |z| < 1. Consequently ∫ ∞ e−yz Γ(a) z Õ az 1 πz s log = − − ζ(s, a) ds. Γ(z + a) a n=1 n(a + n) 2πi C s sin πs

Hence

6

Γ(a) Γ 0(a) 1 log −z − Γ(z + a) Γ(a) 2πi

It can be proved that τ(σ) may be taken to be

1 2 (1

∫ C

πz s ζ(s, a) ds. s sin πs

− σ) when 0 ≤ σ ≤ 1. See Landau [405, §237].

13.6 The asymptotic expansion of log Γ(z + a)

289

Now let D be a semicircle of (large) radius N with centre at s = 32 , the semicircle lying on the right of the line σ = 23 . On this semicircle ζ(s, a) = O(1), |z s | = |z| σ e−t arg s , and so the integrand is O{|z| σ e−π |t |−t arg z }. The constants implied in the symbol O are independent of s and z throughout. Hence if |z| < 1 and −π + δ ≤ arg z ≤ π − δ, where δ is positive, the integrand is O |z| σ e−δ |t | , and hence ∫ πz s ζ(s, a) ds → 0 D s sin πs as N → ∞. It follows at once that, if |arg z | ≤ π − δ and |z| < 1, log

Γ 0(a) 1 Γ(a) = −z + Γ(z + a) Γ(a) 2πi

3 2 +i∞

∫

3 2 −i∞

πz s ζ(s, a) ds. s sin πs

But this integral defines an analytic function of z for all values of |z| if |arg z| ≤ π − δ. Hence, by §5.5 the above equation, proved when |z| < 1, persists for all values of |z| when |arg z| ≤ π − δ. ∫ 23 ±iR πz s Now consider ζ(s, a) ds, where n is a fixed integer and R is going to tend −n− 12 ±iR s sin πs to infinity. By §13.51, the integrand is O z σ e−sR Rτ(σ) , where −n − 21 ≤ σ ≤ 32 ; and hence if the upper signs be taken, or if the lower signs be taken, the integral tends to zero as R → ∞. Therefore, by Cauchy’s theorem, Γ(a) Γ 0(a) 1 log = −z + Γ(z + a) Γ(a) 2πi

∫

−n− 21 +i∞

−n− 12 −i∞

n Õ πz s Rm, ζ(s, a) ds + s sin πs m=−1

where Rm is the residue of the integrand at s = −m. Now, on the new path of integration πz s −n− 21 −δ |t |τ(−n− 12 ) |t| , e s sin πs ζ(s, a) < K z where∫K is independent of z and t, and τ(σ) is the function defined in §13.51. Consequently, ∞ 1 since e−δ |t | |t|τ(−n− 2 ) dt converges, we have −∞

log

n Γ(a) Γ 0(a) Õ 1 = −z + Rm + O(z−n− 2 ), Γ(z + a) Γ(a) m=−1

when |z| is large.

(−1)m z−m ζ(−m, a) Now, when m is a positive integer, Rm = , and so by §13.14, −m m −m 0 (−1) z φm+2 (a) 0 Rm = , where φm (a) denotes the derivative of Bernoulli’s polynomial. m(m + 1)(m + 2) Also R0 is the residue at s = 0 of 1 1 + 61 π 2 s2 + · · · (1 + s log z + · · · ) s

1 2

− a + s ζ 0(0, a) + · · · ,

The Zeta-Function of Riemann

290

and so R0 = =

1 2 1 2

− a log z + ζ 0(0, a) − a log z + log Γ(a) − 12 log(2π),

by §13.21. And, using §13.21 and writing s = S + 1, R−1 is the residue at S = 0 of 1 π2 S2 1 Γ 0(a) 2 − (1 − S + S − · · · ) 1 + + · · · z(1 + S log z + · · · ) − +··· . S 6 S Γ(a) Γ 0(a) + z. Consequently, finally, if |arg z| ≤ π − δ and |z| is large, Γ(a) log Γ(z + a) = z + a − 12 log z − z + 21 log(2π) n 0 Õ (−1)m−1 φm+2 (a) 1 + + O(z−n− 2 ). m m(m + 1)(m + 2)z m=1

Hence R−1 = −z log z + z

In the special case when a = 1, this reduces to the formula found previously in §12.33 for a more restricted range of values of arg z. The asymptotic expansion just obtained is valid when a is not restricted by the inequality 0 < a ≤ 1; but the investigation of it involves the rather more elaborate methods which are necessary for obtaining inequalities satisfied by ζ(s, a) when a does not satisfy the inequality 0 < a ≤ 1. But if, in the formula just obtained, we write a = 1 and then put z + a for z, it is easily seen that, when |arg(z + a)| ≤ π − δ, we have log Γ(z + a + 1) = z + a + 12 log(z + a) − z − a + 21 log(2π) + o(1); subtracting log(z + a) from each side, we easily see that when both |arg(z + a)| ≤ π − δ

and

|arg z| ≤ π − δ,

we have the asymptotic formula log Γ(z + a) = z + a −

1 2

log z − z + 21 log(2π) + o(1),

where the expression which is o(1) tends to zero as |z| → ∞.

13.7 Miscellaneous examples Example 13.1 (Jensen [359]) Shew that ∫ ∞ 2s−1 s s (2 − 1)ζ(s) = +2 s−1 0

1 4

+ y2

−s/2

sin(s arctan 2y)

dy . −1

e2πy

Example 13.2 (Jensen) Shew that ∫ ∞ 2s−1 dy s ζ(s) = −2 (1 + y 2 )−s/2 sin(s arctan y) πy . s−1 e +1 0 Example 13.3 (Barnes) Discuss the asymptotic expansion of log G(z + a), (Example 12.48) by aid of the generalised zeta-function.

13.7 Miscellaneous examples

291

Example 13.4 (Dirichlet [176]) Shew that, if σ > 1, log ζ(s) =

∞ Õ Õ

1 , mpms m=1

p

the summation extending over the prime numbers p = 2, 3, 5, . . . . Example 13.5 Shew that, if σ > 1, ∞

−

ζ 0(s) Õ Λ(n) = , ζ(s) ns n−1

where Λ(n) = 0 when n is not a power of a prime, and Λ(n) = log p when n is a power of a prime p. Example 13.6 (Lerch [429]) Prove that ∫ ∞ ∫ ∞ 1 2 π2 e−x dx 2 = e−x −wx x s−1 dx. s/2 Γ( s ) 2 0 0 w 2 1 + 4x 2 Example 13.7 (Appell [28]) If φ(s, x) =

∞ Õ xn , ns n=1

where |x| < 1, and Re s > 0, shew that φ(s, x) =

1 Γ(s)

∫

∞

0

xz s−1 dz es − x

and, if s < 1, lim (1 − x)1−s φ(s, x) = Γ(1 − s).

x→1

Example 13.8 (Lerch [426]) If x, a, and s be real, and 0 < a < 1, and s > 1, and if φ(x, a, s) =

∞ Õ e2nπix , (a + n)s n=0

shew that φ(x, a, s) =

1 Γ(s)

∫ 0

∞

e−as z s−1 dz 1 − e2πix−z

and φ(x, a, 1 − s) = Γ(s) πi(s/2−2ax) πi {−s/2+2a(1−x)} e φ(−a, x, s) + e φ(a, 1 − x, s) × . (2π)s Example 13.9 (Hardy) By evaluating the residues at the poles on the left of the straight line taken as contour, shew that, if k > 0, and | arg y| < π2 , ∫ k+i∞ 1 −y e = Γ(u)y −u du, 2πi k−i∞

292

The Zeta-Function of Riemann

and deduce that, if k > 21 , 1 2πi

∫

k+i∞

k−i∞

Γ(u) ζ(2u) du = ϑ(x), (πx)u

and thence that, if a is an acute angle, ∫ ∞ cosh 12 at ξ(t) dt = π cos(a/4) − π2 eia/4 1 + 2ϑ(eia ) . 1 2 t +4 0 Example 13.10 (Hardy) By differentiating 2n times under the integral sign in the last result (Example 13.9) and then making a → π/2, deduce from Example 6.17 of Chapter 6 that ∫ ∞ cosh 41 πt 2n (−1)n π π t ξ(t) dt = cos . 1 2n 2 2 8 t +4 0 By taking n large, deduce that there is no number t0 such that ξ(t) is of fixed sign when t > t0 , and thence that ζ(s) has an infinity of zeros on the line σ = 21 . Note Hardy and Littlewood [281] have shewn that the number of zeros on the line σ = 12 for which 0 < t < T is at least O(T) as T → ∞; if the Riemann hypothesis is true, the number is 1 1 + log 2π T log T − T + O(log T); 2π 2π see Landau [405, p. 370].

14 The Hypergeometric Function

14.1 The hypergeometric series We have already (§2.38) considered the hypergeometric series 1 1+

a·b a(a + 1) · b(b + 1) 2 a(a + 1)(a + 2) · b(b + 1)(b + 2) 3 z+ z + z +··· 1·c 1 · 2 · c(c + 1) 1 · 2 · 3 · c(c + 1)(c + 2)

from the point of view of its convergence. It follows from §2.38 and §5.3 that the series defines a function which is analytic when |z| < 1. It will appear later (§14.53) that this function has a branch point at z = 1 and that if a cut 2 (i.e. an impassable barrier) is made from +1 to +∞ along the real axis, the function is analytic and one-valued throughout the cut plane. The function will be denoted by F(a, b; c; z). Many important functions employed in Analysis can be expressed by means of hypergeometric functions. Thus 3 (1 + z)n = F(−n, β; β; −z), log(1 + z) = zF(1, 1; 2; −z), ez = lim F(1, β; 1; z/β). β→∞

Example 14.1.1 Shew that d ab F(a, b; c; z) = F(a + 1, b + 1; c + 1; z). dz c

14.11 The value of F(a, b; c; 1) when Re(c − a − b) > 0 This analysis is due to Gauss. A method more easy to remember but more difficult to justify is given in Example 14.6.2. 1

The name was given by Wallis in 1655 to the series whose nth term is a(a + b)(a + 2b) · · · (a + (n − 1) b).

2

3

Euler used the term hypergeometric in this sense, the modern use of the term being apparently due to Kummer [388, 389]. The plane of the variable z is said to be cut along a curve when it is convenient to consider only such variations in z which do not involve a passage across the curve in question; so that the cut may be regarded as an impassable barrier. It will be a good exercise for the reader to construct a rigorous proof of the third of these results.

293

The Hypergeometric Function

294

The reader will easily verify, by considering the coefficients of x n in the various series, that if 0 ≤ x < 1, then c{c − 1 − (2c − a − b − 1)x}F(a, b; c; x) + (c − a)(c − b)xF(a, b; c + 1; x) = c(c − 1)(1 − x)F(a, b; c − 1; x) ! ∞ Õ n = c(c − 1) 1 + (un − un−1 )x , n=1

where un is the coefficient of x n in F(a, b; c − 1; x). Now make x → 1. By §3.71, the right-hand side tends to zero if 1+

∞ Í

(un −un−1 ) converges

n=1

to zero, i.e. if un → 0, which is the case when Re(c − a − b) > 0. Also, by §2.38 and §3.71, the left-hand side tends to c(a + b − c)F(a, b; c; 1) + (c − a)(c − b)F(a, b; c + 1; 1) under the same condition; and therefore F(a, b; c; 1) =

(c − a)(c − b) F(a, b; c + 1; 1). c(c − a − b)

Repeating this process, we see that ( m−1 ) Ö (c − a + n)(c − b + n) F(a, b; c; 1) = F(a, b; c + m; 1) (c + n)(c − a − b + n) n=0 ) ( m−1 Ö (c − a + n)(c − b + n) lim F(a, b; c + m; 1), = lim m→∞ (c + n)(c − a − b + n) m→∞ n=0 if these two limits exist. Γ(c)Γ(c − a − b) But (§12.13) the former limit is , if c is not a negative integer; and, if Γ(c − a)Γ(c − b) un (a, b, c) be the coefficient of x n in F(a, b; c; x), and m > |c|, we have |F(a, b; c + m) − 1| ≤ ≤

∞ Õ n=1 ∞ Õ

|un (a, b, c + m)| un (|a|, |b|, m − |c|)

n=1

|c| + |a| + |b| − 1, and is a positive decreasing function of m; therefore, since (m − |c|)−1 → 0, we have lim F(a, b; c + m; 1) = 1;

m→∞

and therefore, finally, F(a, b; c; 1) =

Γ(c)Γ(c − a − b) . Γ(c − a)Γ(c − b)

14.2 The differential equation satisfied by F(a, b; c; z)

295

14.2 The differential equation satisfied by F(a, b; c; z) The reader will verify without difficulty, by the methods of §10.3 that the hypergeometric series is an integral valid near z = 0 of the hypergeometric equation (this equation was given by Gauss). z(1 − z)

du d2u + {c − (a + b + 1)z} − abu = 0; 2 dz dz

from §10.3, it is apparent that every point is an ‘ordinary point’ of this equation, with the exception of 0, 1, ∞, and that these are ‘regular points’. Example 14.2.1 Shew that an integral of the equation

d z z +a dz

d d d z +b u− z −α z −β u=0 dz dz dz

is z a F(a + α, b + α; α − β + 1; z).

14.3 Solutions of Riemann’s P-equation by hypergeometric functions In §10.72 it was observed that Riemann’s differential equation 4 d2u 1 − α − α 0 1 − β − β 0 1 − γ − γ 0 du + + + dz 2 z−a z−b z−c dz 0 0 αα (a − b)(a − c) ββ (b − c)(b − a) γγ 0(c − a)(c − b) + + + z−a (z − c) z−c u × = 0, (z − a)(z − b)(z − c) by a suitable change of variables, could be reduced to a hypergeometric equation; and, carrying out the change, we see that a solution of Riemann’s equation is

z−a z−b

α

z−c z−b

γ

(z − a)(c − b) F α + β + γ, α + β + γ; 1 + α − α ; , (z − b)(c − a) 0

0

provided that α − α 0 is not a negative integer; for simplicity, we shall, throughout this section, suppose that no one of the exponent differences α − α 0, β − β 0, γ − γ 0 is zero or an integer, as (§10.32) in this exceptional case the general solution of the differential equation may involve logarithmic terms; the formulae in the exceptional case will be found in Lindelöf’s memoir [435] to which the reader is referred. See also Klein’s lithographed lectures [372]. Now if α be interchanged with α 0, or γ with γ 0, in this expression, it must still satisfy Riemann’s equation, since the latter is unaffected by this change. 4

The constants are subject to the condition α + α0 + β + β0 + γ + γ0 = 1.

The Hypergeometric Function

296

We thus obtain altogether four expressions, namely, γ z−c (c − b)(z − a) F α + β + γ, α + β 0 + γ; 1 + α − α 0; , z−b (c − a)(z − b) α0 γ z−a z−c (c − b)(z − a) 0 0 0 0 u2 = F α + β + γ, α + β + γ; 1 + α − α; , z−b z−b (c − a)(z − b) α γ0 z−a z−c 0 0 0 0 (c − b)(z − a) u3 = F α + β + γ ,α + β + γ ;1 + α − α ; , z−b z−b (c − a)(z − b) α0 γ0 (c − b)(z − a) z−c z−a 0 0 0 0 0 0 F α + β + γ , α + β + γ ; 1 + α − α; , u4 = z−b z−b (c − a)(z − b) u1 =

z−a z−b

α

which are all solutions of the differential equation. Moreover, the differential equation is unaltered if the triads (α, α 0, a), (β, β 0, b), (γ, γ 0, c) are interchanged in any manner. If therefore we make such changes in the above solutions, they will still be solutions of the differential equation. There are five such changes possible, for we may write {b, c, a}, {c, a, b}, {a, c, b}, {c, b, a}, {b, a, c} in turn in place of {a, b, c}, with corresponding changes of α, α 0, β, β 0,γ, γ 0. We thus obtain 4 × 5 = 20 new expressions, which with the original four make altogether twenty-four series. The twenty new solutions may be written down as follows: β α z−a z−b 0 0 (a − c)(z − b) F β + γ + α, β + γ + α; 1 + β − β ; , = z−c z−c (a − b)(z − c) β0 α z−b z−a (a − c)(z − b) = F β 0 + γ + α, β 0 + γ 0 + α; 1 + β 0 − β; , z−c z−c (a − b)(z − c) β α0 z−b z−a 0 0 0 0 (a − c)(z − b) = F β + γ + α ,β + γ + α ;1 + β − β ; , z−c z−c (a − b)(z − c) β0 α0 z−b z−a (a − c)(z − b) F β 0 + γ + α 0, β 0 + γ 0 + α 0; 1 + β 0 − β; = , z−c z−c (a − b)(z − c)

u5

u6

u7 u8

γ β z−c z−b 0 0 (b − a)(z − c) u9 = F γ + α + β, γ + α + β; 1 + γ − γ ; , z−a z−a (b − c)(z − a) γ0 β (b − a)(z − c) z−b z−c u10 = F γ 0 + α + β, γ 0 + α 0 + β; 1 + γ 0 − γ; , z−a z−a (b − c)(z − a)

u11

14.3 Solutions of Riemann’s P-equation γ β0 z−c z−b (b − a)(z − c) = F γ + α + β 0, γ + α 0 + β 0; 1 + γ − γ 0; , z−a z−a (b − c)(z − a)

z−c u12 = z−a

γ0

z−b z−a

β0

297

(b − a)(z − c) F γ + α + β , γ + α + β ; 1 + γ − γ; , (b − c)(z − a) 0

0

0

0

0

0

β z−b (b − c)(z − a) F α + γ + β, α + γ 0 + β; 1 + α − α 0; , z−c (b − a)(z − c) α0 β z−a z−b (b − c)(z − a) 0 0 0 0 u14 = F α + γ + β, α + γ + β; 1 + α − α; , z−c z−c (b − a)(z − c) α β0 z−a z−b 0 0 0 0 (b − c)(z − a) u15 = F α + γ + β ,α + γ + β ;1 + α − α ; , z−c z−c (b − a)(z − c) α0 β0 z−a z−b (b − c)(z − a) 0 0 0 0 0 0 , u16 = F α + γ + β , α + γ + β ; 1 + α − α; z−c z−c (b − a)(z − c) u13 =

z−a z−c

α

γ α z−a z−c 0 0 (a − b)(z − c) F γ + β + α, γ + β + α; 1 + γ − γ ; , = z−b z−b (a − c)(z − b) γ0 α z−c z−a (a − b)(z − c) = F γ 0 + β + α, γ 0 + β 0 + α; 1 + γ 0 − γ; , z−b z−b (a − c)(z − b) γ α0 z−a z−c 0 0 0 0 (a − b)(z − c) F γ + β + α ,γ + β + α ;1 + γ − γ ; , = z−b z−b (a − c)(z − b) γ0 α0 z−c z−a (a − b)(z − c) 0 0 0 0 0 0 = F γ + β + α , γ + β + α ; 1 + γ − γ; , z−b z−b (a − c)(z − b)

u17

u18

u19

u20

β γ z−b z−c 0 0 (c − a)(z − b) = F β + α + γ, β + α + γ; 1 + β − β ; , z−a z−a (c − b)(z − a) β0 γ z−c (c − a)(z − b) z−b 0 0 0 0 F β + α + γ, β + α + γ; 1 + β − β; , = z−a z−a (c − b)(z − a) β γ0 z−b (c − a)(z − b) z−c = F β + α + γ 0, β + α 0 + γ 0; 1 + β − β 0; , z−a z−a (c − b)(z − a) β0 γ0 z−b z−c (c − a)(z − b) 0 0 0 0 0 0 F β + α + γ , β + α + γ ; 1 + β − β; . = z−a z−a (c − b)(z − a)

u21

u22

u23

u24

(z − a)(c − b) respectively, we (z − b)(c − a) obtain 24 solutions of the hypergeometric equation satisfied by F(A, B; C; x). The existence By writing 0, 1−C, A, B, 0, C− A−B, x for α, α 0, β, β 0, γ,γ 0,

The Hypergeometric Function

298

of these 24 solutions was first shewn by Kummer [388, 389]. They are obtained in a different manner in Forsyth [221, Chap. VI].

14.4 Relations between particular solutions of the hypergeometric equation It has just been shewn that 24 expressions involving hypergeometric series are solutions of the hypergeometric equation; and, from the general theory of linear differential equations of the second order, it follows that, if any three have a common domain of existence, there must be a linear relation with constant coefficients connecting those three solutions. If we simplify u1 , u2 , u3 , u4 ; u17 , u18 , u21 , u22 in the manner indicated at the end of §14.3, we obtain the following solutions of the hypergeometric equation with elements A, B, C, x: y1 = F(A, B; C; x), y2 = (−x)1−C F(A − C + 1, B − C + 1; 2 − C; x), y3 = (1 − x)C−A−B F(C − B, C − A; C; x), y4 = (−x)1−C (1 − x)C−A−B F(1 − B, 1 − A; 2 − C; x), y17 = F(A, B; A + B − C + 1; 1 − x), y18 = (1 − x)C−A−B F(C − B, C − A; C − A − B + 1; 1 − x), y21 = (−x)−B F(A, A − C + 1; A − B + 1; x −1 ), y22 = (−x)−A F(B, B − C + 1; B − A + 1; x −1 ). If |arg(1 − x)| < π, it is easy to see from §2.53 that, when |x| < 1, the relations connecting y1 , y2 , y3 , y4 must be y1 = y3 , y2 = y4 , by considering the form of the expansions near x = 0 of the series involved. In this manner we can group the functions u1, . . . , u24 into six sets of four 5 , viz. u1, u3, u13, u15 ; u2, u4, u14, u16 ; u5, u7, u21, u23 ; u6, u8, u22, u24 ; u9, u11, u17, u19 ; u10, u12, u18, u20, such that members of the same set are constant multiples of one another throughout a suitably chosen domain. In particular, we observe that u1, u3, u13, u15 are constant multiples of a function which (by §5.4, §2.53) can be expanded in the form ( ) ∞ Õ (z − a)a 1 + en (z − a)n n=1

when |z − a| is sufficiently small; when arg(z − a) is so restricted that (z − a)α is one-valued, 5

The special formula F(A, 1; C; x) =

x 1 F C − A, 1; C; , 1−x x−1

which is derivable from the relation connecting u1 with u13 , was discovered in 1730 by Stirling [607, Prop. VII].

14.5 Barnes’ contour integrals

299 0

0

0

this solution of Riemann’s equation is usually written P(α) . And P(α ) ; P(β) , P(β ) ; P(γ) , P(γ ) are defined in a similar manner when |z − a|, |z − b|, |z − c| respectively are sufficiently small. To obtain the relations which connect three members of separate sets of solutions is much more difficult. The relations have been obtained by elaborate transformations of the double circuit integrals which will be obtained later in §14.61; but a more simple and singularly elegant method has recently been discovered by Barnes; of his investigation we shall give a brief account.

14.5 Barnes’ contour integrals for the hypergeometric function This appears in [48]. References to previous work on similar topics by Pincherle, Mellin and Barnes are there given. Consider 1 2πi

∫

i∞

−i∞

Γ(a + s)Γ(b + s)Γ(−s) (−z)s ds, Γ(c + s)

where |arg(−z)| < π, and the path of integration is curved (if necessary) to ensure that the poles of Γ(a + s)Γ(b + s), viz. s = −a − n, −b − n (n = 0, 1, 2, . . .) lie on the left of the path and the poles of Γ(−s), viz. s = 0, 1, 2, . . . , lie on the right of the path. It is assumed that a and b are such that the contour can be drawn, i.e. that a and b are not negative integers (in which case the hypergeometric series is merely a polynomial). From §13.6 it follows that the integrand is O |s| a+b−c−1 exp{− arg(−z) Im(s) − π| Im(s)|} as s → ∞ on the contour, and hence it is easily seen (§5.32) that the integrand is an analytic function of z throughout the domain defined by the inequality |argz| ≤ π − δ, where δ is any positive number. Now, taking note of the relation Γ(−s)Γ(1 + s) = −π cosec πs, consider ∫ Γ(a + s)Γ(b + s) π(−z)s 1 ds, 2πi C Γ(c + s)Γ(1 + s) sin πs where C is the semicircle of radius N + 12 on the right of the imaginary axis with centre at the origin, and N is an integer. Now, by §13.6, we have Γ(a + s)Γ(b + s) π(−z)s (−z)s = O(N a+b−c−1 ) · Γ(c + s)Γ(1 + s) sin sπ sin πs as N → ∞, the constant implied in the symbol O being independent of arg s when s is on the semicircle; and, if s = N + 21 eiθ and |z| < 1, we have (−z)s cosec πs = O exp N + 21 cos θ log |z| − N + 12 sin θ arg(−z) − N + 12 π| sin θ| = O exp N + 12 cos θ log |z| − N + 12 δ| sin θ| h n 1 oi 0 ≤ |θ| ≤ 14 π, O exp 2− 2 N + 12 log |z| oi h n = 1 1 π ≤ |θ| ≤ 12 π. O exp −2− 2 δ N + 21 4

The Hypergeometric Function

300

Hence if log |z| is negative (i.e. |z| < 1), the integrand ∫ tends to zero sufficiently rapidly (for all values of θ under consideration) to ensure that → 0 as N → ∞. C

Now ∫

(∫

i∞

− i(N + 12 )

− −i∞

−i∞

+

∫ C

+

∫

i∞

) ,

i(N + 12 )

by Cauchy’s theorem, is equal to −2πi times the sum of the residues of the integrand at the points s = 0, 1, . . . , N. Make N → ∞, and the last three integrals tend to zero when |arg(−z)| ≤ π − δ, and |z| < 1, and so, in these circumstances, ∫ i∞ N Õ Γ(a + n)Γ(b + n) n Γ(a + s)Γ(b + s)Γ(−s) 1 (−z)s ds = lim z , N →∞ 2πi −i∞ Γ(c + s) Γ(c + n)n! n=0 the general term in this summation being the residue of the integrand at s = n. Thus, an analytic function (namely the integral under consideration) exists throughout the domain defined by the inequality |argz| < π, and, when |z| < 1, this analytic function may be represented by the series ∞ Õ Γ(a + n)Γ(b + n) n z . Γ(c + n) n! n=0 The symbol F(a, b; c; z) will, in future, be used to denote this function divided by Γ(a) Γ(b)/Γ(c).

14.51 The continuation of the hypergeometric series To obtain a representation of the function F(a, b; c; z) in the form of series convergent when |z| > 1, we shall employ the integral obtained in §14.5. If D be the semicircle of radius ρ on the left of the imaginary axis with centre at the origin, it may be shewn 6 by the methods of §14.5 that ∫ Γ(a + s)Γ(b + s)Γ(−s) 1 (−z)s ds → 0 2πi D Γ(c + s) as ρ → ∞, provided that |arg(−z)| < π, |z| > 1 and ρ → ∞ in such a way that the lower bound of the distance of D from poles of the integrand is a positive number (not zero). Hence it can be proved (as in the corresponding work of §14.5) that, when |arg(−z)| < π and |z| > 1, ∫ 1 Γ(a + s)Γ(b + s)Γ(−s) (−z)s ds 2πi Γ(c + s) ∞ Õ Γ(a + n)Γ(1 − c + a + n) sin (c − a − n)π = (−z)−a−n Γ(1 + n)Γ(1 − b + a + n) cos nπ sin(b − a − n)π n=0 + 6

∞ Õ sin(c − b − n)π Γ(b + n)Γ(1 − c + b + n) (−z)−b−n, Γ(1 + n)Γ(1 − a + b + n) cos nπ sin(a − b − n)π n=0

In considering the asymptotic expansion of the integrand when |s | is large on the contour or on D, it is simplest to transform Γ(a + s), Γ(b + s), Γ(c + s) by the relation of §12.14.

14.5 Barnes’ contour integrals

301

the expressions in these summations being the residues of the integrand at the points s = −a − n, s = −b − n respectively. It then follows at once on simplifying these series that the analytic continuation of the series, by which the hypergeometric function was originally defined, is given by the equation Γ(a)Γ(b) Γ(a)Γ(b − a) F(a, b; c; z) = (−z)−a F(a, 1 − c + a; 1 − b + a; z−1 ) Γ(c) Γ(c − a) Γ(b)Γ(a − b) + (−z)−b F(b, 1 − c + b; 1 − a + b; z−1 ), Γ(c − b) where |arg(−z)| < π. It is readily seen that each of the three terms in this equation is a solution of the hypergeometric equation (see §14.4). This result has to be modified when a − b is an integer or zero, as some of the poles of Γ(a + s)Γ(b+ s) are double poles, and the right-hand side then may involve logarithmic terms, in accordance with §14.3. Corollary 14.5.1 Putting b = c, we see that, if |arg(−z)| < π, ∫ i∞ 1 −a Γ(a + s)Γ(−s)(−z)s ds, Γ(a)(1 − z) = 2πi −i∞ where (1 − z)−a → 1 as z → 0, and so the value of |arg(1 − z)| which is less than π always has to be taken in this equation, in virtue of the cut (see §14.1) from 0 to +∞ caused by the inequality |arg(−z)| < π.

14.52 Barnes’ lemma Barnes’ lemma states that, if the path of integration is curved so that the poles of Γ(γ − s)Γ(δ − s) lie on the right of the path and the poles of Γ(α + s)Γ(β + s) lie on the left 7 , then ∫ i∞ Γ(α + γ)Γ(α + δ)Γ(β + γ)Γ(β + δ) 1 Γ(α + s)Γ(β + s)Γ(γ − s)Γ(δ − s) ds = . 2πi −i∞ Γ(α + β + γ + δ) Write I for the expression on the left. If C be defined to be the semicircle of radius ρ on the right of the imaginary axis with centre at the origin, and if ρ → ∞ in such a way that the lower bound of the distance of C from the poles of Γ(γ − s)Γ(δ − s) is positive (not zero), it is readily seen that Γ(α + s)Γ(β + s)Γ(γ − s)Γ(δ − s) Γ(α + s)Γ(β + s) = π 2 cosec π(γ − s) cosec π(δ − s) Γ(1 − γ + s)Γ(1 − δ + s) = O[sα+β+γ+δ−2 exp {−2π | Im(s)|}], as |s| → ∞ on the imaginary axis or on C.

∫ →

Hence the original integral converges; and

0 as ρ

→

∞, when

C

Re(α + β + γ + δ − 1) < 0. Thus, as in §14.5, the integral involved in I is −2πi times 7

It is supposed that α, β, γ, δ are such that no pole of the first set coincides with any pole of the second set.

The Hypergeometric Function

302

the sum of the residues of the integrand at the poles of Γ(γ − s)Γ(δ − s); evaluating these residues we get 8 I=

∞ Õ π Γ(a + γ + n)Γ(β + γ + n) Γ (n + 1)Γ(1 + γ − δ + n) sin π(δ − γ) n=0 ∞ Õ Γ(α + δ + n)Γ(β + δ + n) π + . Γ(n + 1)Γ(1 + δ − γ + n) sin π(γ − δ) n=0

And so, using the result of §12.14 freely, by §14.11: π Γ(α + δ)Γ(β + δ) F(α + δ, β + δ; 1 − γ + δ; 1) sin π(γ − δ) Γ(1 − γ + δ) Γ(α + γ)Γ(β + γ) − F(α + γ, β + γ; 1 − δ + γ; 1) Γ(1 − δ + γ) Γ(α + δ)Γ(β + δ) πΓ(1 − α − β − γ − δ) = sin π(γ − δ) Γ(1 − α − γ)Γ(1 − β − γ) Γ(α + γ)Γ(β + γ) − Γ(1 − α − δ)Γ(1 − β − δ) Γ(α + γ)Γ(β + γ)Γ(α + δ)Γ(β + δ) = Γ(α + β + γ + δ) sin π(α + β + γ + δ) sin π(γ − δ) {sin π(α + γ) sin π(β + γ) − sin π(α + δ) sin π(β + δ)} .

I=

But 2 sin π(α + γ) sin π(β + γ) − 2 sin π(α + δ) sin π(β + δ) = cos π(α − β) − cos π(α + β + 2γ) − cos π(α − β) + cos π(α + β + 2δ) = 2 sin π(γ − δ) sin π(α + β + γ + δ). Therefore I=

Γ(α + γ)Γ(β + γ)Γ(α + δ)Γ(β + δ) Γ(α + β + γ + δ)

which is the required result; it has, however, only been proved when Re(α + β + γ + δ − 1) < 0; but, by the theory of analytic continuation, it is true throughout the domain through which both sides of the equation are analytic functions of, say α, and hence it is true for all values of α, β, γ, δ for which none of the poles of Γ(α + s)Γ(β + s), qua function of s, coincide with any of the poles of Γ(γ − s)Γ(δ − s). Corollary 14.5.2 Writing s + k, α − k, β − k, γ + k, δ + k in place of s, α, β, γ, δ, we see that the result is still true when the limits of integration are −k ± i∞, where k is any real constant. 8

These two series converge (§2.38).

14.6 Solution of Riemann’s equation by a contour integral

303

14.53 The connexion between hypergeometric functions of z and of 1 − z We have seen that, if |arg(−z)| < π, ∫ i∞ 1 Γ(a)Γ(b) Γ(a + s)Γ(b + s)Γ(−s) F(a, b; c; z) = (−z)s ds Γ(c) 2πi −i∞ Γ(c + s) ∫ i∞ ∫ −k+i∞ 1 1 = Γ(a + t)Γ(b + t)Γ(s − t)Γ(c − a − b − t) dt 2πi −i∞ 2πi −k−i∞ Γ(−s) (−z)s × ds, Γ(c − a)Γ(c − b) by Barnes’ lemma. If k be so chosen that the lower bound of the distance between the s contour and the t contour is positive (not zero), it may be shewn that the order of the integrations 9 may be interchanged. Carrying out the interchange, we see that if arg(1 − z) be given its principal value, Γ(c − a)Γ(c − b)Γ(a)Γ(b) F(a, b; c; z) Γ(c) 1 = 2πi

∫

Γ(a + t)Γ(b + t)Γ(c − a − b − t) ∫ i∞ 1 s Γ(s − t)Γ(−s)(−z) ds dt 2πi −i∞

−k−i∞

1 = 2πi

−k+i∞

∫

−k+i∞

Γ(a + t)Γ(b + t)Γ(c − a − b − t)Γ(−t)(1 − z)t dt.

−k−i∞

Now, when |arg(1 − z)| < 2π and |1 − z| < 1, this last integral may be evaluated by the methods of Barnes’ lemma (§14.52); and so we deduce that Γ(c − a)Γ(c − b)Γ(a)Γ(b)F(a, b; c; z) = Γ(c)Γ(a)Γ(b)Γ(c − a − b)F (a, b; a + b − c + 1; 1 − z) + Γ(c)Γ(c − a)Γ(c − b)Γ(a + b − c)(1 − z)c−a−b × F(c − a, c − b; c − a − b + 1; 1 − z), a result which shews the nature of the singularity of F(a, b; c; z) at z = 1. This result has to be modified if c − a − b is an integer or zero, as then Γ(a + t)Γ(b + t)Γ(c − a − b − t)Γ(−t) has double poles, and logarithmic terms may appear. With this exception, the result is valid when |arg(−z)| < π, |arg1 − z| < π. Taking |z| < 1, we may make z tend to a real value, and we see that the result still holds for real values of z such that 0 < z < 1.

14.6 Solution of Riemann’s equation by a contour integral We next proceed to establish a result relating to the expression of the hypergeometric function by means of contour integrals. 9

Methods similar to those of §4.51 may be used, or it may be proved without much difficulty that conditions established by Bromwich [102, §177] are satisfied.

The Hypergeometric Function

304

Let the dependent variable u in Riemann’s P-equation (§10.7) be replaced by a new dependent variable I, defined by the relation u = (z − a)α (z − b)β (z − c)γ I. The differential equation satisfied by I is easily found to be d2 I 1 + α − α 0 1 + β − β 0 1 + γ − γ 0 dI + + + dz 2 z−a z−b z−c dz Í (α + β + γ) {(α + β + γ + 1)z + a(α + β 0 + γ 0 − 1)} I = 0, + (z − a)(z − b)(z − c) which can be written in the form Q(z)

dI d2 I − {(λ − 2)Q 0(z) + R(z)} 2 dz dz + 12 (λ − 2)(λ − 1)Q 00(z) + (λ − 1) R 0(z) I = 0,

where λ = 1 − α − β − γ = α 0 + β 0 + γ 0, Q(z) = (z − a)(z − b)(z − c), Õ R(z) = (α 0 + β + γ)(z − b)(z − c). It must be observed that the function I is not analytic at ∞, and consequently the above differential equation in I is not a case of the generalised hypergeometric equation. We shall now shew that this differential equation can be satisfied by an integral of the form ∫ 0 0 0 I = (t − a)α +β+γ−1 (t − b)α+β +γ−1 (t − c)α+β+γ −1 (z − t)−α−β−γ dt, C

provided that C, the contour of integration, is suitably chosen. For, if we substitute this value of I in the differential equation, the condition 10 that the equation should be satisfied becomes ∫ 0 0 0 (t − a)α +β+γ−1 (t − b)α+β +γ−1 (t − c)α+β+γ −1 (z − t)−α−β−γ−2 K dt = 0, C

where K = (λ − 2) Q(z) + (t − z) Q 0(z) + 12 (t − z)2 Q 00(z) +(t − z){R(z) + (t − z)R 0(z)}

Õ = (λ − 2){Q(t) − (t − z)3 } + (t − z){R(t) − (t − z)2 (α 0 + β + γ)} Õ = −(1 + α + β + γ)(t − a)(t − b)(t − c) + (α 0 + β + γ)(t − b)(t − c)(t − z). ∫ dV It follows that the condition to be satisfied reduces to dt = 0, where C dt V = (t − a)α +β+γ (t − b)α+β +γ (t − c)α+β+γ (t − z)−(1+α+β+γ) . 0

10

0

0

The differentiations under the sign of integration are legitimate (§4.2) if the path C does not depend on z and does not pass through the points a, b, c, z; if C be an infinite contour or if C passes through the points a, b, c or z, further conditions are necessary.

14.6 Solution of Riemann’s equation by a contour integral

305

The integral I is therefore a solution of the differential equation, when C is such that V resumes its initial value after t has described C. Now V = (t − a)α +β+γ−1 (t − b)α+β +γ−1 (t − c)α+β+γ −1 (z − t)−α−β−γ U, 0

0

0

where U = (t − a)(t − b)(t − c)(z − t)−1 . Now U is a one-valued function of t; hence, if C be a closed contour, it must be such that the integrand in the integral I resumes its original value after t has described the contour. Hence finally any integral of the type ∫ 0 0 α β γ (z − a) (z − b) (z − c) (t − a)β+γ+α −1 (t − b)γ+α+β −1 C

× (t − c)α+β+γ −1 (z − t)−α−β−γ dt, 0

where C is either a closed contour in the t-plane such that the integrand resumes its initial value after t has described it, or else is a simple curve such that V has the same value at its termini, is a solution of the differential equation of the general hypergeometric function. Note The reader is referred to the memoirs of Pochhammer [527], and Hobson [314], for an account of the methods by which integrals of this type are transformed so as to give rise to the relations of §14.51 and §14.53. Example 14.6.1 To deduce a real definite integral which, in certain circumstances, represents the hypergeometric series. The hypergeometric series F(a, b; c; z) is, as already shewn, a solution of the differential equation defined by the scheme ∞ 1 0 0 a 0 z . P 1 − c b c − a − b If in the integral ∫ 0 z β t γ+α+β −1 α γ β+γ+α0 −1 (z − a) 1 − (z − c) (t − a) 1− b b C 0 × (t − c)α+β+γ −1 (t − z)−α−β−γ dt, which is a constant multiple of that just obtained, we make b → ∞ (without paying attention to the validity of this process), we are led to consider ∫ t a−c (t − 1)c−b−1 (t − z)−a dt. C

Now the limiting form of V in question is t 1−c+a (t − 1)c−b (t − z)−1−a, and this tends ∫ ∞ to zero at t = 1 and t = ∞, provided Re(c) > Re(b) > 0. We accordingly consider t a−c (t − 1)c−b−1 (t − z)−a dt, where z is not 11 positive and greater than 1. In this 1 11

This ensures that the point t = 1/z is not on the path of integration.

The Hypergeometric Function

306

integral, write t = u−1 ; the integral becomes ∫ 1 ub−1 (1 − u)c−b−1 (1 − uz)−a du. 0

We are therefore led to expect that this integral may be a solution of the differential equation for the hypergeometric series. The reader will easily see that if Re(c) > Re(b) > 0, and if arg u = arg(1 − u) = 0, while the branch of 1 − uz is specified by the fact that (1 − uz)−a → 1 as u → 0, the integral just found is Γ(b)Γ(c − b) F(a, b; c; z). Γ(c) This can be proved by expanding 12 (1 − uz)−a in ascending powers of z when |z| < 1 and using §12.41. Example 14.6.2 Deduce the result of (§14.11) from the preceding example.

14.61 Determination of an integral which represents P(α) We shall now shew how an integral which represents the particular solution P(α) (§14.3) of the hypergeometric differential equation can be found. We have seen (§14.6) that the integral ∫ 0 0 I = (z − a)α (z − b)β (z − c)γ (t − a)β+γ+α −1 (t − b)γ+α+β −1 C

× (t − c)α+β+γ −1 (t − z)−α−β−γ dt 0

satisfies the differential equation of the hypergeometric function, provided C is a closed contour such that the integrand resumes its initial value after t has described C. Now the singularities of this integrand in the t-plane are the points a, b, c, z; and after describing the double circuit contour (§12.43) symbolised by (b+, c+, b−, c−) the integrand returns to its original value. Now, if z lie in a circle whose centre is a, the circle not containing either of the points b and c, we can choose the path of integration so that t is outside this circle, and so |z − a| < |t − a| for all points t on the path. Now choose arg(z − a) to be numerically less than π and arg(z − b), arg(z − c) so that they reduce to arg(a − b), arg(a − c) when z → a, the values of arg(a − b), arg(a − c) being fixed. Now fix arg(t − a), arg(t − b), arg(t − c) at the point N at which the path of integration starts and ends; also choose arg(t − z) to reduce to arg(t − a) when z → a. Then n z − a o (z − b)β = (a − b)β 1 + β +··· , o n za −− ab γ γ (z − c) = (a − c) 1 + γ +··· , a−c 12

The justification of this process by (§4.7) is left to the reader.

14.7 Relations between contiguous hypergeometric functions

307

and since we can expand (t − z)−a−β −γ into an absolutely and uniformly convergent series n o a−z +··· , (t − a)−a−β−γ 1 − (α + β + γ) t−a we may expand the integral into a series which converges absolutely. Multiplying up the absolutely convergent series, we get a series of integer powers of z − a multiplied by (z − a)α . Consequently we must have ∫ (b+,c+,b−,c−) 0 I = (a − b)β (a − c)γ P(α) (t − a)β+γ+α −1 N

× (t − b)γ+α+β −1 (t − c)α+β+γ −1 dt. 0

0

We can define P(α ) , P(β) , P(β ), P(γ) , P(γ ) by double circuit integrals in a similar manner. 0

0

0

14.7 Relations between contiguous hypergeometric functions Let P(z) be a solution of Riemann’s equation with argument z, singularities a, b, c, and exponents α, α 0, β, β 0, γ, γ 0. Further, let P(z) be a constant multiple of one of the six 0 0 0 functions P(α) , P(α ) , P(β) , P(β ) , P(γ) , P(γ ) . Let P`+1, m−1 (z) denote the function which is obtained by replacing two of the exponents, ` and m, in P(z) by ` + 1 and m − 1 respectively. Such functions P`+1, m−1 (z) are said to be contiguous to P(z). There are 6 × 5 = 30 contiguous functions, since ` and m may be any two of the six exponents. It was first shewn by Riemann [556] 13 that the function P(z) and any two of its contiguous functions are connected by a linear relation, the coefficients in which are polynomials in z. There will clearly be 12 × 30 × 29 = 435 of these relations. To shew how to obtain them, we shall take P(z) in the form ∫ 0 0 α β γ P(z) = (z − a) (z − b) (z − c) (t − a)β+γ+α −1 (t − b)γ+α+β −1 C

× (t − c)α+β+γ −1 (t − z)α−β−γ dt, 0

where C is a double circuit contour of the type considered in (§14.61). First, since the integral round C of the differential of any function which resumes its initial value after t has described C is zero, we have ∫ d 0 0 0 (t − a)α +β+γ (t − b)α+β +γ−1 (t − c)α+β+γ −1 (t − z)−α−β−γ dt. 0= C dt On performing the differentiation by differentiating each factor in turn, we get (α 0 + β + γ)P + (α + β 0 + γ − 1)Pα0 +1, β0 −1 + (α + β + γ 0 − 1)Pα0 +1, γ0 −1 (α + β + γ) = Pβ+1,γ0 −1 . z−b Considerations of symmetry shew that the right-hand side of this equation can be replaced by (α + β + γ) Pβ0 −1,γ+1 . z−c 13

Gauss had previously obtained 15 relations between contiguous hypergeometric functions.

The Hypergeometric Function

308

These, together with the analogous formulae obtained by cyclical interchange 14 of (a, α, α 0) with (b, β, β 0) and (c, γ, γ 0), are six linear relations connecting the hypergeometric function P with the twelve contiguous functions Pα+1,β0 −1,

Pβ+1,γ0 −1,

Pγ+1,α0 −1, Pα+1,γ0 −1,

Pβ+1,α0 −1,

Pα0 +1,β0 −1,

Pα0 +1,γ0 −1,

Pβ0 +1,γ0 −1, Pβ0 +1,α0 −1,

Pγ0 +1,α0 −1,

Pγ+1,β0 −1, Pγ0 +1,β0 −1 .

Next, writing t − a = (t − b) + (b − a), and using 15 Pα0 −1 to denote the result of writing α − 1 for α 0 in P, we have 0

P = Pα0 −1,β0 +1 + (b − a)Pα0 −1 . Similarly P = Pα0 −1,γ0 +1 + (c − a)Pα0 −1 . Eliminating Pa0 −1 from these equations, we have (c − b)P + (a − c)Pα0 −1,β0 +1 + (b − a)Pα0 −1,γ0 +1 = 0. This and the analogous formulae are three more linear relations connecting P with the last six of the twelve contiguous functions written above. Next, writing (t − z) = (t − a) − (z − a), we readily find the relation P=

1 Pβ+1,γ0 −1 − (z − a)α+1 (z − b)β (z − c)γ z−b ∫

(t − a)β+γ+α −1 (t − b)γ+α+β −1 (t − c)α+β+γ −1 (t − z)−α−β−γ−1 dt, 0

×

0

0

C

which gives the equations (z − a)−1 P − (z − b)−1 Pβ+1,γ0 −1 = (z − b)−1 P − (z − c)−1 Pγ+1,α0 −1 = (z − c)−1 P − (z − a)−1 Pα+1,β0 −1 . These are two more linear equations between P and the above twelve contiguous functions. We have therefore now altogether found eleven linear relations between P and these twelve functions, the coefficients in these relations being rational functions of z. Hence each of these functions can be expressed linearly in terms of P and some selected one of them; that is, between P and any two of the above functions there exists a linear relation. The coefficients in this relation will be rational functions of z, and therefore will become polynomials in z when the relation is multiplied throughout by the least common multiple of their denominators. The theorem is therefore proved, so far as the above twelve contiguous functions are concerned. It can, without difficulty, be extended so as to be established for the rest of the thirty contiguous functions. Corollary 14.7.1 If functions be derived from P by replacing the exponents α, α 0, β, β 0, γ, γ 0 by α + p, α 0 + q, β + r, β 0 + s, γ + t, γ 0 + u, where p, q, r, s, t ,u are integers satisfying the relation p + q + r + s + t + u = 0, then between P and any two such functions there exists a linear relation, the coefficients in which are polynomials in z. 14 15

The interchange is to be made only in the integrands; the contour C is to remain, unaltered. Pa0 −1 is not a function of Riemann’s type since the sum of its exponents at a, b, c is not unity.

14.8 Miscellaneous examples

309

This result can be obtained by connecting P with the two functions by a chain of intermediate contiguous functions, writing down the linear relations which connect them with P and the two functions, and from these relations eliminating the intermediate contiguous functions. Many theorems which will be established subsequently, e.g. the recurrence formulae for the Legendre functions (§15.21), are really cases of the theorem of this article.

14.8 Miscellaneous examples Example 14.1 Shew that F(a, b + 1; c; z) − F(a, b; c; z) =

az F(a + 1, b + 1; c + 1; z). c

Example 14.2 Shew that if a is a negative integer while β and γ are not integers, then the ratio F(α, β; α + β + 1 − γ; 1 − x) F(α, β; γ; x) is independent of x, and find its value. dP d2 P and dz dz 2 linearly in terms of P and contiguous functions, and hence find the linear relation between dP d2 P P, , and 2 , i.e. verify that P satisfies the hypergeometric differential equation. dz dz Example 14.3 If P(z) be a hypergeometric function, express its derivatives

Example 14.4 Shew that F { 41 , 14 ; 1; 4z(1−z)} satisfies the hypergeometric equation satisfied by F( 12 , 12 ; 1; z). Shew that, in the left-hand half of the lemniscate |z(1 − z)| = 41 , these two functions are equal; and in the right-hand half of the lemniscate, the former function is equal to F( 12 , 12 ; 1; 1 − z). Example 14.5 (Gauss) If Fa+ = F(a + 1, b; c; x), Fa− = F(a − 1, b; c; x) determine the 15 linear relations with polynomial coefficients which connect F(a, b; c; x) with pairs of the six functions Fa+ , Fa− , Fb+ , Fb− , Fc+ , Fc− . Example 14.6 Shew that the hypergeometric equation x(x − 1)

d2 y dy − {γ − (α + β + 1)x} + αβy = 0 2 dx dx

is satisfied by the two integrals (supposed convergent) ∫ 1 z β−1 (1 − z)γ−β−1 (1 − xz)−α dz 0

and ∫

1

z β−1 (1 − z)α−γ {1 − (1 − x)z}−α dz.

0

Example 14.7 (Math. Trip. 1896) Shew that, for values of x between 0 and 1, the solution of the equation x(1 − x)

dy d2 y 1 + (α + β + 1)(1 − 2x) − αβy = 0 2 dx 2 dx

The Hypergeometric Function

310

is

α+1 β+1 3 α β 1 2 2 , ; ; (1 − 2x) + B(1 − 2x)F , ; (1 − 2x) , AF 2 2 2 2 2 2

where A, B are arbitrary constants and F(α, β; γ; x) represents the hypergeometric series. Example 14.8 (Hardy) Shew that lim− F(α, β; γ; x)− x→1

k Õ n+γ−α−β n Γ(α + β − γ − n)Γ(γ − α + n)Γ(γ − β + n)Γ(γ) (1 − x) (−1) n !Γ(γ − α)Γ(γ − β)Γ(α)Γ (β) n=0 =

Γ(γ − α − β)Γ(γ) Γ(γ − α)Γ(γ − β)

where k is the integer such that k ≤ Re(α + β − γ) < k + 1. (This specifies the manner in which the hypergeometric function becomes infinite when x → 1− provided that α + β − γ is not an integer.) Example 14.9 (M. J. M. Hill [308]) Shew that, when Re(γ − α − β) < 0, then Γ(γ)nα+β−γ →1 as n → ∞, Sn (α + β − γ)Γ(α)Γ(β) where Sn denotes the sum of the first n terms of the series for F(α, β; γ; 1). Example 14.10 (Appell [29]) Shew that, if y1 , y2 be independent solutions of dy d2 y +P + Qy = 0, 2 dx dx then the general solution of d3 z d2 z dP dz dQ 2 z=0 + 3P + 2P + + 4Q + 4PQ + 2 dx 3 dx 2 dx dx dx is z = Ay1 2 + By1 y2 + cy2 2 , where A, B, C are constants. Example 14.11 (Clausen [148]) Deduce from Example 14.10 above that, if a + b + ∞

(F(a, b; c; x))2 =

Γ(c)Γ(2c − 1) Õ Γ(2a + n)Γ(a + b + n)Γ(2b + n) n x . Γ(2a)Γ(2b)Γ(a + b) n=0 n!Γ(c + n)Γ(2c − 1 + n)

Example 14.12 (Kummer) Shew that, if |x| < 21 and |x(1 − x) | < 41 , 1 1 F 2α, 2β; α + β + ; x = F α, β; α + β + ; 4x(1 − x) . 2 2 Example 14.13 Deduce from Example 14.12 above that Γ(α + β + 21 )Γ( 12 ) 1 1 F 2α, 2β; α + β + ; = . 2 2 Γ(α + 12 )Γ(β + 12 )

1 2

= c,

14.8 Miscellaneous examples

311

Example 14.14 (Watson [647]) Shew that, if ω = e2πi/3 and Re(a) < 1, 1 πi(3a − 1) Γ(2a)Γ(a − 3 ) 2 3(α−1)/2 F(a, 3a − 1 ; 2a; −ω ) = 3 exp , 6 Γ (3a − 1)Γ( 23 ) 1 πi(3a − 1) Γ(2a)Γ(a − 3 ) . F(a, 3a − 1; 2a; −ω) = 33(α−1)/2 exp − 6 Γ(3a − 1)Γ( 23 ) Example 14.15 (Heymann [302]) 1 1 F − n, − n + 2 2

Shew that n 4 3 1 3 1 8 Γ( 3 )Γ(n + 2 ) ;n + ;− = . 2 2 3 9 Γ( 32 )Γ(n + 43 )

Example 14.16 (Cayley [133]. See also Orr [510]) If (1 − x)α+β−γ F(2α, 2β; 2γ; x) = 1 + Bx + C x 2 + Dx 3 + · · · , shew that F(α, β; γ + 21 ; x) F(γ − α, γ − β; γ + 12 ; x) γ γ(γ + 1) =1+ Bx + C x2 1 γ+2 (γ + 12 )(γ + 32 ) γ(γ + 1)(γ + 2) + Dx 3 + · · · . (γ + 12 )(γ + 32 )(γ + 52 ) Example 14.17 (Le Vavasseur) If the function F(α, β, β 0, γ; x, y) be defined by the equation ∫ 1 Γ(γ) 0 0 uα−1 (1 − u)γ−α−1 (1 − ux)−β (1 − uy)−β du, F(α, β, β , γ; x, y) = Γ(α)Γ(γ − α) 0 then shew that between F and any three of its eight contiguous functions F(α ± 1), F(β ± 1), F(β 0 ± 1), F(γ ± 1), there exists a homogeneous linear equation, whose coefficients are polynomials in x and y. Example 14.18 (Math. Trip. 1893) If γ − α − β < 0, shew that, as x → 1− , Γ(γ)Γ(α + β − γ) γ−α−β F(α, β; γ; x) (1 − x) →1, Γ(α)Γ(β) and that, if γ − α − β = 0, the corresponding approximate formula is 1 Γ(α + β) log → 1. F(α, β; γ; x) Γ(α)Γ(β) 1−x Example 14.19 (Pochhammer) Shew that, when |x| < 1, ∫ (x + , 0+ , x − , 0− ) x 1−γ (ν − x)γ−α−1 ν α−1 (1 − ν)−β dν c

= −4e πiγ sin πα sin π(γ − α) ·

Γ(γ − α)Γ(α) F(α, β; γ; x), Γ(γ)

where c denotes a point on the straight line joining the points 0, x, the initial arguments of ν − x and of ν are the same as that of x, and arg(1 − ν) → 0 as ν → 0.

The Hypergeometric Function

312

Example 14.20 (Barnes) If, when |arg(1 − x)| < 2π, 1 K(x) = 2πi

∫

i∞

Γ(−s)Γ

−i∞

1 2

+s

2

(1 − x)s ds,

+s

2

and, when |argx| < 2π, 1 K (x) = 2πi 0

∫

i∞

Γ(−s)Γ

−i∞

1 2

x s ds,

by changing the variable s in the integral or otherwise, obtain the following relations: K(x) = K 0(1 − x), K(1 − x) = K (x), 0

x K(x) = (1 − x)−1/2 K , x − 1 x−1 , K(1 − x) = x −1/2 K x K 0(x) = x −1/2 K 0(1/x), K 0(1 − x) = (1 − x)−1/2 K 0

if

|arg(1 − x)| < π,

if

|argx| < π,

if

|arg(1 − x)| < π,

if

|argx| < π,

if

|argx| < π,

if

|arg(1 − x)| < π.

1 , 1−x

Example 14.21 (Barnes) With the notation of the preceding example, obtain the following results " #2 ∞ Õ Γ( 12 + n) 2K(x) = x n, n! n=0 " #2 ∞ 1 Õ Γ( + n) 2 xn 2πK 0(x) = − n! n=0 1 1 1 × log x − 4 log 2 + 4 − +···− , 1 2 2n when |x| < 1, |argx| < π; and K(x) = ∓i(−x)− 2 K(1/x) + (−x)− 2 K 0(1/x), 1

1

when |arg(−x)| < π, the ambiguous sign being the same as the sign of Im(x). Example 14.22 (Appell [30])

Hypergeometric series in two variables are defined by the

14.8 Miscellaneous examples

313

equations F1 (α; β, β 0; γ; x, y) =

Õ αm+n βm β 0

n

m,n

F2 (α; β, β 0; γ, γ 0; x, y) =

Õ αm+n βm β 0

n

m,n

F3 (α, α 0, β, β 0; γ; x, y) =

m!n!γm+n m!n!γm γn0

Õ αm α 0 βm β 0 n

m,n

n

m!n!γm+n

x m y n, x m y n, x m y n,

Õ αm+n βm+n F4 (α, β; γ, γ 0; x, y) = x m y n, 0 m!n!γ γ m n m,n where αm = α(α + 1) · · · (α + m − l), and

Í m,n

means

∞ Í ∞ Í m=0 n=0

.

Obtain the differential equations ∂ 2 F1 ∂ 2 F1 ∂F1 ∂F1 + y(1 − x) + {γ − (α + β + 1)x} − βy − αβF1 = 0, ∂ x2 ∂ x∂ y ∂x ∂y ∂ 2 F2 ∂F2 ∂F2 ∂ 2 F2 + {γ − (α + β + 1)x} − βy − αβF2 = 0, x(1 − x) 2 − x y ∂x ∂ x∂ y ∂x ∂y ∂ 2 F3 ∂F3 ∂ 2 F3 + {γ − (α + β + 1)x} − αβF3 = 0, x(1 − x) 2 + y ∂x ∂ x∂ y ∂x ∂ 2 F4 ∂ 2 F4 ∂F4 ∂ 2 F4 x(1 − x) 2 − 2xy − y 2 2 + { γ − (α + β + 1)x} ∂x ∂ x∂ y ∂y ∂x ∂F4 − αβF4 = 0, − (α + β + 1)y ∂y x (1 − x)

and four similar equations, derived from these by interchanging x with y and α, β, γ with α 0, β 0, γ 0 when α 0, β 0, γ 0 occur in the corresponding series. Example 14.23 (Hermite [292]) If α is negative, and if α = −ν + a, where ν is an integer and α is positive, shew that ∞ Γ(x) Γ(α) Õ Rn = + G n (x) , Γ(x + α) x+n n=1 where (−1)n (α − 1)(α − 2) · · · (α − n) G(−n), n! x x x 1+ ··· 1+ G(x) = 1 + α−1 α−2 α−ν G(x) − G(−n) G n (x) = . x+n Rn =

The Hypergeometric Function

314

Example 14.24 When α < 1, shew that ∞

∞

Õ Rn Γ(x)Γ(α − x) Õ Rn = − , Γ(α) x + n x − α−n n=1 n=1 where

(−1)n α (α + 1) · · · (α + n − 1) . n! Example 14.25 (Hermite [292]) When α > 1, and ν and a are respectively the integral and fractional parts of α, shew that Rn =

∞

∞

Γ(x)Γ(α − x) Õ G(x)ρn Õ G(x)ρν+n = − Γ(α) x+n x−a−n n=1 n=1 i h ρ ρ1 ρν−1 0 + +···+ , − G(x) x−a x−a−1 x−a−ν+1 where x x x G(x) = 1 − 1− ··· 1− a a+1 a+ν−1 and (−1)n a(a + 1) · · · (a + n − 1) ρn = . n! Example 14.26 (Saalschütz [570]) (A number of similar results are given by Dougall [187].) If fn (x, y, v) = n x(x + 1)(y + v + n − 1)(y + v + n) n x(y + v + n − 1) + −··· , 1− y(x + v) 2 y(y + 1)(x + v)(x + v + 1) 1 where n is a positive integer and n1 , n2 , . . . are binomial coefficients, shew that fn (x, y, v) =

Γ(y)Γ(y − x + n)Γ(x + v)Γ(υ + n) . Γ(y − x)Γ(y + n)Γ(v)Γ(x + v + n)

Example 14.27 (Dixon [183]) If F(α, β, γ; δ, ε; x) = 1 +

α(α + 1)β(β + 1)γ(γ + 1) 2 αβγ x+ x +··· , δε + 1 δ(δ + 1)ε(ε + 1)1 · 2

shew that, when Re(δ + ε − 23 α − 1) > 0, then F(α, α − δ + 1, α − ε + 1; δ, ε; 1) = 2−α

Γ( 21 )Γ(δ)Γ(ε)Γ(δ + ε − 32 α − 1) Γ(δ − 21 α)Γ(ε − 12 α)Γ( 12 + 12 α)Γ(δ + ε − α − 1)

.

Example 14.28 (Morley [475]) Shew that, if Re(α) < 23 , then 3 ∞ πα Γ(1 − 3 α) Õ α(α + 1) · · · (α + n − 1) 2 1+ = cos 3 . 1 n! 2 Γ(1 − α) n=1 2

14.8 Miscellaneous examples

315

Example 14.29 (Dixon [184]) If ∫ 1∫ 1 x i−1 (1 − x) j−1 y l−1 (1 − y)k−1 (1 − x y)m−j−k dx dy = B(i, j, k, l, m), 0

0

shew, by integrating with respect to x, and also with respect to y, that B(i, j, k, l, m) is a symmetric function of i + j, j + k, k + l, l + m, m + i. Deduce that F(α, β, γ; δ, ε; 1) Γ(δ)Γ(ε)Γ(δ + ε − α − β − γ) is a symmetric function of δ, ε, δ + ε − α − β, δ + ε − β − γ, δ + ε − γ − α. For a proof of a special case by Barnes, see [50]. Example 14.30 If Fn = F(−n, α + n; γ; x) x 1−γ (1 − x)γ−α d n γ+n−1 = {x (1 − x)α+n−γ }, γ(γ + 1) · · · (γ + n − 1) dx n shew that, when n is a large positive integer, and 0 < x < 1, Γ(γ) π 1 1 −γ γ−α− 21 2 Fn = cos{(2n + a)φ − (2γ − 1)} + O , (sin φ) (cos φ) 1√ 1 4 nγ− 2 π nγ+ 2 where x = sin2 φ. Note This result is contained in the great memoir by Darboux [163, 164]. For a systematic development of hypergeometric functions in which one (or more) of the constants is large, see [652].

15 Legendre Functions

15.1 Definition of Legendre polynomials Consider the expression (1 − 2zh + h2 )− 2 ; when |2zh − h2 | < 1, it can be expanded in a series of ascending powers of 2zh − h2 . If, in addition, |2zh| + |h| 2 < 1, these powers can be multiplied out and the resulting series rearranged in any manner (§2.52) since the expansion 1 of [1 − {|2zh| + |h| 2 }]− 2 in powers of |2zh| + |h| 2 then converges absolutely. In particular, if we rearrange in powers of h, we get 1

(1 − 2zh + h2 )− 2 = P0 (z) + hP1 (z) + h2 P2 (z) + h3 P3 (z) + · · · , 1

where P0 (z) = 1, P4 (z) =

P1 (z) = z,

1 2 1 (3z − 1), P3 (z) = (5z3 − 3z), 2 2 1 P5 (z) = (63z5 − 70z3 + 15z), 8

P2 (z) =

1 (35z4 − 30z2 + 3), 8

and generally n(n − 1) n−2 n(n − 1)(n − 2)(n − 3) n−4 (2n)! n z − Pn (z) = n z + z −··· 2 (n!)2 2(2n − 1) 2 · 4 · (2n − 1)(2n − 3) m Õ (2n − 2r)! (−1)r n = z n−2r , 2 r!(n − r)!(n − 2r)! r=0 where m = 12 n or 21 (n − 1), whichever is an integer. If a, b and δ be positive constants, b being so small that 2ab + b2 ≤ 1 − δ, the expansion 1 of (1 − 2zh + h2 )− 2 converges uniformly with respect to z and h when |z| ≤ a, |h| ≤ b. The expressions P0 (z), P1 (z), . . . , which are clearly all polynomials in z, are known as Legendre polynomials, Pn (z) being called the Legendre polynomial of degree n. Other names are Legendre coefficients and Zonal Harmonics. They were introduced into analysis in 1784 by Legendre [420]. It will appear later (§15.2) that these polynomials are particular cases of a more extensive class of functions known as Legendre functions. Example 15.1.1 By giving z special values in the expression (1 − 2zh + h2 )− 2 , shew that 1

Pn (1) = 1, P2n+1 (0) = 0,

Pn (−1) = (−1)n, 1 · 3 · · · (2n − 1) P2n (0) = (−1)n . 2 · 4 · · · (2n) 316

15.1 Definition of Legendre polynomials

317

Example 15.1.2 (Legendre) From the expansion 1 iθ 1 · 3 2 2iθ 2 − 12 h e +··· (1 − 2h cos θ + h ) = 1 + he + 2 2·4 1 −iθ 1 · 3 2 −2iθ × 1 + he + h e +··· , 2 2·4 shew that 1 · 3 · · · (2n − 1) 1 · (2n) Pn (cos θ) = 2 cos(n − 2)θ 2 cos nθ + 2 · 4 · · · (2n) 2 · (2n − 1) 1 · 3 · (2n) · (2n − 2) + 2 cos(n − 4)θ + · · · . 2 · 4 · (2n − 1)(2n − 3) Deduce that, if θ be a real angle, 1 · 3 · · · (2n − 1) 1 · 3 · (2n)(2n − 2) 1 · (2n) ·2+ ·2+··· 2+ 2 · 4 · · · 2n 2 · (2n − 1) 2 · 4 · (2n − 1)(2n − 3) = Pn (1),

|Pn (cos θ)| ≤

so that |Pn (cos θ)| ≤ 1. Example 15.1.3 (Clare, 1905) Shew that, when z = − 21 , Pn = P0 P2n − P1 P2n−1 + P2 P2n−2 − · · · + P2n P0 .

15.11 Rodrigues’ formula for the Legendre polynomials [561] It is evident that, when n is an integer, ( n ) n Õ n! d dn 2 (z − 1)n = n (−1)r z2n−2r dz n dz r=0 r!(n − r)! =

m Õ (−1)r r=0

n! (2n − 2r)! n−2r z , r!(n − r)! (n − 2r)!

where m = 12 n or 12 (n − 1), the coefficients of negative powers of z vanishing. From the general formula for Pn (z) it follows at once that 1 dn 2 (z − 1)n ; 2n n! dz n this result is known as Rodrigues’ formula. Pn (z) =

Example 15.1.4 Shew that Pn (z) = 0 has n real roots, all lying between ±1.

15.12 Schläfli’s integral for Pn (z) [579] From the result of §15.11 combined with §5.22, it follows at once that ∫ 1 (t 2 − 1)n Pn (z) = dt, 2πi C 2n (t − z)n+1

Legendre Functions

318

where C is a contour which encircles the point z once counter-clockwise; this result is called Schläfli’s integral formula for the Legendre polynomials.

15.13 Legendre’s differential equation We shall now prove that the function u = Pn (z) is a solution of the differential equation (1 − z2 )

du d2u − 2z + n(n + 1)u = 0, dz2 dz

which is called Legendre’s differential equation for functions of degree n. For, substituting Schläfli’s integral in the left-hand side, we have, by §5.22, dPn (z) d 2 Pn (z) − 2z + n(n + 1)Pn (z) 2 dz dz ∫ (n + 1) (t 2 − 1)n = {−(n + 2)(t 2 − 1) + 2(n + 1)t(t − z)} dt n n+3 2πi C 2 (t − z) ∫ (n + 1) d (t 2 − 1)n+1 = dt, 2πi · 2n C dt (t − z)n+2

(1 − z2 )

and this integral is zero, since (t 2 − 1)n+1 (t − z)−n−2 resumes its original value after describing C when n is an integer. The Legendre polynomial therefore satisfies the differential equation. The result just obtained can be written in the form d 2 dPn (z) (1 − z ) + n(n + 1)Pn (z) = 0. dz dz Note It will be observed that Legendre’s equation is a particular case of Riemann’s equation, defined by the scheme −1 ∞ 1 P 0 n+1 0 z . 0 −n 0 Example 15.1.5 Shew that the equation satisfied by −∞ 1 −1 P −r n + r + 1 −r 0 −n + r 0

d r Pn (z) is defined by the scheme dzr

z .

Example 15.1.6 If z2 = η, shew that Legendre’s differential equation takes the form d2 y 1 1 dy n(n + 1)y + − + = 0. dη2 2η 1 − η dη 4η(1 − η) Shew that this is a hypergeometric equation. Example 15.1.7 Deduce Schläfli’s integral for the Legendre functions, as a limiting case

15.1 Definition of Legendre polynomials

319

of the general hypergeometric integral of §14.6. Hint. Since Legendre’s equation is given by the scheme ∞ 1 −1 P 0 n+1 0 z , 0 −n 0 the integral suggested is ∫ t −n z n+1 (t − z)−n−1 dt (t + 1)n (t − 1)n lim 1 − lim 1 − b→∞ b→∞ b b C ∫ = (t 2 − 1)n (t − z)−n−1 dt, C

taken round a contour C such that the integrand resumes its initial value after describing it; and this gives Schläfli’s integral.

15.14 The integral properties of the Legendre polynomials We shall now shew that ∫

1

Pm (z)Pn (z) dz =

−1

0

(m , n),

2 2n + 1

(m = n).

These two results were given by Legendre in 1784 and 1789. dr u Let {u}r denote r ; then, if r ≤ n, {(z2 − 1)n }r is divisible by (z2 − 1)n−r ; and so, if dz r < n, {(z2 − 1)n }r vanishes when z = 1 and when z = −1. Now, of the two numbers m, n, let m be that one which is equal to or greater than the other. Then, integrating by parts continually, ∫ 1 2 1 (z − 1)m m (z2 − 1)n n dz = (z2 − 1)m m−1 (z2 − 1)n n −1 − −1 ∫ 1 2 (z − 1)m m−1 (z2 − 1)n n+1 dz −1

.. . = (−1)m

∫

1

−1

(z2 − 1)m (z2 − 1)n n+m dz,

, (z − 1) m−2 , . . . vanish at both limits. Now, when m > n, (z2 − 1)n m+n = 0, since differential coefficients of (z2 − 1)n of order higher than 2n vanish; and so, when m is greater than n, it follows from Rodrigues’ formula that ∫ 1 Pm (z)Pn (z) dz = 0.

since (z − 1) 2

m

2

m

m−1

−1

Legendre Functions

320

When m = n, we have, by the transformations just obtained, ∫ 1 ∫ 1 2 d 2n (z2 − 1)n 2n (z2 − 1)n dz (z − 1)n n (z2 − 1)n n dz = (−1)n dz −1 −1 ∫ 1 = (2n)! (1 − z2 )n dz −1 ∫ 1 = 2 · (2n)! (1 − z2 )n dz 0

= 2 · (2n)!

∫

1 2π

sin2n+1 θ dθ

0

= 2 · (2n)!

2 · 4 · · · (2n) , 3 · 5 · · · (2n + 1)

where cos θ has been written for z in the integral; hence, by Rodrigues’ formula, ∫ 1 2 2 · (2n)! (2n ; n!)2 {Pn (z)}2 dz = n 2 = . (2 n!) (2n + 1)! 2n + 1 −1 We have therefore obtained both the required results. Note It follows that, in the language of Chapter 11, the functions n + orthogonal functions for the interval (−1, 1).

1 1/2 2

Pn (z) are normal

Example 15.1.8 (Clare, 1908) Shew that, if x > 0, ∫ 1 √ −1 1 (cosh 2x − z)− 2 Pn (z) dz = 2 n + 12 e−(2n+1)x . −1

Example 15.1.9 (Clare, 1902) If I =

∫

1

Pm (z)Pn (z) dz, then 0

(i) I = 1/(2n + 1) when m = n, (ii) I = 0 when m − n is even, (−1)µ+ν n!m! (iii) I = m+n−1 when n = 2ν + 1, m = 2µ. 2 (n − m)(n + m + 1) (ν!)2 (µ!)2

15.2 Legendre functions Hitherto we have supposed that the degree n of Pn (z) is a positive integer; in fact, Pn (z) has not been defined except when n is a positive integer. We shall now see how Pn (z) can be defined for values of n which are not necessarily integers. An analogy can be drawn from the theory of the Gamma-function. The expression z! as ordinarily defined (viz. as z(z − 1)(z − 2) · · · 2 · 1) has a meaning only for positive integral values of z; but when the Gamma-function has been introduced, z! can be defined to be Γ(z + 1), and so a function z! will exist for values of z which are not integers. Referring to §15.13, we see that the differential equation (1 − z2 )

d2u du − 2z + n(n + 1)u = 0 2 dz dz

15.2 Legendre functions

321

is satisfied by the expression u=

1 2πi

∫ C

(t 2 − 1)n dt, 2n (t − z)n+1

even when n is not a positive integer, provided that C is a contour such that (t 2 −1)n+1 (t −z)−n−2 resumes its original value after describing C. Suppose then that n is no longer taken to be a positive integer. The function (t 2 − 1)n+1 (t − z) has three singularities, namely the points t = 1, t = −1, t = z; and it is clear that after describing a circuit round the point t = 1 counter-clockwise, the function resumes its original value multiplied by e2πi(n+1) ; while after describing a circuit round the point t = z counterclockwise, the function resumes its original value multiplied by e2πi(−n−2) . If therefore C be a contour enclosing the points t = 1 and t = z, but not enclosing the point t = −1, then the function (t 2 − 1)n+1 (t − z)−n−2 will resume its original value after t has described the contour C. Hence, Legendre’s differential equation for functions of degree n, −n−2

(1 − z2 )

du d2u − 2z + n(n + 1)u = 0, dz2 dz

is satisfied by the expression u=

1 2πi

(1+,z+)

∫ A

(t 2 − 1)n dt, 2n (t − z)n+1

for all values of n; the many-valued functions will be specified precisely by taking A on the real axis on the right of the point t = 1 (and on the right of z if z be real), and by taking arg(t − 1) = arg(t + 1) = 0 and | arg(t − z)| < π at A. This expression will be denoted by Pn (z), and will be termed the Legendre function of degree n of the first kind. We have thus defined a function Pn (z), the definition being valid whether n is an integer or not. Note The function Pn (z) thus defined is not a one-valued function of z; for we might take two contours as shewn in the figure, and the integrals along them would not be the same; to

make the contour integral unique, make a cut in the t plane from −1 to −∞ along the real axis; this involves making a similar cut in the z plane, for if the cut were not made, then, as z varied continuously across the negative part of the real axis, the contour would not vary continuously. It follows, by §5.31, that Pn (z) is analytic throughout the cut plane.

Legendre Functions

322

15.21 The recurrence formulae We proceed to establish a group of formulae (which are really particular cases of the relations between contiguous Riemann P-functions which were shewn to exist in §14.7) connecting Legendre functions of different degrees. d Pn (z), we have If C be the contour of §15.2, and writing Pn0 (z) for dz ∫ ∫ 1 (t 2 − 1)n n+1 (t 2 − 1)n 0 Pn (z) = n+1 dt; P (z) = dt. n 2 πi C (t − z)n+1 2n+1 πi C (t − z)n+2 Now

d (t 2 − 1)n+1 2(n + 1)t(t 2 − 1)n (n + 1)(t 2 − 1)n+1 = − , dt (t − z)n+1 (t − z)n+1 (t − z)n+2

and so, integrating, 0=2

t(t 2 − 1)n dt − (t − z)n+1

∫ C

∫ C

(t 2 − 1)n+1 dt. (t − z)n+2

Therefore ∫

1 2n+1 πi

C

(t 2 − 1)n 1 dt = n+2 n (t − z) 2 πi

∫ C

(t 2 − 1)n+1 z dt − n+1 n+2 (t − z) 2 πi

∫ C

(t 2 − 1)n dt. (t − z)n+1

Consequently Pn+1 (z) − zPn (z) =

1 2n+1 πi

∫ C

(t 2 − 1)n dt. (t − z)n

(15.1)

Differentiating 1 , we get 0 (z) − zPn0 (z) − Pn (z) = nPn (z), Pn+1

and so 0 Pn+1 (z) − zPn0 (z) = (n + 1)Pn (z)

This is the first of the required formulae. Next, expanding the equation d t(t 2 − 1)n dt = 0, dt (t − z)n

∫ C

we find that ∫ C

(t 2 − 1)n dt + 2n (t − z)n

∫ C

t 2 (t 2 − 1)n−1 dt − n (t − z)n

∫ C

t(t 2 − 1) dt = 0. (t − z)n+1

Writing (t − 1) + 1 for t and (t − z) + z for t in this equation, we get ∫ 2 ∫ 2 ∫ (t − 1)n (t − 1)n−1 (t 2 − 1)n (n + 1) dt + 2n dt − nz dt = 0. n n n+1 C (t − z) C (t − z) C (t − z) 2

2

Using (15.1), we have at once (n + 1) {Pn+1 (z) − zPn (z)} + nPn−1 (z) − nzPn (z) = 0. 1

The process of differentiating under the sign of integration is readily justified by §4.2.

(15.2)

15.2 Legendre functions

323

That is to say (n + 1)Pn+1 (z) − (2n + 1)zPn (z) + nPn−1 (z) = 0,

(15.3)

a relation 2 connecting three Legendre functions of consecutive degrees. This is the second of the required formulae. We can deduce the remaining formulae from (15.2) and (15.3) thus: Differentiating (15.3), we have 0 0 (z) − (2n + 1)Pn (z) = 0. (z) − zPn0 (z) − n zPn0 (z) − Pn−1 (n + 1) Pn+1 0 Using (15.2) to eliminate Pn+1 (z), and then dividing by 3 n we get 0 (z) = nPn (z). zPn0 (z) − Pn−1

(15.4)

0 0 Pn+1 (z) − Pn−1 (z) = (2n + 1)Pn (z).

(15.5)

Adding (15.2) and (15.4) we get

0 Lastly, writing n − 1 for n in (15.2) and eliminating Pn−1 (z) between the equation so obtained and (15.4), we have

(z2 − 1)Pn0 (z) = nzPn (z) − nPn−1 (z).

(15.6)

The formulae (15.2), (15.3), (15.4), (15.5), (15.6), are called the recurrence formulae. Note The above proof holds whether n is an integer or not, i.e. it is applicable to the general Legendre functions. Another proof which, however, only applies to the case when n is a positive integer (i.e. is only applicable to the Legendre polynomials) is as follows: Write V = (1 − 2hz + h2 )− 2 . Then, equating coefficients 4 of powers of h in the expansions on each side of the equation 1

∂V = (z − h)V, ∂h we have nPn (z) − (2n − 1)zPn−1 (z) + (n − 1)Pn−2 (z) = 0, which is the formula (15.3). (1 − 2hz + h2 )

Similarly, equating coefficients of powers of h in the expansions on each side of the equation ∂V ∂V h = (z − h) , ∂h ∂z we have dPn (z) dPn−1 (z) z − = nPn (z), dz dz which is the formula (15.4). The others can be deduced from these. Example 15.2.1 (Hargreaves) Shew that, for all values of n, d 2 2 {z(Pn 2 + Pn+1 ) − 2Pn Pn+1 } = (2n + 3)Pn+1 − (2n + 1)Pn2 . dz 2 3 4

This relation was given in substance by Lagrange [392] in a memoir on probability. If n = 0, we have P0 (z) = 1, P−1 (z) = 1, and the result (15.4) is true but trivial. The reader is recommended to justify these processes.

Legendre Functions

324

Example 15.2.2 (Trinity, 1900) If Mn (x) =

d dz

n

xz

,

(ze cosech z) z=0

shew that dMn (x) = nMn−1 (x) and dx

∫

1

Mn (x) dx = 0.

−1

Example 15.2.3 (Clare, 1898) Prove that if m and n are integers such that m ≤ n, both being even or both odd, ∫ 1 dPm (z) dPn (z) dz = m(m + 1). dz dz −1 Example 15.2.4 (Math. Trip. 1897) Prove that, if m, n are integers and m ≥ n, ∫ 1 2 (n − 1)n(n + 1)(n + 2) d Pm (z) d 2 Pn (z) dz = {3m(m + 1) − n(n + 1) + 6} 2 2 dz dz 48 −1 × {1 + (−1)n+m }.

15.211 The expression of any polynomial as a series of Legendre polynomials Let fn (z) be a polynomial of degree n in z. Then it is always possible to choose a0, a1, . . . , an so that fn (z) ≡ a0 P0 (z) + a1 P1 (z) + · · · + an Pn (z), for, on equating coefficients of z n, z n−1, . . . on each side, we obtain equations which determine an, an−1, . . . uniquely in turn, in terms of the coefficients of powers of z in fn (z). To determine a0, a1, . . . , an in the most simple manner, multiply the identity by Pr (z), and integrate. Then, by §15.14, ∫ 1 2ar fn (z)Pr (z) dz = , 2r +1 −1 when r = 0, 1, 2, . . . , n; when r > n, the integral on the left vanishes. Example 15.2.5 (Legendre [421]) Given z n = a0 P0 (z) + a1 P1 (z) + · · · + an Pn (z), determine a0 , a1 , . . . , a n . Hint. Equate coefficients of z n on both sides; this gives an = Let In,m =

∫

2n (n!)2 . (2n)!

1

z n Pm (z) dz, so that, by the result just given,

−1

Im,m =

2m+1 (m!)2 . (2m + 1)!

Now when n−m is odd, In,m is the integral of an odd function with limits ±1, and so vanishes; and In,m also vanishes when n − m is negative and even.

15.2 Legendre functions

325

To evaluate In,m when n − m is a positive even integer, we have from Legendre’s equation ∫ 1 ∫ 1 d n m(m + 1) z Pm (z) dz = − z n {(1 − z2 )Pm0 (z)} dz dz −1 −1 ∫ 1 n 1 2 0 = − z (1 − z )Pm (z) −1 + n z n−1 (1 − z2 )Pm0 (z) dz −1

=n z

n−1

2

1

(1 − z )Pm (z) −1 ∫ 1 −n {(n − 1)z n−2 − (n + 1)z n }Pm (z) dz, −1

on integrating by parts twice; and so m(m + 1)In,m = n(n + 1)In,m − n(n − 1)In−2,m . Therefore n(n − 1) In−2,m (n − m)(n + m + 1) n(n − 1) · · · (m + 1) = Im,m, (n − m)(n − 2 − m) · · · 2 · (n + m + 1)(n + m − 1) · · · (2m + 3)

In,m =

by carrying on the process of reduction. Consequently In,m =

2m+1 n!( 12 n + 12 m)! ( 12 n − 12 m)!(n + m + 1)!

,

and so am =

0, when n − m is odd or negative and

(2m + 1)2 + ( 1 n − 1 m)!(n + m + 1)! 2 2 m

n!( 21 n

1 m)! 2

when n − m is even and positive.

(15.7)

Example 15.2.6 Express cos nθ as a series of Legendre polynomials of cos θ when n is an integer. Example 15.2.7 (St John’s, 1899) Evaluate the integrals ∫ 1 ∫ 1 zPn (z)Pn+1 (z) dz, z2 Pn (z)Pn+1 (z) dz. −1

−1

Example 15.2.8 (Trinity, 1894) Shew that ∫ 1 2n(n + 1) (1 − z2 ) {Pn0 (z)}2 dz = . 2n + 1 −1 Example 15.2.9 (St John’s, 1898) Shew that nPn (cos θ) =

n Õ r=1

cos rθPn−r (cos θ).

326

Example 15.2.10 (Trinity, 1895)

Legendre Functions ∫ 1 If un = (1 − z2 )n P2m (z) dz, where m < n, shew that −1

(n − m)(2n + 2m + 1)un = 2n2 un−1 .

15.22 Murphy’s expression of Pn (z) as a hypergeometric function This appears in [482]. Murphy’s result was obtained only for the Legendre polynomials. Since (§15.13) Legendre’s equation is a particular case of Riemann’s equation, it is to be expected that a formula can be obtained giving Pn (z) in terms of hypergeometric functions. To determine this formula, take the integral of §15.2 for the Legendre function and suppose that | 1 − z | < 2; to fix the contour C, let δ be any constant such that 0 < δ < 1, and suppose that z is such that |1 − z| ≤ 2(1 − δ); and then take C to be the circle 5 |1 − t| = 2 − δ. Since 1 − z 2 − 2δ 1 − t ≤ 2 − δ < 1, we may expand (t − z)−n−1 into the uniformly convergent series 6 ( ) 2 z − 1 (n + 1)(n + 2) z − 1 −n−1 −n−1 (t − z) = (t − 1) 1 + (n + 1) +··· . + t−1 2! t−1 Substituting this result in Schläfli’s integral, and integrating term-by-term (§4.7) we get ∫ ∞ Õ (z − 1)r (n + 1)(n + 2) · · · (n + r) (1+, z+) (t 2 − 1)n Pn (z) = dt 2n+1 πi r! (t − 1)n+1+r A r=0 ∞ Õ (z − 1)r (n + 1)(n + 2) · · · (n + r) d r n (t + 1) , = 2n (r!)2 dt r t=1 r=0 by §5.22. Since arg(t + 1) = 0 when t = 1, we get r d n (t + 1) = 2n−r n(n − 1) · · · (n − r + 1), dt r t=1 and so, when |1 − z| ≤ 2(1 − δ) < 2, we have r ∞ Õ (n + 1)(n + 2) · · · (n + r) · (−n)(1 − n) · · · (r − 1 − n) 1 1 Pn (z) = − z (r!)2 2 2 r=0 1 1 = F n + 1, −n; 1; − z . 2 2 This is the required expression; it supplies a reason §14.53 why the cut from −1 to −∞ could not be avoided in §15.2. Corollary 15.2.1 From this result, it is obvious that, for all values of n, Pn (z) = P−n−1 (z). 5 6

This circle contains the points t = 1, t = z. The series terminates if n be a negative integer.

15.2 Legendre functions

327

Note When n is a positive integer, the result gives the Legendre polynomial as a polynomial in 1 − z with simple coefficients. Example 15.2.11 (Trinity, 1907) Shew that, if m be a positive integer, m+1 Γ(2m + n + 2) d Pm+n (z) = m+1 . m+1 dz 2 (m + 1)!Γ(n) z=1 Example 15.2.12 (Murphy) Shew that the Legendre polynomial Pn (cos θ) is equal to (−1)n F n + 1, −n; 1; cos2 (θ/2) , and to cosn (θ/2) × F −n, −n; 1; tan2 (θ/2) .

15.23 Laplace’s integrals for Pn (z) This appears in Laplace’s Mécanique Céleste, [409, Livre XI, Ch. 2.]. For the contour employed in this section, and for some others introduced later in the chapter, we are indebted to Mr J. Hodgkinson. We shall next shew that, for all values of n and for certain values of z, the Legendre function Pn (z) can be represented by the integral (called Laplace’s first integral) ∫ 1 π 1 {z + (z2 − 1) 2 cos φ} n dφ. π 0 (A) Proof applicable only to the Legendre polynomials. When n is a positive integer, we have, by §15.12, ∫ 1 (t 2 − 1)n Pn (z) = n+1 dt, 2 πi C (t − z)(n+1) where C is any contour which encircles the point z counter-clockwise. Take C to be the circle 1 with centre z and radius |z2 − 1| 1/2 , so that, on C, t = z + (z2 − 1) 2 eiφ , where φ may be taken to increase from −π to π. Making the substitution, we have, for all values of z, !n ∫ π 1 1 {z − 1 + (z2 − 1) 2 eiφ }{z + 1 + (z2 − 1) 2 eiφ } 1 i dφ Pn (z) = n+1 1 2 πi −π (z2 − 1) 2 eiφ ∫ π 1 1 = {z + (z2 − 1) 2 cos φ} n dφ 2π −π ∫ 1 π 1 = {z + (z2 − 1) 2 cos φ} n dφ, π 0 since the integrand is an even function of φ. The choice of the branch of the two-valued 1 function (z2 − 1) 2 is obviously a matter of indifference. (B) Proof applicable to the Legendre functions, where n is unrestricted. Make the same substitution as in (A) in Schläfli’s integral defining Pn (z); it is, however, necessary in addition to verify that t = 1 is inside the contour and t = −1 outside it, and it is

Legendre Functions

328

also necessary that we should specify the branch of {z + (z2 − 1) 2 cos φ} n , which is now a many-valued function of φ. The conditions that t = 1, t = −1 should be inside and outside C respectively are that the 1 distances of z from these points should be less and greater than |z2 − 1| 2 . These conditions are both satisfied if |z − 1| < |z + 1|, which gives Re(z) > 0, and so (giving arg z its principal value) we must have | arg z| < 12 π. Therefore ∫ π 1 1 Pn (z) = {z + (z2 − 1) 2 cos φ} n dφ, 2π −π 1

where the value of arg{z + (z2 − 1) 2 cos φ} is specified by the fact that it, being equal to arg(t 2 − 1) − arg(t − z), is numerically less than π when t is on the real axis and on the right of z (see §15.2). 1

Now as φ increases from −π to π, z + (z2 − 1) 2 cos φ describes a straight line in the Argand 1 1 diagram going from z − (z 2 − 1) 2 to z + (z2 − 1) 2 and back again; and since this line does not 1 pass through the origin 7 , arg{z + (z2 − 1) 2 cos φ} does not change by so much as π on the range of integration. 1 Now suppose that the branch of {z + (z2 − 1) 2 cos φ} n which has to be taken is such that it reduces to z n e2kπi n (where k is an integer) when φ = 21 π. Then ∫ e2nkπi π 1 {z + (z2 − 1) 2 cos φ} n dφ, Pn (z) = 2π −π 1

where now that branch of the many-valued function is taken which is equal to z n when φ = 21 π. Now make z → 1 by a path which avoids the zeros of Pn (z); since Pn (z) and the integral are analytic functions of z when | arg z| < 21 π, k does not change as z describes the path. And so we get e2nkπi = 1. Therefore, when | arg z| < 12 π and n is unrestricted, ∫ π 1 1 {z + (z2 − 1) 2 cos φ} n dφ, Pn (z) = 2π −π where arg{z + (z2 − 1) 2 cos φ} is to be taken equal to arg z when φ = 12 π. This expression for Pn (z), which may, again, obviously be written ∫ 1 π 1 {z + (z2 − 1) 2 cos φ} n dφ, π 0 1

is known as Laplace’s first integral for Pn (z). Corollary 15.2.2 From Corollary 15.2.1 it is evident that, when | arg z| < π2 , ∫ 1 π dφ , Pn (z) = n 0 {z + (z2 − 1) 12 cos φ} n+1 a result, due to Jacobi [351], known as Laplace’s second integral for Pn (z). 7

It only does so if z is a pure imaginary; and such values of z have been excluded.

15.2 Legendre functions

329

Example 15.2.13 Obtain Laplace’s first integral by considering ∫ π ∞ Õ 1 hn {z + (z2 − 1) 2 cos φ} n dφ, n=0

0

and using Example 6.2.1. Example 15.2.14 Shew, by direct differentiation, that Laplace’s integral is a solution of Legendre’s equation. Example 15.2.15 (Binet) If s < 1, |h| < 1 and (1 − 2h cos θ + h2 )−s =

∞ Õ

bn cos nθ,

n=0

shew that bn =

2 sin sπ π

∫

1

0

h n x n+s−1 dx . (1 − x)s (1 − xh2 )s

Example 15.2.16 When z > 1, deduce Laplace’s second integral from his first integral by the substitution {z − (z2 − 1) 2 cos θ}{z + (z2 − 1) 2 cos φ} = 1. 1

1

Example 15.2.17 By expanding in powers of cos φ, shew that for a certain range of values of z, ∫ 1 π n 1 − n 1 2 n n −2 {z + (z − 1) 2 cos φ} dφ = z F − , ; 1; 1 − z . π 0 2 2 Example 15.2.18 Shew that Legendre’s equation is defined by the scheme 0 P − 21 n 1 + 1n 2 2

1 2

∞ + 21 n − 21 n

1 0 0

ξ ,

where z = 12 (ξ 2 + ξ − 2 ). 1

1

15.231 The Mehler–Dirichlet integral for Pn (z) This comes from Dirichlet [174] and Mehler [465]. Another expression for the Legendre function as a definite integral may be obtained in the following way: For all values of n, we have, by the preceding theorem, ∫ 1 π Pn (z) = {z + (z2 − 1)1/2 cos φ} n dφ. π 0 In this integral, replace the variable φ by a new variable h, defined by the equation h = z + (z2 − 1)1/2 cos φ, and we get ∫ z+(z 2 −1)1/2 i Pn (z) = h n (1 − 2hz + h2 )−1/2 dh; π z−(z 2 −1)1/2

Legendre Functions

330

the path of integration is a straight line, arg h is determined by the fact that h = z when φ = 21 π, and (1 − 2hz + h2 )−1/2 = −i (z2 − 1)1/2 sin φ. Now let z = cos θ; then ∫ ei θ i Pn (cos θ) = h n (1 − 2hz + h2 )−1/2 dh. π e−i θ Now (θ being restricted so that − π2 < θ < π2 when n is not a positive integer) the path of integration may be deformed 8 into that arc of the circle |h| = 1 which passes through h = 1, and joins the points h = e−iθ , h = eiθ , since the integrand is analytic throughout the region between this arc and its chord 9 . Writing h = eiφ we get ∫ 1 θ e(n+1/2)iφ dφ , Pn (cos θ) = π −θ (2 cos φ − 2 cos θ)1/2 and so Pn (cos θ) =

2 π

θ

∫ 0

cos(n + 21 )φ dφ ; {2(cos φ − cos θ)}1/2

it is easy to see that the positive value of the square root is to be taken. This is known as Mehler’s simplified form of Dirichlet’s integral. The result is valid for all values of n. Example 15.2.19 Prove that, when n is a positive integer, ∫ sin(n + 21 )φ dφ 2 π Pn (cos θ) = . π θ {2(cos θ − cos φ)}1/2 (Write π − θ for θ and π − φ for φ in the result just obtained.) Example 15.2.20 Prove that 1 Pn (cos θ) = 2πi

∫

h n dh , (h2 − 2h cos θ + 1)1/2

the integral being taken along a closed path which encircles the two points h = e±iθ , and a suitable meaning being assigned to the radical. Note Hence (or otherwise) prove that, if θ lie between 61 π and 56 π,

Pn (cos θ) =

cos(nθ + φ)

+

12 cos(nθ + 3φ) 2(2n + 3) (2 sin θ) 32

(2 sin θ) 4 2 · 4 · · · 2n , π 3 · 5 · · · (2n + 1) cos(nθ + 5φ) 12 · 32 + · · · · · · 5 + 2 · 4 · (2n + 3)(2n + 5) (2 sin θ) 2 1 2

where φ denotes 12 θ − 14 π.

Shew also that the first few terms of the series give an approximate value of Pn (cos θ) for

8

9

If θ be complex and Re cos θ > 0 the deformation of the contour presents slightly greater difficulties. The reader will easily modify the analysis given to cover this case. The integrand is not analytic at the ends of the arc but behaves like (h − e±i θ )−1/2 near them; but if the region be indented §6.23 at e±i θ and the radii of the indentations be made to tend to zero, we see that the deformation is legitimate.

15.3 Legendre functions of the second kind

331

all values of θ between 0 and π which are not nearly equal to either 0 or π. And explain how this theorem may be used to approximate to the roots of the equation Pn (cos θ) = 0. (See Heine [287, vol. I, p. 171]; Darboux [161].)

15.3 Legendre functions of the second kind We have hitherto considered only one solution of Legendre’s equation, namely Pn (z). We proceed to find a second solution. We have seen (§15.2) that Legendre’s equation is satisfied by ∫ (t 2 − 1)n (t − z)−n−1 dt, taken round any contour such that the integrand returns to its initial value after describing it. Let D be a figure-of-eight contour formed in the following way: let z be not a real number between ±1; draw an ellipse in the t-plane with the points ±1 as foci, the ellipse being so small that the point t = z is outside. Let A be the end of the major axis of the ellipse on the right of t = 1. Let the contour D start from A and describe the circuits (1− , −1+ ), returning to A (cf. §12.43), and lying wholly inside the ellipse. Let | arg z| ≤ π and let | arg(z −t)| → arg z as t → 0 on the contour. Let arg(t + 1) = arg(t − 1) = 0 at A. Then a solution of Legendre’s equation valid in the plane (cut along the real axis from 1 to −∞) is ∫ (t 2 − 1)n dt 1 , Q n (z) = 4i sin nπ D 2n (z − t)n+1 if n is not an integer. When Re(n + 1) > 0, we may deform the path of integration as in §12.43, and get ∫ 1 1 Q n (z) = n+1 (1 − t 2 )n (z − t)−n−1 dt 2 −1 (where arg(1 − t) = arg(1 + t) = 0); this will be taken as the definition of Q n (z) when n is a positive integer or zero. When n is a negative integer (= −m − 1) Legendre’s differential equation for functions of degree n is identical with that for functions of degree m, and accordingly we shall take the two fundamental solutions to be Pm (z), Q m (z). We call Q n (z) the Legendre function of degree n of the second kind.

15.31 Expansion of Q n (z) as a power series We now proceed to express the Legendre function of the second kind as a power series in z−1 . We have, when the real part of n + 1 is positive, ∫ 1 1 (1 − t 2 )n (z − t)−n−1 dt. Q n (z) = n+1 2 −1 Suppose that |z| > 1. Then the integrand can be expanded in a series uniformly convergent

Legendre Functions

332

with regard to t, so that −n−1 t (1 − t 2 )n 1 − dt z −1 ( ) ∫ 1 ∞ r Õ t (n + 1)(n + 2) · · · (n + r) 1 2 n (1 − t ) 1 + dt = n+1 n+1 2 z z r! −1 r=1 "∫ # ∫ ∞ 1 Õ (n + 1) · · · (n + 2s) 1 1 2 n 2 n 2s (1 − t ) dt + (1 − t ) t dt , = n n+1 2 z 2s!z2s 0 0 s=1

Q n (z) =

1 n+1 2 z n+1

∫

1

where r = 2s, the integrals arising from odd values of r vanishing. Writing t 2 = u we get without difficulty, from §12.41, 1 1 1 1 3 −2 π 2 Γ(n + 1) 1 F n + , n + 1; n + ; z . Q n (z) = n+1 2 2 2 2 2 Γ(n + 23 ) z n+1 The proof given above applies only when the real part of (n + 1) is positive (see §4.5); but a similar process can be applied to the integral ∫ 1 1 2 Q n (z) = (t − 1)n (z − t)−n−1 dt, 4i sin nπ D 2n ∫ the coefficients being evaluated by writing (t 2 − 1)n t r dt in the form D

enπi

∫

(1−)

(1 − t 2 )n t r dt + enπi

∫

0

(−1+)

(1 − t 2 )n t r dt;

0

and then, writing t = u and using §12.43, the same result is reached, so that the formula 1 1 π 2 Γ(n + 1) 1 1 1 3 1 F Q n (z) = n+1 n + , n + 1; n + ; 2 Γ(n + 32 ) z n+1 2 2 2 2 z2 2

is true for unrestricted values of n (negative integer values excepted) and for all values 10 of z, such that |z| > 1, | arg z| < π. Example 15.3.1 Shew that, when n is a positive integer, ∫ ∞ (−2)n n! d n 2 n 2 −n−1 (z − 1) (v − 1) dv . Q n (z) = (2n)! dz n z It is easily verified that Legendre’s equation can be derived from the equation d2w dw + 2(n − 1)z + 2nw = 0, 2 dz dz dnw by differentiating n times and writing u = . Two independent solutions of this equation dz n are found to be ∫ ∞ (z2 − 1)n and (z2 − 1)n (v 2 − 1)−n−1 dv. (1 − z2 )

z 10

When n is a positive integer it is unnecessary to restrict the value of arg z.

15.3 Legendre functions of the second kind

It follows that

333

∫ ∞ dn 2 n 2 −n−1 (z − 1) (v − 1) dv dz n z

is a solution of Legendre’s equation. As this expression, when expanded in ascending powers of z−1 , commences with a term in z−n−1 , it must be a constant multiple 11 of Q n (z); and on comparing the coefficient of z−n−1 in this expression with the coefficient of z−n−1 in the expansion of Q n (z), as found above, we obtain the required result. Example 15.3.2 Shew that, when n is a positive integer, the Legendre function of the second kind can be expressed by the formula ∫ ∞∫ ∞∫ ∞ ∫ ∞ Q n (z) = 2n n! ··· (v 2 − 1)−n−1 (dv)n+1 . z

v

v

v

Example 15.3.3 Shew that, when n is a positive integer, ∫ ∞ n Õ 2n · n! n−t (−z) v t (v 2 − 1)−n−1 dv. Q n (z) = t!(n − t)! z t=0 This result can be obtained by applying the general integration theorem ∫ ∞ ∫ ∞∫ ∞∫ ∞ ∫ ∞ n Õ (−z)n−t v t f (v) dv ··· f (v) (dv)n+1 = t!(n − t)! z z v v v t=0 to the preceding result.

15.32 The recurrence formulae for Q n (z) The functions Pn (z) and Q n (z) have been defined by means of integrals of precisely the same form, namely ∫ (t 2 − 1)n (t − z)−n−1 dt, taken round different contours. It follows that the general proof of the recurrence formulae for Pn (z), given in §15.21, is equally applicable to the function Q n (z); and hence that the Legendre function of the second kind satisfies the recurrence formulae 0 Q n+1 (z) − zQ n0 (z) = (n + 1)Q n (z),

(n + 1)Q n+1 (z) − (2n + 1)zQ n (z) + nQ n−1 (z) = 0, 0 zQ n0 (z) − Q n−1 (z) = nQ n (z), 0 0 Q n+1 (z) − Q n−1 (z) = (2n + 1)Q n (z),

(z2 − 1)Q n0 (z) = nzQ n (z) − nQ n−1 (z). Example 15.3.4 Shew that Q0 (z) = 11

1 z+1 log , 2 z−1

Q1 (z) =

Pn (z) contains positive powers of z when n is an integer.

1 z+1 z log − 1, 2 z−1

Legendre Functions

334

and deduce that Q2 (z) =

z+1 3 1 P2 (z) log − z 2 z−1 2

and that Q n (z) 1 z+1 = log − Pn (z) 2 z−1

1

.

12

z−

22

3z − 5z −

32 (n − 1)2 . 7z − . . − (2n − 1)z

Example 15.3.5 Shew by the recurrence formulae that, when n is a positive integer 12 , 1 z+1 − Q n (z) = fn−1 (z), Pn (z) log 2 z−1 where fn−1(z) consists of the positive (and zero) powers of z in the expansion of z+1 1 P (z) log z−1 in descending powers of z. 2 n Note This example shews the nature of the singularities of Q n (z) at ±1, when n is an integer, which make the cut from −1 to +1 necessary. For the connexion of the result with the theory of continued fractions, see Gauss [233], and Frobenius [226]; the formulae of Example 15.3.4 are due to them.

15.33 The Laplacian integral for Legendre functions of the second kind This formula was first given by Heine [287, p. 147]. It will now be proved that, when Re(n + 1) > 0, ∫ ∞n o −n−1 1 Q n (z) = dθ, z + (z2 − 1) 2 cosh θ 0

where arg{z + (z2 − 1) cosh θ} has its principal value when θ = 0, if n be not an integer. First suppose that z > 1. In the integral of §15.3, viz. ∫ 1 1 Q n (z) = n+1 (1 − t 2 )n (z − t)−n−1 dt, 2 −1 1 2

write t= 12

eθ (z + 1)1/2 − (z − 1)1/2 , eθ (z + 1)1/2 + (z − 1)1/2

If −1 < z < 1, it is apparent from these formulae that Q n (z + 0i) − Q n (z − 0i) = −πiPn (z). It is convenient to define Q n (z) for such values of z to be 12 Q n (z + 0i) + 12 Q n (z − 0i). The reader will observe that this function satisfies Legendre’s equation for real values of z.

15.3 Legendre functions of the second kind

335

so that the range (−1, 1) of real values of t corresponds to the range (−∞, ∞) of real values of θ. It then follows (as in (15.1)) by straightforward substitution that ∫ o −n−1 1 ∞n 1 z + (z2 − 1) 2 cosh θ dθ Q n (z) = 2 ∫ ∞−∞ n o −n−1 1 = z + (z2 − 1) 2 cosh θ dθ, 0

since the integrand is an even function of θ. Note To prove the result for values of z not comprised in the range of real values greater than 1, we observe that the branch points of the integrand, qua function of z, are at the points ±1 and at points where z + (z2 − 1)1/2 cosh θ vanishes; the latter are the points at which z = ± coth θ. Hence Q n (z) and ∫ ∞ 1 {z + (z2 − 1) 2 cosh θ}−n−1 dθ 0 13

are both analytic at all points of the plane when cut along the line joining the points z = ±1. By the theory of analytic continuation the equation proved for positive values of 1 z − 1 persists for all values of z in the cut plane, provided that arg {z + (z2 − 1) 2 cosh θ} is given a suitable value, namely that one which reduces to zero when z − 1 is positive. The integrand is one-valued in the cut plane [and so is Q n (z)] when n is a positive integer; 1 but arg{z + (z2 − 1) 2 cosh θ} increases by 2π as arg z does so, and therefore if n be not a positive integer, a further cut has to be made from z = −1 to z = −∞. These cuts give the necessary limitations on the value of z; and the cut when n is not an integer ensures that 1 1 1 arg{z + (z2 − 1) 2 } = 2 arg{(z + 1) 2 + (z − 1) 2 } has its principal value. Example 15.3.6 Obtain this result for complex values of z by taking the path of integration to be a certain circular arc before making the substitution eθ (z + 1) 2 − (z − 1) 2 1

t=

1

eθ (z + 1) 2 + (z − 1) 2 1

1

,

where θ is real. Example 15.3.7 (Trinity, 1893) Shew that, if z > 1 and coth a = z, ∫ a 1 Q n (z) = {z − (z2 − 1) 2 cosh u} n du, 0

where arg{z − (z − 1) cosh u} = 0. 2

1 2

15.34 Neumann’s formula for Q n (z), when n is an integer This appears in F. Neumann [490]. When n is a positive integer, and z is not a real number between 1 and −1, the function Q n (z) is expressed in terms of the Legendre function of the first kind by the relation ∫ dy 1 1 Pn (y) , Q n (z) = 2 −1 z−y 13

It is easy to shew that the integral has a unique derivative in the cut plane.

Legendre Functions

336

which we shall now establish. When |z| > 1 we can expand the integrand in the uniformly convergent series Pn (y)

∞ Õ ym . z m+1 m=0

Consequently 1 2

∫

∞

1

Pn (y) −1

1 Õ −m−1 dy = z z − y 2 m=0

∫

1

y m Pn (y) dy.

−1

The integrals for which m − n is odd or negative vanish (15.7); and so ∫ ∫ ∞ 1 Õ −n−2m−1 1 n+2m 1 1 dy = Pn (y) z y Pn (y) dy 2 −1 z − y 2 m=0 −1 ∞

1 Õ −n−2m−1 2n+1 (n + 2m)!(n + m)! z = 2 m=0 m!(2n + 2m + 1)! 2n (n!)2 −n−1 n 1 n 3 −2 = z F + , + 1; n + ; z (2n + 1)! 2 2 2 2 = Q n (z), by §15.31. The theorem is thus established for the case in which |z| > 1. Since each side of the equation ∫ dy 1 1 Pn (y) Q n (z) = 2 −1 z−y represents an analytic function, even when |z| is not greater than unity, provided that z is not a real number between −1 and +1, it follows that, with this exception, the result is true (§5.5) for all values of z. The reader should notice that Neumann’s formula apparently expresses Q n (z) as a onevalued function of z, whereas it is known to be many-valued (Example 15.3.4). The reason for the apparent discrepancy is that Neumann’s formula has been established when the z-plane is cut from −1 to +1, and Q n (z) is one-valued in the cut plane. Example 15.3.8 Shew that, when −1 ≤ Re z ≤ 1, | Q n (z) | ≤ | Im z | −1 ; and that for other values of z, | Q n (z) | does not exceed the larger of | z − 1 | −1 and | z + 1 | − 1 . Example 15.3.9 Shew that, when n is a positive integer, Q n (z) is the coefficient of h n in the expansion of h−z . (1 − 2hz + h2 )−1/2 arccosh (z2 − 1)1/2 Hint. For |h| sufficiently small, ∞ Õ n=0

∫ ∫ ∞ Õ h n 1 Pn (y) dy 1 1 (1 − 2hy + h2 )−1/2 dy = 2 −1 z − y 2 −1 (z − y) n=0 h−z = (1 − 2hz + h2 )−1/2 arccosh . (z2 − 1)1/2

h n Q n (z) =

15.4 Heine’s development of (t − z)−1

337

This result has been investigated by Heine [287, vol. I, p. 134] and Laurent [411].

15.4 Heine’s development of (t − z)−1 as a series of Legendre polynomials in z This appears in [286]. We shall now obtain an expansion which will serve as the basis of a general class of expansions involving Legendre polynomials. The reader will readily prove by induction from the recurrence formulae (2m + 1)tQ m (t) − (m + 1)Q m+1 (t) − mQ m−1 (t) = 0, (2m + 1)zPm (z) − (m + 1)Pm+1 (z) − mPm−1 (z) = 0, that n

Õ n+1 1 = (2m + 1)Pm (z)Q m (t) + {Pn+1 (z)Q n (t) − Pn (z)Q n+1 (t)}. t − z m=0 t−z Using Laplace’s integrals, we have Pn+1 (z)Q n (t) − Pn (z)Q n+1 (t) =

1 π

∫

π

∫

∞

{z + (z2 − 1) 2 cos φ} n 1

{t + (t 2 − 1) 2 cosh u} n+1 1

0

0

× [z + (z2 − 1) 2 cos φ − {t + (t 2 − 1) 2 cosh u}−1 ] dφ du. 1

1

Now consider z + (z2 − 1) 21 cos φ . t + (t 2 − 1) 21 cosh u Let cosh a, sinh a be the semi-major axes of the ellipses with foci ±1 which pass through z and t respectively. Let θ be the eccentric angle of z; then z = cosh(a + iθ), |z ± (z − 1) cos φ| = | cosh(a + iθ) ± sinh(a + iθ) cos φ| 2

1 2

= {cosh2 α − sin2 θ + (cosh2 a − cos2 θ) cos2 φ ± 2 sinh a cosh α cos φ} 2 . 1

This is a maximum for real values of φ when cos φ = ∓1; and hence |z ± (z2 − 1) 2 cos φ| 2 ≤ 2 cosh2 a − 1 + 2 cosh a(cosh2 a − 1) 2 = exp(2α). 1

1

Similarly | t + (t 2 − 1) 2 cosh u | ≤ exp a. Therefore 1

|Pn+1 (z)Q n (t) − Pn (z)Q n+1 (t)| ≤ π −1 exp{n(a − α)}

∫

π

∫

∞

V dφ du, 0

0

where z + (z 2 − 1) 21 cos φ 1 |V | = + | {t + (t 2 − 1) 2 cosh u} | −2 . 1 t + (t 2 − 1) 2 cosh u Therefore |Pn+1 (z)Q n (t) − Pn (z)Q n+1 (t)| → 0, as n → ∞, provided a x > 1, prove that y+1 (x + 1) (y − 1) ∫ 1 ∞ Õ dz 1 2 Pn (x)Q n (y). = {(x + 1)(y − 1)} 1 {(1 − z2 )(1 − k 2 z2 )} 2 n=0 Í

Hint. Substitute Laplace’s integrals on the right and integrate with regard to φ. Example 15.4.3 (Frobenius [226]) Shew that ∞

1 (x + 1)(y − 1) Õ (2n + 1)Q n (x)Q n (y). log = 2(y − x) (x − 1)(y + 1) n=0

15.5 Ferrers’ associated Legendre functions Pnm (z) and Q m n (z) We shall now introduce a more extended class of Legendre functions. If m be a positive integer and −1 < z < 1, n being unrestricted (when n is a positive integer it is unnecessary to restrict the value of arg z), the functions m d m Pn (z) m 2 m/2 d Q n (z) , Q (z) = (1 − z ) n dz m dz m will be called Ferrers’ associated Legendre functions of degree n and order m of the first and second kinds respectively. (Ferrers writes Tnm (z) for Pnm (z).) It may be shewn that these functions satisfy a differential equation analogous to Legendre’s equation. For, differentiate Legendre’s equation

Pnm (z) = (1 − z2 )m/3

(1 − z2 )

d2 y dy − 2z + n(n + 1)y = 0 2 dz dz

Legendre Functions

340

m times and write v for

dm y . We obtain the equation dz m

(1 − z2 )

d2v dv − 2z(m + 1) + (n − m)(n + m + 1)v = 0. dz 2 dz

Write w = (1 − z2 )m/2 v, and we get d2w dw m2 (1 − z ) 2 − 2z + n(n + 1) − w = 0. dz dz 1 − z2 2

This is the differential equation satisfied by Pnm (z) and Q m n (z). Note From the definitions given above, several expressions for the associated Legendre functions may be obtained. Thus, from Schläfli’s formula we have ∫ (1+,z+) (n + 1)(n + 2) · · · (n + m) m 2 m/2 Pn (z) = (1 − z ) (t 2 − 1)n (t − z)−n−m−1 dt, 2n+1 πi A where the contour does not enclose the point t = −1. Further, when n is a positive integer, we have, by Rodrigues’ formula, (1 − z2 )m/2 d n+m (z2 − 1)n . 2n n! dz n+m Shew that Legendre’s associated equation is defined by the

Pnm (z) = Example 15.5.1 (Olbricht) scheme

∞ 1 10 1 P 2m n + 1 2m − 1 m −n − 1 m 2 2

1 2

−

1 z 2

.

15.51 The integral properties of the associated Legendre functions The generalisation of the theorem of §15.14 is the following: When n, r, m are positive integers and n > m, r > m, then ∫

1

−1

Pnm (z)Prm (z) dz

0 (r , n), = 2 (n + m)! (r = n). 2n + 1 (n − m)!

To obtain the first result, multiply the differential equations for Pnm (z), Prm (z) by Prm (z), Pnm (z) respectively and subtract; this gives dP m (z) dP m (z) d (1 − z2 ) Prm (z) n − Pnm (z) r dz dz dz + (n − r)(n + r + 1)Prm (z)Pnm (z) = 0. On integrating between the limits −1, +1, the result follows when n and r are unequal, since the expression in square brackets vanishes at each limit. To obtain the second result, we observe that dP m (z) Pnm+1 (z) = (1 − z2 )1/2 n + mz(1 − z2 )−1/2 Pnm (z); dz

15.6 Hobson’s definition of the associated Legendre functions

341

squaring and integrating, we get m 2 ∫ 1 ∫ 1" m+1 2 dP m (z) dPn (z) 2 + 2mzPnm (z) n Pn (z) dz = (1 − z ) dz dz −1 −1 m2 z 2 m 2 + P (z) dz 1 − z2 n ∫ 1 ∫ 1 m 2 dP m (z) d =− Pnm (z) (1 − z2 ) n dz − m Pn (z) dz dz dz −1 −1 ∫ 1 2 2 m z {Pnm (z)}2 dz, + 2 −1 1 − z on integrating the first two terms in the first lines on the right by parts. If now we use the differential equation for Pnm (z) to simplify the first integral in the second line, we at once get ∫ 1 ∫ 1 m+1 2 {Pn (z)} dz = (n − m)(n + m + 1) {Pnm (z)}2 dz. −1

−1

By repeated applications of this result we get ∫ 1 {Pnm (z)}2 dz = (n − m + 1)(n − m + 2) · · · n −1

× (n + m)(n + m − 1) · · · (n + 1)

∫

1

{Pn (z)}2 dz, −1

and so ∫

1

{Pnm (z)}2 dz =

−1

2 (n + m)! . 2n + 1 (n − m)!

(15.8)

15.6 Hobson’s definition of the associated Legendre functions So far it has been taken for granted that the function (1 − z2 )m/2 which occurs in Ferrers’ definition of the associated functions is purely real; and since, in the more elementary physical applications of Legendre functions, it usually happens that −1 < z < 1, no complications arise. But as we wish to consider the associated functions as functions of a complex variable, it is undesirable to introduce an additional cut in the z-plane by giving arg(1 − z) its principal value. Accordingly, in future, when z is not a real number such that −1 < z < 1, we shall follow Hobson in defining the associated functions by the equations Pnm (z) = (z2 − 1)m/2

d m Pn (z) dz m

2 m/2 Qm n (z) = (z − 1)

d m Q n (z) , dz m

where m is a positive integer, n is unrestricted and arg z, arg(z + 1), arg(z − 1) have their principal values. When m is unrestricted, Pnm (z) is defined by Hobson to be m/2 1 z+1 F −n, n + 1; 1 − m; 12 − 21 z ; Γ(1 − m) z − 1

Legendre Functions

342

and Barnes has given a definition of Q m n (z) from which the formula 1 sin(n + m)π Γ(n + m + 1)Γ 2 (z2 − 1)m/2 m Q n (z) = sin nπ z n+m+1 2n+1 Γ(n + 32 ) ×F n2 + m2 + 1, n2 + m2 + 12 ; n + 32 ; z−2 may be obtained. Throughout this work we shall take m to be a positive integer.

15.61 Expression of Pnm (z) as an integral of Laplace’s type If we make the necessary modification in the Schläfli integral of §15.5, in accordance with the definition of §15.6, we have ∫ (1+,z+) (n + 1)(n + 2) · · · (n + m) 2 m m/2 Pn (z) = (z − 1) (t 2 − 1)n (t − z)−n−m−1 dt. 2n+1 πi A Write t = z + (z2 − 1)1/2 eiφ , as in §15.23; then Pnm (z) =

(n + 1)(n + 2) · · · (n + m) 2 (z − 1)m/2 2π

∫

2π+α

α

{z + (z2 − 1)1/2 cos φ} n dφ, {(z2 − 1)1/2 eiφ } m

where α is the value of φ when t is at A, so that | arg(z2 − 1)1/2 + α| < π. Now, as in §15.23, the integrand is a one-valued periodic function of the real variable φ with period 2π, and so ∫ (n + 1)(n + 2) · · · (n + m) π m Pn (z) = {z + (z2 − 1)1/2 cos φ} n e−miφ dφ. 2π −π Since {z + (z2 − 1)1/2 cos φ} n is an even function of φ, we get, on dividing the range of integration into the parts (−π, 0) and (0, π), ∫ (n + 1)(n + 2) · · · (n + m) π {z + (z2 − 1)1/2 cos φ} n cos mφ dφ. Pnm (z) = π 0 The ranges of validity of this formula, which is due to Heine (according as n is or is not an integer), are precisely those of the formula of §15.23. Example 15.6.1 Shew that, if | arg z| < 21 π, Pnm (z) = (−1)m

n(n − 1) · · · (n − m + 1) π

∫ 0

π

{z +

cos mφ dφ , − 1)1/2 cos φ} n+1

(z2

where the many-valued functions are specified as in §15.23.

15.7 The addition-theorem for the Legendre polynomials This appears in Legendre [421, vol. II, p. 262–269]. An investigation of the theorem based on physical reasoning will be given subsequently (§18.4).

15.7 The addition-theorem for the Legendre polynomials

343

Let z = x x 0 − (x 2 − 1)1/2 (x 2 − 1)1/2 cos ω, where x, x 0, ω are unrestricted complex numbers. Then we shall shew that n Õ (n − m)! m Pn (x)Pnm (x 0) cos mω. Pn (z) = Pn (x)Pn (x 0) + 2 (−1)m (n + m)! m=1 First let Re(x 0) > 0, so that x + (x 2 − 1)1/2 cos(ω − φ) x 0 + (x 02 − 1)1/2 cos φ is a bounded function of φ in the range 0 < φ < 2π. If M be its upper bound and if |h| < M −1, then ∞ Õ {x + (x 2 − 1)1/2 cos(ω − φ)} n hn {x 0 + (x 02 − 1)1/2 cos φ} n+1 n=0 converges uniformly with regard to φ, and so (§4.7) ∫ π ∞ Õ {x + (x 2 − 1)1/2 cos(ω − φ)} n hn dφ {x 0 + (x 02 − 1)1/2 cos φ} n+1 −π n=0 ∫ πÕ ∞ h n {x + (x 2 − 1)1/2 cos(ω − φ)} n = dφ {x 0 + (x 02 − 1)1/2 cos φ} n+1 −π n=0 ∫ π dφ = . 0 + (x 02 − 1)1/2 cos φ − h{x + (x 2 − 1)1/2 cos(ω − φ)} x −π Now, by a slight modification of Example 6.2.1 it follows that ∫ π dφ 2π = 2 , 2 − C 2 )1/2 A + B cos φ + C sin φ (A − B −π where that value of the radical is taken which makes | A − (A2 − B2 − C 2 )1/2 | < |(B2 + C 2 )1/2 |. Therefore ∫ π

dφ + − cos φ − h{x + (x 2 − 1)1/2 cos(ω − φ)} −π 2π = 0 2 2 1/2 2 [(x − hx) − {(x − 1) − h(x − 1)1/2 cos ω}2 − {h(x 2 − 1)1/2 sin ω}2 ]1/2 2π = ; (1 − 2hz + h2 )1/2 x0

(x 02

1)1/2

and when h → 0, this expression has to tend to 2πP0 (x 0) by §15.23. Expanding in powers of h and equating coefficients, we get ∫ π 1 {x + (x 2 − 1)1/2 cos(ω − φ)} n Pn (z) = dφ. 2π −π {x 0 + (x 02 − 1)1/2 cos φ } n+1 Now Pn (z) is a polynomial of degree n in cos ω, and can consequently be expressed in the

Legendre Functions

344

n Õ 1 A0 + Am cos mω, where the coefficients A0, A1, . . . , An are independent of ω; to 2 m=1 determine them, we use Fourier’s rule (§9.12), and we get ∫ 1 π Am = Pn (z) cos mω dω π −π ∫ π ∫ π 1 {x + (x 2 − 1)1/2 cos(ω − φ)} n cos mω dφ dω = 2 2π −π −π {x 0 + (x 02 − 1)1/2 cos φ} n+1 ∫ π ∫ π 1 {x + (x 2 − 1)1/2 cos(ω − φ)} n cos mω = 2 dω dφ 2π −π −π {x 0 + (x 02 − 1)1/2 cos φ} n+1 ∫ π ∫ π 1 {x + (x 2 − 1)1/2 cos ψ} n cos m(φ + ψ) = 2 dψ dφ, 2π −π −π {x 0 + (x 02 − 1)1/2 cos φ} n+1

form

on changing the order of integration, writing ω = φ + ψ and changing the limits for ψ from ±π − φ to ±π. Now ∫ π {x + (x 2 − 1)1/2 cos ψ} n sin mψ dψ = 0, −π

since the integrand is an odd function; and so, by §15.61, ∫ π cos mφ Pnm (x) n! Am = dφ π(n + m)! −π {x 0 + (x 02 − 1)1/2 cos φ} n+1 (n − m)! m = 2(−1)m P (x)Pnm (x 0). (n + m)! n Therefore, when | arg z 0 | < 21 π, Pn (z) = Pn (x)Pn (x 0) + 2

n Õ

(−1)m

m=1

(n − m)! m P (x)Pnm (x 0) cos mω. (n + m)! n

But this is a mere algebraical identity in x, x 0 and cos ω (since n is a positive integer) and so is true independently of the sign of Re(x 0). The result stated has therefore been proved. The corresponding theorem with Ferrers’ definition is Pn {x x 0 + (1 − x 2 )1/2 (1 − x 02 )1/2 cos ω} n Õ (n − m)! m = Pn (x)Pn (x 0) + 2 Pn (x)Pnm (x 0) cos mω. (n + m)! m=1

15.71 The addition theorem for the Legendre functions Let x, x be two constants, real or complex, whose arguments are numerically less than 12 π; and let (x ± 1)1/2 , (x 0 ± 1)1/2 be given their principal values; let ω be real and let 0

z = x x 0 − (x 2 − 1)1/2,

(x 02 − 1)1/2 cos ω.

Then we shall shew that, if | arg z| < 12 π for all values of the real variable ω, and n be not

15.7 The addition-theorem for the Legendre polynomials

345

a positive integer, Pn (z) = Pn (x)Pn (x 0) + 2

∞ Õ

(−1)m

m=1

Γ(n − m + 1) m P (x)Pnm (x 0) cos mω. Γ(n + m + 1) n

Let cosh α, cosh α be the semi-major axes of the ellipses with foci ±1 passing through x, x respectively. Let β, β 0 be the eccentric angles of x, x 0 on these ellipses so that π π π π − | sinh 12 ξ 0 |, the argument of the denominators does not change when φ increases by 2π; for similar reasons, the arguments of the first and third numerators increase by 2π, and the argument of the second does not change; therefore the circle contains the points t = 1, t = z, and not t = −1, so it is a possible contour. Making these substitutions it is readily found that ∫ π 1 {x + (x 2 − 1)1/2 cos(ω − φ)} n Pn (z) = dφ, 2π −π {x 0 + (x 02 − 1)1/2 cos φ} n+1 and the rest of the work follows the course of §15.7 except that the general form of Fourier’s theorem has to be employed. 15

This follows from the fact that cos β0 > 0.

Legendre Functions

346

Example 15.7.1 (Heine [287], Neumann [489]) Shew that, if n be a positive integer, Q n {x x 0 + (x 2 − 1)1/2 (x 02 − 1)1/2 cos ω} ∞ Õ −m 0 = Q n (x)Pn (x 0) + 2 Qm n (x)Pn (x ) cos mω, m=1

when ω is real, Re(x 0) ≥ 0, and |(x 0 − 1)(x + 1)| < |(x − 1)(x 0 + 1)|.

15.8 The function Cnν (z) A function connected with the associated Legendre function Pnm (z) is the function Cnν (z), which for integral values of n is defined to be the coefficient of h n in the expansion of (1 − 2hz + h2 )−ν in ascending powers of h. This function has been studied by Gegenbauer [240]. It is easily seen that Cnν (z) satisfies the differential equation d 2 y (2ν + 1)z dy n(n + 2ν) + 2 − 2 y = 0. dz2 z − 1 dz z −1 For all values of n and ν, it may be shewn that we can define a function, satisfying this equation, by a contour integral of the form ∫ (1 − z2 )1/2−ν C

(1 − t 2 )n+ν−1/2 dt, (t − z)n+1

where C is the contour of §15.2; this corresponds to Schläfli’s integral. The reader will easily prove the following results: (I) When n is a integer Cnν (z) =

(−2)n ν(ν + 1) · · · (ν + n − 1) dn 1 (1 − z2 )1/2−ν n {(1 − z2 )n+ν− 2 }; n!(2n + 2ν − 1)(2n + 2ν − 2) · · · (n + 2ν) dz

since Pn (z) = Cn1/2 (z), Rodrigues’ formula is a particular case of this result. (II) When r is an integer, r+ 1

Cn−r2 (z) =

dr 1 Pn (z), (2r − 1)(2r − 3) · · · 3 · 1 dzr

whence r+ 1

Cn−r2 (z) =

(z2 − 1)−r/3 Pr (z). (2r − 1) · (2r − 3) · · · 3 · 1 n

The last equation gives the connexion between the functions Cnν (z) and Pnr (z).

15.9 Miscellaneous examples

347

(III) Modifications of the recurrence formulae for Pn (z) are the following: n ν C (z) = 0, 2ν n n + 2ν ν ν+1 Cnν+1 (z) − zCn−1 (z) = Cn (z), 2ν ν dCn (z) ν+1 = 2νCn−1 (z), dz ν ν−1 nCnν (z) = (n − 1 + 2ν)zCn−1 (z) − 2ν(1 − z2 )Cn−2 (z).

ν+1 ν+1 Cn−1 (z) − Cn−2 (z) −

15.9 Miscellaneous examples The functions involved in Examples 15.1–15.30 are Legendre polynomials. Example 15.1 (Math. Trip. 1898) Prove that when n is a positive integer, Pn (z) =

n Õ (n + p)!(−1) p {(1 − z) p + (−1)n (1 + z) p }. 2 2 p+1 (n − p)!p! 0

∫ Example 15.2 (Math. Trip. 1896)

1

z(1 − z2 )

Prove that

m − n = ±1, and determine its value in these cases.

−1

dPn dPm dz is zero unless dz dz

Example 15.3 (Math. Trip. 1899) Shew (by induction or otherwise) that when n is a positive integer, ∫ 1 (2n + 1) Pn 2 (z) dz = 1−

z zPn 2

2 − 2z(P1 2 + P2 2 + · · · + Pn−1 ) + 2(P1 P2 + P2 P3 + · · · + Pn−1 Pn ).

Example 15.4 (Clare, 1906) Shew that zPn0 (z) = nPn (z) + (2n − 3)Pn−2 (z) + (2n − 7)Pn−4 (z) + · · · . Example 15.5 (Math. Trip. 1904) Shew that z

2

Pn00(z)

= n(n − 1)Pn (z) +

p Õ

(2n − 4r + 1){r(2n − 2r + 1) − 2}Pn−2r (z),

r=1

where p = 21 n or 12 (n − 1). Example 15.6 (Trin. Coll. Dublin) Shew that the Legendre polynomial satisfies the relation ∫ z ∫ z d 2 Pn (z2 − 1)2 2 = n(n − 1)(n + 1)(n + 2) dz Pn (z) dz. dz 1 1 Example 15.7 (Peterhouse, 1905) Shew that ∫ 1 z2 Pn+1 (z)Pn−1 (z) dz = 0

n(n + 1) . (2n − 1)(2n + 1)(2n + 3)

Legendre Functions

348

Example 15.8 (Peterhouse, 1907) Shew that the values of ∫ 1 (1 − z2 )2 Pm000(z)Pn0 (z) dz −1

are as follows: 1. 8n(n + 1) when m − n is positive and even, 2. −2n(n2 − 1)(n − 2)/(2n + 1) when m = n, 3. 0 for other values of m and n. Example 15.9 (Math. Trip. 1907) Shew that n Õ

n! cosr θPr (cos θ). r!(n − r)! r=0 ∫ π Example 15.10 (Clare, 1903) Shew, by evaluating Pn (cos θ) dθ (Example 15.3.1), and sin θPn (sin θ) = n

(−1)r

0

then integrating by parts, that ∫

1

Pn (µ) arcsin µ dµ =

−1

when n is even

0

1 · 3 · · · (n − 2) π 2 · 4 · · · (n + 1)

2 when n is odd.

Example 15.11 (Adams [9]) If m and n be positive integers, and m ≤ n, shew by induction that m Õ Am−r Ar An−r 2n + 2m − 4r + 1 Pn+m−2r (z), Pm (z)Pn (z) = An+m−r 2n + 2m − 2r + 1 r=0 where Am =

1 · 3 · 5 · · · (2m − 1) . m!

Example 15.12 By expanding in ascending powers of u shew that Pn (z) =

(−1)n d n 2 (u + z2 )−1/2, n! dz n

where u2 is to be replaced by (1 − z2 ) after the differentiation has been performed. Example 15.13 (Heun [299]) Shew that Pn (z) can be expressed as a constant multiple of a determinant in which all elements parallel to the auxiliary diagonal are equal (i.e. all elements are equal for which the sum of the row-index and column-index is the same); the determinant containing n rows, and its elements being 1 1 1 1 1 z, − , z, − , z, . . . , z. 3 3 5 5 2n − 1 Example 15.14 (Silva) Shew that, if the path of integration passes above t = 1, ∫ 2 ∞ {z(1 − t 2 ) − 2t(1 − z2 )1/2 } n Pn (z) = dt. πi 0 (1 − t 2 )n+1

15.9 Miscellaneous examples

349

Example 15.15 (Math. Trip. 1893) By writing cot θ 0 = cot θ − h cosec θ and expanding sin θ 0 in powers of h by Taylor’s theorem, shew that Pn (cos θ) =

d n (sin θ) (−1)n cosecn+1 θ . n! d(cot θ)n ∞ Í

Example 15.16 (Glaisher [247]) By considering 1 Pn (z) = √ n! π

∫

n=0

∞

e

−(1−z 2 )t 2

−∞

h n Pn (z), shew that

d − dz

n

e−z

2 2

t

dt.

Example 15.17 (Math. Trip. 1894) The equation of a nearly spherical surface of revolution is r = 1 + α{P1 (cos θ) + P3 (cos θ) + · · · + P2n−1 (cos θ)}, where α is small; shew that if α2 be neglected the radius of curvature of the meridian is 1+α

n−1 Õ

{n(4m + 3) − (m + 1)(8m + 3)}P2m+1 (cos θ).

m=0

Example 15.18 (Trinity, 1894) The equation of a nearly spherical surface of revolution is r = α {1 + ε Pn (cos θ)}, where ε is small. Shew that if ε 3 be neglected, its area is 1 2 n2 + n + 2 2 4πα 1 + ε . 2 2n + 1 Example 15.19 (Routh [564]) Shew that, if k is an integer and (1 − 2hz + h2 )−k/2 =

∞ Õ

αn Pn (z),

n=0

then αn =

1 (k−3) 1 hn 2 2 (k−3) (2n + 1) ∂ 2 2 ∂ h + x −n+k/2−2 y n+k/2−2, ∂x ∂y (1 − h2 )k−2 1 · 3 · 5 · · · (k − 2)

where x and y are to be replaced by unity after the differentiations have been performed. Example 15.20 (Catalan) Shew that ∫ 1 2 1 {Pn (x)Pn−1 (z) − Pn−1 (x)Pn (z)} dx = − , n −1 z − x ∞ Õ 1 d 1 1 Pn (z) Pn−1 (z) + Pn+1 (z) = −1. 2n + 1 dz n n + 1 n=1 Example 15.21 Let x 2 + y 2 + z2 = r 2 , z = µr, the numbers involved being real, so that −1 < µ < 1. Shew that (−1)n r n+1 ∂ n 1 Pn (µ) = , n! ∂z n r

Legendre Functions

350

where r is to be treated as a function of the independent variables x, y, z in performing the differentiations. Example 15.22 With the notation of Example 15.3.4, shew that r−z (−1)n r n+1 ∂ n 1 log , Q n (µ) = n! ∂z n 2r r+z (−1)n r n+3 ∂ n 1 (n + 1)Pn (µ) + µPn0 (µ) = . n! ∂z n r 3 Example 15.23 Shew that, if |h| and |z| are sufficiently small, ∞

Õ 1 − h2 = (2n + 1)h n Pn (z). (1 − 2hz + h2 )3/2 n=0 Example 15.24 (Math. Trip. 1894) Prove that Pn+1 (z)Q n−1 (z) − Pn−1 (z)Q n+1 (z) =

2n + 1 z. n(n + 1)

Example 15.25 (Bauer) If the arbitrary function f (x) can be expanded in the series f (x) =

∞ Õ

αn Pn (x),

n=0

converging uniformly in a domain which includes the point x = 1, shew that the expansion of the integral of this function is ∫ x ∞ Õ 1 αn−1 αn+1 f (x) dx = −α0 − α1 + − Pn (x). 3 2n − 1 2n + 3 1 n=1 Example 15.26 (Bauer [58]) Determine the coefficients in Neumann’s expansion of eαz in a series of Legendre polynomials. Example 15.27 (Catalan) Deduce from Example 15.25 that 2 ∞ π Õ 1 · 3 · 5 · · · (2n − 1) {P2n+1 (z) − P2n−1 (z)}. arcsin z = 2 0 2 · 4 · 6 · · · 2n Example 15.28 (Schläfli; Hermite [293]) Shew that 1 z+1 1 Q n (z) = log · Pn (z)−{Pn−1 (z)P0 (z) + Pn−2 (z)P1 (z) 2 z−1 2 1 1 + Pn−3 (z)P2 (z) + · · · + P0 (z)Pn−1 (z)}. 3 n Example 15.29 (Math. Trip. 1898) Shew that 1 dn z+1 1 z+1 2 n (z − 1) log − Pn (z) log . Q n (z) = n n 2 n! dz z−1 2 z−1 Prove also that Q n (z) =

1 z+1 Pn (z) log − fn−1 (z), 2 z−1

15.9 Miscellaneous examples

351

where fn−1 (z) = =

2n − 5 2n − 9 2n − 1 Pn−1 (z) + Pn−3 (z) + Pn−5 (z) + · · · 1·n 3(n − 1) 5(n − 2) z−1 z−1 2 k n + (k n − 1) n(n+1) + k n − 1 − 1 n(n−1)(n+1)(n+2) 2 2 2 1

+ kn − 1 −

2

1 2

−

1 3

2

n(n−1)(n−2)(n+1)(n+2)(n+3) 12 22 32

1 2

z−1 3 2

2

+···

,

where k n = 1 + + + · · · + The first of these expressions for fn−1 (z) was given by Christoffel [145] and he also gives a generalisation of Example 15.28; the second was given by Stieltjes [298, p. 59]. 1 2

1 3

1 . n

Example 15.30 Shew that the complete solution of Legendre’s differential equation is ∫ ∞ dt y = APn (z) + BPn (z) , 2 (t − 1){Pn (t)}2 z the path of integration being the straight line which when produced backwards passes through the point t = 0. Example 15.31 (Schläfli) Shew that {z + (z2 − 1)1/2 }α =

∞ Õ

Bm Q2m−α−1 (z),

m=0

where Bm = −

α(α − 2m + 12 ) Γ(m − 12 )Γ(m − α − 12 ) . 2π m! Γ(m − α + 1)

Example 15.32 Shew that, when Re(n + 1) > 0, ∫ ∞ h−n−1 dh , and Q n (z) = 2 1/2 z+(z 2 −1)1/2 (1 − 2hz + h ) ∫ z+(z 2 −1)1/2 h n dh Q n (z) = . (1 − 2hz + h2 )1/2 0 Example 15.33 (Hobson) Shew that Qm n (z)

=e

mπi

Γ(n + 1) Γ(n − m + 1)

∫ 0

∞

{z +

cosh mu du , − 1)1/2 cosh u} n+1

(z2

where Re(n + 1) > m. Example 15.34 Obtain the expansion of Pn (z) when | arg z| < π as a series of powers of 1/z, when n is not an integer, namely tan nπ {Q n (z) − Q−n−1 (z)} π 2n Γ(n + 12 ) n 1−n n 1 1 = , − , − n, 2 z F 2 2 2 z Γ(n + 1)Γ( 21 ) 1 −n−1 2 Γ(−n − 2 ) −n−1 n n+1 3 1 + z F + 1, ,n + , 2 . 2 2 2 z Γ(−n)Γ( 12 )

Pn (z) =

Legendre Functions

352

[This is most easily obtained by the method of §14.51.] Example 15.35 (Olbricht) Shew that the differential equation for the associated Legendre function Pnm (z) is defined by the schemes 16

P

0

∞

1

− 12 n

m

− 12 n

1 n 2

+

−m

1 2

1 n 2

+

1/2

z+(z 2 −1) z−(z 2 −1)1/2 1 2

, P

0

∞

1

− 21 n

1 m 2

0

1 n 2

+

1 2

− 12 m

1 1−z 2

1 2

.

Example 15.36 Shew that the differential equation for Cnν (z) is defined by the scheme P

−1 ∞ − 21 − ν n + 2ν 0 −n

1 2

1 −ν 0

z .

Example 15.37 (Math. Trip. 1896) Prove that, if ys =

s (2n + 1)(2n + 3) · · · (2n + 2s − 1) 2 s d Pn , (z − 1) n(n2 − 1)(n2 − 4) · · · {n2 − (s − 1)2 }(n + s) dz s

then 2 (2n + 1) 2n + 3 Pn + Pn−2, 2n − 1 2n − 1 3(2n + 3) 3(2n + 5) (2n + 3)(2n + 5) y3 = Pn+3 − Pn+1 + Pn−1 − Pn−3, 2n − 1 2n − 3 (2n − 1)(2n − 3) y2 = Pn+2 −

and find the general formula. Example 15.38 (Math. Trip. 1901) Shew that " cos{(n + 12 )θ − 14 π + 21 mπ} Γ(n + m + 1) 2 Pnm (cos θ) = √ (2 sin θ)1/2 π Γ(n + 23 ) +

3 3 1 (12 − 4m2 ) cos{(n + 2 )θ − 4 π + 2 mπ} 2(2n + 3) (2 sin θ)3/2

# 5 5 1 (12 − 4m2 )(32 − 4m2 ) cos{(n + 2 )θ − 4 π + 2 mπ} + +··· , 2 · 4 · (2n + 3)(2n + 5) (2 sin θ)5/2 obtaining the ranges of values of m, n and θ for which it is valid. Example 15.39 (Macdonald [445, 447]) Shew that the values of n, for which Pn−m (cos θ) vanishes, decrease as θ increases from 0 to π when m is positive; and that the number of real zeros of Pn−m (cos θ) for values of θ between −π and π is the greatest integer less than n − m + 1. 16

See also Example 15.5.1.

15.9 Miscellaneous examples

353

Example 15.40 (Legendre) Obtain the formula ∫ π −1/2 1 1 − 2h{cos ω cos φ + sin ω sin φ cos(θ 0 − θ)} + h2 dθ 2π −π ∞ Õ = h n Pn (cos ω)Pn (cos φ). n=0

Example 15.41 (Trinity, 1893) If f (x) = x 2 for x ≥ 0, and f (x) = −x 2 for x < 0, shew that, if f (x) can be expanded into a uniformly convergent series of Legendre polynomials in the range (−1, 1), the expansion is ∞ Õ 3 1 · 3 · · · (2r − 3) 4r + 3 f (x) = P1 (x) − (−1)r P2r+1 (x). 4 4 · 6 · 8 · · · 2r 2r + 4 r=1 Example 15.42 (Gegenbauer [241]) If

∞ Í 1 = h n Cnν (z), shew that (1 − 2hz + h2 )ν n=0

Cnν {x x1 − (x 2 − 1)1/2 (x12 − 1)1/2 cos φ} n λ 2 Γ(2ν − 1) Õ λ 4 Γ(n − λ + 1){Γ(ν + λ)} (2ν + 2λ − 1) = (−1) {Γ(ν)}2 λ=0 Γ(n + 2ν + λ) ν− 1

ν+λ ν+λ (x1 )Cλ 2 (cos φ) (x)Cn−λ × (x 2 − 1) 2 λ (x1 2 − 1) 2 λ Cn−λ ∫ e1 Example 15.43 (Pincherle [522]) If σn (z) = (t 3 − 3tz + 1)−1/2 t n dt, where e1 is the 1

least root of t 3 − 3tz + 1 = 0, shew that

1

0

(2n + 1)σn+1 − 3(2n − 1)zσn−1 + 2(n − 1)σn−2 = 0, and 4(4z3 − 1)σn000 + 144z2 σn00 − z(12n2 − 24n − 291)σn0 − (n − 3)(2n − 7)(2n + 5)σn = 0, where σn000 =

d 3 σn (z) , etc. dz 3

Example 15.44 (Pincherle [520]) If (h3 − 3hz + 1)−1/2 =

∞ Í n=0

Rn (z)h n, shew that

2(n + 1)Rn+1 − 3(2n + 1)zRn + (2n − 1)Rn−2 = 0, 0 nRn + Rn−2 − zRn0 = 0,

and 4(4z3 − 1)Rn000 + 96z2 Rn00 − z(12n2 + 24n − 91)Rn0 − n(2n + 3)(2n + 9)Rn = 0, where Rn000 =

d 3 Rn , etc. dz 3

Example 15.45 (Schendel [575])

If An (x) =

1 dn {(x 2 − 1)n (x − 1)}, obtain 2n n!(x − 1) dx n

the recurrence formula (n + 1)(2n − 1)An (x) − {(4n2 − 1)x + 1}An−1 (x) + (n − 1)(2n + 1)An−2 (x) = 0.

Legendre Functions

354

Example 15.46 If n is not negative and m is a positive integer, shew that the equation (x 2 − 1)

dy d2 y + (2n + 2)x = m(m + 2n + 1)y 2 dx dx

has the two solutions Km (x) = (x 2 − 1)−n

dm 2 (x − 1)m+n, dx m

Lm (x) = (x 2 − 1)−n

∫

1

−1

(t 2 − 1)n Km (t) dt, x−t

when x is not a real number such that −1 ≤ x ≤ 1. Example 15.47 (Clare, 1901) Prove that {1 − hx − (1 − 2hx + h )

∞ Õ

n h n+m 1 d n+m x 2 − 1 } = m(x − 1) . (n + m)! n dx n+m 2 n=m

2 1/2 m

2

m

∞ (m + α)n Í x m, shew that m! m=0 n d αt+xs t ) Fα,n (x) = (e = e x Pn (x, α), dt n t=0

Example 15.48 (Trinity, 1905) If Fα,n (x) =

where Pn (x, α) is a polynomial of degree n in x; and deduce that d Pn (x, α). dx Example 15.49 (Léauté) If Fn (x) be the coefficient of z n in the expansion of Pn+1 (x, α) = (x + α)Pn (x, α) + x

ehz

2hz e xz − e−kz

in ascending powers of z, so that F0 (x) = 1,

F1 (x) = x,

F2 (x) =

3x 2 − h2 , . . ., 6

shew that: 1. Fn (x) is a homogeneous polynomial of degree n in x and h; dFn (x) 2. = Fn−1 (x) for n ≥ 1; ∫ dx k

Fn (x) dx = 0 for n ≥ 1;

3. −k

4. If y = α0 F0 (x) + α1 F1 (x) + α2 F2 (x) + · · · , where α0 , α1 , α2, . . . are real constants, then dr y the mean-value of r in the interval from x = −h to x = +h is αr . dx Example 15.50 (Appell) If Fn (x) be defined as in the preceding example, shew that, when −h < x < h, 2h2m πx 1 2πx 1 3πx F2m (x) = (−1)m 2m cos − 2m cos + 2m cos +··· , π h 2 h 3 h 2m+1 πx 1 2πx 1 3πx m 2h F2m+1 (x) = (−1) sin − 2m+1 sin + 2m+1 sin +··· . π 2m+1 h 2 h 3 h

16 The Confluent Hypergeometric Function

16.1 The confluence of two singularities of Riemann’s equation We have seen (§10.8) that the linear differential equation with two regular singularities only can be integrated in terms of elementary functions; while the solution of the linear differential equation with three regular singularities is substantially the topic of Chapter 14. As the next type in order of complexity, we shall consider a modified form of the differential equation which is obtained from Riemann’s equation by the confluence of two of the singularities. This confluence gives an equation with an irregular singularity (corresponding to the confluent singularities of Riemann’s equation) and a regular singularity corresponding to the third singularity of Riemann’s equation. The confluent equation is obtained by making c → ∞ in the equation defined by the scheme 0 ∞ c 1 P 2 + m −c c − k z . 1 − m 0 k 2 The equation in question is readily found to be d 2 u du k + + + 2 dz dz z

1 4

! − m2 u = 0. z2

(16.1)

We modify this equation by writing u = e− 2 z Wk,m (z) and obtain as the equation 1 ( ) 1 d 2W 1 k 4 − m2 + − + + W = 0. (16.2) dz2 4 z z2 1

The reader will verify that the singularities of this equation are at 0 and ∞, the former being regular and the latter irregular; and when 2m is not an integer, two integrals of equation (16.2) which are regular near 0 and valid for all finite values of z are given by the series ( ) 1 +m−k ( 12 + m − k)( 32 + m − k) 2 2 1/2+m − 21 z Mk,m (z) = z e 1+ z+ z +··· , 1!(2m + 1) 2!(2m + 1)(2m + 2) ( ) 1 −m−k ( 12 − m − k)( 32 − m − k) 3 2 1/2−m − 12 z Mk,−m (z) = z e 1+ z+ z +··· . 1!(1 − 2m) 2!(1 − 2m)(2 − 2m) 1

This equation was given by Whittaker [671], for Wk , m (z).

355

The Confluent Hypergeometric Function

356

These series obviously form a fundamental system of solutions. Note Series of the type above have been considered by Kummer [389, p. 139] and more recently by Jacobsthal [355] and Barnes [49]; the special series in which k = 0 had been investigated by Lagrange in 1762–1765 [396, vol. I, p. 480]. In the notation of Kummer, modified by Barnes, they would be written 1 F1 12 ± m − k; ±2m + 1; z ; the reason for d2 y dy discussing solutions of equation (16.2) rather than those of the equation z 2 − (z − ρ) − dz dz ay = 0, of which 1 F1 (a; ρ; z) is a solution, is the greater appearance of symmetry in the formulae, together with a simplicity in the equations giving various functions of Applied Mathematics (see §16.2) in terms of solutions of equation (16.2).

16.11 Kummer’s formulae (I) We shall now shew that, if 2m is not a negative integer, then z−1/2−m Mk,m (z) = (−z)−1/2−m M−k,m (−z); that is to say, ( e−z

( 1 + m − k)( 32 + m − k) 2 +m−k z+ 2 z +··· 1+ 1!(2m + 1) 2!(2m + 1)(2m + 2)

=1−

1 2

)

+m+k ( 1 + m + k)( 32 + m + k) 2 z+ 2 z −··· . 1!(2m + 1) 2!(2m + 1)(2m + 2) 1 2

For, replacing e−z by its expansion in powers of z, the coefficient of z n in the product of absolutely convergent series on the left is (−1)n F n!

1 2

(−1)n Γ(2m + 1)Γ(m + 12 + k + n) + m − k, −n; 2m + 1; 1 = , n! Γ(m + 12 + k)Γ(2m + 1 + n)

by §14.11, and this is the coefficient of z n on the right (the result is still true when m + 12 + k is a negative integer, by a slight modification of the analysis of §14.11); we have thus obtained the required result. This will be called Kummer’s first formula. (II) The equation ) ( ∞ 2p Õ z , M0,m (z) = z1/2+m 1 + 24p p!(m + 1)(m + 2) · · · (m + p) p=1 valid when 2m is not a negative integer, will be called Kummer’s second formula. To prove it we observe that the coefficient of z n+m+1/2 in the product 2m+1/2 e−z/2 1 F1 (m + 12 ; 2m + 1; z), of which the second and third factors possess absolutely convergent expansions, is (§3.73) ( 12 + m)( 23 + m) · · · (n − m + 12 ) F −n, −2m − n; −n + 12 − m; 12 n!(2m + 1)(2m + 2) · · · (2m + n) ( 1 + m)( 23 + m) · · · (n − m + 12 ) = 2 F − 12 n, −m − 12 n; −n + 12 − m; 1 , n!(2m + 1)(2m + 2) · · · (2m + n)

16.1 The confluence of two singularities of Riemann’s equation

357

by Kummer’s relation (see Chapter 14, Examples 14.12 and 14.13) F(2α, 2β; α + β + 12 ; x) = F{α, β; α + β + 21 ; 4x(1 − x)}, valid when 0 ≤ x ≤ 21 ; and so the coefficient of z n+m+1/2 (by §14.11) is ( 12 + m)( 32 + m) · · · (n − m + 12 ) Γ(−n + 12 − m)Γ( 12 ) n!(2m + 1)(2m + 2) · · · (2m + n) Γ( 12 − m − 12 n)Γ( 12 − 21 n) =

Γ( 12 − m)Γ( 12 ) n!(2m + 1)(2m + 2) · · · (2m + n)Γ( 12 − m − 12 n)Γ( 12 − 12 n)

,

and when n is odd this vanishes; for even values of n (= 2p) it is Γ( 21 − m)(− 12 )(− 32 ) · · · ( 12 − p) (2p)!22p (m + 12 )(m + 23 ) · · · (m + p − 12 )(m + 1)(m + 2) · · · (m + p)Γ( 12 − m − p) 1 · 3 · · · (2p − 1) 1 = = 4p . 3p (2p)! 2 (m + 1)(m + 2) · · · (m + p) 2 · p!(m + 1)(m + 2) · · · (m + p)

16.12 Definition of the function Wk,m (z) The function Wk,m (z) was defined by means of an integral in this manner by Whittaker [671]. The solutions Mk,±m (z) of equation (16.2) of §16.1 are not, however, the most convenient to take as the standard solutions, on account of the disappearance of one of them when 2m is an integer. The integral obtained by confluence from that of §14.6, when multiplied by a constant multiple of ez/2 , is 2 ∫ (0+) 1 1 1 Wk,m (z) = − Γ k + 12 − m e−z/2 z k (−t)−k− 2 +m (1 + t/z)k− 2 +m e−t dt. 2πi ∞ It is supposed that arg z has its principal value and that the contour is so chosen that the point t = −z is outside it. The integrand is rendered one-valued by taking | arg(−t)| ≤ π and taking that value of arg(1 + t/z) which tends to zero as t → 0 by a path lying inside the contour. Under these circumstances it follows from §5.32 that the integral is an analytic function of z. To shew that it satisfies equation (16.2), write ∫ (0+) 1 1 v= (−t)−k− 2 +m (1 + t/z)k− 2 +m e−t dt; ∞ 3

and we have without difficulty 1 d2v 2k dv 4 − m2 + k(k − 1) + − 1 + v dz2 z dz z2 ∫ (k − 21 + m) (0+) d −k+1/2+m (1 + t/z)k−3/2+m e−t dt =− t 2 z dt ∞ = 0, 2 3

A suitable contour has been chosen and the variable t of §14.6 replaced by −t. The differentiations under the sign of integration are legitimate by Corollary 4.4.1.

The Confluent Hypergeometric Function

358

since the expression in braces tends to zero as t → +∞; and this is the condition that e−z/2 z k v should satisfy (16.2). Accordingly the function Wk,m (z) defined by the integral 1 Γ k+ − 2πi

1 2

−m e

(0+)

∫

− 12 z k

(−t)−k− 2 +m (1 + t/z)k− 2 +m e−t dt 1

1

z

∞

is a solution of the differential equation (16.2). The formula for Wk,m (z) becomes nugatory when k − 12 − m is a negative integer. To overcome this difficulty, we observe that whenever Re k − 12 − m ≤ 0 and k − 12 − m is not an integer, we may transform the contour integral into an infinite integral, after the manner 1 of §12.22; and so, when Re k − 2 − m ≤ 0, 1

e− 2 z z k Wk,m (z) = 1 Γ( 2 − k + m)

∫

∞

t −k− 2 +m (1 + t/z)k− 2 +m e−t dt. 1

1

0

This formula suffices to define Wk,m (z) in the critical cases when m + 12 − k is a positive integer, and so Wk,m (z) is defined for all values of k and m and all values of z except negative real values 4 . Example 16.1.1 Solve the equation b c d2u + a + u=0 + dz 2 z z2 in terms of functions of the type Wk,m (z), where a, b, c are any constants.

16.2 Expression of various functions by functions of the type Wk,m (z) It has been shewn 5 that various functions employed in Applied Mathematics are expressible by means of the function Wk,m (z); the following are a few examples: (I) The Error function 6 which occurs in connexion with the theories of Probability, Errors of Observation, Refraction and Conduction of Heat is defined by the equation ∫ ∞ 2 Erfc(x) = e−t dt, x

where x is real. 4

5 6

When z is real and negative, Wk , m (z) may be defined to be either Wk , m (z + 0i) or Wk , m (z − 0i), whichever is more convenient. Whittaker [671]; this paper contains a more complete account than is given here. This name is also applied to the function Erf(x) =

∫ 0

x

√ e−t dt = 2

π − Erfc(x). 2

16.2 Expression of various functions by functions of the type Wk,m (z)

359

Writing t = x 2 (w 2 − 1) and then w = s/x in the integral for W− 41 , 41 (x 2 ), we get ∫ ∞ −1/2 −t 2 − 21 − 21 x 2 W− 14 , 14 (x ) = x e 1 + t/x 2 e dt ∫0 ∞ 1 2 3 2 = 2x 2 e− 2 x e x (1 − w 2 ) dw ∫ 1∞ 1 2 1 2 e−s ds, = 2x 2 e 2 x x

and so the error function is given by the formula Erfc(x) = 21 x − 2 e− 2 x W− 14 , 14 (x 2 ). 1

1

2

Other integrals which occur in connexion with the theory of Conduction of Heat, e.g. ∫

b

e−t

2

−x 2 /t 2

dt, can be expressed in terms of error functions, and so in terms of Wk,m

a

functions. Example 16.2.1 of x.

Shew that the formula for the error function is true for complex values

(II) The Incomplete Gamma-function, studied by Legendre and others 7 , is defined by the equation ∫ x γ(n, x) = t n−1 e−t dt. 0

By writing t = s − x in the integral for W 21 (n−1), 12 n (x), the reader will verify that γ(n, x) = Γ(n) − x 2 (n−1) e− 2 x W 12 (n−1), 12 n (x). 1

1

(III) The Logarithmic-integral function, which has been discussed by Euler and others 8 , is of considerable importance in the higher parts of the Theory of Prime Numbers; see Landau [405]. It is defined, when | arg(− log z)| < π, by the equation ∫ z dt . li(z) = 0 log t On writing s − log z = u and then u = − log t in the integral for W− 12 ,0 (− log z), it may be verified that li(z) = −(− log z)− 2 z 2 W− 21 ,0 (− log z). 1

1

It will appear later that Weber’s Parabolic Cylinder functions (§16.5) and Bessel’s Circular Cylinder functions (Chapter 17) are particular cases of the Wk,m function. Other functions of like nature are given in the Miscellaneous Examples at the end of this chapter. 7 8

Legendre [421, vol. I, p. 339]; Hočevar [324]; Schlömilch [584]; Prym [545]. Euler [201]; Soldner [597]; Bessel [69]; Laguerre [397]; Stieltjes [605].

The Confluent Hypergeometric Function

360

Note The error function has been tabulated by Encke [195], and Burgess [107]. The logarithmic-integral function has been tabulated by Bessel and by Soldner. Jahnke & Emde [356], and Glaisher [253], should also be consulted.

16.3 The asymptotic expansion of Wk,m (z), when |z| is large From the contour integral by which Wk,m (z) was defined, it is possible to obtain an asymptotic expansion for Wk,m (z) valid when | arg z| < π. For this purpose, we employ the result given in Chapter 5, Example 5.6, that λ λt λ(λ − 1) · · · (λ − n + 1) t n t =1+ +···+ + Rn (t, z), 1+ z 1z n! zn where

λ ∫ t/z λ(λ − 1) · · · (λ − n) t Rn (t, z) = 1+ un (1 + u)−λ−1 du. n! z 0

Substituting this in the formula of §16.12 and integrating term-by-term, it follows from the result of §12.22 that ( m2 − (k − 12 )2 {m2 − (k − 12 )2 }{m2 − (k − 32 )2 } 1 + Wk,m (z) = e− 2 z z k 1 + 1!z 2!z2 {m2 − (k − 12 )2 }{m2 − (k − 23 )2 } · · · {m2 − (k − n + 21 )2 } n!z n ) ∫ ∞ 1 1 + t −k− 2 +m Rn (t, z)e−t dt Γ(−k + 21 + m) 0 provided that n be taken so large that Re n − k − 21 + m > 0. Now, if | arg z| ≤ π − α and |z| > 1, then ) 1 ≤ | (1 + t/z) | ≤ 1 + t Re(z) ≥ 0 , |1 + t/z| ≥ sin α Re(z) ≤ 0 +··· +

and so 9 ∫ λ(λ − 1) · · · (λ − n) (1 + t) |λ| |Rn (t, z)| ≤ (sin α) |λ| n!

|t/z |

un (1 + u) |λ| du.

0

Therefore λ(λ − 1) · · · (λ − n) (1 + t) |λ| n+1 |λ| −1 |Rn (t, z)| < (sin α) |λ| |t/z| (1 + t) (n + 1) , n! since 1 + u < 1 + t. Therefore, when |z| > 1, ∫ ∞ ∫ ∞ 1 1 −k− 2 +m −t −k+ 12 +m+n 2|λ| −n−1 −t t Rn (t, z)e dt = O t (1 + t) |z| e dt Γ(−k + 12 + m) 0 0 = O(z −n−1 ), 9

It is supposed that λ is real; the inequality has to be slightly modified for complex values of λ.

16.4 Contour integrals of the Mellin–Barnes type for Wk,m (z)

361

since the integral converges. The constant implied in the symbol O is independent of arg z, but depends on α, and tends to infinity as α → 0. That is to say, the asymptotic expansion of Wk,m (z) is given by the formula 1

Wk,m (z) ∼ e− 2 z z k ( ) 2 ∞ Õ m − (k − 21 )2 m2 − (k − 32 )2 · · · m2 − (k − n + 12 )2 × 1+ n!z n n=1 for large values of |z| when | arg z| ≤ π − α < π.

16.31 The second solution of the equation for Wk,m (z) The differential equation (16.2) of §16.1 satisfied by Wk,m (z) is unaltered if the signs of z and k are changed throughout. Hence, if | arg(−z)| < π, W−k,m (−z) is a solution of the equation. Since, when | arg z| < π, Wk,m (z) = e−z/2 z k 1 + O z−1 , whereas, when | arg(−z)| < π, W−k,m (−z) = ez/2 (−z)−k 1 + O z−1 , the ratio Wk,m (z)/W−k,m (−z) cannot be a constant, and so Wk,m (z) and W−k,m (−z) form a fundamental system of solutions of the differential equation.

16.4 Contour integrals of the Mellin–Barnes type for Wk,m (z) Consider now e−z/2 z k I= 2πi

∫

i∞

−i∞

Γ(s)Γ(−s − k − m + 12 )Γ(−s − k + m + 12 ) Γ(−k − m + 12 )Γ(−k + m + 12 )

z s ds,

(16.3)

where | arg z| < 23 π, and neither of the numbers k ± m + 21 is a positive integer or zero 10 ; the contour has loops if necessary so that the poles of Γ(s) and those of Γ −s − k − m + 12 × Γ −s − k + m + 21 are on opposite sides of it. It is easily verified, by §13.6, that, as s → ∞ on the contour, Γ(s)Γ −s − k − m + 21 Γ −s − k + m + 12 = O(e−3π |s |/2 |s| −2k−1/2 ), and so the integral represents a function of z which is analytic at all points 11 in the domain | arg z| ≤ 23 π − α < 32 π. Now choose N so that the poles of Γ −s − k − m + 12 Γ −s − k + m + 12 are on the right of the line Re(s) = −N − 21 ; and consider the integral taken round the rectangle whose corners are ±iξ, −N − 12 ± iξ, where ξ is positive 12 and large. The reader will verify that, 10 11 12

In these cases the series of §16.3 terminates and Wk , m (z) is a combination of elementary functions. The integral is rendered one-valued when Re(z) < 0 by specifying arg z. The line joining ±iξ may have loops to avoid poles of the integrand as explained above.

The Confluent Hypergeometric Function

362

when | arg z| ≤ 32 π − α, the integrals

∫

−N − 12 −iξ

∫

−N − 12 +iξ

and −iξ

tend to zero as ξ → ∞; and

iξ

so, by Cauchy’s theorem, ∫ 1 1 e−z/2 z k i∞ Γ(s)Γ(−s − k − m + 2 )Γ(−s − k + m + 2 ) s z ds 2πi Γ(−k − m + 12 )Γ(−k + m + 12 ) −i∞ ) ( N ∫ −N − 21 +i∞ 1 1 Õ Γ(s)Γ(−s − k − m + )Γ(−s − k + m + ) 1 1 2 2 = e− 2 z z k z s ds , Rn + 1 1 1 2πi Γ(−k − m + )Γ(−k + m + ) −i∞ −N − 2 n=0 2 2 where Rn is the residue of the integrand at s = −n. Write s = −N − 12 + it, and the modulus of the last integrand is 1 |z| −N − 2 O e−α |t | |t| N −2k , ∫ where the constant implied in the symbol O is independent of z. Since

±∞

e−α |t | |t| N −2k dt

converges, we find that ( I=e

−z/2 k

z

N Õ

) Rn + O(|z|

−N −1/2

) .

n=0

But, on calculating the residue Rn , we get Rn =

Γ(n − k − m + 21 )Γ(n − k + m + 12 )

(−1)n z−n n!Γ(−k − m + 12 )Γ(−k + m + 12 ) 2 m − (k − 12 )2 m2 − (k − 23 )2 · · · m2 − (k − n + 12 )2 , = n!z n

and so I has the same asymptotic expansion as Wk,m (z). Further, I satisfies the differential equation for Wk,m (z); for, on substituting ∫ i∞ Γ(s)Γ −s − k − m + 12 Γ −s − k + m + 12 z s ds −i∞

for v in the expression (given in §16.12) z2

d2v dv dv + 2k z + k − m − 12 k + m − 12 v − z2 , 2 dz dz dz

we get ∫

i∞

Γ(s)Γ −s − k − m + 32 Γ −s − k + m + 32 z s ds −i∞ ∫ ∞i − Γ(s + 1)Γ −s − k − m + 12 Γ −s − k + m + 12 z s+1 ds −∞i ∫ i∞ ∫ 1+i∞ 3 = − Γ(s)Γ −s − k − m + Γ −s − k + m + 32 z s ds. 2 −i∞ 1−i∞

Since there are no poles of the last integrand between the contours, and since the integrand

16.4 Contour integrals of the Mellin–Barnes type for Wk,m (z)

363

tends to zero as |s| → ∞, s being between the contours, the expression under consideration vanishes, by Cauchy’s theorem; and so I satisfies the equation for Wk,m (z). Therefore I = AWk,m (z) + BW−k,m (−z), where A and B are constants. Making |z| → ∞ when Re(z) > 0 we see, from the asymptotic expansions obtained for I and W∓k,m (±z), that A = 1, B = 0. Accordingly, by the theory of analytic continuation, the equality I = Wk,m (z) persists for all values of z such that | arg z| < π; and, for values 13 of arg z such that π ≤ | arg z| < 23 π, Wk,m (z) may be defined to be the expression I. Example 16.4.1 Shew that 1

e− 2 z Wk,m (z) = 2πi

∫

i∞

−i∞

Γ(s − k)Γ(−s − m + 12 )Γ(−s + m + 12 ) Γ(−k − m + 12 )Γ(−k + m + 12 )

z s ds,

taken along a suitable contour. Example 16.4.2 Obtain Barnes’ integral for Wk,m (z) by writing ∫ i∞ Γ(s)Γ(−s − k − m + 12 ) s −s 1 z t ds 2πi −i∞ Γ(−k − m + 12 ) for (1 + t/z)k− 2 +m in the integral of §16.12 and changing the order of integration. 1

16.41 Relations between Wk,m (z) and Mk,±m (z) If we take the expression F(s) ≡ Γ(s)Γ −s − k − m +

1 2

Γ −s − k + m +

1 2

which occurs in Barnes’ integral for Wk,m (z), and write it in the form π 2 Γ(s) , Γ(s + k + m + 12 )Γ(s + k − m + 12 ) cos(s + k + m)π cos(s + k − m)π we see, by §13.6, that, when Re(s) ≥ 0, we have, as |s| → ∞, F(s) = O exp −s − 12 − 2k log s + s sec(s + k + m)π sec(s + k − m)π. ∫ 3 Hence, if | arg z| < 2 π, F(s)z s ds, taken round a semicircle on the right of the imaginary axis, tends to zero as the radius of the semicircle tends to infinity, provided the lower bound of the distance of the semicircle from the poles of the integrand is positive (not zero). Therefore Í 1 e− 2 z z k · ( R 0 ) Wk,m (z) = − , Γ(−k − m + 12 )Γ(−k + m + 12 ) 13

It would have been possible, by modifying the path of integration in §16.3, to have shewn that that integral could be made to define an analytic function when | arg z | < 3π/2. But the reader will see that it is unnecessary to do so, as Barnes’ integral affords a simpler definition of the function.

The Confluent Hypergeometric Function

364

Í where R 0 denotes the sum of the residues of F(s) at its poles on the right of the contour (cf. §14.5) which occurs in equation (16.3) of §16.4. Evaluating these residues we find without difficulty that, when | arg z| < 23 π, and 2m is not an integer 14 , Wk,m (z) =

Γ(−2m) Γ(2m) Mk,m (z) + 1 Mk,−m (z). − m − k) Γ( 2 + m − k)

Γ( 12

Example 16.4.3 (Barnes) Shew that, when | arg(−z)| < 23 π and 2m is not an integer, W−k,m (−z) =

Γ(2m) Γ(−2m) M−k,m (−z) + 1 M−k,−m (−z). Γ( 12 − m + k) Γ( 2 + m + k)

These results are given in the notation explained in §16.1. Example 16.4.4 When − 12 π < arg z < 32 π; and − 23 π < arg(−z) < 12 π, shew that Mk,m (z) =

Γ(2m + 1) ( 12 +m+k)πi Γ(2m + 1) kπi Wk,m (z). e W−k,m (−z) + 1 e 1 Γ( 2 + m − k) Γ( 2 + m + k)

Example 16.4.5 (Barnes) Obtain Kummer’s first formula (§16.11) from the result ∫ i∞ 1 z n e−z = Γ(n − s)z s ds. 2πi −i∞

16.5 The parabolic cylinder functions. Weber’s equation 1 Consider the differential equation satisfied by w = z− 2 Wk,− 14 21 z2 ; it is ( ) ( ) 1 3 1 d 1 d(wz 2 ) 1 2k 1 4 + − + 2 + 4 wz 2 = 0; z dz z dz 4 z z this reduces to 1 2 d2w + 2k − z w = 0. dz 2 4 Therefore the function Dn (z) = 2 2 n+ 4 z− 2 W 12 n+ 41 ,− 41 1

1

1

1 2 z 2

satisfies the differential equation d 2 Dn (z) + n+ dz2

1 2

− 41 z2 Dn (z) = 0.

Accordingly Dn (z) is one of the functions associated with the parabolic cylinder in harmonic analysis (see Weber [655] and Whittaker [670]); the equation satisfied by it will be called Weber’s equation. 14

When 2m is an integer some of the poles are generally double poles, and their residues involve logarithms of z. The result has not been proved when k − 21 ± m is a positive integer or zero, but may be obtained for such values of k and m by comparing the terminating series for Wk , m (z) with the series for Mk ,±m (z).

16.5 The parabolic cylinder functions. Weber’s equation

365

From §16.41, it follows that 1

when |arg z|

arg z > 41 π, we can assign to −z and −iz arguments between ± 43 π; and arg(−z) = arg z − π, arg(−iz) = arg z − 12 π; and then, applying the asymptotic expansion of §16.5 to Dn (−z) and D−n−1 (−iz), we see that, if 54 π > arg z > 41 π, n(n − 1) n(n − 1)(n − 2)(n − 3) 1 2 Dn (z) ∼e− 4 z z n 1 − + − · · · 2z2 2 · 4z4 √ 2π nπi 14 z 2 −n−1 (n + 1)(n + 2) (n + 1)(n + 2)(n + 3)(n + 4) − e e z 1+ + + · · · . Γ(−n) 1 · 2z2 2 · 4z4 This formula is not inconsistent with that of §16.5 since in their common range of validity, 1 2 viz. 14 π < arg z < 34 π, e 2 z z−2n−1 is o(z−m ) for all positive values of m. To obtain a formula valid in the range − 41 π > arg z > − 54 π, we use the formula √ 2π − 12 (n+1)πi D−n−1 (iz), Dn (z) = e−nπi Dn (−z) + e Γ(−n)

and we get an asymptotic expansion which differs from that which has just been obtained only in containing e−nπi in place of enπi . Since Dn (z) is one-valued and one or other of the expansions obtained is valid for all values of arg z in the range −π ≤ arg z ≤ π, the complete asymptotic expansion of Dn (z) has been obtained.

16.6 A contour integral for Dn (z) ∫

(0+)

Consider

e−zt− 2 t (−t)−n−1 dt, where | arg(−t)| ≤ π; it represents a one-valued analytic 1 2

∞

function of z throughout the z-plane (§5.32) and further 2 ∫ (0+) ∫ (0+) d d −zt− 12 t 2 d −zt− 12 t 2 −n−1 − z + n e (−t) dt = {e (−t)−n } dt = 0, 2 dz dz dt ∞ ∞ the differentiations under the sign of integration being easily justified; accordingly the integral 1 2 satisfies the differential equation satisfied by e 4 z Dn (z); and therefore ∫ (0+) 1 2 − 14 z 2 e e−zt− 2 t (−t)−n−1 dt = aDn (z) + bD−n−1 (iz), ∞

where a and b are constants. Now, if the expression on the right be called En (z), we have ∫ (0+) ∫ (0+) 1 2 − 12 t 2 −n−1 0 (−t) dt, En (0) = e− 2 t (−t)−n dt. En (0) = e ∞

∞

16.7 Properties of Dn (z) when n is an integer

367

To evaluate these integrals, which are analytic functions of n, we suppose first that Re(n) < 0; then, deforming the paths of integration, we get ∫ ∞ 1 2 En (0) = −2i sin(n + 1)π e− 2 t t −n−1 dt ∫ ∞0 = 2−n/2i sin nπ e−u u−n/2−1 du 0 = 2−n/2i sin(nπ)Γ − n2 . Similarly En0 (0) = −2(1−n)/2i sin(nπ)Γ( 21 (1 − n)). Both sides of these equations being analytic functions of n, the equations are true for all values of n; and therefore b = 0,

a=

Γ( 12 − 12 n) Γ( 21 )2n/2

1 2− 2 n i sin(nπ)Γ − n2 = 2iΓ(−n) sin nπ.

Therefore Γ(n + 1) − 14 z 2 Dn (z) = − e 2πi

∫

(0+) 1 2

e−zt− 2 t (−t)−n−1 dt.

∞

16.61 Recurrence formulae for Dn (z) Form the equation ∫

(0+)

o d n −zt− 12 t 2 e (−t)−n−1 dt dt ∞ ∫ (0+) 1 2 = −z(−t)−n−1 + (−t)−n + (n + 1)(−t)−n−2 e−zt− 2 t dt,

0=

∞

after using §16.6, we see that Dn+1 (z) − zDn (z) + nDn−1 (z) = 0. Further, by differentiating the integral of §16.6, it follows that Dn0 (z) + 12 zDn (z) − nDn−1 (z) = 0. Example 16.6.1 Obtain these results from the ascending power series of §16.5.

16.7 Properties of Dn (z) when n is an integer When n is an integral, we may write the integral of §16.6 in the form 1 2 ∫ (0+) 1 2 n!e− 4 z e−zt− 2 t Dn (z) = − dt. 2πi (−t)n+1 If now we write t = v − z, we get 1 2 ∫ (z+) 1 2 e− 2 v n!e 4 z dv 2πi (v − z)n+1 n 1 2 d 1 2 = (−1)n e 4 z n e− 2 z , dz

Dn (z) = (−1)n

The Confluent Hypergeometric Function

368

a result due to Hermite [289]. Also, if m and n be unequal integers, we see from the differential equations that 00 Dn (z)Dm (z) − Dm (z)Dn00(z) + (m − n)Dm (z)Dn (z) = 0,

and so ∫

∞

(m − n) −∞

∞ 0 Dm (z)Dn (z) dz = Dn (z)Dm (z) − Dm (z)Dn0 (z) −∞ = 0,

by the expansion of §16.5 in descending powers of z (which terminates and is valid for all values of arg z when n is a positive integer). Therefore if m and n are unequal positive integers ∫

∞

Dm (z)Dn (z) dz = 0.

−∞

On the other hand, when m = n, we have (n + 1)

∫

∞

{Dn (z)} dz = 2

−∞

∫

∞

−∞

0 Dn (z) Dn+1 (z) + 12 zDn+1 (z) dz

= [Dn (z)Dn+1 (z)]∞ −∞ ∫ ∞ 1 zDn (z)Dn+1 (z) − Dn+1 (z)Dn0 (z) dz + 2 −∞ ∫ ∞ {Dn+1 (z)}2 dz, = −∞

on using the recurrence formula, integrating by parts and then using the recurrence formula again. It follows by induction that ∫

∞

{Dn (z)} dz = n! 2

−∞

= n!

∞

∫

{D0 (z)}2 dz

∫−∞∞

1 2

e− 2 z dz

−∞

= (2π) 2 n!, 1

by Corollary 12.1.1 and §12.2. It follows at once that if, for a function f (z), an expansion of the form f (z) = a0 D0 (z) + a1 D1 (z) + · · · + an Dn (z) + · · · exists, and if it is legitimate to integrate term-by-term between the limits −∞ and ∞, then an =

∫

1 1

(2π) 2 n!

∞

Dn (t) f (t) dt. −∞

16.8 Miscellaneous examples

369

16.8 Miscellaneous examples Example 16.1 Shew that, if the integral is convergent, then Γ(2m + 1)z m+ 2 2−2m Γ( 12 + m + k)Γ( 12 + m − k) ∫ 1 1 1 1 × (1 + u)− 2 +m−k (1 − u)− 2 +m+k e 2 zu du. 1

Mk,m (z) =

−1

Example 16.2 Shew that Mk,m (z) = z 2 +m e−z/2 lim F 1

ρ→∞

1 2

+ m − k, 12 + m − k + ρ; 2m + 1; z/ρ .

Example 16.3 Obtain the recurrence formulae Wk,m (z) = z 2 Wk− 21 ,m− 12 (z) +

− k + m Wk−1,m (z), 1 Wk,m (z) = z 2 Wk− 21 ,m+ 12 (z) + 12 − k − m Wk−1,m (z), n 2o 0 Wk−1,m (z). (z) = k − 12 z Wk,m (z) − m2 − k − 12 zWk,m 1

1 2

Example 16.4 Prove that Wk,m (z) is the integral of an elementary function when either of the numbers k − 21 ± m is a negative integer. Example 16.5 Shew that, by a suitable change of variables, the equation (a2 + b2 x)

dy d2 y + (a1 + b1 x) + (a0 + b0 x)y = 0 dx 2 dx

can be brought to the form ξ

dη d2η + (c − ξ) − aη = 0. dξ 2 dξ

Derive this equation from the equation for F(a, b; c; x) by writing x = ξ/b and making b → ∞. Example 16.6 Shew that the cosine integral of Schlömilch and Besso [71], defined by the equation ∫ ∞ cos t dt, Ci(z) = t z is equal to 1 − 12 12 iz+ 41 πi 1 1 1 1 z e W− 12 ,0 (−iz) + z− 2 e− 2 iz− 4 πi W− 12 ,0 (iz). 2 2 Shew also that Schlömilch’s function, defined in [582] by the equations ∫ ∞ ∫ ∞ −u e −ν −zt ν−1 z S(ν, z) = (1 + t) e dt = z e du, uν 0 z is equal to zν/2−1 ez/2W− 12 ν, 12 − 12 ν (z).

370

The Confluent Hypergeometric Function

Example 16.7 Express in terms of Wk,m functions the two functions ∫ ∞ −t ∫ z e sin t dt, Ei(z) ≡ dt. Si(z) ≡ t t z 0 The results of Examples 16.8, 16.9, 16.10 below were communicated to us by Mr. Bateman. Example 16.8 Shew that Sonine’s polynomial [598], defined by the equation Tmn (z) =

zn z n−1 − n!(m + n)!0! (n − 1)!(m + n − 1)!1! z n−2 −··· , + (n − 2)!(m + n − 2)! 2!

is equal to 1 1 z− 2 (m+1) ez/2Wn+ 12 m+ 12 , 21 m (z). n!(m + n)! Example 16.9 Shew that the function φm (z) defined by Lagrange [394] in 1762–1765 and by Abel [5, p. 284] as the coefficient of h m in the expansion of (1 − h)−1 e−hz/(1−k) is equal to (−1)m −1/2 z/2 z e Wm+ 12 ,0 (z). m! Example 16.10 Shew that the Pearson–Cunningham function [517, 158], ωn,m (z), defined as ( 1 (n + 21 m)(n − 12 m) e−z (−z)n− 2 m 1 − z Γ(n − 12 m + 1) ) (n + 21 m)(n + 12 m − 1)(n − 12 m)(n − 21 m − 1) −··· , + 2!z2 is equal to 1

(−1)n− 2 m 1 z− 2 (m+1) e−z/2Wn+ 12 , 12 m (z). 1 Γ(n − 2 m + 1) Example 16.11 (Whittaker) Shew that, if | arg z| < 14 π and | arg(1 + t)| ≤ π, ∫ Γ( 21 n + 1) (−1+) 1 z 2 t 1 1 e 4 (1 + t)− 2 n−1 (1 − t) 2 (n−1) dt. Dn (z) = 1 − 2 (n−1) 2 πi −∞ Example 16.12 Shew that, if n be not a positive integer and if | arg z| < 43 π, then ∫ 1 1 1 − 14 z 2 i∞ Γ( 2 t − 2 n)Γ(−t) √ t−n−2 t Dn (z) = e ( 2) z dt, 2πi Γ(−n) −i∞ ∫ (0−) and that this result holds for all values of arg z if the integral be , the contours enclosing ). the poles of Γ(−t) but not those of Γ( t−n 2

∞

16.8 Miscellaneous examples

371

Example 16.13 Shew that, if | arg a| < 12 π, ∫ (0+) 1 π 3/2 2n/2−m e πi(m− 2 ) 1 2 e( 4 −a)z z m Dn (z) dz = 1 Γ(−m)Γ( 12 m − 12 n + 1)a 2 (m+1) x 1 n m+1 m−n ; + 1; 1 − ×F − , . 2 2 2 2a Example 16.14 (Watson) Deduce from Example 16.13 that, if the integral is convergent, then ∫ x √ −1−m 3 2 e− 4 z z m Dm+1 (z) dz = 2 Γ(m + 1) sin (1 − m) π4 . 0

Example 16.15 (Watson) Shew that, if n be a positive integer, and if ∫ ∞ 1 2 En (x) = e− 4 z (z − x)−1 Dn (z) dz, −∞

√ 1 2 then En (x) = ±ie∓nπi 2πΓ(n + 1)e− 4 x D−n−1 (∓ix), the upper or lower signs being taken according as the imaginary part of x is positive or negative. Example 16.16 (Adamoff) Shew that, if n be a positive integer, ∫ x cos µ n+2 − 21 41 x 2 n −2u 2 Dn (x) = (−1) 2 (2π) e u e (2xu) du, sin 0 where µ is 21 n or 21 (n − 1), whichever is an integer, and the cosine or sine is taken as n is even or odd. Example 16.17 (Adamoff) Shew that, if n be a positive integer, r 2 √ n+1 14 x 2 − 12 n µ Dn (x) = (−1) n e e (J1 + J2 − J3 ), π where ∫ ∞ √ cos −n(v−1)2 J1 = e (xv n) dv, sin −∞ ∫ ∞ √ cos J2 = σ(v) (xv n) dv, sin 0 ∫ 0 √ cos −n(v−1)2 J3 = e (xv n) dv, sin −∞ and σ(v) = e 2 n(1−v ) v n − e−n(v−1) . 1

2

2

Example 16.18 (Adamoff) With the notation of the preceding examples, shew that, when x is real, √ 1 1 1 2 cos − − x J1 = π 2 n 2 e 4 (xv n) dv sin while J3 satisfies both the inequalities |J3 |

21 π or < − 12 π. When n is an integer (17.6) reduces at once to Bessel’s integral, and (17.7) does so when we make use of the equation Jn (z) = (−1)n J−n (z), which is true for integer values of n.

Bessel Functions

382

Equation (17.6), as already stated, is due to Schläfli [576, p. 148], and equation (17.7) was given by Sonine [598]. These trigonometric integrals for the Bessel functions may be regarded as corresponding to Laplace’s integrals for the Legendre functions. For (see Example 17.1.6) Jm (z) satisfies the confluent form (obtained by making n → ∞) of the equation for Pnm (1 − z2 /2n2 ). But Laplace’s integral for this function is a multiple of ∫ 0

π

( ) 1/2 n 2 2 2 z z 1 − + 1− 2 −1 cos φ cos mφ dφ 2 2n 2n n ∫ π iz −2 = 1 + cos φ + O(n ) cos mφ dφ. n 0

The limit of the integrand as n → ∞ is eiz cos φ cos mφ, and this exhibits the similarity of Laplace’s integral for Pnm (z) to the Bessel–Schläfli integral for Jm (z). ∫ π 1 Example 17.2.9 (Callandreau [112]) From the formula J0 (x) = e−ix cos φ dφ, by a 2π −π change of order of integration, shew that, when n is a positive integer and cos θ > 0, ∫ ∞ 1 e−x cos θ J0 (x sin θ)x n dx. Pn (cos θ) = Γ(n + 1) 0 Example 17.2.10 Shew that, with Ferrers’ definition of Pnm (cos θ), ∫ ∞ 1 Pnm (cos θ) = e−x cos θ Jm (x sin θ)x n dx Γ(n − m + 1) 0 when n and m are positive integers and cos θ > 0. See Hobson [313].

17.24 Bessel functions whose order is half an odd integer We have seen (§17.2) that when the order n of a Bessel function Jn (z) is half an odd integer, the difference of the roots of the indicial equation at z = 0 is 2n, which is an integer. We now shew that, in such cases, Jn (z) is expressible in terms of elementary functions. For 1/2 1/2 z2 2z z4 2 1− sin z, J1/2 (z) = + −··· = π 2·3 2·3·4·5 πz and therefore (§17.211) if k is a positive integer 1

Jk+1/2 (z) =

(−1)k (2z)k+ 2

dk sin z . z d(z2 )k

π2 On differentiating out the expression on the right, we obtain the result that 1

Jk+1/2 (z) = Pk sin z + Q k cos z, where Pk , Q k are polynomials in z− 2 . 1

Example 17.2.11 Shew that J− 12 (z) =

2 πz

1/2 cos z.

17.3 Hankel’s contour integral for Jn (z)

383

Example 17.2.12 Prove by induction that if k be an integer and n = k + 12 , then 1/2 2 Jn (z) = πz " ) ( Õ (−1)r (4n2 − 12 )(4n2 − 32 ) · · · {4n2 − (4r − 1)2 } × cos z − 12 nπ − 41 π 1 + (2r) ! 26r z2r r=1 # Õ (−1)r (4n2 − 12 )(4n2 − 32 ) · · · {4n2 − (4r − 3)2 } + sin z − 12 nπ − 41 π , (2r − 1)! 26r−3 z2r−1 r=1 the summations being continued as far as the terms with the vanishing factors in the numerators. cos z dk k+ 12 Example 17.2.13 Shew that z is a solution of Bessel’s equation for z d(z2 )k Jk+ 12 (z). d 2m+1 y + y = 0 is dz 2m+1 2m n o Õ 1 1 cp J−m− 21 (2a p z 2 ) + iJm+ 21 (2a p z 2 ) , 1

Example 17.2.14 (Lommel) Shew that the solution of z m+ 2 y = z 2 m+ 4 1

1

p=0

where c0, c1, . . . , c2m are arbitrary and a0, a1, . . . , a2m are the roots of a2m+1 = i.

17.3 Hankel’s contour integral for Jn (z) This appears in [273]. Consider the integral ∫ (1+,−1−) 1 y = zn (t 2 − 1)n− 2 cos(zt) dt, A

where A is a point on the right of the point t = 1, and arg(t − 1) = arg(t + 1) = 0 at A; the contour may conveniently be regarded as being in the shape of a figure of eight. We shall shew that this integral is a constant multiple of Jn (z). It is easily seen that the 1 integrand returns to its initial value after t has described the path of integration; for (t − 1)n− 2 1 is multiplied by the factor e(2n−1)πi after the circuit (1+) has been described, and (t + 1)n− 2 is multiplied by the factor e−(2n−1)πi after the circuit (−1−) has been described. ∞ Í (−1)r (zt)2r 2 Since (t − 1)n−1/2 converges uniformly on the contour, we have (2r)! r=0

y=

∫ ∞ Õ (−1)r z n+2r r=0

(2r)!

(1+,−1−)

t 2r (t 2 − 1)n−1/2 dt, A

(see §4.7). To evaluate these integrals, we observe firstly that they are analytic functions of n for all values of n, and secondly that, when Re (n + 12 ) > 0, we may deform the contour into the circles |t − 1| = δ, |t + 1| = δ and the real axis joining the points t = ±(1 − δ) taken twice,

Bessel Functions

384

and then we may make δ → 0; the integrals round the circles tend to zero and, assigning to t − 1 and t + 1 their appropriate arguments on the modified path of integration, we get, if arg(1 − t 2 ) = 0 and t 2 = u, ∫ (1+,−1−) t 2τ (t 2 − 1)n−1/2 dt A ∫ 1 ∫ −1 (n−1/2)πi 2τ 2 n−1/2 −(n−1/2)πi =e t (1 − t ) dt + e t 2τ (1 − t 2 )n−1/2 dt 1 −1 ∫ 1 t 2τ (1 − t 2 )n−1/2 dt = −4i sin n − 21 π 0 ∫ 1 = −2i sin n − 12 π uτ−1/2 (1 − u)n−1/2 du 0 = 2i sin n + 12 πΓ r + 12 Γ n + 21 /Γ(n + r + 1). Since the initial and final expressions are analytic functions of n for all values of n, it 1 follows from §5.5 that this equation, proved when Re n + 2 > 0, is true for all values of n. Accordingly ∞ Õ (−1)r z n+2r 2i sin(n + 12 ) πΓ(r + 12 )Γ(n + 12 ) (2r)!Γ(n + r + 1) r=0 n+1 1 = 2 i sin n + 2 πΓ n + 21 Γ 12 Jn (z),

y=

on reduction. Accordingly, when Γ

1 2

−n

−1

, 0, we have ∫ Γ( 21 − n)( 12 z)n (1+,−1−) 2 Jn (z) = (t − 1)n−1/2 cos(zt) dt. 2πiΓ 12 A Corollary 17.3.1 When Re n + 21 > 0, we may deform the path of integration, and obtain the result ∫ 1 z n 2 Jn (z) = (1 − t 2 )n−1/2 cos(zt) dt 1 1 Γ n + 2 Γ 2 −1 n ∫ π/2 2 2z sin2n φ cos(z cos φ) dφ. = Γ n + 12 Γ 12 0 Example 17.3.1 Shew that, when Re n + 21 > 0, ∫ π z n 2 Jn (z) = e±iz cos φ sin2n φ dφ. Γ n + 12 Γ 21 0 Example 17.3.2 Obtain the result Jn (z) =

Γ n

z n 2 + 21 Γ

∫ 1 2

π

cos(z cos φ) sin2n φ dφ,

0

when Re (n) > 0, by expanding in powers of z and integrating (§4.7) term-by-term.

17.4 Connexion between Bessel coefficients and Legendre functions

385

Example 17.3.3 Shew that when − 12 < n < 12 , Jn (z) has an infinite number of real zeros. Hint. Let z = m + 21 π where m is zero or a positive integer; then by the corollary above Jn mπ + 21 π =

zn 2n−1 Γ n + 12 Γ

1 1 2

u 2 0

− u1 + u2 − · · · + (−1)m um ,

where ∫ 2r +1 2m+1 ur = (1 − t 2 )n−1/2 cos m + 12 πt dt 2r −1 2m+1 ( ) 2 n−1/2 ∫ 1/(m+ 12 ) 2r − 1 = 1− t+ sin m + 12 πt dt, 2m + 1 0 so, since n − 12 < 0, um > um−1 > um−2 > · · · , and hence Jn mπ + 12 π has the sign of (−1)m . This method of proof, for n = 0, is due to Bessel. Example 17.3.4 Shew that if n be real, Jn (z) has an infinite number of real zeros; and find an upper limit to the numerically smallest of them. Hint. Use Example 17.3.3 combined with §17.22.

17.4 Connexion between Bessel coefficients and Legendre functions We shall now establish a result due to Heine [287], which renders precise the statement of §17.11, Example 17.1.6 concerning the expression of Bessel coefficients as limiting forms of hypergeometric functions. The apparently different result given in [287] is due to the difference between Heine’s associated Legendre function and Ferrers’ function. When | arg(1 ± z)| < π, n is unrestricted and m is a positive integer, it follows by differentiating the formula of §15.22 that, with Ferrers’ definition of Pnm (z), Γ(n + m + 1) 1 1 (1 − z) 2 m (1 + z) 2 m · m ! Γ(n − m + 1) ×F −n + m, n + 1 + m; m + 1; 12 − 21 z , and so, if | arg z| < 12 π, | arg 1 − 41 z2 /n2 | < π, we have m/2 Γ(n + m + 1)z m n−m z2 z2 Pnm 1 − 2 = m 1− 2 2n 2 · m ! Γ(n − m + 1) 4n Pnm (z) =

2m

×F −n + m, n + 1 + m; m + 1; 14 z2 n−2 . Now make n → +∞ (n being positive, but not necessarily integral), so that, if δ = n−1 , δ → 0 continuously through positive values. Then Γ(n + m + 1)n−m → 1, Γ(n − m + 1)nm by §13.6, and 1 − z2 /4n2

m/2

→ 1.

Bessel Functions

386

Further, the (r + 1)th term of the hypergeometric series is (1 − mδ)(1 + mδ + rδ) 1 − (m + 1)2 δ2 1 − (m + 2)2 δ2 · · · 1 − (m + r − 1)2 δ2 r (−1) (m + 1)(m + 2) · · · (m + r) · r! z 2r × ; 2 this is a continuous function of δ and the series of which this is the (r + 1)th term is easily seen to converge uniformly in a range of values of δ including the point δ = 0; so, by §3.32, we have r ∞ (−1)r 2z zm Õ z2 −m m lim n Pn 1 − 2 = m n→∞ 2n 2 · m! r=0 (m + 1)(m + 2) · · · (m + r) r ! = Jm (z), which is the relation required. Example 17.4.1 Shew that h z i lim n−m Pnm cos = Jm (z). n→∞ n The special case of this when m = 0 was given by Mehler [464]; see also [465]. Example 17.4.2 Shew that Bessel’s equation is the confluent form of the equations defined by the schemes 0 ∞ n ic P −n −ic

1 2 1 2

c + ic − ic

z

0 ∞ c 1 n 0 z ,e P 2 −n 3 − 2ic 2ic − 1 2 0 ∞ c2 P 1 1 n (c − n) 0 z2 2 2 1 1 − n − (c + n) n + 1 2 2

is

, ,

the confluence being obtained by making c → ∞.

17.5 Asymptotic series for Jn (z) when |z| is large We have seen (§17.212) that Jn (z) =

z−1/2 22n+1/2 e

1 1 2 (n+ 2 )πi

Γ(n + 1)

M0,n (2iz),

where it is supposed that | arg z| < π, − 12 π < arg(2iz) < 32 π. But for this range of values of z M0,n (2iz) =

Γ(2n + 1) (n+ 12 )πi Γ(2n + 1) W0,n (2iz) + e W0,n (−2iz) Γ 12 + n Γ 12 + n

by Example 16.4.4, if − 23 π < arg(−2iz) < 12 π; and so, when | arg z| < π, n 1 1 o 1 − 21 (n+ 12 )πi 2 (n+ 2 ) πi W Jn (z) = e (2iz) + e W (−2iz) . 0,n 0,n (2πz)1/2

17.5 Asymptotic series for Jn (z) when |z| is large

387

But, for the values of z under consideration, the asymptotic expansion of W0,n (±2iz) is (4n2 − 12 ) (4n2 − 12 )(4n2 − 32 ) + ±··· e∓iz 1 ± 8iz 2!(8iz)2 ) (±1)r 4n2 − 12 4n2 − 32 · · · 4n2 − (2r − 1)2 −r + O(z ) , + r!(8iz)r and therefore, combining the series, the asymptotic expansion of Jn (z), when |z| is large and | arg z| < π, is 12 h

nπ π cos z − − 2 4 ( 2 ) ∞ r Õ (−1) 4n − 12 4n2 − 32 · · · 4n2 − (4r − 1)2 × 1+ (2r)!26r z2r r=1 # ∞ nπ π Õ (−1)r 4n2 − 12 4n2 − 32 · · · 4n2 − (4r − 3)2 + sin z − − 2 4 r=1 (2r − 1)!26r−3 z2r−1 21 h i 2 nπ π nπ π = cos z − − · Un (z) − sin z − − · Vn (z) , πz 2 4 2 4

2 Jn (z) ∼ πz

where Un (z), −Vn (z) have been written in place of the series. The reader will observe that if n is half an odd integer these series terminate and give the result of Example 17.2.12. Even when z is not very large, the value of Jn (z) can be computed with great accuracy from this formula. Thus, for all positive values of z greater than 8, the first three terms of the asymptotic expansion give the value of J0 (z) and J1 (z) to six places of decimals. This asymptotic expansion was given by Poisson [530, p. 350] (for n = 0) and by Jacobi (for n = 0) and by Jacobi [353, p. 94] (for general integral values of n) for real values of z. Complex values of z were considered by Hankel [273] and several subsequent writers. The method of obtaining the expansion here given is due to Barnes [49]. Asymptotic expansions for Jn (z) when the order n is large have been given by Debye [168] and Nicholson [495]. An approximate formula for Jn (nx) when n is large and 0 < x < 1, namely n √ o x n exp n 1 − x 2 n on , √ (2πn)1/2 (1 − x 2 )1/4 1 + 1 − x 2 was obtained by Carlini in 1817 in a memoir reprinted in Jacobi [354, vol. VII, pp. 189–245]. The formula was also investigated by Laplace [409, vol. V, 1827] in 1827, on the hypothesis that x is purely imaginary. A more extended account of researches on Bessel functions of large order is given in [676]. Example 17.5.1 By suitably modifying Hankel’s contour integral (§17.3), shew that, when

Bessel Functions

388

| arg z| < 21 π and Re n + Jn (z) =

Γ n+

1 1 2

1 2

> 0, " 1

(2πz) 2

e

n− 12 iu du 1+ e u 2z 0 n− 21 # ∫ ∞ iu 1 1 1 +e−i(z− 2 nπ− 4 π) du ; e−u un− 2 1 − 2z 0

i(z− 12 nπ− 21 π)

∫

∞

−u n− 21

and deduce the asymptotic expansion of Jn (z) when |z| is large and | arg z| < 21 π. Hint. Take the contour to be the rectangle whose corners are ±1, ±1 + iN, the rectangle being indented 1 at ±1, and make N → ∞; the integrand being (1 − t 2 )n− 2 eist . Example 17.5.2 Shew that, when | arg z| < π2 and Re n + 12 > 0, 2n+1 z n Jn (z) = 1 Γ n + 12 π 2

∫

π/2

1 e−2z cot φ cosn− 2 φ cosec2n+1 φ sin z − n − 12 φ dφ.

0

Hint. Write u = 2z cot φ in the preceding example. Example 17.5.3 (Schafheitlin, [573]) Shew that, if | arg z| < 21 π and Re n + 12 > 0, then ∫ ∞ ∫ ∞ 1 1 1 1 Aeiz z n v n− 2 (1 + iv)n− 2 e−2vz dv + Be−iz z n v n− 2 (1 − iv)n− 2 e−2vz dv 0

0

is a solution of Bessel’s equation. Further, determine A and B so that this may represent Jn (z).

17.6 The second solution of Bessel’s equation when the order is an integer We have seen in §17.2 that, when the order n of Bessel’s differential equation is not an integer, the general solution of the equation is αJn (z) + βJ−n (z), where α and β are arbitrary constants. When, however, n is an integer, we have seen that Jn (z) = (−1)n J−n (z), and consequently the two solutions Jn (z) and J−n (z) are not really distinct. We therefore require in this case to find another particular solution of the differential equation, distinct from Jn (z), in order to have the general solution. We shall now consider the function Yn (z) = 2πenπi

Jn (z) cos nπ − J−n (z) , sin 2nπ

which is a solution of Bessel’s equation when 2n is not an integer. The introduction of this function Yn (z) is due to Hankel [273, p. 472]. When n is an integer, Yn (z) is defined by the

17.6 The second solution of Bessel’s equation

389

limiting form of this equation, namely Yn (z) = lim 2πe(n+ε)πi ε→0

Jn+ε (z) cos(nπ + επ) − J−n−ε (z) sin 2(n + ε)π

2πenπi {(−1)n Jn+ε (z) − J−n−ε (z)} ε→0 sin 2επ = lim ε −1 {Jn+ε (z) − (−1)n J−n−ε (z)} . = lim

ε→0

To express Yn (z) in terms of Wk,m functions, we have recourse to the result of §17.5, which gives o ε −1 hn 12 (n+ε+ 12 )πi 1 1 Yn (z) = lim e W0,n+ε (2iz) + e− 2 (n+ε+ 2 )πi W0,n+ε (−2iz) 1 e→0 (2πz) 2 n 1 oi 1 1 1 −(−1)n e 2 (−n−ε+ 2 )πi W0,n+ε (2iz) + e− 2 (−n−ε+ 2 )πi W0,n+ε (−2iz) , remembering that Wk,m = Wk,−m . Hence, since 1 lim W0,n+ε (2iz) = W0,n (2iz), we have e→0

1 o π 2 n ( 12 n+ 43 )πi 1 3 W0,n (2iz) + e−( 2 n+ 4 )πi W0,n (−2iz) . Yn (z) = e 2z

This function (n being an integer) is obviously a solution of Bessel’s equation; it is called a Bessel function of the second kind. Another function (also called a function of the second kind) was first used by Weber [657, p. 148] and by Schläfli [578, p. 17]. It is defined by the equation Jn (z) cos nπ − J−n (z) Yn (z) cos nπ , = sin nπ πenπi or by the limits of these expressions when n is an integer. This function which exists for all values of n is taken as the canonical function of the second kind by Nielsen [500], and formulae involving it are generally (but not always) simpler than the corresponding formulae involving Hankel’s function. The asymptotic expansion for Yn (z), corresponding to that of §17.5 for Jn (z), is that, when |arg z| < π and n is an integer, 1/2 2 sin z − 12 nπ − 14 π Un (z) + cos z − 21 nπ − 14 π Vn (z) , Yn (z) ∼ πz Yn (z) =

where Un (z) and Vn (z) are the asymptotic expansions defined in §17.5, their leading terms being 1 and (4n2 − 1)/8z respectively. Example 17.6.1 (Hankel) Prove that dJn (z) dJ−n (z) − (−1)n , dn dn where n is made an integer after differentiation. Yn (z) =

Example 17.6.2 Shew that if Yn (z) be defined by the equation of Example 17.6.1, it is a solution of Bessel’s equation when n is an integer. 1

This is most easily seen from the uniformity of the convergence with regard to ε of Barnes’ contour integral (§16.4) for W0, n+ε (2iz).

Bessel Functions

390

17.61 The ascending series for Yn (z) The series of §17.6 is convenient for calculating Yn (z) when |z| is large. To obtain a convenient series for small values of |z|, we observe that, since the ascending series for J±(n+ε) (z) are uniformly convergent series of analytic functions 2 of ε, each term may be expanded in powers of ε and this double series may then be arranged in powers of ε (§§5.3, 5.4). Accordingly, to obtain Yn (z), we have to sum the coefficients of the first power of ε in the terms of the series ∞ ∞ Õ Õ (−1)r ( 12 z)−n+2r−ε (−1)r ( 21 z)n+2r+ε − (−1)n . r!Γ(n + ε + r + 1) r!Γ(−n − ε + r + 1) r=0 r=0

Now, if s be a positive integer or zero and t a negative integer, the following expansions in powers of ε are valid: z n+ε+2r z n+2r 1 = 1 + ε log z + · · · , 2 2 2 0 1 1 Γ (s + 1) = 1−ε +··· Γ(s + ε + 1) Γ(s + 1) Γ(s + 1) ! ) ( s Õ 1 −1 = m +··· , 1 − ε −γ + Γ(s + 1) m=1 1 sin(t + ε)π =− Γ(−t − ε) = (−1)t+1 εΓ(−t) + · · · , Γ(t + ε + 1) π where γ is Euler’s constant (§12.1). Accordingly, picking out the coefficient of ε, we see that "∞ # ∞ z Õ Õ (−1)r ( 2z )n+2r (−1)r ( 2z )−n+2r n Yn (z) = log + (−1) 2 r=0 r! Γ(n + r + 1) r! Γ(−n + r + 1) r=0 ! ∞ n+r Õ Õ (−1)r ( 2z )n+2r 1 + γ− r! Γ(n + r + 1) m r=0 m=1 ! ∞ n−r r z −n+2r Õ Õ (−1) ( 2 ) 1 n + (−1) γ− r! Γ(−n + r + 1) m r=n m=1 + (−1)n

n−1 Õ (−1)r ( z )−n+2r 2

r!

r=0

(−1)r−n+1 Γ(n − r),

and so Yn (z) =

∞ Õ (−1)r ( z )n+2r 2

r=0

r! (n + r)! −

n+r r Õ 1 Õ 1 2 log + 2γ − − 2 m m=1 m m=1

z

n−1 z −n+2r Õ ( ) (n − r − 1)! 2

r=0 2

(

The proof of this is left to the reader.

r!

.

)

17.7 Bessel functions with purely imaginary argument

391

When n is an integer, fundamental solutions 3 of Bessel’s equations, regular near z = 0, are Jn (z) and Yn (z) or Yn (z). Karl Neumann [486, p. 41] took as the second solution the function Y (n) (z) defined by the equation Y (n) (z) =

1 Yn (z) + Jn (z)(log 2 − γ); 2

but Yn (z) and Yn (z) are more useful for physical applications. Example 17.6.3 Shew that the function Yn (z) satisfies the recurrence formulae nYn (z) =

1 z (Yn+1 (z) + Yn−1 (z)) 2

and Yn0(z) =

1 (Yn−1 (z) − Yn+1 (z)) . 2

Shew also that Hankel’s function Yn (z) and Neumann’s function Y (n) (z) satisfy the same recurrence formulae. Note. These are the same as the recurrence formulae satisfied by Jn (z). Example 17.6.4 (Schläfli [576]) Shew that, when |arg z| < 12 π, ∫ π ∫ ∞ πYn (z) = sin(z sin θ − nθ) dθ − e−z sinh θ enθ + (−1)n e−nθ dθ. 0

0

Example 17.6.5 Shew that Y (0) (z) = J0 (z) log z + 2 J2 (z) − 12 J4 (z) + 13 J6 (z) − · · · .

17.7 Bessel functions with purely imaginary argument The function 4 ∞ Õ ( 2z )n+2r In (z) = i Jn (iz) = r!(n + r)! r=0 −n

is of frequent occurrence in various branches of applied mathematics; in these applications z is usually positive. The reader should have no difficulty in obtaining the following formulae: (i) (ii) (iii) (iv) (v) 3

4

In−1 (z) − In+1 (z) =

2n I (z). z n d n n {z I (z)} = z I (z). n n−1 dz d −n −n {z I (z)} = z In+1 (z). n dz 2 d In (z) 1 dIn (z) n2 + − 1 + 2 In (z) dz 2 z dz z

When Re n +

1 2

= 0.

zn > 0, In (z) = n 1 2 Γ( 2 )Γ(n + 12 )

∫

π

cosh(z cos φ) sin2n φ dφ.

0

Euler [202, pp. 187, 233] gave a second solution (involving a logarithm) of the equation in the special cases n = 0, n = 1. This notation was introduced by Basset [51, p. 17]; in 1886 he had defined In (z) as i n Jn (iz); [52].

Bessel Functions

392

(vi)

When − 32 π < arg z < 12 π, the asymptotic expansion of In (z) is " # ∞ 2 2 2 2 2 2 Õ ez {4n − 1 }{4n − 3 } · · · {4n − (2r − 1) } In (z) ∼ 1+ (−1)r r! 23r zr (2πz)1/2 r=1 # " 1 ∞ Õ e−(n+ 2 )πi e−z {4n2 − 12 }{4n2 − 32 } · · · {4n2 − (2r − 1)2 } , + 1+ r! 23r zr (2πz)1/2 r=1

the second series being negligible when |arg z| < 12 π. The result is easily seen to be valid 1 1 over the extended range − 32 π < arg z < 23 π if we write e±(n+ 2 )πi for e−(n+ 2 )πi , the upper or lower sign being taken according as arg z is positive or negative.

17.71 Modified Bessel functions of the second kind When n is a positive integer or zero, I−n (z) = In (z); to obtain a second solution of the modified Bessel equation (iv) above, we define 5 the function Kn (z) for all values of n by the equation 1/2 π cos nπ W0,n (2z), Kn (z) = 2z so that Kn (z) = π2 (I−n (z) − In (z)) cot nπ. Whether n be an integer or not, this function is a solution of the modified Bessel equation, and when |arg z| < 32 π it possesses the asymptotic expansion # " 1/2 ∞ Õ {4n2 − 12 }{4n2 − 32 } · · · {4n2 − (2r − 1)2 } π −z Kn (z) ∼ e cos(nπ) 1 + 2z r! 23r zr r=1 for large values of |z|. When n is an integer, Kn (z) is defined by the equation π Kn (z) = lim {I−n−ε (z) − In+ε (z)} cot πε, ε→0 2 which gives (cf. §17.61) ( ) ∞ n+r r Õ ( 2z )n+2r z 1Õ 1 1Õ 1 Kn (z) = − log + γ − − r!(n + r)! 2 2 m=1 m 2 m=1 m r=0 n−1

+ 5

1 Õ z −n+2r (−1)n−r (n − r − 1)! 2 r=0 2 r!

The notation K n (z) was used by Basset [52, p. 11] to denote a function which differed from the function now defined by the omission of the factor cos nπ, and Basset’s notation has since been used by various writers, notably Macdonald. The object of the insertion of the factor is to make In (z) and K n (z) satisfy the same recurrence formulae. Subsequently Basset [51, p. 19] used the notation K n (z) to denote a slightly different function, but the latter usage has not been followed by other writers. The definition of K n (z) for integral values of n which is given here is due to Gray and Mathews [259, p. 68], and is now common (see Example 17.40), but the corresponding definition for non-integral values has the serious disadvantage that the function vanishes identically when 2n is an odd integer. The function was considered by Riemann [555] and Hankel [273, p. 498].

17.8 Neumann’s expansion

393

as an ascending series. Example 17.7.1 Shew that Kn (z) satisfies the same recurrence formulae as In (z).

17.8 Neumann’s expansion of an analytic function in a series of Bessel coefficients We shall now consider the expansion of an arbitrary function f (z), analytic in a domain including the origin, in a series of Bessel coefficients, in the form f (z) = a0 J0 (z) + a1 J1 (z) + a2 J2 (z) + · · · , where α0, α1, α2, . . . are independent of z. This appears in Neumann [487] and Kapteyn [365]. Assuming the possibility of expansions of this type, let us first consider the expansion of 1/(t − z); let it be 1 = O0 (t)J0 (z) + 2O1 (t)J1 (z) + 2O2 (t)J2 (z) + · · · , t−z where the functions On (t) are independent of z. We shall now determine conditions which On (t) must satisfy if the series on the right is to be a uniformly convergent series of analytic functions; by these conditions On (t) will be determined, and it will then be shewn that, if On (t) is so determined, then the series on the right actually converges to the sum 1/(t − z) when |z| < |t|. ∂ 1 ∂ + = 0, we have Since ∂t ∂z t − z O00 (t)J0 (z)

+2

∞ Õ

On0 (t)Jn (z)

+

O0 (t)J00(z)

+2

so that, on replacing

On (t)Jn0 (z) ≡ 0,

n=1

n=1

2Jn0 (z)

∞ Õ

by Jn−1 (z) − Jn+1 (z), we find

∞ Õ 0 2On (t) + On+1 (t) − On−1 (t) Jn (z) = 0. O00 (t) + O1 (t) J0 (z) + n=1

Accordingly the successive functions O1 (t), O2 (t), O3 (t), . . . are determined by the recurrence formulae O1 (t) = −O0 0(t),

On+1 (t) = On−1 (t) − 2On0 (t),

and, putting z = 0 in the original expansion, we see that O0 (t) is to be defined by the equation O0 (t) = 1/t. These formulae shew without difficulty that On (t) is a polynomial of degree n in 1/t. We shall next prove by induction that On (t), so defined, is equal to ∫ on n oni √ √ 1 ∞ −tu hn e u + u2 + 1 + u − u2 + 1 du 2 0 when Re (t) > 0. For the expression is obviously equal to O0 (t) or O1 (t) when n is equal to 0

394

Bessel Functions

or 1 respectively; and ∫ ∫ ∞ n o n−1 on √ √ 1 ∞ −tu n d u ± u2 + 1 e du − e−tu u ± u2 + 1 du 2 0 dt 0 ∫ ∞ n o n o √ √ n−1 1 = e−tu u ± u2 + 1 1 + 2u2 ± 2u u2 + 1 du 2 0 ∫ o n+1 √ 1 ∞ −tu n = u ± u2 + 1 du, e 2 0 whence the induction is obvious. Writing u = sinh θ, we see that, according as n is even or odd, see [321, §§79, 264], on n o n i cosh √ √ 1 hn 2 2 u+ u +1 + u− u +1 = nθ sinh 2 n(n − 1) sinhn−2 θ = 2n−1 sinhn θ + 2(2n − 2) n(n − 1)(n − 2)(n − 3) + sinhn−4 θ + · · · , 2 · 4(2n − 2)(2n − 4) and hence, when Re (t) > 0, we have on integration, t2 2n−1 n ! t4 On (t) = n+1 1 + + +··· , t 2 (2n − 2) 2 · 4 (2n − 2) (2n − 4) the series terminating with the term in t n or t n−1 ; now, whether Re(t) be positive or not, On (t) is defined as a polynomial in 1/t; and so the expansion obtained for On (t) is the value of On (t) for all values of t. Example 17.8.1 Shew that, for all values of t, ∫ ∞ hn on n oni √ √ 1 e−x x + x 2 + t 2 + x − x 2 + t 2 dx, On (t) = n+1 2t 0 and verify that the expression on the right satisfies the recurrence formulae for On (t).

17.81 Proof of Neumann’s expansion The method of §17.8 merely determined the coefficients in Neumann’s expansion of 1/(t − z), on the hypothesis that the expansion existed and that the rearrangements were legitimate. To obtain a proof of the validity of the expansion, we observe that ( 2z )n 2n−1 n ! {1 + θ n } , On (t) = n+1 {1 + φn } , Jn (z) = n! t where θ n → 0, φn → 0 as n → ∞, when z and t are fixed. Hence the series ∞ Õ O0 (t)J0 (z) + 2 On (t)Jn (z) ≡ F(z, t) n=1

is comparable with the geometrical progression whose general term is z n /t n+1 , and this progression is absolutely convergent when |z| < |t|, and so the expansion for F(z, t) is absolutely convergent (§2.34) in the same circumstances.

17.8 Neumann’s expansion

395

Again if |z| ≤ r, |t| ≥ R, where r < R, the series is comparable with the geometrical progression whose general term is r n /Rn+1 , and so the expansion for F(z, t) converges uniformly throughout the domains |z| ≤ r and |t| ≥ R by §3.34. Hence, by §5.3, term-byterm differentiations are permissible, and so

∞ ∞ Õ Õ ∂ ∂ On (t) Jn0 (z) + On0 (t)Jn (z) + O0 (t) J00(z) + 2 F(z, t) = O00 (t) J0 (z) + 2 ∂t ∂z n=1 n=1 ∞ Õ 0 = O00 (t) + O1 (t) J0 (z) + 2On (t) + On+1 (t) − On−1 (t) Jn (z) n=1

= 0, by the recurrence formulae. ∂ ∂ + Since F(z, t) = 0, it follows that F(z, t) is expressible as a function of t − z; and ∂t ∂z since F(0, t) = O0 (t) = 1/t, it is clear that F(z, t) = 1/(t − z). It is therefore proved that ∞ Õ 1 On (t) Jn (z), = O0 (t) J0 (z) + 2 t−z n=1

(17.8)

provided that |z| < |t|. Hence, if f (z) be analytic when |z| ≤ r, we have, when |z| < r, ∫

f (t) dt t−z ) ( ∫ ∞ Õ 1 On (t)Jn (z) dt f (t) O0 (t)J0 (z) + 2 = 2πi n=1 ∫ ∞ 1 Õ Jn (z) On (t) f (t) dt, = J0 (z) f (0) + πi n=1

1 f (z) = 2πi

by §4.7, the paths of integration being the circle |t| = r; and this establishes the validity of Neumann’s expansion when |z| < r and f (z) is analytic when |z| ≤ r. Example 17.8.2 (K. Neumann) Shew that cos z = J0 (z) − 2J2 (z) + 2J4 (z) − · · · , sin z = 2J1 (z) − 2J3 (z) + 2J5 (z) − · · · . Example 17.8.3 (K. Neumann) Shew that z n 2

=

∞ Õ (n + 2r)(n + r − 1)! r=0

r!

Jn+2r (z).

Bessel Functions

396

Example 17.8.4 (W. Kapteyn) Shew that, when |z| < |t|, ∫ ∞ ∞ ∞ n on Õ Õ √ Jn (z) O0 (t)J0 (z) + 2 On (t)Jn (z) = t −n−1 e−x x + x 2 + t 2 dx 0

n=−∞

n=1

=

∞

∫ 0

1 = t =

−x

e t n+1

∞ Õ n=−∞

∞

∫

on n √ Jn (z) x + x 2 + t 2 dx

zx exp

0

t

− x dx

1 . t−z

17.82 Schlömilch’s expansion of an arbitrary function in a series of Bessel coefficients of order zero Schlömilch [581] (see also Chapman [142]) has given an expansion of quite a different character from that of Neumann. His result may be stated thus: Any function f (x), which has a continuous differential coefficient with limited total fluctuation for all values of x in the closed range (0, π), may be expanded in the series f (x) = a0 + a1 J0 (x) + a2 J0 (2x) + a3 J0 (3x) + · · · , valid in this range, where ∫ π/2 ∫ 1 π u f 0(u sin θ) dθ du, a0 = f (0) + π 0 0 ∫ ∫ π/2 2 π an = u cos nu f 0(u sin θ) dθ du π 0 0

(n > 0).

Schlömilch’s proof is substantially as follows: Let F(x) be the continuous solution of the integral equation ∫ 2 π/2 f (x) = F(x sin φ) dφ. π 0 Then (§11.81) F(x) = f (0) + x

∫

π/2

f 0(x sin θ) dθ.

0

In order to obtain Schlömilch’s expansion, it is merely necessary to apply Fourier’s theorem to the function F(x sin φ). We thus have ( ∫ ) ∫ ∞ ∫ 2 π/2 1 π 2Õ π f (x) = dφ F(u) du + cos nu cos(nx sin φ)F(u) du π 0 π 0 π n=1 0 ∫ ∞ ∫ 1 π 2Õ π = cos nu F(u)J0 (nx) du, F(u) du + π 0 π n=1 0 the interchange of summation and integration being permissible by §4.7 and §9.44.

17.9 Tabulation of Bessel functions

397

In this equation, replace F(u) by its value in terms of f (u). Thus we have ∫ ∫ π/2 1 π 0 f (0) + u f (x) = f (u sin θ) dθ du π 0 0 ∫ ∞ ∫ π/2 ∞ 2Õ 0 + J0 (nx) cos nu f (0) + u f (u sin θ) dθ du, π n=1 0 0 which gives Schlömilch’s expansion. Example 17.8.5 (Math. Trip. 1895) Shew that, if 0 ≤ x ≤ π, the expression 1 1 π2 − 2 J0 (x) + J0 (3x) + J0 (5x) + · · · 4 9 25 is equal to x; but that, if π ≤ x ≤ 2π, its value is

√ x + 2π arccos(π/x) − 2 x 2 − π 2,

where arccos(π/x) is taken between 0 and π3 . Find the value of the expression when x lies between 2π and 3π.

17.9 Tabulation of Bessel functions Hansen used the asymptotic expansion (§17.5) to calculate tables of Jn (x) which are given by Lommel in [441]. Meissel [466] tabulated J0 (x) and J1 (x) to 12 places of decimals from x = 0 to x = 15.5, while the British Assoc. Report (1909), p. 33, gives tables by which Jn (x) and Yn (x) may be calculated when x > 10. Tables of J 31 (x), J 32 (x), J− 31 (x), J− 32 (x) are given by Dinnik [171]. Tables of the second solution of Bessel’s equation have been given by the following writers: B. A. Smith [594] (see also [469]), Aldis [16], Airey [14]. The functions In (x) have been tabulated in the British Assoc. Reports, (1889) p. 28, (1893) p. 223, (1896) p. 98, (1907) p. 94; also by Aldis √ [15], by Isherwood [338] and by E. Anding [18]. Tables of Jn (x i), a function employed in the theory of alternating currents in wires, have been given in the British Assoc. Reports, 1889, 1893, 1896 and 1912; by Kelvin [627], by Aldis [16] and by Savidge [572]. Formulae for computing the zeros of J0 (z) were given by Stokes [610] and the 40 smallest zeros were tabulated by Willson and Peirce [682]. The roots of an equation involving Bessel functions were computed by Kalähne [364]. A number of tables connected with Bessel functions are given in British Assoc. Reports, 1910–1914, and also by Jahnke & Emde [356].

17.10 Miscellaneous examples Example 17.1 (K. Neumann) Shew that cos(z sin θ) = J0 (z) + 2J2 (z) cos 2θ + 2J4 (z) cos 4θ + · · · , sin(z sin θ) = 2J1 (z) sin θ + 2J3 (z) sin 3θ + 2J5 (z) sin 5θ + · · · Example 17.2 By expanding each side of the equations of Example 17.1 in powers of sin θ, express z n as a series of Bessel coefficients.

Bessel Functions

398

Example 17.3 By multiplying the expansions for exp and considering the terms independent of t, shew that

z 2

(t − 1/t) and exp − 2z (t − 1/t)

{J0 (z)}2 + 2 {J1 (z)}2 + 2 {J2 (z)}2 + 2 {J3 (z)}2 + · · · = 1. √ Deduce that, for the Bessel coefficients, |J0 (z)| ≤ 1, |Jn (z)| ≤ 1/ 2, for n ≥ 1, when z is real. ∫ 1 π k k Example 17.4 (Bourget [94]) If Jm (z) = 2 cosk u cos(mu − z sin u) du (this function π 0 reduces to a Bessel coefficient when k is zero and m an integer), shew that Jmk (z) =

∞ Õ 1 z p N−m,k,p , p! 2 p=0

where N−m,k,p is the Cauchy number defined by the equation ∫ π 1 N−m,k,p = e−miu (eiu + e−iu )k (eiu − e−iu ) p du. 2π −π Shew further that; k−1 k−1 Jmk (z) = Jm−1 (z) + Jm+1 (z), k k (z) − Jm+1 (z) . and zJmk+2 (z) = 2mJmk+1 (z) − 2(k + 1) Jm−1

Example 17.5 (Bourget) If v and M are connected by the equations M = E − e sin E,

cos v =

cos E − e , 1 − e cos E

where |e| < 1,

shew that k ∞ ∞ Õ Õ 1 1 v = M + 2(1 − e ) e Jm k (me) sin mM, 2 m m=1 k=0 2

1 2

where Jmk (z) is defined as in Example 17.4. Example 17.6 (Math. Trip. 1893) Prove that, if m and n are integers, cnm 2 2 21 ∂ m z n, Pn (cos θ) = n Jm (x + y ) r ∂z where z = r cos θ, x 2 + y 2 = r 2 sin2 θ, and cnm is independent of z. Example 17.7 Shew that the solution of the differential equation ( 2 2 0 2) d 2 y φ 0 dy 1 φ0 1 d φ0 1 ψ 00 1 d ψ 00 ψ − + − − + + ψ 2 − ν 2 + 41 y dz 2 φ dz 4 φ 2 dz φ 4 ψ0 2 dz ψ 0 ψ = 0, where φ and ψ are arbitrary functions of z, is 1 φψ 2 y= {AJν (ψ) + BJ−ν (ψ)}. ψ0

17.10 Miscellaneous examples

399

Example 17.8 (Trinity, 1908) Shew that ∫ x 1 J0 (x) + J1 (x) + J3 (x) + J5 (x) + · · · = {J0 (t) + J1 (t)} dt − 1 . 2 0 Example 17.9 (Schläfli [576] and Schönholzer [588]) Shew that Jµ (z)Jν (z) =

∞ Õ n=0

(−1)n Γ(µ + ν + 2n + 1)( 2z )µ+ν+2n n!Γ(µ + n + 1)Γ (ν + n + 1)Γ(µ + ν + n + 1)

for all values of µ and ν. Example 17.10 (Math. Trip. 1899) Shew that, if n is a positive integer and m + 2n + 1 is positive, ∫ x ∫ x m m+1 2 (m − 1) x Jn+1 (x)Jn−1 (x) dx = x {Jn+1 (x)Jn−1 (x) − Jn (x)} + (m + 1) x m Jn2 (x) dx. 0

0

Example 17.11 Shew that J3 (z) + 3

d 3 J0 (z) dJ0 (z) = 0. +4 dz dz 3

Example 17.12 Shew that Jn+1 (z) = Jn (z)

1−

1 z/n(n + 1) 2 1 2 ( 2 z) /(n + 1)(n + ( 12 z)2 /(n + 2)(n

1−

. 2) + 3)

. 1 − ..

Example 17.13 (Lommel) Shew that J−n (z) Jn−1 (z) + J−n+1 (z)Jn (z) = Example 17.14 If

2 sin nπ . πz

Jn+1 (z) be denoted by Q n (z), shew that zJn (z) dQ n (z) 1 2(n + 1) = − Q n (z) + z{Q n (z)}2 . dz z z

Example 17.15 (K. Neumann) Shew that, if R2 = r 2 + r12 − 2rr1 cos θ and r1 > r > 0, J0 (R) = J0 (r)J0 (r1 ) + 2 Y0 (R) = J0 (r) Y0 (r1 ) + 2

∞ Õ n=1 ∞ Õ

Jn (r) Jn (r1 ) cos nθ, Jn (r)Yn (r1 ) cos nθ.

n=1

Example 17.16 (K. Neumann) Shew that, if Re (n + 12 ) > 0, ∫ 12 π π J2n (2z cos θ) dθ = {Jn (z) }2 . 2 0

Bessel Functions

400

Example 17.17 (Math. Trip. 1896) Shew how to express z2n J2n (z) in the form AJ2 (z) + BJ0 (z), where A and B are polynomials in z; and prove that √ √ √ √ 3J6 ( 30) + 5J2 ( 30) = 0. J4 ( 6) + 3J0 ( 6) = 0, Example 17.18 Shew that, if α , β and n > −1, ∫ x d d 2 2 (α − β ) x Jn (αx)Jn (βx) dx = x Jn (αx) Jn (βx) − Jn (βx) Jn (αx) , dx dx 0 2 ∫ x d 2 2 2 2 2 2 2α x{Jn (αx)} dx = (α x − n ){Jn (ax)} + x Jn (αx) . dx 0 Example 17.19 (Lommel [441]) Prove that, if n > −1, and Jn (α) = Jn (β) = 0 while α , β, ∫ 1 ∫ 1 1 x Jn (αx)Jn (βx) dx = 0, and x{Jn (αx)}2 dx = {Jn+1 (α)}2 . 2 0 0 Hence prove that, when n > −1, the roots of Jn (x) = 0, other than zero, are all real and unequal. Hint. If α could be complex, take β to be the conjugate complex. Example 17.20 Let x 1/2 f (x) have an absolutely convergent integral in the range 0 ≤ x ≤ 1; let H be a real constant and let n ≥ 0. Then, if k1, k2, . . . denote the positive roots of the equation k −n {k Jn0 (k) + H Jn (k)} = 0, shew that, at any point x for which 0 < x < 1 and f (x) satisfies one of the conditions of §9.43, f (x) can be expanded in the form f (x) =

∞ Õ

Ar Jn (kr x),

r=1

where Ar =

∫

−1 ∫

1

x{Jn (kr x)}2 dx

1

x f (x) Jn (kr x) dx.

0

0

In the special case when H = −n, k 1 is to be taken to be zero, the equation determining k 1, k 2, . . . being Jn+1 (k) = 0, and the first term of the expansion is A0 x n where ∫ 1 A0 = (2n + 2) x n+1 f (x) dx. 0

Discuss, in particular, the case when H is infinite, so that Jn (k) = 0, shewing that ∫ 1 −2 Ar = 2{Jn+1 (kr )} x f (x) Jn (kr x) dx. 0

Note This result is due to Hobson [318]; see also W. H. Young [686]. The formal expansion was given with H infinite (when n = 0) by Fourier and (for general values of n) by Lommel; proofs were given by Hankel and Schläfli. The formula when H = −n was given incorrectly by Dini [170], the term A0 x n being printed as A0 , and this error was not corrected by Nielsen. See Bridgeman [98] and Chree [144]. The expansion is usually called the Fourier–Bessel expansion.

17.10 Miscellaneous examples

401

Example 17.21 (Clare, 1900) Prove that, if the expansion α2 − x 2 = A1 J0 (λ1 x) + A2 J0 (λ2 x) + · · · exists as a uniformly convergent series when −α ≤ x ≤ α, where λ1, λ2, . . . are the positive roots of J0 (λα) = 0, then An = 8{αλn 3 J1 (λn α) }−1 . Example 17.22 (Math. Trip. 1906) If k 1, k 2, . . . are the positive roots of Jn (kα) = 0, and if x n+2 =

∞ Õ

Ar Jn (kr x),

r=1

this series converging uniformly when 0 ≤ x ≤ α, then Ar =

2a n−1 dJn (kr α) (4n + 4 − α2 kr 2 ) ÷ . kr 2 da

Example 17.23 (Sonine [598]) Shew that ∫ π2 x n−m Jn (x) = n−m−1 Jm (x sin θ) cos2n−2m−1 θ sinm+1 θ dθ 2 Γ(n − m) 0 when n > m > −1. Example 17.24 (Nicholson [497]) Shew that, if σ > 0, 3/2 3/2 ∫ ∞ 2σ πσ 1/2 2σ 3 + J−1/3 . cos(t − σt) dt = √ J1/3 3/2 3 33/2 0 3 3 Example 17.25 (Math. Trip. 1904) If m be a positive integer and u > 0, deduce from Bessel’s integral formula that ∫ ∞ e−x sinh u Jm (x) dx = e−mu sech u. 0

Example 17.26 (Sonine [598]) Prove that, when x > 0, ∫ ∫ 2 ∞ 2 ∞ J0 (x) = sin(x cosh t) dt, Y0 (x) = − cos(x cosh t) dt. π 0 π 0 Hint. Take the contour of §17.1 to be the imaginary axis indented at the origin and a semicircle on the left of this line. Example 17.27 (Weber [656]) Shew that ∫

(

∞

x −1 J0 (xt) sin x dx =

0

π 2

arccosec t

0 0, 12 ∫ π 2 2 1 Jm (z sin θ) sinm+1 θdθ = z− 2 Jm+ 12 (z). π 0 Example 17.32 (Weber [654]; Math. Trip. 1898) Shew that, if 2n + 1 > m > −1, ∫ ∞ Γ( 21 m + 12 ) x −n+m Jn (ax) dx = 2−n+m a n−m−1 . Γ(n − 12 m + 12 ) 0 Example 17.33 (Lommel) Shew that ∞

z Õ 2p + 1 = {Jp+ 12 (z)}2 . π p=0 2

17.10 Miscellaneous examples

403

Example 17.34 (Math. Trip. 1894) In the equation n2 d 2 y 1 dy + + 1 + 2 + y = 0, dz2 z dz z n is real; shew that a solution is given by cos(n log z) − where um denotes

m Í r=1

∞ Õ

(−1)m z2m cos(um − n log z)

m=1

22m m! (1 + n2 ) 2 (4 + n2 ) 2 · · · (m2 + n2 ) 2 1

1

1

,

arctan(n/r).

Example 17.35 (Cauchy [128]; Nicholson [496]) Shew that, when n is large 1 − 13 2 1 Jn (n) = 2− 3 3− 6 π −1 Γ n + O(n−1 ). 3 Example 17.36 (Mehler [464]) Shew that ∫ K0 (x) =

∞

0

t J0 (t x) dt. 1 + t2

Example 17.37 (Math. Trip. 1900) Shew that eλ cos θ = 2n−1 Γ(n)

∞ Õ (n + k)Ckn (cos θ)λ−n In+k (λ). k=0

Example 17.38 (Sonine [598]) Shew that, if ∫ ∞ W= Jm (ax)Jm (bx)Jm (cx)x 1−m dx, 0

a, b, c being positive, and m is a positive integer or zero, then

W=

0

Í a−m b −mc −m {2 b2 c2 1 1 3m−1 2 2 π Γ m+ 2

−

Í

4 m− 12

a }

0

if if

(a − b)2 > c2, (a + b)2 > c2 > (a − b)2,

if

(a + b)2 > c2 .

Example 17.39 (Macdonald [448]) Shew that, if n > −1, m > − 21 and ∫ ∞ W= Jn (ax)Jn (bx)Jm (cx)x 1−m dx, 0

a, b, c being positive, then

W=

0 if (a − b)2 > c2,

(2π) a m−1 bm−1 c−m (1 − µ2 ) − 21

1 π 2

− 12

e a m−1 bm−1 c−m sin(m−n)π π

(m− 12 )πi

where µ = (a2 + b2 − c2 )/2ab, and µ1 = −µ.

(µ21 − 1)

1 4 (2m−1)

1 4 (2m−1)

1 2 −m n− 12

P

Q

1 2 −m n− 12

(µ) if (a + b)2 > c2 > (a − b)2,

(µ1 ) if c2 > (a + b)2,

Bessel Functions

404

Example 17.40 (Math. Trip. 1898; Basset [52]) Shew that, if Re (m + 12 ) > 0, ∫ π zm Im (z) = m cosh(z cos φ) sin2m φ dφ, 2 Γ(m + 21 )Γ( 12 ) 0 and, if | arg z| < 21 π, zm Γ

Km (z) =

1 2

cos mπ

2m Γ(m +

1 ) 2

∫

∞

e−z cosh φ sinh2m φ dφ.

0

Prove also that (2z)m Km (z) = √ Γ m + 12 cos mπ π

∫

∞

(u2 + z2 )−m− 2 cos u du. 1

0

Hint. The first integral may be obtained by expanding in powers of z and integrating termby-term. To obtain the second, consider ∫ (1+, −1+1) 1 zm e−st (t 2 − 1)m− 2 dt, ∞

where initially arg(t − 1) = arg(t + 1) = 0. Take |t| > 1 on the contour, expand (t 2 − 1)m− 2 in descending powers of t, and integrate term-by-term. The result is 1

2ie2mπi sin(2mπ)Γ(2m)2−m Γ(1 − m)I−m (z). Also, deforming the contour by flattening it, the integral becomes ∫ ∞ ∫ 2mπi m 2mπi m −st 2 m− 12 2ie z sin 2mπ dt + 2ie z cos mπ e (t − 1)

1

1

e−st (1 − t 2 )m− 2 dt;

−1

1

and consequently I−m (z) − Im (z) =

21−m sin(mπ)z m Γ 21 Γ(m + 12 )

∫

∞ 1

e−st (t 2 − 1)m− 2 dt.

1

Example 17.41 (K. Neumann) Shew that On (z) satisfies the differential equation d 2 On (z) 3 dOn (z) n2 − 1 + On (z) = gn, + 1− dz 2 z dz 2 where ( gn =

z−1 nz −2

if n is even if n is odd.

Example 17.42 (K. Neumann) If f (z) be analytic throughout the ring-shaped region bounded by the circles c, C whose centres are at the origin, establish the expansion f (z) =

1 1 α0 J0 (z) + α1 J1 (z) + α2 J2 (z) + · · · + β0 O0 (z) + β1 O1 (z) + β2 O2 (z) + · · · , 2 2

where 1 αn = πi

∫ f (t)On (t) dt, C

1 βn = πi

∫ f (t)Jn (t) dt. C

17.10 Miscellaneous examples

405

Example 17.43 (Math. Trip. 1905) Shew that, if x and y are positive, ∫ ∞ −βx e−ir e J0 (k y)k dk = , β r 0 p √ √ where r = x 2 + y 2 and β = k 2 − 1 or i 1 − k 2 , according as k > 1 or k < 1. Example 17.44 Shew that, with suitable restrictions on n and on the form of the function f (x), ∫ ∞ ∫ ∞ 0 0 0 0 f (x) = Jn (t x) t f (x )Jn (t x )x dx dt. 0

0

Note A proof with an historical account of this important theorem is given by Nielsen [500, p. 360–363]. It is due to Hankel, but (in view of the result of §9.7) it is often called the Fourier–Bessel integral. Example 17.45 (K. Neumann) If C be any closed contour, and m and n are integers, shew that ∫ ∫ ∫ Jm (z)Jn (z) dz = Om (z)On (z) dz = Jm (z)On (z) dz = 0, C

C

C

unless C contains the origin and m = n; in which case the first two integrals are still zero, but the third is equal to πi (or 2πi, if m = 0) if C encircles the origin once counter-clockwise. Example 17.46 (K. Neumann) Shew that, if a p,q =

(−1) p , p! q!

and if n be a positive integer, then z−2n =

n Õ

an−m,n+m−1 O2m−1 (z),

m=1

while z1−2n = an−1,n−1 O0 (z) + 2

n−1 Õ

am−1,n+m−1 O2m (z).

m=1

Example 17.47 (K. Neumann) If n Õ 22m (m!)2 n2 (n2 − 12 )(n2 − 22 ) · · · (n2 − (m − 1)2 ) Ωn (y) = , (2m)! y 2m+2 m=0 shew that (y 2 − x 2 )−1 = Ω0 (y)J02 (x) + 2

∞ Õ

Ωn (y)Jn2 (x)

n=1

when the series on the right converges. Example 17.48 (Macdonald [446]) Shew that, if c > 0, Re (n) > −1, and Re (a ± b)2 > 0, then ∫ c+∞i 1 Jn (a)Jn (b) = t −1 exp{(t 2 − a2 − b2 )/(2t)} In (ab/t) dt. 2πi c−∞i

Bessel Functions

406

Example 17.49 (Gegenbauer [239]) Deduce from Example 17.48, or otherwise prove, that (a2 + b2 − 2ab cos θ)− 2 n Jn {(a2 + b2 − 2ab cos θ) 2 } ∞ Õ n = 2 Γ(n) (m + n)a−n b−n Jm+n (a)Jm+n (b)Cmn (cos θ). 1

1

m=0

Example 17.50 (Schafheitlin [573]; Math. Trip. 1903) Shew that ∫ y= Jm (t) Jn tz1/2 t k−1 dt C

satisfies the equation n2 d2 y k 1 dy y 2 2 + + + k −m + =0 2 dz z z − 1 dz z 4z(z − 1) if kt k Jm (t) Jn tz1/2 − t k+1 Jm0 (t) Jn tz1/2 + z1/2 t k+1 Jm (t)Jm0 tz1/2 resumes its initial value after describing the contour. Deduce that, when 0 < z < 1, ∫ ∞ 1 1 Γ(a)z 2 (γ−1) Ja−β (t) Jγ−1 tz 2 t a+β−γ dt = −a−β F(a, β; γ; z). 2γ Γ(1 − β)Γ(γ) 0

18 The Equations of Mathematical Physics

18.1 The differential equations of mathematical physics The functions which have been introduced in the preceding chapters are of importance in the applications of mathematics to physical investigations. Such applications are outside the province of this book; but most of them depend essentially on the fact that, by means of these functions, it is possible to construct solutions of certain partial differential equations, of which the following are among the most important:

(I) Laplace’s equation ∂ 2V ∂ 2V ∂ 2V + + 2 = 0, ∂ x2 ∂ y2 ∂z which was originally introduced in a memoir [408, p. 252] on Saturn’s rings. If (x, y, z) be the rectangular coordinates of any point in space, this equation is satisfied by the following functions which occur in various branches of mathematical physics: (i) (ii) (iii) (iv)

The gravitational potential in regions not occupied by attracting matter. The electrostatic potential in a uniform dielectric, in the theory of electrostatics. The magnetic potential in free aether, in the theory of magnetostatics. The electric potential, in the theory of the steady flow of electric currents in solid conductors. (v) The temperature, in the theory of thermal equilibrium in solids. (vi) The velocity potential at points of a homogeneous liquid moving irrotationally, in hydrodynamical problems.

Notwithstanding the physical differences of these theories, the mathematical investigations are much the same for all of them: thus, the problem of thermal equilibrium in a solid when the points of its surface are maintained at given temperatures is mathematically identical with the problem of determining the electric intensity in a region when the points of its boundary are maintained at given potentials.

(II) The equation of wave motions ∂ 2V ∂ 2V ∂ 2V 1 ∂ 2V + + = . ∂ x2 ∂ y2 ∂z2 c2 ∂t 2 407

408

The Equations of Mathematical Physics

This equation is of general occurrence in investigations of undulatory disturbances propagated with velocity c independent of the wave length; for example, in the theory of electric waves and the electro-magnetic theory of light, it is the equation satisfied by each component of the electric or magnetic vector; in the theory of elastic vibrations, it is the equation satisfied by each component of the displacement; and in the theory of sound, it is the equation satisfied by the velocity potential in a perfect gas.

(III) The equation of conduction of heat 1 ∂V ∂ 2V ∂ 2V ∂ 2V + + 2 = . 2 2 ∂x ∂y ∂z k ∂t This is the equation satisfied by the temperature at a point of a homogeneous isotropic body; the constant k is proportional to the heat conductivity of the body and inversely proportional to its specific heat and density.

(IV) Two-dimensional wave motion A particular case of the preceding equation, when the variable z is absent, is ∂ 2V ∂ 2V 1 ∂ 2V + = . ∂ x2 ∂ y2 c2 ∂t 2 This is the equation satisfied by the displacement in the theory of transverse vibrations of a membrane; the equation also occurs in the theory of wave motion in two dimensions.

(V) The equation of telegraphy ∂V ∂ 2V ∂ 2V + K R . = ∂t 2 ∂t ∂ x2 This is the equation satisfied by the potential in a telegraph cable when the inductance L, the capacity K, and the resistance R per unit length are taken into account. It would not be possible, within the limits of this chapter, to attempt an exhaustive account of the theories of these and the other differential equations of mathematical physics; but, by considering selected typical cases, we shall expound some of the principal methods employed, with special reference to the uses of the transcendental functions. LK

18.2 Boundary conditions A problem which arises very frequently is the determination, for one of the equations of §18.1, of a solution which is subject to certain boundary conditions; thus we may desire to find the temperature at any point inside a homogeneous isotropic conducting solid in thermal equilibrium when the points of its outer surface are maintained at given temperatures. This amounts to finding a solution of Laplace’s equation at points inside a given surface, when the value of the solution at points on the surface is given.

18.3 A general solution of Laplace’s equation

409

A more complicated problem of a similar nature occurs in discussing small oscillations of a liquid in a basin, the liquid being exposed to the atmosphere; in this problem we are given, effectively, the velocity potential at points of the free surface and the normal derivate of the velocity potential where the liquid is in contact with the basin. The nature of the boundary conditions, necessary to determine a solution uniquely, varies very much with the form of differential equation considered, even in the case of equations which, at first sight, seem very much alike. Thus a solution of the equation ∂ 2V ∂ 2V + =0 ∂ x2 ∂ y2 (which occurs in the problem of thermal equilibrium in a conducting cylinder) is uniquely determined at points inside a closed curve in the x–y-plane by a knowledge of the value of V at points on the curve; but in the case of the equation ∂ 2V 1 ∂ 2V − =0 ∂ x 2 c2 ∂t 2 (which effectively only differs from the former in a change of sign), occurring in connexion with transverse vibrations of a stretched string, where V denotes the displacement at time t at distance x from the end of the string, it is physically evident that a solution is determined ∂V uniquely only if both V and are given for all values of x such that 0 ≤ x ≤ l, when t = 0 ∂t (where l denotes the length of the string). Physical intuitions will usually indicate the nature of the boundary conditions which are necessary to determine a solution of a differential equation uniquely; but the existence theorems, which are necessary from the point of view of the pure mathematician, are usually very tedious and difficult (see Forsyth [220, §§216–220], where an apparently simple problem is discussed).

18.3 A general solution of Laplace’s equation It is possible to construct a general solution of Laplace’s equation in the form of a definite integral (see Whittaker [672]). This solution can be employed to solve various problems involving boundary conditions. Let V(x, y, z) be a solution of Laplace’s equation which can be expanded into a power series in three variables valid for points of (x, y, z) sufficiently near a given point (x0, y0, z0 ). Accordingly we write x = x0 + X,

y = y0 + Y,

z = z0 + Z;

and we assume the expansion V = a0 + a1 X + b1Y + c1 Z + a2 X 2 + b2Y 2 + c2 Z 2 + 2d2Y Z + 2e2 Z X + 2 f2 XY + · · · , it being supposed that this series is absolutely convergent whenever |X | 2 + |Y | 2 + |Z | 2 ≤ a, where a is some positive constant (the functions of applied mathematics satisfy this condition). If this expansion exists, V is said to be analytic at (x0, y0, z0 ). It can be proved by the

The Equations of Mathematical Physics

410

methods of §§3.7, 4.7 that the series converges uniformly throughout the domain indicated and may be differentiated term-by-term with regard to X,Y or Z any number of times at points inside the domain. If we substitute the expansion in Laplace’s equation, which may be written ∂ 2V ∂ 2V ∂ 2V + + = 0, ∂ X 2 ∂Y 2 ∂ Z 2 and equate to zero (§3.73) the coefficients of the various powers of X,Y and Z, we get an infinite set of linear relations between the coefficients, of which a2 + b2 + c2 = 0 may be taken as typical. There are 21 n(n − 1) of these relations 1 between the 12 (n + 2)(n + 1) coefficients of the terms of degree n in the expansion of V, so that there are only 12 (n + 2)(n + 1) − 12 n(n − 1) = 2n + 1 independent coefficients in the terms of degree n in V. Hence the terms of degree n in V must be a linear combination of 2n + 1 linearly independent particular solutions of Laplace’s equation, these solutions being each of degree n in X,Y and Z. To find a set of such solutions, consider (Z + iX cos u + iY sin u)n ; it is a solution of Laplace’s equation which may be expanded in a series of sines and cosines of multiples of u, thus: n n Õ Õ hm (X,Y, Z) sin mu, gm (X,Y, Z) cos mu + m=1

m=0

the functions gm (X,Y, Z) and hm (X,Y, Z) being independent of u. The highest power of Z in gm (X,Y, Z) and hm (X,Y, Z) is Z n−m and the former function is an even function of Y , the latter an odd function, hence the functions are linearly independent. They therefore form a set of 2n + 1 functions of the type sought. Now by Fourier’s rule 2 (§9.12) ∫ π πgm (X,Y, Z) = (Z + iX cos u + iY sin u)n cos mu du, ∫−ππ πhm (X,Y, Z) = (Z + iX cos u + iY sin u)n sin mu du, −π

and so any linear combination of the 2n + 1 solutions can be written in the form ∫ π (Z + iX cos u + iY sin u)n fn (u) du, −π

where fn (u) is a rational function of eiu . Now it is readily verified that, if the terms of degree n in the expression assumed for V 1

2

If ar , s , t (where r + s + t = n) be the coefficient of X r Y s Z t in V , and if the terms of degree n − 2 in ∂2 V ∂2 V ∂2 V + + 2 2 ∂X ∂Y ∂Z 2 be arranged primarily in powers of X and secondarily in powers of Y, the coefficient ar , s , t does not occur in any term after X r −2Y s Z t (or X r Y s−2 Z t if r = 0 or 1), and hence the relations are all linearly independent. 2π must be written for π in the coefficient of g0 (X, Y , Z).

18.3 A general solution of Laplace’s equation

411

be written in this form, the series of terms under the integral sign converges uniformly if |X | 2 + |Y | 2 + |Z | 2 be sufficiently small, and so (§4.7) we may write ∫ πÕ ∞ V= (Z + iX cos u + iY sin u)n fn (u)du. −π n=0

But any expression of this form may be written ∫ π V= F(Z + iX cos u + iY sin u, u) du, −π

where F is a function such that differentiations with regard to X,Y or Z under the sign of integration are permissible. And, conversely, if F be any function of this type, V is a solution of Laplace’s equation. This result may be written ∫ π V= f (z + ix cos u + iy sin u, u) du, −π

on absorbing the terms −z0 − ix0 cos u − iy0 sin u into the second variable; and, if differentiations under the sign of integration are permissible, this gives a general solution of Laplace’s equation; that is to say, every solution of Laplace’s equation which is analytic throughout the interior of some sphere is expressible by an integral of the form given. This result is the three-dimensional analogue of the theorem that V = j(x + iy) + g(x − iy) is the general solution of ∂ 2V ∂ 2V + = 0. ∂ x2 ∂ y2 Remark 18.3.1 A distinction has to be drawn between the primitive of an ordinary differential equation and general integrals of a partial differential equation of order higher than the first. For a discussion of general integrals of such equations, see Forsyth [219, chapter 12]. Two apparently distinct primitives are always directly transformable into one another by d2 y + y = 0, we means of suitable relations between the constants; thus in the case of dx 2 can obtain the primitive C sin(x + ε) from A cos x + B sin x by defining C and ε by the equations C sin ε = A, C cos ε = B. On the other hand, every solution of Laplace’s equation is expressible in each of the forms ∫ π ∫ π f (x cos t + y sin t + iz, t) dt, g(y cos u + z sin u + ix, u) du; −π

−π

but if these are known to be the same solution, there appears to be no general analytical relation, connecting the functions f and g, which will directly transform one form of the solution into the other. Example 18.3.1 Shew that the potential of a particle of unit mass at (a, b, c) is ∫ π 1 du 2π −π (z − c) + i(x − a) cos u + i(y − b) sin u

The Equations of Mathematical Physics

412

at all points for which z > c. Example 18.3.2 Shew that a general solution of Laplace’s equation of zero degree in x, y, z is ∫ π log(x cos t + y sin t + iz)g(t) dt, −π

∫

π

if

g(t) dt = 0. Express the solutions

−π

r 2 = x 2 + y2 + z2 .

r+z x and log in this form, where z+r r−z

Example 18.3.3 Shew that, in the case of the equation p1/2 + q1/2 = x + y (where p = ∂z/∂ x, q = ∂z/∂ y), integrals of Charpit’s subsidiary equations (see Forsyth [221, chapter 9]), are (i) p1/2 − x = y − q1/2 = a, (ii) p = q + a2 . Deduce that the corresponding general integrals are derived from (i) z = 31 (x + a)3 + 31 (y − a)3 + F(a), 0 = (x + a)2 − (y − a)2 + F 0(a); (ii) 4z = 31 (x + y)3 + 2a2 (x − y) − a4 (x + y)−1 + G(a), 0 = 4a(x − y) − 4a3 (x + y)−1 + G 0(a) and thence obtain a differential equation determining the function G(a) in terms of the function F(a) when the two general integrals are the same.

18.31 Solutions of Laplace’s equation involving Legendre functions If an expansion for V, of the form assumed in §18.3, exists when x0 = y0 = z0 = 0, we have seen that we can express V as a series of expressions of the type ∫ π (z + ix cos u + iy sin u)n cos mu du, −π ∫ π (z + ix cos u + iy sin u)n sin mu du, −π

where n and m are integers such that 0 ≤ m ≤ n. We shall now examine these expressions more closely. If we take polar coordinates, defined by the equations x = r sin θ cos φ,

y = r sin θ sin φ,

z = r cos θ,

18.3 A general solution of Laplace’s equation

413

we have ∫

π

(z + ix cos u + iy sin u)n cos mu du ∫ π n =r {cos θ + i sin θ cos(u − φ)} n cos mu du −π ∫ π−φ = rn {cos θ + i sin θ cos ψ} n cos m(φ + ψ) dψ −π−φ ∫ π n =r {cos θ + i sin θ cos ψ} n cos m(φ + ψ) dψ −π ∫ π n = r cos mφ {cos θ + i sin θ cos ψ} n cos mψ dψ,

−π

−π

since the integrand is a periodic function of ψ and (cos θ + i sin θ cos ψ)n sin mψ is an odd function of ψ. Therefore (§15.61), with Ferrers’ definition of the associated Legendre function, ∫ π 2πi m n ! n m r Pn (cos θ) cos mφ. (z + ix cos u + iy sin u)n cos mu du = (n + m) ! −π Similarly ∫

π

(z + ix cos u + iy sin u)n sin mu du =

−π

2πi m · n ! n m r Pn (cos θ) sin mφ. (n + m) !

Therefore r n Pnm (cos θ) cos mφ and r n Pnm (cos θ) sin mφ are polynomials in x, y, z and are particular solutions of Laplace’s equation. Further, by §18.3, every solution of Laplace’s equation, which is analytic near the origin, can be expressed in the form ) ( n ∞ Õ Õ (m) (m) m n (An cos mφ + Bn sin mφ)Pn (cos θ) . r An Pn (cos θ) + V= m=1

n=0

Any expression of the form An Pn (cos θ) +

n Õ m (m) (A(m) n cos mφ + Bn sin mφ)Pn (cos θ), m=1

where n is a positive integer, is called a surface harmonic of degree n; a surface harmonic of degree n multiplied by r n is called a solid harmonic (or a spherical harmonic) of degree n. The curves on a unit sphere (with centre at the origin) on which Pn (cos θ) vanishes are n parallels of latitude which divide the surface of the sphere ninto zones, and so Pn (cos θ) is cos o called (see §15.1) a zonal harmonic; and the curves on which mφ · Pnm (cos θ) vanishes sin are n − m parallels of latitude and 2m meridians, which divide the surface of the sphere into quadrangles whose angles are right angles, and so these functions are called tesseral harmonics. A solid harmonic of degree n is evidently a homogeneous polynomial of degree n in x, y, z and it satisfies Laplace’s equation.

The Equations of Mathematical Physics

414

It is evident that, if a change of rectangular coordinates 3 is made by rotating the axes about the origin, a solid harmonic (or a surface harmonic) of degree n transforms into a solid harmonic (or a surface harmonic) of degree n in the new coordinates. Spherical harmonics were investigated with the aid of Cartesian coordinates by W. Thomson in 1862, see [626] and Thomson and Tait [628, pp. 171–218]; they were also investigated independently in the same manner at about the same time by Clebsch [149]. Example 18.3.4 If coordinates r, θ, φ are defined by the equations x = r cos θ,

y = (r 2 − 1)1/2 sin θ cos φ,

z = (r 2 − 1)1/2 sin θ sin φ,

shew that Pnm (r)Pnm (cos θ) cos mφ satisfies Laplace’s equation.

18.4 The solution of Laplace’s equation which satisfies assigned boundary conditions at the surface of a sphere We have seen (§18.31) that any solution of Laplace’s equation which is analytic near the origin can be expanded in the form V(r, θ, φ) = ) ( n ∞ Õ Õ (m) m (A(m) r n An Pn (cos θ) + n cos mφ + Bn sin mφ)Pn (cos θ) ; m=1

n=0

and, from §3.7, it is evident that if it converges for a given value of r, say a, for all values of θ and φ such that 0 ≤ θ ≤ π, −π ≤ φ ≤ π, it converges absolutely and uniformly when r < a. To determine the constants, we must know the boundary conditions which V must satisfy. A boundary condition of frequent occurrence is that V is a given bounded integrable function of θ and φ, say f (θ, φ), on the surface of a given sphere, which we take to have radius a, and V is analytic at points inside this sphere. (m) We then have to determine the coefficients An, A(m) n , Bn from the equation f (θ, φ) =

∞ Õ n=0

( a

n

An Pn (cos θ) +

n Õ

) (A(m) n

cos mφ +

Bn(m)

sin mφ)Pnm (cos θ)

.

m=1

Assuming that this series converges uniformly throughout the domain 0 ≤ θ ≤ π, −π ≤ φ ≤ π, (this is usually the case in physical problems), multiplying by Pnm (cos θ)

n cos o

mφ,

sin

integrating term-by-term (§4.7) and using the results of §§15.14 and 15.51 on the integral 3

Laplace’s operator

∂2 ∂2 ∂2 + + 2 is invariant for changes of rectangular axes. 2 2 ∂x ∂y ∂z

18.4 The solution of Laplace’s equation

415

properties of Legendre functions, we find that ∫ π∫ π 2 (n + m)! (m) A , f (θ 0, φ 0)Pnm (cos θ 0) cos mφ 0 sin θ 0 dθ 0 dφ 0 = πa n 2n + 1 (n − m)! n ∫−ππ ∫0 π (n + m)! (m) 2 · B , f (θ 0, φ 0)Pnm (cos θ 0) sin mφ 0 sin θ 0 dθ 0 dφ 0 = πa n 2n + 1 (n − m)! n −π 0 ∫ π∫ π 2 f (θ 0, φ 0)Pn (cos θ 0) sin θ 0 dθ 0 dφ 0 = 2πa n An . 2n + 1 −π 0 Therefore, when r < a, ( ∫ ∫ ∞ Õ 2n + 1 r n π π V(r, θ, φ) = f (θ 0, φ 0) Pn (cos θ)Pn (cos θ 0) 4π a −π 0 n=0 ) n Õ (n − m)! m Pn (cos θ)Pnm (cos θ 0) cos m(φ − φ 0) sin θ 0 dθ 0 dφ 0 . +2 (n + m)! m=1 The series which is here integrated term-by-term converges uniformly when r < a, since the expression under the integral sign is a bounded function of θ, θ 0, φ, φ 0, and so (§4.7) ( ∫ π∫ π ∞ r n Õ 4πV(r, θ, φ) = f (θ 0, φ 0) (2n + 1) Pn (cos θ)Pn (cos θ 0) a −π 0 n=0 ) n Õ (n − m)! m +2 Pn (cos θ)Pnm (cos θ 0) cos m(φ − φ 0) sin θ 0 dθ 0 dφ 0 . (n + m)! m=1 Now suppose that we take the line (θ, φ) as a new polar axis and let (θ 10 , φ10 ) be the new coordinates of the line whose old coordinates were (θ 0, φ 0); we consequently have to replace Pn (cos θ) by 1 and Pnm (cos θ) by zero; and so we get ∫ π∫ π ∞ r n Õ 4πV(r, θ, φ) = f (θ 0, φ 0) (2n + 1) Pn (cos θ 10 ) sin θ 10 dθ 10 dφ10 a −π 0 n=0 ∫ π∫ π ∞ r n Õ = Pn (cos θ 10 ) sin θ 0 dθ 0 dφ 0 . f (θ 0, φ 0) (2n + 1) a −π 0 n=0 If, in this formula, we make use of the result of Example 15.23 of Chapter 15 we get ∫ π∫ π a(a2 − r 2 ) sin θ 0 dθ 0 dφ 0 , 4πV(r, θ, φ) = f (θ 0, φ 0) 3 (r 2 − 2ar cos θ 10 + a2 ) 2 −π 0 and so V(r, θ, φ) =

a(a2 − r 2 ) 4π

∫

π

−π

∫

π

f (θ 0, φ 0) sin θ 0 dθ 0 dφ 0 3

0

[r 2 − 2ar {cos θ cos θ 0 + sin θ sin θ 0 cos(φ − φ 0)} + a2 ] 2

.

In this compact formula the Legendre functions have ceased to appear explicitly. The last formula can be obtained by the theory of Green’s functions. For properties of such functions the reader is referred to Thomson and Tait [628].

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Remark 18.4.1 From the integrals for V(r, θ, φ) involving Legendre functions of cos θ 10 and of cos θ, cos θ 0 respectively, we can obtain a new proof of the addition theorem for the Legendre polynomial. For let ( χn (θ 0, φ 0) = Pn (cos θ 10 )− Pn (cos θ)Pn (cos θ 0) +2

n Õ (n − m)!

(n + m)! m=1

) Pn m (cos θ)Pn m (cos θ 0) cos m(φ − φ 0) ,

and we get, on comparing the two formulae for V(r, θ, φ), ∫ π∫ π ∞ r n Õ χn (θ 0, φ 0) sin θ 0 dθ 0 dφ 0 . 0= f (θ 0, φ 0) (2n + 1) a −π 0 n=0 If we take f (θ 0, φ 0) to be a surface harmonic of degree n, the term involving r n is the only one which occurs in the integrated series; and in particular, if we take f (θ 0, φ 0) = χn (θ 0, φ 0), we get ∫ π∫ π { χn (θ 0, φ 0)}2 sin θ 0 dθ 0 dφ 0 = 0. −π

0

Since the integrand is continuous and is not negative it must be zero; and so χn (θ 0, φ 0) ≡ 0; that is to say we have proved the formula Pn (cos θ 10 ) = Pn (cos θ)Pn (cos θ 0) + 2

n Õ (n − m)!

(n + m)! m=1

Pnm (cos θ)Pnm (cos θ 0) cos m(φ − φ 0),

wherein it is obvious that cos θ 10 = cos θ cos θ 0 + sin θ sin θ 0 cos(φ − φ 0), from geometrical considerations. We have thus obtained a physical proof of a theorem proved elsewhere (§15.7) by purely analytical reasoning. (The absence of the factor (−1)m which occurs in §15.7 is due to the fact that the functions now employed are Ferrers’ associated functions.) Example 18.4.1 Find the solution of Laplace’s equation analytic inside the sphere r = 1 which has the value sin 3θ cos φ at the surface of the sphere. Solution. 158 r 3 P31 (cos θ) cos φ − 51 r P11 (cos θ) cos φ. Example 18.4.2 Let fn (r, θ, φ) be equal to a homogeneous polynomial of degree n in x, y, z. Shew that ∫ π∫ π fn (a, θ, φ)Pn {cos θ cos θ 0 + sin θ sin θ 0 cos(φ − φ 0)}a2 sin θ dθ dφ −π

0

=

4πa2 fn (a, θ 0, φ 0). 2n + 1

Hint. Take the direction (θ 0, φ 0) as a new polar axis.

18.5 Laplace’s equation and Bessel coefficients

417

18.5 Solutions of Laplace’s equation which involve Bessel coefficients A particular case of the result of §18.3 is that ∫ π ek(z+ix cos u+iy sin u) cos mu du −π

is a solution of Laplace’s equation, k being any constant and m being any integer. Taking cylindrical-polar coordinates (ρ, φ, z) defined by the equations x = ρ cos φ,

y = ρ sin φ,

the above solution becomes ∫ π ∫ π kz ikρ cos(u−φ) kz e e cos mu du = e eikρ cos v cos m(v + φ) · dv −π −π ∫ π kz = 2e eikρ cos v cos mv cos mφ dv 0 ∫ π kz = 2e cos(mφ) eikρ cos v cos mv dv, 0

and so, using Example 17.1.3 we see that 2πi e cos(mφ)Jm (k ρ) is a solution of Laplace’s equation analytic near the origin. Similarly, from the expression ∫ π ek(z+ix cos u+iy sin u) sin mu du, m kz

−π

where m is an integer, we deduce that 2πi m ekz sin(mφ)Jm (k ρ) is a solution of Laplace’s equation.

18.51 The periods of vibration of a uniform membrane This is based upon Euler [205], Poisson [533] and Bourget [95]. For a detailed discussion of vibrations of membranes, see also Rayleigh [550]. The equation satisfied by the displacement V at time t of a point (x, y) of a uniform plane membrane vibrating harmonically is 1 ∂ 2V ∂ 2V ∂ 2V + = , ∂ x2 ∂ y2 c2 ∂t 2 where c is a constant depending on the tension and density of the membrane. The equation can be reduced to Laplace’s equation by the change of variable given by z = cti. It follows, from §18.5, that expressions of the form sin sin Jm (k ρ) (mφ) (ckt) cos cos satisfy the equation of motion of the membrane. Take as a particular case a drum, that is to say a membrane with a fixed circular boundary of radius R. Then one possible type of vibration is given by the equation V = Jm (k ρ) cos mφ cos ckt,

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provided that V = 0 when ρ = R; so that we have to choose k to satisfy the equation Jm (k R) = 0. This equation to determine k has an infinite number of real roots (Example 17.3.3), k 1, k 2, k 3, . . . say. A possible type of vibration is then given by V = Jm (kr ρ) cos mφ cos ckr t

(r = 1, 2, 3, . . .).

This is a periodic motion with period 2π/(ckr ); and so the calculation of the periods depends essentially on calculating the zeros of Bessel coefficients (see §17.9). Example 18.5.1 The equation of motion of air in a circular cylinder vibrating perpendicularly to the axis OZ of the cylinder is 1 ∂ 2V ∂ 2V ∂ 2V + = 2 2, 2 2 ∂x ∂y c ∂t V denoting the velocity potential. If the cylinder have radius R, the boundary condition is ∂V that = 0 when ρ = R. Shew that the determination of the free periods depends on finding ∂ρ the zeros of Jm0 (ζ) = 0.

18.6 A general solution of the equation of wave motions It may be shewn 4 by the methods of §18.3 that a general solution of the equation of wave motions ∂ 2V ∂ 2V ∂ 2V 1 ∂ 2V + + 2 = 2 2 2 2 ∂x ∂y ∂z c ∂t is V=

∫

π

−π

∫

π

f (x sin u cos v + y sin u sin v + z cos u + ct, u, v) du dv,

−π

where f is a function (of three variables) of the type considered in §18.3. Regarding an integral as a limit of a sum, we see that a physical interpretation of this equation is that the velocity potential V is produced by a number of plane waves, the disturbance represented by the element f (x sin u cos v + y sin u sin v + z cos u + ct, u, v) δu δv being propagated in the direction (sin u cos v, sin u sin v, cos u) with velocity c. The solution therefore represents an aggregate of plane waves travelling in all directions with velocity c.

18.61 Solutions of the equation of wave motions which involve Bessel functions We shall now obtain a class of particular solutions of the equation of wave motions, useful for the solution of certain special problems. In physical investigations, it is desirable to have the time occurring by means of a factor 4

See the paper previously cited [672, p. 342–345] or [646].

18.6 A general solution of the equation of wave motions

419

sin ckt or cos ckt, where k is constant. This suggests that we should consider solutions of the type ∫ π∫ π V= eik(x sin u cos v+y sin u sin v+z cos u+ct) f (u, v) du dv. −π

0

Physically this means that we consider motions in which all the elementary waves have the same period. Now let the polar coordinates of (x, y, z) be (r, θ, φ) and let (ω, ψ) be the polar coordinates of the direction (u, v) referred to new axes such that the polar axis is the direction (θ, φ), and the plane ψ = 0 passes through OZ; so that cos ω = cos θ cos u + sin θ sin u cos(φ − v), sin u sin(φ − v) = sin ω sin ψ. Also, take the arbitrary function f (u, v) to be Sn (u, v) sin u, where Sn denotes a surface harmonic in u, v of degree n; so that we may write Sn (u, v) = S¯n (θ, φ; ω, ψ), where (§18.31) S¯n is a surface harmonic in ω, ψ of degree n. We thus get ∫ π∫ π ikct V=e eikr cos ω S¯n (θ, φ; ω, ψ) sin ω dω dψ. −π

0

Now we may write (§18.31) S¯n (θ, φ; ω, ψ) =An (θ, φ)Pn (cos ω) n Õ (m) An (θ, φ) cos mψ + Bn(m) (θ, φ) sin mψ Pnm (cos ω), + m=1

A(m) n (θ, φ)

where An (θ, φ), and Bn(m) (θ, φ) are independent of ψ and ω. Performing the integration with respect to ψ, we get ∫ π ikct V = 2πe An (θ, φ) eikr cos ω Pn (cos ω) sin ω dω 0 ∫ 1 = 2πeikct An (θ, φ) eikr µ Pn (µ) dµ −1 ∫ 1 1 dn 2 (µ − 1)n dµ, = 2πeikct An (θ, φ) eikr µ n n 2 n! dµ −1 by Rodrigues’ formula (§15.11); on integrating by parts n times and using Hankel’s integral (Corollary 17.3.1), we obtain the equation ∫ 1 2π ikct n V = n e An (θ, φ)(ikr) eikr µ (1 − µ2 )n dµ 2 n −1 = (2π) 2 i n eikct (kr)− 2 Jn+ 21 (kr)An (θ, φ), 3

1

and so V is a constant multiple of eikct r − 2 Jn+ 12 (kr)An (θ, φ). Now the equation of wave motions is unaffected if we multiply x, y, z and t by the same constant factor, i.e. if we multiply r and t by the same constant factor leaving θ and φ 1

420

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unaltered; so that An (θ, φ) may be taken to be independent of the arbitrary constant k which multiplies r and t. Hence lim eikct r − 2 k −n− 2 Jn+ 12 (kr)An (θ, φ) 1

1

k→0

is a solution of the equation of wave motions; and therefore r n An (θ, φ) is a solution (independent of t) of the equation of wave motions, and is consequently a solution of Laplace’s equation; it is, accordingly, permissible to take An (θ, φ) to be any surface harmonic of degree n; and so we obtain the result that n cos o n cos o r −1/2 Jn+ 12 (kr)Pnm (cos θ) (mφ) (ckt) sin sin is a particular solution of the equation of wave motions.

18.611 Application of §18.61 to a physical problem The solution just obtained for the equation of wave motions may be used in the following manner to determine the periods of free vibration of air contained in a rigid sphere. The velocity potential V satisfies the equation of wave motions and the boundary condition ∂V = 0 when r = a, where a is the radius of the sphere. Hence is that ∂r n cos o n cos o (ckt) (mφ) V = r −1/2 Jn+ 12 (kr)Pnm (cos θ) sin sin gives a possible motion if k is so chosen that d −1/2 {r Jn+ 12 (kr)}r=a = 0. dr This equation determines k; on using §17.24, we see that it may be written in the form tan ka = fn (ka), where fn (ka) is a rational function of ka. In particular the radial vibrations, in which V is independent of θ and φ, are given by taking n = 0; then the equation to determine k becomes simply tan ka = ka; and the pitches of the fundamental radial vibrations correspond to the roots of this equation.

18.7 Miscellaneous examples Example 18.1 If V be a solution of Laplace’s equation which is symmetrical with respect to OZ, and if V = f {z} on OZ, shew that if f (ζ) be a function which is analytic in a domain of values (which contains the origin) of the complex variable ζ, then ∫ 1 π 1 f {z + i(x 2 + y 2 ) 2 cos φ} dφ V= π 0 at any point of a certain three-dimensional region.

18.7 Miscellaneous examples

421

Deduce that the potential of a uniform circular ring of radius c and of mass M lying in the plane XOY with its centre at the origin is ∫ M π 2 1 1 [c + {z + i(x 2 + y 2 ) 2 cos φ}2 ]− 2 dφ. π 0 Example 18.2 (Dougall) If V be a solution of Laplace’s equation, which is of the form emiφ F(ρ, z), where (ρ, φ, z) are cylindrical coordinates, and if this solution is approximately equal to ρm emiφ f (z) near the axis of z, where f (ζ) is of the character described in Example 18.1, shew that ∫ π m ! ρm emiφ f (z + i ρ cos t) sin2m t dt. V= Γ(m + 21 )Γ( 12 ) 0 Example 18.3 (Forsyth [217]) If u be determined as a function of x, y and z by means of the equation Ax + By + Cz = 1, where A, B, C are functions of u such that A2 + B2 + C 2 = 0, shew that (subject to certain general conditions) any function of u is a solution of Laplace’s equation. Example 18.4 (Sylvester [616]) A, B are two points outside a sphere whose centre is C. A layer of attracting matter on the surface of the sphere is such that its surface density σP at P is given by the formula σP → (AP · BP)−1 . Shew that the total quantity of matter is unaffected by varying A and B so long as C A · CB ˆ are unaltered; and prove that this result is equivalent to the theorem that the surface and ACB integral of two harmonics of different degrees taken over the sphere is zero. Example 18.5 (Appell [32]) Let V(x, y, z) be the potential function defined analytically as due to particles of masses λ + i µ, λ − i µ at the points (a + ia 0, b + ib0, c + ic 0) and (a − ia 0, b − ib0, c − ic 0) respectively. Shew that V(x, y, z) is infinite at all points of a certain real circle, and if the point (x, y, z) describes a circuit intertwined once with this circle the initial and final values of V(x, y, z) are numerically equal, but opposite in sign. Example 18.6 Find the solution of Laplace’s equation analytic in the region for which a < r < A, it being given that on the spheres r = a and r = A the solution reduces to ∞ Õ

cn Pn (cos θ),

n=0

∞ Õ

Cn Pn (cos θ),

n=0

respectively. Example 18.7 (Trinity, 1893) Let O 0 have coordinates (0, 0, c), and let ˆ = θ, POZ

POˆ 0 Z = θ 0,

PO = r,

PO 0 = r 0 .

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422

Shew that Pn (cos θ 0) = r 0n+1

(

Pn (cos θ) c 2 Pn+2 (cos θ) (cos θ) + (n + 1) c Pn+1 + (n+1)(n+2) + · · · o, r n+1 n r n+2 2! r n+3 r P1 (cos θ) (n+1)(n+2) r 2 P2 (cos θ) 1 n (−1) c n+1 + (n + 1) c n+2 + +··· 2! c n+3

if r > c if r < c.

Obtain a similar expansion for r 0n Pn0 (cos θ). Example 18.8 (St John’s, 1899) At a point (r, θ, φ) outside a uniform oblate spheroid whose semi-axes are a, b and whose density is ρ, shew that the potential is 1 m2 P2 (cos θ) m4 P4 (cos θ) 2 4πρa b − + −··· , 3r 3 · 5 r3 5·7 r5 where m2 = a2 − b2 and r > m. Obtain the potential at points for which r < m, Example 18.9 (Bauer [57]) Shew that ∞ 1 Õ 1 eir cos θ = 12 π 2 i n (2n + 1)r − 2 Pn (cos θ)Jn+ 12 (r). n=0

Example 18.10 (Lamé) Shew that if x±iy = h cos h(ξ±iη), the equation of two-dimensional wave motions in the coordinates ξ and η is h2 ∂ 2V ∂ 2V ∂ 2V 2 2 + = (cosh ξ − cos η) . ∂ξ 2 ∂η2 c2 ∂t 2 Note Examples 18.10, 18.11, 18.12 and 18.14 are most easily proved by using Lamé’s result, [400], that if (λ, µ, ν) be orthogonal coordinates for which the line-element is given by the formula (δx)2 + (δy)2 + (δz)2 = (H1 δλ)2 + (H2 δµ)2 + (H3 δν)2, Laplace’s equation in these coordinates is ∂ H2 H3 ∂V ∂ H3 H1 ∂V ∂ H1 H2 ∂V + + = 0. ∂λ H1 ∂λ ∂ µ H2 ∂ µ ∂ν H3 ∂ν A simple method (due to W. Thomson [625]) of proving this result, by means of arguments of a physical character, is reproduced by Lamb [398, §111]. Analytical proofs, based on Lamé’s proof, are given by Bertrand [68, p. 181–187] and Goursat [256, p. 155–159]; and a most compact proof is due to Neville [491]. Another proof is given by Heine [287, p. 303–306]. Example 18.11 (Niven [503]) Let x = (c +r cos θ) cos φ, y = (c +r cos θ) sin φ, z = r sin θ; shew that the surfaces for which r, θ, φ respectively are constant form an orthogonal system; and shew that Laplace’s equation in the coordinates r, θ, φ is ∂ 2V ∂ ∂V 1 ∂ ∂V r r(c + r cos θ) + (c + r cos θ) + = 0. ∂r ∂r r ∂θ ∂θ c + r cos θ ∂φ2 Example 18.12 (Hicks [303]) Let P have cartesian coordinates (x, y, z) and polar coordinates (r, θ, φ). Let the plane POZ meet the circle x 2 + y 2 = k 2 , z = 0 in the points α, γ; and let α Pˆγ = ω, log(Pα/Pγ) = σ.

18.7 Miscellaneous examples

423

Shew that Laplace’s equation in the coordinates σ, ω, φ is ∂ ∂V ∂V sinh σ ∂ sinh σ + ∂σ cosh σ − cos ω ∂σ ∂ω cosh σ − cos ω ∂ω ∂ 2V 1 = 0; + sinh σ(cosh σ − cos ω) ∂φ2 and shew that a solution is m V = (cosh σ − cos ω) 2 cos nω cos mφPn− 1 (cosh σ). 1

2

Example 18.13 Shew that (R + ρ − 2Rρ cos φ + c ) 2

2 − 21

2

=

∫ 1 ∞ Õ e− 2 mπi m=0

π

∞

∫

π

dk

0

e−ck Jm (k ρ)eik R cos n cos mu du,

−π

and deduce an expression for the potential of a particle in terms of Bessel functions. Example 18.14 (Lamé) Shew that if a, b, c are constants and λ, µ, ν are confocal coordinates, defined as the roots of the equation in ε y2 z2 x2 + + = 1, a2 + ε b2 + ε c2 + ε then Laplace’s equation may be written ∂ ∂V ∂ ∂V ∂ ∂V ∆λ (µ − ν) ∆λ + ∆µ (ν − λ) ∆µ + ∆ν (λ − µ) ∆ν = 0, ∂λ ∂λ ∂µ ∂µ ∂ν ∂ν p where ∆λ = (a2 + λ)(b2 + λ)(c2 + λ). Example 18.15 (Bateman [53]) Shew that a general solution of the equation of wave motions is ∫ π V= F(x cos θ + y sin θ + iz, y + iz sin θ + ct cos θ, θ) dθ. −π

Example 18.16 If U = f (x, y, z, t) be a solution of 1 ∂U ∂ 2U ∂ 2U ∂ 2U + + , = a2 ∂t ∂ x2 ∂ y2 ∂z2 prove that another solution of the equation is 2 x y z 1 x + y2 + z2 − 32 U=t f , , , − exp − . t t t t 4a2 t Example 18.17 (Bateman [53]) Shew that a general solution of the equation of wave motions, when the motion is independent of φ, is ∫ π f (z + i ρ cos θ, ct + ρ sin θ) dθ −π ∫ b∫ π a + z + ct cos θ + arcsin h F(a, θ) dθ da, ρ sin θ 0 −π where ρ, φ, z are cylindrical coordinates and a, b are arbitrary constants.

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424

Example 18.18 (Bateman [54]) If V = f (x, y, z) is a solution of Laplace’s equation, shew that 2 r − a2 r 2 + a2 az 1 f , , V= 1 2(x − iy) 2i(x − iy) x − iy (x − iy) 2 is another solution. Example 18.19 (Bateman [54]) If U = f (x, y, z, t) is a solution of the equation of wave motions, shew that another solution is x 1 y r2 − 1 r2 + 1 U= f , , , . z − ct z − ct z − ct 2(z − ct) 2c(z − ct) Example 18.20 (Bateman [54]) If l = x − iy, m = z + iω, n = x 2 + y 2 + z2 + ω2 and λ = x + iy, µ = z − iω, ν = −1, so that lλ + mµ + nν = 0, shew that any homogeneous solution, of degree zero, of ∂ 2U ∂ 2U ∂ 2U ∂ 2U + + + =0 ∂ x2 ∂ y2 ∂z2 ∂ω2 satisfies ∂ 2U ∂ 2U ∂ 2U + + = 0; ∂l∂λ ∂m∂ µ ∂n∂ν and obtain a solution of this equation in the form −α −α0

l λ

−β0 −γ −γ0

m µ −β

n ν

a, b, c P α, β, γ, ζ , α 0, β 0, γ 0

where lλ = (b − c)(ζ − a), mµ = (c − a)(ζ − b), nν = (a − b)(ζ − c). Example 18.21 (Blades [75]) Note. The functions introduced in this example and the next are known as internal and external spheroidal harmonics respectively. If (r, θ, φ) are spheroidal coordinates, defined by the equations x = c(r 2 + 1) 2 sin θ cos φ, 1

y = c(r 2 + 1) 2 sin θ sin φ, 1

z = cr cos θ,

where x, y, z are rectangular coordinates and c is a constant, shew that, when n and m are integers, ∫ π x cos t + y sin t + iz n cos o Pn (mt) dt c sin −π n cos o (n − m)! m = 2π Pn (ir)Pnm (cos θ) (mφ). (n + m)! sin Example 18.22 (Jeffery [358]) With the notation of Example 18.21, shew that, if z , 0, ∫ π n cos o x cos t + y sin t + iz n cos o (n − m)! m (mt) dt = 2π Q n (ir)Pnm (cos θ) (mφ). Qn sin c sin (n + m)! −π

18.7 Miscellaneous examples

425

Example 18.23 (Donkin [186]; Hobson [322]) Prove that the most general solution of Laplace’s equation which is of degree zero in x, y, z is expressible in the form x + iy x − iy V= f +F , r+z r+z where f and F are arbitrary functions.

19 Mathieu Functions

19.1 The differential equation of Mathieu The preceding five chapters have been occupied with the discussion of functions which belong to what may be generally described as the hypergeometric type, and many simple properties of these functions are now well known. In the present chapter we enter upon a region of Analysis which lies beyond this, and which is, as yet, only very imperfectly explored. The functions which occur in Mathematical Physics and which come next in order of complication to functions of hypergeometric type are called Mathieu functions; these functions are also known as the functions associated with the elliptic cylinder. They arise from the equation of two-dimensional wave motion, namely ∂2 V ∂2 V 1 ∂2 V + = . ∂ x2 ∂ y2 c2 ∂t 2 This partial differential equation occurs in the theory of the propagation of electromagnetic waves; if the electric vector in the wave-front is parallel to OZ and if E denotes the electric force, while (Hx , Hy , 0) are the components of magnetic force, Maxwell’s fundamental equations are ∂Hy ∂Hx 1 ∂E = − , 2 c ∂t ∂x ∂y

∂Hx ∂E =− , ∂t ∂y

∂Hy ∂E = , ∂t ∂x

c denoting the velocity of light; and these equations give at once ∂2 E ∂2 E 1 ∂2 E = + . 2 2 c ∂t ∂ x2 ∂ y2 In the case of the scattering of waves, propagated parallel to OX, incident on an elliptic cylinder for which OX and OY are axes of a principal section, the boundary condition is that E should vanish at the surface of the cylinder. The same partial differential equation occurs in connexion with the vibrations of a uniform plane membrane, the dependent variable being the displacement perpendicular to the membrane; if the membrane be in the shape of an ellipse with a rigid boundary, the boundary condition is the same as in the electromagnetic problem just discussed. The differential equation was discussed by Mathieu [457, p. 137] in 1868 in connexion with the problem of vibrations of an elliptic membrane in the following manner: Suppose that the membrane, which is in the plane XOY when it is in equilibrium, is vibrating with 426

19.1 The differential equation of Mathieu

427

frequency p. Then, if we write V = u(x, y) cos(pt + ε), the equation becomes ∂ 2 u ∂ 2 u p2 + + u = 0. ∂ x 2 ∂ y 2 c2 Let the foci of the elliptic membrane be (±h, 0, 0), and introduce new real variables. (The introduction of these variables is due to Lamé, who called ξ the thermometric parameter. They are more usually known as confocal coordinates. See Lamé [404, 1ere Leçon].) Let ξ, η, defined by the complex equation x + iy = h cosh(ξ + iη), so that x = h cosh ξ cos η, y = h sinh ξ sin η. The curves, on which ξ or η is constant, are evidently ellipses or hyperbolas confocal with the boundary; if we take ξ ≥ 0 and −π < η ≤ π, to each point (x, y, 0) of the plane corresponds one and only one 1 value of (ξ, η). The differential equation for u transforms into 2 ∂ 2 u ∂ 2 u h 2 p2 + + 2 (cosh2 ξ − cos2 η) u = 0. ∂ξ 2 ∂η2 c If we assume a solution of this equation of the form u = F(ξ) G(η), where the factors are functions of ξ only and of η only respectively, we see that 1 d 2 F(ξ) h2 p2 1 d 2 G(η) h2 p2 2 2 + cosh ξ = − − cos η . F(ξ) dξ 2 c2 G(η) dη2 c2 Since the left-hand side contains ξ but not η, while the right-hand side contains η but not ξ, F(ξ) and G(η) must be such that each side is a constant, A, say, since ξ and η are independent variables. We thus arrive at the equations 2 2 2 2 h p d 2 G(η) h p d 2 F(ξ) 2 2 + cosh ξ − A F(ξ) = 0, − cos η − A G(η) = 0. dξ 2 c2 dη2 c2 By a slight change of independent variable in the former equation, we see that both of these equations are linear differential equations, of the second order, of the form d2u + (a + 16q cos 2z)u = 0, dz 2 where a and q are constants. (Their actual values are a = A − h2 p2 /(2c2 ), q = h2 p2 /(32c2 ); the factor 16 is inserted to avoid powers of 2 in the solution.) It is obvious that every point (infinity excepted) is a regular point of this equation. This is the equation which is known as Mathieu’s equation and, in certain circumstances (§19.2), particular solutions of it are called Mathieu functions. 1

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This may be seen most easily by considering the ellipses obtained by giving ξ various positive values. If the ellipse be drawn through a definite point (ξ , η) of the plane, η is the eccentric angle of that point on the ellipse. A proof of this result, due to Lamé, is given in numerous textbooks; see the Note to Example 18.10, Chapter 18.

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Mathieu Functions

19.11 The form of the solution of Mathieu’s equation In the physical problems which suggested Mathieu’s equation, the constant a is not given a priori, and we have to consider how it is to be determined. It is obvious from physical considerations in the problem of the membrane that u(x, y) is a one-valued function of position, and is consequently unaltered by increasing η by 2π; and the condition 3 G(η +2π) = G(η) is sufficient to determine a set of values of a in terms of q. And it will appear later (§§19.4, sub:19.41) that, when a has not one of these values, the equation G(η + 2π) = G(η) is no longer true. When a