An Introduction to Special Functions [1st ed.] 3319413449, 978-3-319-41344-0, 978-3-319-41345-7

Table of contents :
Front Matter....Pages i-viii
Picard’s Theorems....Pages 1-14
The Weierstrass Factorization Theorem....Pages 15-25
Entire Functions of Finite Order....Pages 27-38
Bernoulli Numbers and Polynomials....Pages 39-48
Summation Formulae....Pages 49-66
The Euler Gamma-Function....Pages 67-92
Linear Differential Equations....Pages 93-113
Hypergeometric Functions....Pages 115-163
Back Matter....Pages 165-168

Citation preview

UNITEXT 102

Carlo Viola

An Introduction to Special Functions

UNITEXT - La Matematica per il 3+2 Volume 102

Editor-in-chief A. Quarteroni Series editors L. Ambrosio P. Biscari C. Ciliberto M. Ledoux W.J. Runggaldier

More information about this series at http://www.springer.com/series/5418

Carlo Viola

An Introduction to Special Functions

123

Carlo Viola Department of Mathematics University of Pisa Pisa Italy

ISSN 2038-5722 ISSN 2038-5757 (electronic) UNITEXT - La Matematica per il 3+2 ISBN 978-3-319-41344-0 ISBN 978-3-319-41345-7 (eBook) DOI 10.1007/978-3-319-41345-7 Library of Congress Control Number: 2016944325 Mathematics Subject Classification (2010): 30-01, 30A10, 33B15, 33C05, 33C15, 34M03, 65B15 © Springer International Publishing Switzerland 2016 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. Printed on acid-free paper This Springer imprint is published by Springer Nature The registered company is Springer International Publishing AG Switzerland

Preface

These lecture notes stem from a course that I gave at the doctoral school in mathematics at Pisa University during the academic year 2013–2014. Their primary purpose is to expound some material well suited for a second course on analytic functions of one complex variable, after a first elementary course dealing with the basic concepts in this theory, such as the residue theorem, Cauchy’s integral formula, Taylor and Laurent series expansions, poles and essential singularities, branch points, etc. These basic subjects are the only background assumed in this book; I have made a serious attempt to avoid more advanced prerequisites, sometimes at the cost of choosing slightly longer but more elementary proofs of the theorems. As the title suggests, however, the topics included have been especially chosen to provide the reader with the main notions and results in the theory of functions of one complex variable leading to a rigorous treatment of some special functions: in the first place, the Euler gamma function, which can be considered the most frequently used non-elementary mathematical function, and, secondly, the most important in the family of hypergeometric functions, namely the Euler–Gauss hypergeometric function 2F1 and the Kummer confluent hypergeometric function 1F1. These functions are indispensable tools in ‘higher calculus’ and are often encountered in all branches of pure and applied mathematics. I have tried to treat, on a solid ground, the basic material concerning these functions in the hope of providing a reasonably detailed exposition of their main properties. The content of these notes is classical, and therefore there is nothing original, except perhaps the selection of the subject matter. Naturally, I have borrowed much from many excellent books, some of which are listed in the bibliography, and it would be impossible to acknowledge in detail all my indebtedness to other authors. Special thanks are due to A. Perelli, who kindly read a preliminary draft of these lecture notes and made valuable remarks and suggestions. Pisa, Italy May 2016

Carlo Viola

v

Contents

1 Picard’s Theorems . . . . . . . . . . . . . 1.1 Borel–Carathéodory’s Theorem . 1.2 Schottky’s Theorem . . . . . . . . . 1.3 Picard’s First Theorem . . . . . . . 1.4 Picard’s Second Theorem . . . . .

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1 1 3 10 11

Weierstrass Factorization Theorem Mittag-Leffler’s Series . . . . . . . . . . Infinite Products . . . . . . . . . . . . . . Weierstrass’ Products . . . . . . . . . . .

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15 15 19 21

3 Entire Functions of Finite Order . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Order of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Hadamard’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

27 27 31

4 Bernoulli Numbers and Polynomials . 4.1 Euler’s Factorization of sin z . . . . 4.2 Bernoulli Numbers . . . . . . . . . . 4.3 Bernoulli Polynomials . . . . . . . .

2 The 2.1 2.2 2.3

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5 Summation Formulae . . . . . . . . . . . . . . . . . . 5.1 Stirling’s Formula for n!. . . . . . . . . . . . . . 5.2 Partial Summation and Euler’s Constant . . . 5.3 The Euler–MacLaurin Summation Formula. 5.4 Kronecker’s Summation Formula. . . . . . . .

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67 67 70 76 82 88

6 The 6.1 6.2 6.3 6.4 6.5

Euler Gamma-Function. . Eulerian Integrals . . . . . . Euler’s Limit Formula and Stirling’s Formula . . . . . . The Psi-Function . . . . . . . Binet’s Integral Formulae .

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........... ........... Consequences . ........... ........... ...........

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viii

7 Linear Differential Equations . . . . . . . 7.1 Cauchy’s Method . . . . . . . . . . . . 7.2 Singular Points of the Coefficients . 7.3 Fuchs’ Theorem . . . . . . . . . . . . .

Contents

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8 Hypergeometric Functions . . . . . . . . . . . . . . . . 8.1 Totally Fuchsian Differential Equations . . . . 8.2 The Hypergeometric Differential Equation . . 8.3 Euler’s Integral Representation of 2 F1 . . . . . 8.4 2 F1 as a Function of the Parameters. . . . . . . 8.5 Linear Transformations . . . . . . . . . . . . . . . 8.6 The Confluent Hypergeometric Function 1 F1 . 8.7 The Confluent Hypergeometric Equation . . . 8.8 Mellin–Barnes’ Integral Representations . . . .

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115 115 116 120 127 129 136 139 155

Bibliography. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

Chapter 1

Picard’s Theorems

Our main goal in this chapter is to give an elementary proof of Picard’s first and second theorems, which we base upon Schottky’s theorem (Theorem 1.2). We begin with a proof of Borel–Carathéodory’s inequalities (1.1) and (1.2), which yield useful  upper bounds for | f (z)| and  f (n) (z) (n ≥ 1) in terms of Re f (z). In the proof of Schottky’s Theorem 1.2 we shall employ the inequality (1.1), while (1.2) will be used in Chap. 3, in the proof of Hadamard’s Theorem 3.4.

1.1 Borel–Carathéodory’s Theorem Theorem 1.1 (Borel–Carathéodory) Let 0 < r < R, let z 0 ∈ C, and let f (z) be a function regular1 in the closed disc |z − z 0 | ≤ R. Then max | f (z)| ≤

|z−z 0 |=r

2r R −r

max Re f (z) +

|z−z 0 |=R

R +r | f (z 0 )|. R −r

(1.1)

A similar upper bound holds for the derivatives of f (z) : if max Re f (z) ≥ 0,

|z−z 0 |=R

these notes, a function f (z) of a complex variable z is said to be regular at a point z ∗ ∈ C if it is holomorphic (i.e., satisfies the Cauchy–Riemann equations) in an open neighbourhood of z ∗ . The function f (z) is regular in a set S ⊂ C if it is regular at every point of S.

1 Throughout

© Springer International Publishing Switzerland 2016 C. Viola, An Introduction to Special Functions, UNITEXT - La Matematica per il 3+2 102, DOI 10.1007/978-3-319-41345-7_1

1

2

1 Picard’s Theorems

then, for n = 1, 2, . . . ,   max  f (n) (z) ≤

|z−z 0 |=r

 2n+2 n! R  max Re f (z) + | f (z )| . 0 (R − r )n+1 |z−z0 |=R

(1.2)

Proof By a translation, we may plainly assume z 0 = 0. The theorem is obviously true if f (z) is constant. Otherwise, we denote for brevity A = max Re f (z), |z|=R

f (z) and is regular in |z| ≤ R, the maximum of  f (z)we  first assume f (0) = 0. Since e e  = eRe f (z) in |z| ≤ R is attained only on the border |z| = R. Therefore

A > Re f (0) = 0. The function ϕ(z) : =

f (z) 2 A − f (z)

is regular in |z| ≤ R because the real part of the denominator is >0. Let f (z) = u + iv with u, v ∈ R. From −2 A + u ≤ u ≤ 2 A − u we get |ϕ(z)|2 =

u2 + v 2 ≤ 1. (2 A − u)2 + v 2

Since ϕ(0) = f (0) = 0, the function ϕ(z)/z is regular in |z| ≤ R, whence  ϕ(z)     ≤ max  |z|=R z

 ϕ(z)  1    ≤ .  z R

In particular, for |z| = r , max |ϕ(z)| ≤ |z|=r

r . R

Hence for any z with |z| = r we get | f (z)| = 2 A

r/R 2 Ar |ϕ(z)| ≤ 2A = , |1 + ϕ(z)| 1 − r/R R −r

(1.3)

that is (1.1) in the present case f (0) = 0. If f (0) = 0, we apply (1.3) to the function f (z) − f (0). We obtain | f (z) − f (0)| ≤

 2r 2r  max Re( f (z) − f (0)) = max Re f (z) − Re f (0) . R − r |z|=R R − r |z|=R

1.1 Borel–Carathéodory’s Theorem

3

Therefore, for |z| = r , | f (z)| − | f (0)| ≤ | f (z) − f (0)| ≤

 2r  max Re f (z) + | f (0)| , R − r |z|=R

and (1.1) follows. If max Re f (z) ≥ 0, (1.1) yields |z|=R

max | f (z)| ≤ |z|=r

 R +r  max Re f (z) + | f (0)| . R − r |z|=R

(1.4)

For any z such that |z| = r , by Cauchy’s integral formula we have f

(n)

n! (z) = 2πi

 |t−z|=(R−r )/2

f (t) dt. (t − z)n+1

(1.5)

Since |t| ≤ |z| + |t − z| = r + (R − r )/2 = (R + r )/2, (1.4) yields | f (t)| ≤

max

| f (τ )| ≤

R + 21 (R + r ) 

R − 21 (R + r )   4R max Re f (τ ) + | f (0)| . < R − r |τ |=R |τ |=(R+r )/2

max Re f (τ ) + | f (0)|



|τ |=R

Thus, by (1.5),  n! 4R  max Re f (τ ) + | f (0)| ((R − r )/2)n R − r |τ |=R  2n+2 n! R  max = Re f (τ ) + | f (0)| . (R − r )n+1 |τ |=R

 (n)   f (z) ≤



1.2 Schottky’s Theorem Lemma 1.1 Let R0 and M be positive constants. Let ϕ() be a real function such that 0 ≤ ϕ() ≤ M

(1.6)

for 0 <  < R0 . Assume there exists a constant C > 0 such that ϕ() ≤

 C ϕ(R) 2 (R − )

(1.7)

4

1 Picard’s Theorems

for any , R satisfying 0 <  < R < R0 . Then, for any  such that 0 <  < R0 , 44 C 2 . (R0 − )4

ϕ() ≤

Proof Let 0 <  < 1 < R0 . From (1.6) and (1.7) with R = 1 we get ϕ() ≤

 C C ϕ(1 ) ≤ M 1/2 . 2 (1 − ) (1 − )2

Thus, for 0 <  < 1 < 2 < R0 , C M 1/2 , (2 − 1 )2

ϕ(1 ) ≤ whence, again by (1.7) with R = 1 , ϕ() ≤

 C C ϕ(1 ) ≤ (1 − )2 (1 − )2



C (2 − 1 )2

1/2 M 1/4 .

Iterating this process we get, by induction on n, C ϕ() ≤ (1 − )2



C (2 − 1 )2

1/2

 ···

C (n − n−1 )2

1/2n−1 M 1/2

n

(1.8)

for any 0 <  < 1 < 2 < · · · < n < R0 . For any  with 0 <  < R0 we choose 1 =

 + R0 1 + R0 n−1 + R0 , 2 = , . . . , n = . 2 2 2

Then (1 − )2 =

2 2 (R0 − )2 2 = (R0 − ) , . . . , ( −  2 = (R0 − ) , , ( −  ) ) n 2 1 n−1 22 24 22n

whence, by (1.8), 4C ϕ() ≤ (R0 − )2



1/2n−1 4n C n ··· M 1/2 (R0 − )2 1+1/2+···+1/2n−1 C n M 1/2 . (R0 − )2

42 C (R0 − )2 

= 41+2(1/2)+···+n(1/2

n−1

)

1/2



1.2 Schottky’s Theorem

5

For n → +∞ we get 1+1/2+1/22 +··· C (R0 − )2  (1−1/2)−1  2 C C −2 4 = 4(1−1/2) = 4 . (R0 − )2 (R0 − )2

ϕ() ≤ 41+2(1/2)+3(1/2

2

)+···





We now proceed to a proof of Schottky’s theorem, which implies a strong constraint for a function regular in a disc and omitting two distinct values. Theorem 1.2 (Schottky) Let the function f (z) be regular and satisfy f (z) = 0 and f (z) = 1 for all z in the disc |z| ≤ R0 . Then, for any z, R such that |z| ≤ R < R0 , | f (z)| ≤ exp

K R04 , (R0 − R)4

where K > 0 is a constant depending only on f (0). Proof Let g1 (z) = log f (z),

g2 (z) = log(1 − f (z)),

where the logarithms take their principal values at z = 0, i.e., −π < Im g1 (0) = arg f (0) ≤ π,

−π < Im g2 (0) = arg(1 − f (0)) ≤ π, (1.9)

and are defined by continuity in the disc |z| ≤ R0 . By the assumption f (z) = 0, 1, in the disc |z| ≤ R0 the functions g1 (z) and g2 (z) have no branch points, and hence are regular and = 2kπi (k ∈ Z). For any r such that 0 < r ≤ R0 , let M1 (r ) = max |g1 (z)|, |z|=r

M2 (r ) = max |g2 (z)|, |z|=r

M(r ) = max{M1 (r ), M2 (r )}, and G 1 (r ) = − min Re g1 (z) = max log |z|=r

|z|=r

1 . | f (z)|

We apply Borel–Carathéodory’s inequality (1.1) to the function −g1 (z). We get M1 () ≤

2 r + G 1 (r ) + |g1 (0)| r − r −

(1.10)

6

1 Picard’s Theorems

for any , r such that 0 <  < r < R < R0 . If G 1 (r ) > 1,

(1.11)

let z 0 be such that |z 0 | = r and G 1 (r ) = log

1 . | f (z 0 )|

(1.12)

Then | f (z 0 )| = e−G 1 (r ) < e−1
1/2, whence ∞

| f (z 0 )|n =

n=1

| f (z 0 )| < 2 | f (z 0 )| < 1. 1 − | f (z 0 )|

It follows that, by (1.14), |g2 (z 0 ) − 2kπi|
1, whence log |g2 (0) − 2kπi| > 0. From (1.16), (1.22) and the maximum principle we get, for k = 0, |h(0)| ≤ log |g2 (0) − 2kπi| + π ≤ log(|g2 (0)| + 2|k|π) + π ≤ log(|g2 (0)| + 1 + M2 (R)) + π ≤ log(2M2 (R) + 1) + π. Combining (1.23) and (1.24) we obtain, for any value of the integer k,   |h(0)| ≤ log(2M2 (R) + 1) +  log |g2 (0)|  + π,

(1.24)

8

1 Picard’s Theorems

whence, by (1.19) and (1.21),    2R

2 log(2M2 (R) + 1) +  log |g2 (0)|  + π . R −r

max |h(z)| ≤ |z|=r

(1.25)

By (1.20) and (1.25), 2R R −r 2R < R −r 4R < R −r

G 1 (r ) ≤

  

2 log(2M2 (R) + 1) +  log |g2 (0)|  + π + log 2   

2 log(2M2 (R) + 1) +  log |g2 (0)|  + 2π

   log(2M2 (R) + 1) +  log |g2 (0)|  + π .

The last inequality is proved under the assumption (1.11), but is obviously true if G 1 (r ) ≤ 1. Therefore, by (1.10),     2r  4R

log(2M2 (R) + 1) +  log |g2 (0)|  + π + |g1 (0)| r − R −r (1.26)   

8Rr log(2M2 (R) + 1) + |g1 (0)| +  log |g2 (0)|  + π . < (R − r )(r − )

M1 () ≤

We have log(2M2 (R) + 1) ≤

log(3M2 (R)) = log M2 (R) + log 3, if M2 (R) ≥ 1 log 3, if M2 (R) < 1

= log+ M2 (R) + log 3, where log+ x = max{log x, 0} for x > 0. Clearly log+ x < log(2M2 (R) + 1)
0, in the punctured disc  P z0 , : = {z ∈ C  0 < |z − z 0 | < }

(1.29)

the function f (z) is regular and has a Laurent series expansion f (z) =

+∞

an (z − z 0 )n

(1.30)

n=−∞

with an = 0 for infinitely many n < 0. We require the following Lemma 1.2 (Weierstrass) Let z 0 ∈ C be an isolated essential singularity for f (z). For any ε > 0, the set f (Pz0 ,ε ) of the values taken by f (z) in the punctured disc Pz0 ,ε = {z ∈ C  0 < |z − z 0 | < ε} is dense in C. Proof By contradiction. Assume there exist  > 0, η > 0 and a ∈ C such that f (z) is regular in the punctured disc (1.29) and satisfies | f (z) − a| ≥ η for all z ∈ Pz0 , . Then in Pz0 , the function g(z) : = ( f (z) − a)−1 is regular and satisfies |g(z)| = Denote by g(z) =

1 1 ≤ . | f (z) − a| η +∞ n=−∞

bn (z − z 0 )n

(1.31)

(1.32)

12

1 Picard’s Theorems

the Laurent series expansion of g(z) in Pz0 , . Then, for any n ∈ Z and any δ such that 0 < δ < ,  g(z) 1 bn = dz. 2πi (z − z 0 )n+1 |z−z 0 |=δ

For n = −(k + 1) with k ≥ 0 we get, by (1.31),   1     |bn | = b −(k+1) =  2πi ≤



  g(z)(z − z 0 ) dz  k

|z−z 0 |=δ

δ δ k+1 · 2πδ = →0 2πη η k

(δ → 0),

whence b−1 = b−2 = b−3 = . . . = 0, and (1.32) is a Taylor series. Thus g(z) is regular at z 0 , whence f (z) = a + 1/g(z) either is regular or has a pole at z 0 , contradicting  the assumption that z 0 is an essential singularity for f (z). Theorem 1.4 (Picard’s second theorem) Let z 0 ∈ C be an isolated essential singularity for f (z). For any ε > 0, the set f (Pz0 ,ε ) of the values taken by f (z) in the punctured disc Pz0 ,ε = {z ∈ C  0 < |z − z 0 | < ε} is either C, or C \ {a} for an exceptional a ∈ C depending on f and on the essential singularity z 0 , but independent of ε. Proof By contradiction. Assume there exist  > 0 and a, b ∈ C, with a = b, such that f (z) is regular in the punctured disc (1.29) and satisfies f (z) = a and f (z) = b for all z ∈ Pz0 , . We claim that there exists a sequence of positive numbers rν , with  > r1 > r2 > r3 > · · · → 0, such that | f (z)| ≤ C for |z − z 0 | = rν

(ν = 1, 2, 3, . . . ),

(1.33)

where C is a constant independent of ν. For convenience, by a translation we may assume z 0 = 0. Then, by replacing f (z) with f (z) − a , b−a we may also assume  = 1, a = 0 and b = 1. By Lemma 1.2 there exists a sequence z 1 , z 2 , z 3 , . . . ∈ P0,1 such that |z 1 | > |z 2 | > |z 3 | > · · · → 0 and | f (z ν ) − c1 | < c2

(ν = 1, 2, 3, . . . ),

(1.34)

where c1 and c2 are any fixed real numbers satisfying c1 > 2 and 0 < c2 < c1 − 2 (for instance, we may take c1 = 3 and c2 = 21 ).

1.4 Picard’s Second Theorem

13

Let rν = |z ν | (ν = 1, 2, 3, . . . ), and let w = log z for z ∈ P0,1 . Then z = ew is a one-to-one mapping of the half strip  S : = {w = u + iv  u < 0, −π < v ≤ π} onto P0,1 , and in the half-plane u < 0 the function F(w) : = f (ew ) is regular and periodic with period 2πi. Let wν = log z ν (ν = 1, 2, 3, . . . ) with the principal value of the logarithm, whence wν = u ν + iv ν ∈ S and u ν = log |z ν | → −∞. Thus for any sufficiently large ν we have u ν < −4π. For such ν, the function

f ν (t) : = F(wν + t) = f (ewν +t )

is regular in the disc |t| ≤ 4π and  satisfies4 f ν (t) = 0, 1. By Theorem 1.2 with R =  

2π and R0 = 4π we get f ν (t) ≤ exp(2 K ν ) for |t| ≤ 2π, where the constant K ν depends only on f ν (0) = f (z ν ). Moreover, with the notation √ used in the proof of

/ K ∗ , so that an upper Theorem 1.2, we may take K ν = 218 K 02 with K 0 = 1 + K bound for K ν with an absolute constant is obtained from an upper bound for    

= max{|g1 (0)| +  log |g2 (0)| , |g2 (0)| +  log |g1 (0)| } + π + log 3 K and a lower bound for K ∗ = max{|g1 (0)|, |g2 (0)|}, and hence from upper and lower bounds with positive absolute constants for    f ν (0) = log2 | f (z ν )| + arg2 f (z ν ) |g1 (0)| =  log and    f ν (0)) = log2 |1 − f (z ν )| + arg2 (1 − f (z ν )), |g2 (0)| =  log(1 − where, in accordance with (1.9), arg denotes the principal argument. By virtue of (1.34), K ν does not exceed an absolute constant. Therefore we get |F(w)| ≤ C, with an absolute constant C, in the disc |w − wν | ≤ 2π, and a fortiori for w = u + iv satisfying u = u ν = Re wν = log |z ν | and |v| ≤ π. It follows that | f (z)| ≤ C for |z| = |z ν | = rν , i.e., (1.33).

14

1 Picard’s Theorems

By the same method used in the proof of Lemma 1.2, (1.33) easily implies that f (z) is regular at z 0 . For, denoting by (1.30) the Laurent expansion of f (z) in Pz0 , , we have  f (z) 1 an = dz (n ∈ Z ; ν = 1, 2, 3, . . . ). 2πi (z − z 0 )n+1 |z−z 0 |=rν

Then, for n = −(k + 1) with k ≥ 0, one gets by (1.33)   1     |an | = a −(k+1) =  2πi ≤

Crνk 2π



  f (z)(z − z 0 ) dz  k

|z−z 0 |=rν

· 2πrν = Crνk+1 → 0

(ν → ∞)

whence a−1 = a−2 = a−3 = . . . = 0, and (1.30) is a Taylor series, i.e., f (z) is regular at z 0 . This contradicts the assumption that z 0 is an essential singularity for f (z), and therefore completes the proof of Picard’s second theorem.  Since the ε in the statement of Theorem 1.4 is arbitrarily small, it follows that for any b ∈ C, or for any b ∈ C \ {a} if the exceptional a exists, there are infinitely many z ∈ Pz0 ,ε such that f (z) = b.

Chapter 2

The Weierstrass Factorization Theorem

2.1 Mittag-Leffler’s Series A function f (z) is meromorphic in an open set A ⊂ C if it is regular in A except for a finite or infinite sequence z 1 , z 2 , . . . ∈ A of poles of f (z) (of any multiplicities). For short, in what follows a function meromorphic in the whole C will simply be called meromorphic. If f (z) is meromorphic with only finitely many poles z 1 , z 2 , . . . , z N of respective multiplicities μ1 , μ2 , . . . , μ N , for any n = 1, 2, . . . , N let the Laurent series expansion of f (z) in the neighbourhood of z n be denoted by f (z) =

∞ 

ak(n) (z − z n )k +

μn 

k=0

k=1

bk(n) . (z − z n )k

Then G(z) : = f (z) −

μn N   n=1 k=1

bk(n) (z − z n )k

is plainly an entire function. Conversely, for any entire function G(z), the function f (z) = G(z) +

μn N   n=1 k=1

bk(n) (z − z n )k

(2.1)

is meromorphic, with finitely many poles z 1 , . . . , z N of respective multiplicities μ1 , . . . , μ N . If a meromorphic function f (z) has infinitely many poles z 1 , z 2 , . . . , in general the decomposition similar to (2.1) with an entire function G(z) does not hold, because the series © Springer International Publishing Switzerland 2016 C. Viola, An Introduction to Special Functions, UNITEXT - La Matematica per il 3+2 102, DOI 10.1007/978-3-319-41345-7_2

15

16

2 The Weierstrass Factorization Theorem μn ∞   n=1 k=1

bk(n) (z − z n )k

(2.2)

is not expected to converge. However, Mittag-Leffler has shown that (2.2) can be suitably modified, so that the series thus obtained is totally convergent in any compact subset of C not containing the poles z 1 , z 2 , . . . . We recall that a series of functions ∞ 

u n (z)

n=1

is said to be totally convergent in a set K ⊂ C, if there exists a sequence of constants cn > 0 such that |u n (z)| ≤ cn for all n = 1, 2, . . . and for all z ∈ K , with ∞ 

cn < +∞.

n=1

By Weierstrass’ test, if

∞ 

u n (z) is totally convergent in K , it is absolutely and

n=1

uniformly convergent in K . In view of subsequent applications we prove a special case of Mittag-Leffler’s theorem, corresponding to the most important case where the infinitely many poles z 1 , z 2 , . . . are all simple. Theorem 2.1 (Mittag-Leffler) Let z 1 , z 2 , . . . → ∞ be a sequence of distinct complex numbers satisfying 0 < |z 1 | ≤ |z 2 | ≤ . . . . Let m 1 , m 2 , . . . be any sequence of non-zero complex numbers. Then there exists a (not unique) sequence p1 , p2 , . . . of non-negative integers, depending only on the sequences (z n ) and (m n ), such that the series f (z) : =

∞   z  pn m n zn z − zn n=1

(2.3)

is totally convergent, and hence absolutely and uniformly convergent, in any compact set K ⊂ C \ {z 1 , z 2 , . . . }. Thus the function f (z) is meromorphic, with simple poles z 1 , z 2 , . . . having respective residues m 1 , m 2 , . . . .

2.1 Mittag-Leffler’s Series

17

Proof We choose a sequence of real numbers 0 < r1 ≤ r2 ≤ . . . → +∞ satisfying rn < |z n | (n = 1, 2, . . . ). In the disc |z| ≤ rn we have  m  |m n | |m n | n   ≤  ≤  z − zn |z n | − |z| |z n | − rn

(2.4)

and z rn   < 1.   ≤ zn |z n | Let ε1 + ε2 + . . . be any convergent series of positive constants εn . For every n, let pn be any non-negative integer such that  r  pn εn n (|z n | − rn ). < |z n | |m n |

(2.5)

Since rn /|z n | < 1, (2.5) is satisfied for any sufficiently large pn . Then in the disc |z| ≤ rn we get by (2.4) and (2.5)  z  pn m   r  pn |m | n  n n  < εn .  ≤  zn z − zn |z n | |z n | − rn

(2.6)

Take any compact set K ⊂ C \ {z 1 , z 2 , . . . }, and choose an integer N such that K is contained in the disc |z| ≤ r N . Let  z  pn m  n   Mn = max  . z∈K zn z − zn Then, by (2.6), for any z ∈ K we have  z  pn m  n     ≤ zn z − zn



Mn if n < N εn if n ≥ N .

Since the series of constants M1 + . . . + M N −1 + ε N + ε N +1 + . . . converges, (2.3) is totally convergent in K .  Remark 2.1 Fix any z ∈ C, and let n be such that |z| < |z n |. Then 2 |z n | > |z| + |z n |, whence z 2 |z| 2 |z|   ≤ ,   ≤ zn |z| + |z n | |z − z n | and  z  pn +1  z  pn m  n     |m n |   ≤ 2 |z|  . zn zn z − zn

18

2 The Weierstrass Factorization Theorem

It follows that the sequence ( pn ) in Theorem 2.1 is such that  z  pn +1   |m n |   < +∞ for every z ∈ C. zn n=1

∞ 

(2.7)

Conversely, any sequence ( pn ) of non-negative integers satisfying (2.7) is such that the series (2.3) is totally convergent in any compact set K ⊂ C \ {z 1 , z 2 , . . . }. For, if z ∗ ∈ C is such that |z ∗ | > max |z|, for any n satisfying |z n | > |z ∗ | + 1 we have z∈K

|z ∗ | |z n | = + 1 < |z ∗ | + 1, |z n | − |z ∗ | |z n | − |z ∗ | whence  1 |z ∗ | + 1 1   z ∗  < = 1+  . |z n | − |z ∗ | |z n | |z ∗ | z n Thus, for any z ∈ K ,  z  pn m   z  pn |m |  z  pn +1  1  n  n   ∗  ∗ ≤ 1+ |m n |   .   ≤   zn z − zn z n |z n | − |z ∗ | |z ∗ | zn By (2.7), the series (2.3) is totally convergent in K . Remark 2.2 If g(z) is any meromorphic function with infinitely many poles z 1 , z 2 , . . . , all simple and with respective residues m 1 , m 2 , . . . , then for any sequence ( pn ) of non-negative integers satisfying (2.7) the function G(z) : = g(z) − f (z), where f (z) is given by (2.3), is plainly entire. Thus g(z) has the Mittag-Leffler series expansion ∞   z  pn m n g(z) = G(z) + , zn z − zn n=1 where G(z) is an entire function. Remark 2.3 The function f (z) =

∞   z  pn m n zn z − zn n=1

in (2.3) has the Taylor series expansion around z = 0 given by f (z) = −

∞    k=0

n pn ≤k

m n z n−(k+1)

zk

(2.8)

2.1 Mittag-Leffler’s Series

19

with radius of convergence |z 1 |. For, in the disc |z| ≤ r1 < |z 1 | we have f (z) = −

∞ 

mn

n=1

z pn p +1

zn n

∞ ∞   1 zk = − mn . k+1 1 − z/z n z n=1 k= pn n

(2.9)

Since, by (2.5), ∞  ∞  n=1 k= pn

|m n |

∞ ∞     1 |z|k |z| pn  z  pn |m n | = = |m |   n |z n |k+1 |z n | pn +1 1 − |z|/|z n | z n |z n | − |z| n=1 n=1 ∞  ∞   rn  pn |m n | ≤ ≤ εn < +∞, |z n | |z n | − rn n=1 n=1

we may interchange the sums on the right-hand side of (2.9). Therefore f (z) = −

∞ 

zk



m n z n−(k+1) ,

n pn ≤k

k=0

and the radius of convergence of this Taylor series is the distance from the origin to the closest singular point of f (z), i.e., to the pole z 1 .

2.2 Infinite Products Let an ∈ C, an = 0 (n = 1, 2, . . . ). The infinite product



an is defined by

n ∞ 

an = lim

n→∞

n=1

n 

ak = A.

(2.10)

k=1

The infinite product (2.10) converges if the limit A exists and A = 0, ∞. If A = 0 or A = ∞, the infinite product is said to be divergent to zero or, respectively, to infinity. If (2.10) converges, then lim an = lim

n→∞

n→∞

 n k=1

ak

n−1 

ak

= A/A = 1.

k=1

Let an (z) = 0 (n = 1, 2, . . . ) be

a sequence of functions defined for z in a an (z) is uniformly convergent to a function set K ⊂ C. The infinite product n

f (z) = 0, ∞ in K if, for n → ∞, the sequence of partial products

20

2 The Weierstrass Factorization Theorem

f n (z) =

n 

ak (z)

k=1

converges uniformly to f (z) in K , i.e., if for any ε > 0 there exists n 0 such that | f (z) − f n (z)| < ε for all n > n 0 and for all z ∈ K . Lemma 2.1 Let the functions u n (z) (n = 1, 2, . . . ) be regular in a compact set K ⊂ C, and let the series ∞  u n (z) n=1

be totally convergent in K . Then the infinite product ∞ 

 exp u n (z) = exp

 ∞

n=1

u n (z)

n=1

is uniformly convergent in K . Proof For any z 1 , z 2 ∈ C we have 1 1 e z1 − e z2 = (z 1 − z 2 ) + (z 12 − z 22 ) + (z 13 − z 23 ) + . . . 2! 3!   1 1 = (z 1 − z 2 ) 1 + (z 1 + z 2 ) + (z 12 + z 1 z 2 + z 22 ) + . . . , 2! 3! whence   1 1 |e z1 − e z2 | ≤ |z 1 − z 2 | 1 + (|z 1 | + |z 2 |) + (|z 1 | + |z 2 |)2 + . . . 2! 3! ≤ |z 1 − z 2 | e|z1 |+|z2 | . Therefore     ∞ N    exp  u (z) − exp u (z) n n   n=1

(2.11)

n=1

   N ∞ N    ∞   ≤  u n (z) − u n (z) exp |u n (z)| + |u n (z)| n=1

n=1

n=1

n=1

n=1

    N ∞   ∞    ≤  u n (z) − u n (z) exp 2 |u n (z)| . n=1

n=1

2.2 Infinite Products

21

Let ∞ 

|u n (z)| = U (z),

n=1

let   M = exp 2 max U (z) , z∈K

and for any ε > 0 let N0 be such that   N   ∞  ε  u n (z) − u n (z) <  M n=1 n=1 for all N > N0 and all z ∈ K . Then, by (2.11),       ∞ N N   ∞      exp  u n (z) − exp u n (z)  ≤ M  u n (z) − u n (z) < ε.  n=1

n=1

n=1



n=1

2.3 Weierstrass’ Products Lemma 2.2 Let f (z) be a meromorphic function. Let z 1 , z 2 , . . . = 0 be the poles of f (z), all simple with respective residues m 1 , m 2 , . . . ∈ Z. Then the function z ϕ(z) : = exp

f (t) dt

(2.12)

0

is meromorphic. The zeros (resp. poles) of ϕ(z) are the points z n such that m n > 0 (resp. m n < 0), and the multiplicity of z n as a zero (resp. pole) of ϕ(z) is m n (resp. −m n ). Proof First we remark that ϕ(z) is a one-valued function, because if γ and γ are any two paths of endpoints 0 and z not passing through the poles z n , by the residue theorem we get   f (t) dt = f (t) dt + 2πi R, γ

γ

22

2 The Weierstrass Factorization Theorem

where R is the sum of residues of f (t) at the poles between γ and γ , each residue being taken with + or − sign according to the mutual position of γ and γ around the corresponding pole. Since m 1 , m 2 , . . . ∈ Z, we get R ∈ Z. Therefore 

 f (t) dt = e2πi R exp

exp γ

 f (t) dt = exp

γ

f (t) dt. γ

Plainly ϕ(z) is regular and = 0 in C \ {z 1 , z 2 , . . . }. The  z function f 1 (z) : = f (z) − m 1 /(z − z 1 ) is regular in C \ {z 2 , z 3 , . . . }, whence exp 0 f 1 (t) dt is regular and = 0 in C \ {z 2 , z 3 , . . . }. Thus  z

z ϕ(z) = exp

f (t) dt = exp 0

z = exp

z f 1 (t) dt + m 1

0

0

dt t − z1



  z  = (z − z 1 )m 1 ϕ1 (z), f 1 (t) dt · exp m 1 log 1 − z1

0

z where ϕ1 (z) = (−z 1 )−m 1 exp 0 f 1 (t) dt is regular and = 0 in C \ {z 2 , z 3 , . . . }. By  the same argument for z 2 , z 3 , . . . the lemma follows. For any sequence z 1 , z 2 , . . . ∈ C, either finite or satisfying lim z n = ∞, from Theorem 2.1 and Lemma 2.2 we deduce the existence of a meromorphic function with zeros and poles at z 1 , z 2 , . . . with arbitrary multiplicities m 1 , m 2 , . . . . This yields the following Corollary 2.1 Every meromorphic function is the quotient of two entire functions. Proof Let g(z) be meromorphic, and let z 1 , z 2 , . . . = 0 be the poles of g(z) with respective multiplicities m 1 , m 2 , . . . . Let f (z) be the function (2.3) and let ϕ(z) be the function (2.12). By applying Theorem 2.1 and Lemma 2.2, we see that ϕ(z) is entire with zeros z 1 , z 2 , . . . of respective multiplicities m 1 , m 2 , . . . . Then the product g(z)ϕ(z) has no poles, and therefore is an entire function h(z), whence g(z) = h(z)/ϕ(z) with h(z) and ϕ(z) entire functions. If g(z) has a pole at z = 0 of multiplicity m, then  g (z) : = z m g(z) is regular at z = 0, whence  g (z) = h(z)/ϕ(z) with entire h(z) and ϕ(z), and g(z) = h(z)/(z m ϕ(z))  with entire h(z) and z m ϕ(z). Theorem 2.2 (Weierstrass) Let F(z) be meromorphic, and regular and = 0 at z = 0. Let z 1 , z 2 , . . . be the zeros and poles of F(z) with respective multiplicities |m 1 |, |m 2 |, . . . , where m n > 0 if z n is a zero and m n < 0 if z n is a pole of F(z). Then there exist integers p1 , p2 , . . . ≥ 0 and an entire function G(z) such that F(z) = e

G(z)

 n

  pn 1  z k z m n , 1− exp m n zn k zn k=1

(2.13)

2.3 Weierstrass’ Products

23

where the product converges uniformly in any compact set K ⊂ C \ {z 1 , z 2 , . . . }. In (2.13) one can take any sequence of integers p1 , p2 , . . . ≥ 0 satisfying (2.7). Proof Let f (z) be the function (2.3) with integer exponents p1 , p2 , . . . ≥ 0 satisfying (2.7), and let ϕ(z) be the function (2.12). By Theorem 2.1 and Lemma 2.2 ϕ(z) is meromorphic, with zeros z n of multiplicities m n if m n > 0, and with poles z n of multiplicities |m n | if m n < 0. Thus F(z) and ϕ(z) have the same zeros and poles multiplicities, whence F(z)/ϕ(z) is entire and = 0. Therefore

with the same  log F(z)/ϕ(z) = G(z) is an entire function, and F(z) = e G(z) ϕ(z).

(2.14)

By uniform convergence of (2.3) on a path of endpoints 0 and z not containing z 1 , z 2 , . . . , term-by-term integration of (2.3) from 0 to z is allowed. Thus from (2.12) we get z    p t n mn ϕ(z) = exp dt zn t − zn n

(2.15)

0

=

 n

z  mn m n (t/z n ) pn − 1  dt exp + t − zn z n t/z n − 1 0

z  pn  mn m n   t k−1  = dt exp + t − zn z n k=1 z n n 0

pn   1  z k z m n = exp log 1 − + mn zn k zn n k=1  p n  1  z k  z m n . 1− exp m n = zn k zn n k=1 



Then (2.13) follows from (2.14). Moreover, for any z ∈ K we can choose the integration path from 0 to z of length bounded by a constant depending only on K , whence, by Theorem 2.1, the series z    t  pn m n dt zn t − zn n 0

is totally convergent in K . Hence, by Lemma 2.1, the product in (2.13) is uniformly convergent in K . 

24

2 The Weierstrass Factorization Theorem

If the meromorphic function F(z) has a zero of multiplicity m or a pole of multiplicity −m at z = 0, then z −m F(z) is regular and = 0 at z = 0, and therefore can be represented by (2.13). Thus in this case the Weierstrass factorization formula (2.13) becomes   pn  z m n 1  z k m G(z) . (2.16) 1− exp m n F(z) = z e zn k zn n k=1 Corollary 2.2 The Taylor series expansion with centre z = 0 of the entire function G(z) in (2.13) is G(z) = log F(0) +

∞   k=1

k  k−1   1 d F (z) −k z . (2.17) + m n zn (k − 1)! dz k−1 F(z) z=0 k n pn 0, we write f (z)  g(z)

(z → z 0 ),

(3.1)

where z 0 can be either finite or infinity, to mean that there exists a constant C > 0, independent of z, such that | f (z)| ≤ Cg(z) for all z in a suitable neighbourhood of z 0 . Thus (3.1) is synonymous with f (z) = O(g(z)) (z → z 0 ), where O is Landau’s asymptotic symbol. If the constant C depends on some parameters λ, μ, . . . , we may write f (z) λ,μ,... g(z), or f (z) = Oλ,μ,... (g(z)). Similarly, given two sequences an ∈ C and bn > 0, we write an  bn to mean an = O(bn ), i.e., that there exists a constant C > 0 independent of n such that, for any n, |an | ≤ Cbn . If f (z) and g(z) are both > 0, Vinogradov’s notation g(z)  f (z) means f (z)  g(z), and similarly for sequences an , bn > 0.

3.1 Order of a Function We define the order of an entire function F(z) as the infimum α of the exponents A > 0 such that (3.2) F(z)  exp(|z| A ) (z → ∞), and we denote α = ord F(z). © Springer International Publishing Switzerland 2016 C. Viola, An Introduction to Special Functions, UNITEXT - La Matematica per il 3+2 102, DOI 10.1007/978-3-319-41345-7_3

27

28

3 Entire Functions of Finite Order

If there exist no A > 0 satisfying (3.2), we define ord F(z) = +∞. From the above definition it follows that F(z) ε exp(|z| A+ε )

for every ε > 0

(3.3)

if and only if ord F(z) ≤ A. Lemma 3.1 If ord F1 (z) = α1 and ord F2 (z) = α2 , then ord (F1 (z) + F2 (z)) ≤ max{α1 , α2 } and ord (F1 (z)F2 (z)) ≤ max{α1 , α2 }. Proof Let α = max{α1 , α2 }. Then F1 (z) ε exp(|z|α+ε ) and F2 (z) ε exp(|z|α+ε ), whence

and

|F1 (z) + F2 (z)| ≤ |F1 (z)| + |F2 (z)| ε exp(|z|α+ε )

|F1 (z)F2 (z)| = |F1 (z)| |F2 (z)| ε exp(2|z|α+ε ) ε exp(|z|α+2ε ).



Remark 3.1 Since |z|n  exp(|z|ε ), from Lemma 3.1 it follows that, for any polynomial P(z), ord P(z) = 0. If deg P(z) then ord e P(z) = n. To see this, note  P(z)  =Ren,P(z) n   =e ≤ e|P(z)|  exp(|z|n+ε ), so that that |P(z)|  |z| , whence e P(z) n n−1 ≤ n. If P(z) = a0 z + a1 z + . . . + an (a0 = 0), and if z → ∞ along ord e one of the n half-lines defined by arg z = − n1 arg a0 , whence arg(a0 z n ) = arg a0 + n arg z = 0 so that a0 z n > 0, then Re P(z) − |z|n−ε = Re(a0 z n ) + O(|z|n−ε ) = |a0 ||z|n + O(|z|n−ε ) → +∞. This shows that  P(z)  e  = lim sup exp(Re P(z) − |z|n−ε ) = +∞, lim sup n−ε ) z→∞ exp(|z| z→∞ whence ord e P(z) ≥ n. We aim at relating the order of an entire function with the density of its zeros. To do this, we require the following Lemma 3.2 Let 0 < r < R, and let f (z) be regular in the disc |z − z 0 | ≤ R and not identically zero. Let N be the number of zeros of f (z) in the disc |z − z 0 | ≤ r , counted with multiplicity. Then | f (z 0 )| ≤

 r N R

max | f (z)|,

|z−z 0 |=R

(3.4)

whence, if f (z 0 ) = 0, max | f (z)| 1 |z−z 0 |=R log . N≤ log(R/r ) | f (z 0 )|

(3.5)

3.1 Order of a Function

29

Proof The inequality (3.5) is an obvious consequence of (3.4) by taking logarithms. In order to prove (3.4), by a translation we may assume z 0 = 0. Also, by replacing f (z) with  f (z) : = f (Rz), we may assume R = 1, since  f (z) is regular in |z| ≤ 1 and has N zeros in |z| ≤ r/R. Let z 1 , . . . , z N be the zeros of f (z) in |z| ≤ r , where a zero of multiplicity m is repeated m times, and let g(z) = f (z)

N  1 − zn z n=1

z − zn

.

Since f (z n ) = 0, g(z) is regular in |z| ≤ 1. Moreover, each of the Blaschke factors (1 − z n z)/(z − z n ) maps the circumference |z| = 1 onto itself, because for any ζ with |ζ| = 1 we get        −iϑ − ζ   1 − ζ eiϑ    = eiϑ   e   eiϑ − ζ  = 1.  eiϑ − ζ  Thus, for |t| ≤ 1,

|g(t)| ≤ max |g(z)| = max | f (z)|, |z|=1

|z|=1

whence   N  N      t − zn   t − zn      | f (t)| = |g(t)| | f (z)|  1 − z t  ≤ max  1 − z t . |z|=1 n

n=1

n=1

n

In particular, for t = 0, | f (0)| ≤



max | f (z)| |z|=1

N 

|z n | ≤ r N max | f (z)|. |z|=1

n=1



This proves (3.4).

Theorem 3.1 Let F(z) be an entire function not identically zero, and let N (r ) denote the number of zeros of F(z) in the disc |z| ≤ r , counted with multiplicity. If α = ord F(z) < +∞, then, for any ε > 0,

N (r ) ε r α+ε

(r → +∞).

Proof Assume F(0) = 0. By (3.5) with z 0 = 0, R = 2r , we obtain N (r ) ≤

max |F(z)| 1 |z|=2r log . log 2 |F(0)|

(3.6)

30

3 Entire Functions of Finite Order

Since

  max |F(z)|  exp (2r )α+ε ,

|z|=2r

we get

N (r )  (2r )α+ε  r α+ε .

If z = 0 is a zero of F(z)of multiplicity m, we apply (3.6) to the function z −m F(z)  −m  which plainly satisfies ord z F(z) = α. Let z 1 , z 2 , . . . → ∞ be a sequence of non-zero complex numbers. We define the exponent of convergence of the sequence (z n ) as the infimum β of the exponents B > 0 such that ∞ |z n |−B < +∞. (3.7) n=1

If

∞

|z n |−B = +∞ for every B > 0, the exponent of convergence of (z n ) is +∞.

n=1

Theorem 3.2 If an entire function F(z) satisfying F(0) = 0 has order α, and if the sequence z 1 , z 2 , . . . of the zeros of F(z) has exponent of convergence β, then β ≤ α. Proof From (3.6) with r = |z n | we get n  |z n |α+ε , whence |z n |−(α+2ε)  n −(α+2ε)/(α+ε) = n −(1+ε1 ) , for a suitable ε1 > 0. Therefore ∞

|z n |−(α+2ε) 

n=1

∞

n −(1+ε1 ) < +∞.

n=1

By (3.7) we get β ≤ α.



Theorem 3.3 If the zeros z 1 , z 2 , . . . of an entire function F(z) such that F(0) = 0 have exponent of convergence β < +∞, the Weierstrass factorization formula (2.13) can be written in the form F(z) = e G(z)

p  z 1  z k 1− exp , zn k zn n k=1

(3.8)

where p ≥ 0 is an integer independent of n. Proof For any integer p such that p + 1 > β we get n

|z n |−( p+1) < +∞,

(3.9)

3.1 Order of a Function

31

whence, for every z ∈ C,

 z  p+1 < +∞,   zn n

i.e., the constant sequence pn = p satisfies (2.18).



For a sequence z 1 , z 2 , . . . → ∞ of non-zero complex numbers with exponent of convergence β < +∞, we use the standard notation p   z 1  z k z exp , p := 1 − , E zn zn k zn k=1

(3.10)

and, if p is the least integer satisfying (3.9), we call p    z z 1  z k 1− exp E , p = zn zn k zn n n k=1

(3.11)

the Weierstrass canonical product for the sequence (z n ).

3.2 Hadamard’s Theorem In accordance with Theorems 3.2 and 3.3, for any entire function F(z) satisfying F(0) = 0 and α = ord F(z) < +∞ we shall use the factorization formula (3.8) with the canonical product (3.11). For such a function F(z), the following theorem yields strong information on the entire function G(z) appearing in (3.8). Theorem 3.4 (Hadamard) Let F(z) = e G(z)

p  1  z k z 1− exp zn k zn n k=1

(3.12)

be an entire function of order α < +∞, where the product is the canonical product for the zeros z n of F(z). Then the entire function G(z) in (3.12) is a polynomial of degree not exceeding α. Proof Let ν = α (the integer part of α). By the Taylor series expansion of G(z), the theorem is equivalent to proving that G (ν+1) (z) = 0 identically. Since p is the least integer satisfying (3.9) we have p ≤ β, where β is the exponent of convergence of the sequence (z n ). By Theorem 3.2 we get p ≤ α, whence p ≤ ν = α. Therefore p dν+1 1  z k = 0. dz ν+1 k=1 k z n

32

3 Entire Functions of Finite Order

Thus, taking the logarithm of (3.12) and then differentiating ν + 1 times we obtain dν F (z) 1 (ν+1) = G (z) − ν! , dz ν F(z) (z − z)ν+1 n n

(3.13)

because term-by-term differentiation is allowed by (2.15) and by uniform convergence of the Mittag-Leffler series (2.3). For any R > 0 we define ϕ R (z) : =

z −1 F(z)   1− , F(0) |z |≤R zn

(3.14)

n

and ψ R (z) : = log ϕ R (z),

(3.15)

where we take the principal logarithm at z = 0, i.e., ψ R (0) = log 1 = 0. Plainly ϕ R (z) is an entire function, and satisfies ϕ R (z) = 0 for |z| ≤ R. Thus ψ R (z) is regular in the disc |z| ≤ R. Also, from (3.14) and (3.15) we get ψ R (z) = log F(z) − log F(0) −



log(z n − z) +

|z n |≤R

Therefore ψ (ν+1) (z) = R



log z n .

|z n |≤R

1 dν F (z) + ν! , ν dz F(z) (z n − z)ν+1 |z |≤R n

and (3.13) yields 1 dν F (z) + ν! ν dz F(z) (z n − z)ν+1 n 1 = ψ (ν+1) (z) + ν! . R (z − z)ν+1 n |z |>R

G (ν+1) (z) =

(3.16)

n

Since |1 − z/z n | ≥ |z|/|z n | − 1 ≥ 1 for |z| = 2R and |z n | ≤ R, we get by (3.14) |ϕ R (z)| ≤

  |F(z)| ε exp (2R)α+ε |F(0)|

(3.17)

for |z| = 2R, hence also for |z| ≤ 2R by the maximum principle. From (3.15) and (3.17) we obtain Re ψ R (z) = log |ϕ R (z)| ε R α+ε

for |z| ≤ R.

(3.18)

3.2 Hadamard’s Theorem

33

Moreover, again by the maximum principle, from ψ R (0) = 0 we get max Re ψ R (z) ≥ 0.

|z|=R

Thus we can apply Borel–Carathéodory’s inequality (1.2) to the function ψ R (z) with r = R/2 and z 0 = 0. By (3.18) we obtain   max ψ (ν+1) (z) ν R

R max Re ψ R (z) ε,ν R α−ν−1+ε . (R/2)ν+2 |z|=R

|z|=R/2

Hence for |z| ≤ R/2 we get, by (3.16), G (ν+1) (z) ε,ν R α−ν−1+ε +

|z n |>R

ε,ν R α−ν−1+ε +



1 |z n − z|ν+1

(3.19)

|z n |−(ν+1) ,

|z n |>R

because |z n − z| ≥ |z n | − |z| ≥ |z n | − R/2 > 21 |z n |. Since ν = α we have α < ν + 1, whence lim R α−ν−1+ε = 0 R→∞

for small ε > 0. Moreover, by Theorem 3.2, β ≤ α < ν + 1, whence

sufficiently |z n |−(ν+1) < +∞. Thus n

lim

R→∞



|z n |−(ν+1) = 0.

|z n |>R

For any fixed z ∈ C, (3.19) holds for all R ≥ 2|z|. Since G (ν+1) (z) is independent of  R, making R → ∞ in (3.19) we obtain G (ν+1) (z) = 0. For any polynomial P(z) of degree n, e P(z) is a non-vanishing entire function, of order n by Remark 3.1 above. The converse of this statement is also true, in the following form. Corollary 3.1 Let F(z) be an entire function satisfying F(z) = 0 for all z ∈ C, and let α = ord F(z) < +∞. Then α ∈ N and F(z) = e G(z) , where G(z) is a polynomial of degree α. Proof We use a simplified version of the proof of Theorem 3.4, where the product  on the right-hand side of (3.12) is empty. Let ν = α, let n

34

3 Entire Functions of Finite Order

ψ(z) = log

F(z) , F(0)

where we take the principal logarithm at z = 0, i.e., ψ(0) = log 1 = 0, and let G(z) = log F(z) = ψ(z) + log F(0). Then G (ν+1) (z) = ψ (ν+1) (z) =

dν F (z) . dz ν F(z)

(3.20)

Since F(z) = 0 for all z ∈ C, G(z) and ψ(z) are entire functions. For any R > 0, by the maximum principle we have max Re ψ(z) ≥ Re ψ(0) = 0.

|z|=R

Thus, by Borel–Carathéodory’s inequality (1.2) with r = R/2 and z 0 = 0,   max ψ (ν+1) (z) 

|z|=R/2

R max Re ψ(z). (R/2)ν+2 |z|=R

(3.21)

Since α = ord F(z), for |z| ≤ R and for any ε > 0 we get    F(z)   α+ε    ,  F(0)  ε exp R    F(z)   ε R α+ε . Re ψ(z) = log  F(0) 

whence

Therefore, by (3.20) and (3.21),     max G (ν+1) (z) = max ψ (ν+1) (z) ε R α−ν−1+ε .

|z|≤R/2

|z|≤R/2

Hence, for any fixed z ∈ C,  (ν+1)  G (z) ε R α−ν−1+ε for all R ≥ 2|z|. Since ν = α we have α < ν + 1, whence α − ν − 1 + ε < 0 for sufficiently small ε > 0. For R → ∞ we obtain G (ν+1) (z) = 0 identically. The Taylor expansion of G(z) implies that G(z) is a polynomial of degree μ ≤ ν = α. By Remark 3.1, F(z) = e G(z) has order μ, whence α = μ ≤ α. Thus α = μ = α ∈ N, and deg G(z) = μ = α.  By (2.15) and Lemma 2.2 the Weierstrass canonical product (3.11) is an entire function, the order of which is given by the following

3.2 Hadamard’s Theorem

35

Theorem 3.5 Let z 1 , z 2 , . . . → ∞ be any sequence of non-zero complex numbers with exponent of convergence β < +∞. The canonical product p   z  z 1  z k 1− exp E , p = zn zn k zn n n k=1

is an entire function of order β. Proof Let α be the order of the canonical product (3.11). By Theorem 3.2 we know that β ≤ α. Thus we must prove that α ≤ β, i.e., by (3.3),   z E , p ε exp(|z|β+ε ) zn n for every ε > 0. For this purpose, it suffices to prove that         z z    E , p  = log  E , p  ε |z|β+ε log  z z n

n

(z → ∞),

(3.22)

n

n

because from (3.22) with ε/2 in place of ε one gets     z E , p  ε |z|β+ε/2 = o(|z|β+ε ) ≤ ϑ|z|β+ε log  zn n with 0 < ϑ < 1, whence     z    E , p  ≤ exp(ϑ|z|β+ε ) = o exp(|z|β+ε ) .  z n n If |z/z n | ≤

1 2

we have

    z  log  E , p  ≤ z n

     log E z , p  =   z n

  p    1  z k   log 1 − z +  zn k zn  k=1

  ∞ ∞    1  z k   z k ≤ =  −   k z n  k= p+1 z n k= p+1  z  p+1   z  p+1  1 1     1 + + 2 + ... = 2  , ≤   zn 2 2 zn whence, by (3.9), |z n |≥2|z|

    z  log  E , p  ≤ 2 |z| p+1 |z n |−( p+1)  |z| p+1 . z n

|z n |≥2|z|

(3.23)

36

3 Entire Functions of Finite Order

Since p is the least integer satisfying (3.9), we have p ≤ β ≤ p + 1. If β = p + 1, (3.23) yields     z  log  E , p   |z|β . z n |z |≥2|z| n

If β < p + 1, for any ε satisfying 0 < ε < p + 1 − β we get by (3.23)     z  log  E , p  ≤ 2 |z|β+ε |z n |−( p+1) |z| p+1−β−ε z n |z |≥2|z| |z |≥2|z| n

n

≤

2 |z| |z n |−(β+ε) ε |z|β+ε . 2 p+1−β−ε |z |≥2|z| β+ε

n

In either case

|z n

If |z/z n | >

    z log  E , p  ε |z|β+ε . zn |≥2|z|



1 2

(3.24)

we have

  p    z  z 1  z k  log  E , p  = log 1 −  + Re zn zn k zn k=1 p  z     ≤ log 1 +   + zn k=1

 z  p+ε   ε   . zn

⎧  z ε ⎪ ⎨   if p = 0  z k zn       z p  ⎪ zn ⎩   if p ≥ 1 zn

Therefore

    z log  E , p  ε |z| p+ε |z n |−( p+ε) z n |z n | max{ p, q} we get  k−1  d F (z) 1 + z n−k = 0, (k − 1)! dz k−1 F(z) z=0 n whence (3.28).



Chapter 4

Bernoulli Numbers and Polynomials

4.1 Euler’s Factorization of sin z By a standard application of the Weierstrass–Hadamard factorization formula (3.12) we prove the following Theorem 4.1 (Euler) The entire function sin(πz)/πz satisfies  ∞   z2 sin(πz) = 1− 2 πz n n=1

(4.1)

for all z ∈ C. Proof By Remark 3.1, the exponential functions e±πi z have order 1. Thus, by Lemma 3.1, the entire function F(z) : =

1 π2 z 2 sin(πz) = (eπi z − e−πi z ) = 1 − + ... πz 2πi z 3!

has order α ≤ 1 and satisfies F(0) = 1,

F  (0) = 0.

(4.2)

Since eπi z = e−πi z if and only if e2πi z = 1, i.e., 2πi z = log 1 = 2kπi with k ∈ Z, the zeros of F(z) are z = ±1, ±2, ±3, . . . with exponent of convergence β = 1. By Theorem 3.2 we have β = 1 ≤ α, whence α = β = 1. Also, the least integer p satisfying (3.9) for |z n | = n is plainly p = 1. Hence, in the present case, (3.12) becomes  ∞  ∞    z2 z  z/n  z  −z/n sin(πz) G(z) G(z) 1− 2 , = e e e 1− 1+ = e F(z) = πz n n n n=1 n=1 © Springer International Publishing Switzerland 2016 C. Viola, An Introduction to Special Functions, UNITEXT - La Matematica per il 3+2 102, DOI 10.1007/978-3-319-41345-7_4

39

40

4 Bernoulli Numbers and Polynomials

where G(z) is a polynomial of degree q ≤ α = p = 1 by Theorem (3.4). Thus, by (3.26) and (4.2), F  (0) G(z) = log F(0) + z = 0, F(0) 

and (4.1) follows. Since the zeros of the entire function cos(πz) =

π2 z 2 1 πi z (e + e−πi z ) = 1 − + ... 2 2!

are z = ±(n − 1/2) (n = 1, 2, 3, . . . ), the same method used in the proof of Theorem 4.1 yields the factorization ∞   cos(πz) = 1− n=1

 4z 2 . (2n − 1)2

The Riemann zeta-function is defined by ζ(s) : =

∞ 

n −s

(s ∈ C, Re s > 1).

(4.3)

n=1

As Riemann proved in his celebrated paper, published in 1859, on the average distribution of prime numbers, ζ(s) can be analytically continued in the whole C to a meromorphic function of s, regular in C \ {1} and having a simple pole with residue 1 at s = 1. Here our aim is to prove Euler’s formulae (4.16) for the values ζ(2k) =

∞ 

n −2k

(k = 1, 2, 3, . . . )

(4.4)

n=1

of the zeta-function at even positive integers 2k. We apply (3.28) to the function (4.1). The logarithmic derivative of (4.1) is 1 d log sin(πz) − log z − log π = π cot(πz) − , dz z and the genus of (4.1) is g = max{ p, q} = 1. Thus (3.28) yields ∞  n=1

n

−h

∞  + (−n)−h = − n=1

h−1   d 1 1 π cot(πz) − (h − 1)! dz h−1 z z=0 (h = 2, 3, 4, . . . ). (4.5)

4.1 Euler’s Factorization of sin z

41

For odd h = 2k + 1 the two sides of (4.5) vanish; hence (4.5) gives no information on ζ(2k + 1). For even h = 2k, (4.5) becomes ζ(2k) =

∞  n=1

n

−2k

2k−1   d π 1 1 − cot(πz) = (2k − 1)! dz 2k−1 2z 2 z=0 (k = 1, 2, 3, . . . ). (4.6)

This formula is equivalent to the Taylor series expansion ∞  π 1 − cot(πz) = ζ(2k) z 2k−1 2z 2 k=1

(|z| < 1).

(4.7)

4.2 Bernoulli Numbers We can express (4.6) in a different form by means of the Bernoulli numbers, which allow to write the Laurent series expansion of cot z at z = 0. Definition 4.1 The Bernoulli numbers are defined by

Bn =

z dn n z dz e − 1

 (n = 0, 1, 2, . . . ), z=0

i.e., ∞  z zn = B n ez − 1 n! n=0

(|z| < 2π).

(4.8)

Since we have identically

ez

z z −z z + = −z − , −1 2 e −1 2

the function z/(e z − 1) + z/2 is even and takes the value 1 at z = 0. Hence (4.8) yields B0 = 1,

1 B1 = − , 2

B2k+1 = 0 (k = 1, 2, 3, . . . ).

(4.9)

42

4 Bernoulli Numbers and Polynomials

From (4.8) we get, for |z| < 2π, 1 =

 ∞ n=0

Bn

zn n!



ez − 1 = z

 ∞

Bn

n=0

=

zn n!

  ∞ n=1

∞  n−1  n=1

k=0

z n−1 n!



 Bk z n−1 . k! (n − k)!

Thus, for n ≥ 2, n−1  k=0

Bk = 0, k! (n − k)!

whence, multiplying by n!, n−1    n

k

k=0

Bk = 0

(n = 2, 3, 4, . . . ).

(4.10)

(n = 0, 2, 3, 4, . . . ),

(4.11)

Adding Bn , (4.10) yields n    n k=0

k

Bk = Bn

which can symbolically be written as Bn = (1 + B)n

(n = 1),

where in the binomial expansion of (1 + B)n the symbolic powers B k are replaced by Bk . Note that, for n = 1, instead of (4.11) we have, by (4.9), 1    1 k=0

k

Bk =

1 = −B1 . 2

(4.12)

From (4.10) we get the recurrence formula Bn−1

      n 1 n B1 + . . . + = − 1+ Bn−2 n 1 n−2

(n = 2, 3, 4, . . . ), (4.13)

which shows that Bn ∈ Q for all n ≥ 0, and yields, for the first few B2k ,

4.2 Bernoulli Numbers

43

1 1 1 1 , B4 = − , B6 = , B8 = − , 6 30 42 30 5 691 7 3617 , B12 = − , B14 = , B16 = − ,.... = 66 2730 6 510

B2 = B10

(4.14)

By (4.8) and (4.9), in the disc |z| < π we get the Laurent expansion ∞ ei z + e−i z 1 2i z 1 (2i z)n e2i z + 1 = i + = i + = i B n ei z − e−i z e2i z − 1 z e2i z − 1 z n=0 n!   ∞ ∞  1  (2z)2k z 2k−1 1 1 − iz + = + , = i+ (−1)k B2k (−1)k 22k B2k z (2k)! z k=1 (2k)! k=1

cot z = i

whence ∞  π 1 z 2k−1 − cot(πz) = (−1)k−1 (2π)2k B2k 2z 2 2 (2k)! k=1

(|z| < 1).

(4.15)

Comparing (4.7) with (4.15) we obtain Euler’s formulae ζ(2k) = (−1)k−1

(2π)2k B2k 2 (2k)!

(k = 1, 2, 3, . . . ).

(4.16)

By (4.14) and (4.16), ζ(2) =

π4 π6 π8 π2 , ζ(4) = , ζ(6) = , ζ(8) = , .... 6 90 945 9450

Euler’s formulae (4.16) show that (−1)k−1 B2k > 0 for k ≥ 1, i.e., B2 , B4 , B6 , . . . have alternate signs. Moreover, writing (4.16) in the form (−1)k−1 B2k =

2 (2k)! ζ(2k), (2π)2k

(4.17)

we get 2

(2k)! π 2 (2k)! k−1 < (−1) B ≤ 2k (2π)2k 3 (2π)2k

(k = 1, 2, 3, . . . ),

because 1 < ζ(2k) ≤ ζ(2) =

π2 . 6

(4.18)

44

4 Bernoulli Numbers and Polynomials

Since π 2 (2k)! (2k + 2)! ≤ 2 2k 3 (2π) (2π)2k+2 if and only if 3(k + 1)(2k + 1) ≥ π 4 , i.e., if and only if k ≥ 4, by (4.18) we obtain −B8 < B10 < −B12 < . . . . Thus from the values (4.14) we see that |B2k | decreases for k ≤ 3 and increases for k ≥ 3. An asymptotic formula for the growth of |B2k | will be given in (5.13). Since B2k ∈ Q, (4.16) shows that, for each k = 1, 2, 3, . . . , ζ(2k) is a rational multiple of π 2k and hence is transcendental by Lindemann’s theorem (1882) on the transcendence of π. For each k = 1, 2, 3, . . . , ζ(2k + 1) is conjectured to be transcendental, but very little has been proved so far about the arithmetical nature of ζ(2k + 1). Apéry (1979) proved that ζ(3) ∈ / Q, Rivoal (2000) proved that there exist infinitely many positive integers k such that ζ(2k + 1) ∈ / Q (although Rivoal’s theorem is ineffective, i.e., for no explicitly given integer k ≥ 2 the irrationality of ζ(2k + 1) has been proved), and Zudilin (2004) proved that at least one of the four numbers ζ(5), ζ(7), ζ(9) and ζ(11) is irrational.

4.3 Bernoulli Polynomials Definition 4.2 The Bernoulli polynomials are defined by

Bn (x) =

∂ n z ex z ∂z n e z − 1

 (n = 0, 1, 2, . . . ), z=0

i.e., ∞  z ex z zn = B (x) n ez − 1 n! n=0

(|z| < 2π).

(4.19)

From the generating function (4.8) of the Bernoulli numbers we get, for |z| < 2π, z ex z = ez − 1

 ∞

Bn

n=0

zn n!

  ∞ n=0

xn

zn n!

 =

∞  n    n n=0

k=0

k

 Bk x n−k

zn , n!

whence Bn (x) =

n    n k=0

k

Bk x n−k

(n = 0, 1, 2, . . . ),

(4.20)

4.3 Bernoulli Polynomials

45

which can symbolically be written as Bn (x) = (x + B)n . Thus, by (4.20), B0 (x) = 1, B2 (x) = x 2 − x + 16 , B4 (x) = x 4 − 2 x 3 + x 2 − ···············

B1 (x) = x − 21 , B3 (x) = x 3 − 23 x 2 + 1 , B5 (x) = x 5 − 25 x 4 + 30 ···············

1 2 5 3

x, x3 −

1 6

x,

From (4.20) we get Bn (0) = Bn

(4.21)

and Bn (1) =

n    n

k

k=0

Bk ,

whence, by (4.9), (4.11) and (4.12), Bn (1) =

Bn if n = 1 = (−1)n Bn . 1 if n = 1 2

(4.22)

The Bernoulli polynomials satisfy several symmetry properties and functional equations, which are easily obtained from (4.19) or (4.20). By (4.19) we have, for any x, y ∈ C and for |z| < 2π, ∞  n=0

Bn (x + y)

z e x z e yz zn = z = n! e −1

 ∞

 zn n! n=0     n ∞ n   n n−k z Bk (x) y , = k n! n=0 k=0

Bn (x)

n=0

zn n!

  ∞

yn

whence the identity Bn (x + y) =

n    n k=0

k

Bk (x) y n−k .

(4.23)

46

4 Bernoulli Numbers and Polynomials

In particular, by (4.22), Bn (1 + x) =

n    n k=0

k

Bk (1) x n−k =

n    n k=0 k=1

k

Bk x n−k +

=

n n−1 x 2

n    n k=0

k

Bk x n−k + n x n−1 .

Therefore, by (4.20), Bn (x + 1) − Bn (x) = n x n−1 .

(4.24)

Differentiating (4.24) we get, for n ≥ 1,  Bn (x + 1) − Bn (x) = n(n − 1) x n−2 = n Bn−1 (x + 1) − Bn−1 (x) , whence Bn (x + 1) − n Bn−1 (x + 1) = Bn (x) − n Bn−1 (x). Thus the polynomial Bn (x) − n Bn−1 (x) has period 1, and therefore is constant. From (4.20) and (4.21) we obtain Bn (0)



 n Bn−1 = n Bn−1 (0). n−1

=

Hence Bn (x) − n Bn−1 (x) is identically zero, and we get the differentiation formula Bn (x) = n Bn−1 (x)

(n = 1, 2, 3, . . . ).

(4.25)

Another interesting consequence of (4.24) is, for any n, M, N ∈ N with M ≤ N , Bn+1 (N + 1) − Bn+1 (M) =

N  



Bn+1 (k + 1) − Bn+1 (k) =

k=M

N 

(n + 1) k n ,

k=M

whence the formula for the sum of consecutive nth powers: N  k=M

kn =

Bn+1 (N + 1) − Bn+1 (M) . n+1

(4.26)

4.3 Bernoulli Polynomials

47

For M = 1, (4.26) yields 1 N (N + 1) 1 B2 (N + 1) − B2 = (N 2 + N ) = , 2 2 2 1 1 3 3 2 1  12 + 22 + . . . + N 2 = B3 (N + 1) = N + N + N 3 3 2 2 N (N + 1)(2N + 1) = , 6 1 4 1 B4 (N + 1) − B4 = N + 2N 3 + N 2 13 + 23 + . . . + N 3 = 4 4 N 2 (N + 1)2 , = 4  1 1 5 5 1  14 + 24 + . . . + N 4 = B5 (N + 1) = N5 + N4 + N3 − N 5 5 2 3 6 N (N + 1)(2N + 1)(3N 2 + 3N − 1) , = 30

1 + 2 + ... + N =

etc. Remarkably, from the above formulae we see that 13 + 23 + . . . + N 3 = (1 + 2 + . . . + N )2 . From (4.22) and (4.23) we get Bn (1 − x) =

n    n k=0

k

Bk (1)(−x)

n−k

n    n (−1)k Bk · (−1)n−k x n−k = k k=0 n    n Bk x n−k , = (−1)n k k=0

and (4.20) yields the symmetry property of Bn (x) around x = 1/2 : Bn (1 − x) = (−1)n Bn (x).

(4.27)

By (4.23), Bn (x + 1) =

n    n k=0

k

Bk (x) =

n−1    n k=0

k

Bk (x) + Bn (x),

whence, by (4.24), n−1    n k=0

k

Bk (x) = n x n−1 .

(4.28)

48

4 Bernoulli Numbers and Polynomials

Moreover, by (4.19), for |z| < 2π we have ∞ ∞ m−1 m−1 m−1     z e(x+k/m)z zn   k k  zn = = Bn x + Bn x + n! k=0 m m n! ez − 1 n=0 k=0 n=0 k=0

=

m−1 (z/m) e x z z e x z   z/m k z ex z ez − 1 = m z/m = z e z z/m e − 1 k=0 e −1 e −1 e −1

= m

∞ 

Bn (mx)

n=0

∞  z n Bn (mx) (z/m)n = . n! n! m n−1 n=0

Hence we get the multiplication formula for the Bernoulli polynomials: Bn (mx) = m n−1

m−1  k=0

 k Bn x + m

(n = 0, 1, 2, . . . ; m = 1, 2, . . . ).

(4.29)

Chapter 5

Summation Formulae

5.1 Stirling’s Formula for n! For n ∈ N let π/2 sinn x dx. Sn : = 0

Integrating by parts we get, for n ≥ 2,   π/2 + (n − 1) cos2 x sinn−2 x dx Sn = − sinn−1 x cos x π/2

0

0

π/2   = (n − 1) 1 − sin2 x sinn−2 x dx = (n − 1)Sn−2 − (n − 1)Sn , 0

whence the recurrence formula Sn =

n−1 Sn−2 n

(n ≥ 2).

(5.1)

Since S0 = π/2 and S1 = 1, by repeated application of (5.1) we obtain S2n =

(2n − 1)!! π , (2n)!! 2

S2n+1 =

(2n)!! , (2n + 1)!!

© Springer International Publishing Switzerland 2016 C. Viola, An Introduction to Special Functions, UNITEXT - La Matematica per il 3+2 102, DOI 10.1007/978-3-319-41345-7_5

(5.2)

49

50

5 Summation Formulae

with the usual notation (2n − 1)!! = 1 · 3 · 5 · · · (2n − 1), (2n)!! = 2 · 4 · 6 · · · (2n) for the semifactorial N !! of a positive integer N . For 0 ≤ x ≤ π/2 we have 0 ≤ sin x ≤ 1, whence sin x ≥ sin2 x ≥ sin3 x ≥ . . . . Therefore, integrating from 0 to π/2, S1 ≥ S2 ≥ S3 ≥ . . . . Hence, by (5.2), 1 ≥

1 1 S2n+1 2n S2n+1 = 1− = 1+O . ≥ = S2n S2n−1 2n + 1 2n + 1 n

Thus

1 S2n+1 , = 1+O S2n n

i.e., by (5.2),

(2n)!! (2n − 1)!!

2

n 1  2k 2k π 1 = = +O . 2n + 1 2k − 1 2k + 1 2 n k=1

(5.3)

For n → ∞ we get Wallis’ formula: ∞  2 2 4 4 6 6 π 4k 2 = = ··· . 2 2 4k − 1 1 3 3 5 5 7 k=1

Moreover, by (5.3),

(2n)!! (2n − 1)!!

2 =

  1    1  π (2n + 1) 1 + O = πn 1 + O . 2 n n

(5.4)

  2 By Taylor’s formula, (1 + x)1/2 = 1 + 21 x + 1/2 x + . . . = 1 + O(x) (x → 0). 2 Hence   1 1/2 1 1+O (n → ∞). = 1+O n n Also (2n − 1)!! (2n)!! = (2n)! and (2n)!! = 2n n!, whence (2n)!!2 22n n!2 (2n)!! = = . (2n − 1)!! (2n)! (2n)!

5.1 Stirling’s Formula for n!

51

Thus from (5.4) we get  1  √  22n n!2 = πn 1 + O (2n)! n

(n → ∞).

(5.5)

We can now prove the following Theorem 5.1 (Stirling’s ‘elementary’ formula) For n → ∞ we have  1 √ 1 log n! = n + log n − n + log 2π + O 2 n

(5.6)

and √

n! =

2πn

 n n  e

1+O

 1  n

.

(5.7)

Proof Plainly (5.6) and (5.7) are equivalent since, by Taylor’s formula, exp O( n1 ) = 1 + O( n1 ) and log(1 + O( n1 )) = O( n1 ). We prove (5.6). Let k ≥ 1 be an integer. Then k+1/2 

log x dx =

1/2 

 log(k − x) + log(k + x) dx =

0

k−1/2

(5.8)

0

1/2 =

1/2 log(k 2 − x 2 ) dx



x2 log k 2 + log 1 − 2 dx = log k − Rk , k

0

where

1/2 1 x2 Rk = − log 1 − 2 dx = O 2 k k

(5.9)

0

    since − log 1 − x 2 /k 2 = O 1/k 2 uniformly for 0 ≤ x ≤ 1/2. Thus the series ∞ Rk converges. From (5.8) we get k=1

log n! =

n

k=1

k+1/2  n ∞ ∞

log k = log x dx + Rk − Rk . k=1 k−1/2

k=1

k=n+1

(5.10)

52

5 Summation Formulae

By (5.9)

∞



∞ 1 , k2 k=n+1

Rk = O

k=n+1

and clearly

 ∞

1 dx 1 < = . 2 2 k x n k=n+1 ∞

n

Therefore, by (5.10), n+1/2 

log n! =

log x dx + C1 + O 1/2

 = n+  = n+  = n+

1 n

 1 1 1 log n + − n + C2 + O 2 2 n  1 1  1  log n + log 1 + − n + C2 + O 2 2n n 1  1  1  1 log n + , + O 2 − n + C2 + O 2 2n n n

for suitable constants C1 and C2 . It follows that 1  1 log n − n + C + O , log n! = n + 2 n

(5.11)

where C is a constant. To compute C we take the logarithm of (5.5). We get 2n log 2 + 2 log n! − log(2n)! = log

√

π + (1/2) log n + O(1/n).

Here we replace log n! and log(2n)! by the quantities obtained from the right-hand side of (5.11). This easily yields C = (1/2) log 2 + log

√

π + O(1/n).

Since C is constant, for n → ∞ we get C = log Then (5.6) follows from (5.11) and (5.12).

√

2π.

(5.12) 

In Theorem 6.4 and Corollary 6.2 we shall generalize and improve (5.6) by proving Stirling’s formula for the Euler gamma-function, with an asymptotic expansion of the remainder term.

5.1 Stirling’s Formula for n!

53

From (4.17) and (5.7) we obtain an asymptotic formula for the Bernoulli numbers B2k . Since ∞

n −2k
λ1 be a real number, and let f (t) be a complex valued function, continuously differentiable in the real interval λ1 ≤ t ≤ x. Let a1 , a2 , a3 , . . . be an arbitrary sequence of complex numbers. Then

n λn ≤x

an f (λn ) =

n λn ≤x

an f (x) −

x λ1

n λn ≤t



an f  (t) dt.

(5.14)

54

5 Summation Formulae

Proof Let N be such that λ N ≤ x < λ N +1 . Then

an f (x) −

n λn ≤x

=

an f (λn ) =

n λn ≤x

N x

N

  an f (x) − f (λn )

n=1

an f  (t) dt

n=1 λ n

λ2 =

λ3



a1 f (t) dt + λ1

x



a1 f (t) dt + · · · + λ2

a1 f  (t) dt

λN

λ3



x

a2 f (t) dt + · · · +

+ λ2

a2 f  (t) dt

λN

+ ················· x + a N f  (t) dt λN

λ2 =

λ3  a1 f (t) dt + (a1 + a2 ) f  (t) dt + · · ·

λ1

λ2

x +

(a1 + a2 + · · · + a N ) f  (t) dt

λN

=

x λ1



an f  (t) dt.

n λn ≤t

 The meaning of Theorem 5.2 is that, by introducing twice on the right-hand side of (5.14) suitable information on the partial sum an , one gets information on the full sum a n f (λn ). We give an instance of this by proving an asymptotic formula 1/n, which also allows us to define Euler’s constant γ. for the sum n≤x

We recall that x denotes the integer part of the real number x, and {x} = x − x

denotes the fractional part of x. In Theorem 5.2 we choose λn = n, an = 1 (n = 1, 2, 3, . . . ), and f (t) = 1/t. By (5.14) we get

5.2 Partial Summation and Euler’s Constant

55

x x x

x

t

1 t

1 dt x

1 1 2 = dt = + + n x t x t2 n=1 n=1 n=1 1

1

{x} + = 1− x

x

1 {t} − 2 dt. t t

1

Since {t}/t 2 < 1/t 2 , the integral

∞ 1

({t}/t 2 ) dt converges. Hence

 x

{x} 1 {t} = log x − +γ + dt, n x t2 n=1 ∞

(5.15)

x

∞ 2 where 1 ({t}/t  ∞ ) dt 2is Euler’s constant. For x > 0 we have {x}/x < 1/x  ∞γ = 1 − 2 and x ({t}/t ) dt < x dt/t = 1/x. By (5.15) we obtain the asymptotic formula x

1

1 = log x + γ + O n x n=1

(x → +∞).

(5.16)

Since log x = log x + log(1 + {x}/ x ) = log x + O(1/ x ), writing x = N we get from (5.16) N 1

1 = log N + γ + O n N n=1

(N → +∞),

(5.17)

whence ∞ γ = 1−

{t} dt = lim N →∞ t2

1

N n=1

1 − log N n

= 0. 577215 . . . .

(5.18)

We shall prove two further integral representations ((5.23) and (5.24)) of Euler’s constant γ. First of all we remark that, for any positive integer N and for 0 < t < N , 1+

t t t2 t2 t < 1+ + + + . . . < 1 + + ..., N N 2! N 2 N N2

i.e., 1+

 t t −1 < et/N < 1 − , N N

56

5 Summation Formulae

whence (1 + t/N ) N < et and (1 − t/N ) N < e−t . Therefore 0 < e

−t



 t N t N −t t 1−e 1− − 1− = e N N  2 N

t < e−t 1 − 1 − 2 . N 

(5.19)

Comparing the graphs of the functions y = (1 − x) N and y = 1 − N x we see that 1 − N x < (1 − x) N for all 0 < x ≤ 1/N . Substituting x = t 2 /N 2 we get  √ t2 t 2 N for 0 < t ≤ N , < 1− 1− 2 N N √ and (5.20) is trivially true for N < t < N . Hence, by (5.19) and (5.20),  t N t 2 −t e for 0 < t < N . < 0 < e−t − 1 − N N

(5.20)

(5.21)

With the substitution 1 − t/N = u we get N 0

1 1    t  N dt 1 − uN 1− 1− 1 + u + u 2 + · · · + u N −1 du = du = N t 1−u 0

0

= 1+

1 1 1 + + ··· + . 2 3 N

Thus

N N N 

1 1 − e−t t  N dt e−t − 1 − = + dt n N t t n=1 0

0

N 1 N N −t  1 − e−t dt e t  N dt −t e − 1− + dt + − dt. = N t t t t 0

0

1

1

It follows that

N 1 N −t N 

1 1 − e−t e t  N dt −t e − 1− − log N = + dt − dt. n N t t t n=1 0

0

1

(5.22)

5.2 Partial Summation and Euler’s Constant

57

By (5.21),

N N  t  N dt 1 −t e − 1− 0 < te−t dt < N t N 0

0


0 and gcd(m, n) = 1). However, if α ∈ R, α > 0, unless otherwise specified one conventionally takes arg α = 0 in (6.2), so that αβ = exp(β log α) with the elementary real value of log α. In particular this yields  β α  = | exp(β log α)| = exp Re(β log α)   = exp (Re β) log α = αRe β

© Springer International Publishing Switzerland 2016 C. Viola, An Introduction to Special Functions, UNITEXT - La Matematica per il 3+2 102, DOI 10.1007/978-3-319-41345-7_6

(6.3) (α > 0).

67

68

6 The Euler Gamma-Function

  This applies to t z−1 = exp (z − 1) log t in (6.1), where t ∈ (0, +∞), as well as to the power n −s = exp(−s log n) (n = 1, 2, . . . ) in the definition (4.3) of the Riemann zeta-function. By (6.3) we have |e−t t z−1 | = e−t t Re z−1 , whence for any z ∈ C with Re z > 0 the integral (6.1) converges absolutely both at t = 0 and at t = +∞. Moreover +∞ 1 +∞ −t z−1 −t Re z−1 |e t | dt = e t dt + e−t t Re z−1 dt. 0

0

(6.4)

1

For z in any vertical strip a ≤ Re z ≤ b with a > 0 and b < +∞, we have e−t t Re z−1 ≤ t a−1 in the first integral on the right-hand side of (6.4), and e−t t Re z−1 ≤ e−t t b−1 in the second integral. Since 1 t

a−1

+∞ dt < +∞ for a > 0 and e−t t b−1 dt < +∞,

0

1

the decomposition (6.4) shows that the integral (6.1) converges uniformly in any strip 0 < a ≤ Re z ≤ b and hence is a regular function of z in the strip. Since a > 0 can be taken arbitrarily small, and b arbitrarily large, Γ (z) is regular in the open half-plane Re z > 0. For Re z > 0 we have, by (6.3), |t z | = t Re z → 0

(t → 0+).

Thus from (6.1) we get, integrating by parts, 1 1  −t z +∞ e t + Γ (z) = t=0 z z

+∞ 1 e−t t z dt = Γ (z + 1), z

(6.5)

0

whence the first functional equation for the gamma-function: Γ (z + 1) = z Γ (z).

(6.6)

Since Γ (z + 1) is regular for Re z > −1, (6.5) yields the analytic continuation of Γ (z) in the half-plane Re z > −1 and shows that Γ (z) has a simple pole at z = 0 with residue +∞ e−t dt = 1. Γ (1) = 0

Plainly this argument can be iterated. By n-fold application of (6.5) we get

6.1 Eulerian Integrals

Γ (z) =

69

1 1 Γ (z + 1) = Γ (z + 2) = · · · z z(z + 1) =

(6.7)

1 Γ (z + n), z(z + 1) · · · (z + n − 1)

which yields the analytic continuation of Γ (z) for Re z > −n. Since n is arbitrary, we conclude that Γ (z) is analytically continued in the whole C to a meromorphic function, regular in C \ {0, −1, −2, . . . } and having simple poles at z = 0, −1, −2, . . . . Moreover, changing n to n + 1 in (6.7), Γ (z) =

Γ (z + n + 1) 1 · , z + n z(z + 1) · · · (z + n − 1)

whence Res Γ (z) =

z=−n

(−1)n Γ (1) = (−n)(−n + 1) · · · (−1) n!

(n = 0, 1, 2, . . . ).

(6.8)

Also, by repeated application of (6.6) we get Γ (n + 1) = n(n − 1) · · · 2 · 1 · Γ (1), i.e., Γ (n + 1) = n! (n = 0, 1, 2, . . . ). (6.9) The Euler beta-function is a function of two complex variables, defined by 1 B(x, y) =

t x−1 (1 − t) y−1 dt

(x, y ∈ C, Re x > 0, Re y > 0)

(6.10)

0

and called by Legendre eulerian integral of the first kind. The beta-function can easily be related to the gamma-function, as follows. With the substitution t = u/(1 + u) we get ∞ B(x, y) = 0

u x−1 du. (1 + u)x+y

(6.11)

For any fixed u ∈ (0, +∞), with the substitution (1 + u)v = ϑ we obtain ∞ 0

e−(1+u)v v x+y−1 dv =

∞ 0

e−ϑ

 ϑ x+y−1 dϑ 1+u 1+u

1 = (1 + u)x+y

∞ 0

e−ϑ ϑx+y−1 dϑ =

Γ (x + y) . (1 + u)x+y

70

6 The Euler Gamma-Function

Multiplying by u x−1 and integrating in 0 < u < +∞, by (6.11) we find ∞

∞ du

0

e

−(1+u)v

∞ u

v

x−1 x+y−1

dv = Γ (x + y)

0

0

u x−1 du (1 + u)x+y

(6.12)

= Γ (x + y)B(x, y), where the repeated integral on the left-hand side of (6.12) can be inverted by absolute convergence. For any fixed v ∈ (0, +∞), with the substitution uv = τ we get ∞ e

−uv

∞ u

x−1

du =

0

e−τ

0

 τ x−1 dτ = v −x Γ (x). v v

Hence, multiplying by e−v v x+y−1 and integrating in 0 < v < +∞, ∞

∞ dv

0

e−(1+u)v u x−1 v x+y−1 du

0

∞ = Γ (x)

e−v v x+y−1 v −x dv = Γ (x)Γ (y).

0

Comparing with (6.12) we conclude that B(x, y) =

Γ (x)Γ (y) , Γ (x + y)

(6.13)

which yields the analytic continuation of B(x, y) outside Re x > 0 and Re y > 0.

6.2 Euler’s Limit Formula and Consequences In this section we prove further important representations of Γ (z) and functional equations, which we base upon Euler’s limit formula, stated in the following Theorem 6.1 (Euler) For any z ∈ C \ {0, −1, −2, . . . }, Γ (z) = lim

n→∞

n z n! . z(z + 1)(z + 2) · · · (z + n)

Proof We first assume Re z > 0. From (6.1) we get, for positive integer n,

(6.14)

6.2 Euler’s Limit Formula and Consequences

71

n  t n z−1 1− lim t dt n→∞ n

(6.15)

0

n = lim

e

n→∞

  n

 t n z−1 −t e − 1− t dt − dt n

−t z−1

t

0

0

 n

 t n z−1 −t e − 1− t = Γ (z) − lim dt. n→∞ n 0

By (5.21),  n

 n 2 n n     t t 1 −t z−1 −t Re z−1  e − 1− t e t dt  ≤ dt = e−t t Re z+1 dt  n n n 0

0

0

∞ 1 ≤ e−t t Re z+1 dt → 0 n

(n → ∞).

0

Therefore, by (6.15), n  t n z−1 Γ (z) = lim 1− t dt. n→∞ n

(6.16)

0

With the substitution t/n = τ we get n 

t n z−1 1− t dt = n z n

0

1 (1 − τ )n τ z−1 dτ .

(6.17)

0

By repeated integrations by parts, since Re z > 0, 1 (1 − τ ) τ n

z−1

0

n dτ = z

1 (1 − τ )

n−1

0

= ··· =

n n−1 τ dτ = z z+1

n! z(z + 1) · · · (z + n − 1)

1 (1 − τ )n−2 τ z+1 dτ

z

1 τ z+n−1 dτ = 0

0

n! . z(z + 1) · · · (z + n)

72

6 The Euler Gamma-Function

Thus, by (6.17), n  t n z−1 n z n! , 1− t dt = n z(z + 1) · · · (z + n) 0

and Euler’s limit formula (6.14) follows from (6.16). If Re z ≤ 0, let k ∈ N be such that −(k + 1) < Re z ≤ −k with z = −k. From (6.7) we get 1 Γ (z + k + 1). Γ (z) = z(z + 1) · · · (z + k) Since Re(z + k + 1) > 0, by the previous argument we have 1 n z+k+1 n! lim z(z + 1) · · · (z + k) n→∞ (z + k + 1)(z + k + 2) · · · (z + k + n + 1)

 n z n! n k+1 · = lim n→∞ z(z + 1) · · · (z + n) (z + n + 1) · · · (z + n + k + 1) n z n! n k+1 lim = lim n→∞ z(z + 1) · · · (z + n) n→∞ (n + z + 1) · · · (n + z + k + 1) n z n! . = lim n→∞ z(z + 1) · · · (z + n)

Γ (z) =

 Corollary 6.1 (Euler) For any z ∈ C \ {0, −1, −2, . . . }, Γ (z) =

∞ 1  1 z  z −1 1+ 1+ . z n=1 n n

(6.18)

Proof Since N −1 2 3 4 N 1 n + 1 = = ··· = N, 1+ n n 1 2 3 N −1 n=1

N −1  n=1

from (6.14) we get Nz N →∞ z(1 + z)(1 + z/2) · · · (1 + z/N )

N  N −1  1 z  z −1 1 lim 1+ 1+ = z N →∞ n=1 n n=1 n

Γ (z) = lim

(6.19)

6.2 Euler’s Limit Formula and Consequences

73

=

N  1 z  z −1 1 1+ 1+ lim z N →∞ n=1 n n

=

∞ 1 z  z −1 1  1+ 1+ . z n=1 n n

 From (6.6) and (6.18), z Γ (z) Γ (1 − z) = Γ (1 + z) Γ (1 − z) (6.20) ∞ 1 1+z  1 1−z  1 + z −1  1 − z −1 1  1+ 1+ 1+ 1+ = 2 1 − z n=1 n n n n

 ∞ 1 2 z 2 −1 1 2  1  1 + 1 + = − 2 1 − z 2 n=1 n n n

 ∞ −1 z2 1 1− = 2 1 − z n=1 (n + 1)2  ∞

z 2 −1 1− 2 = . n n=1 By (4.1) and (6.20),

πz , sin(πz)

z Γ (z) Γ (1 − z) =

whence the second functional equation for the gamma-function: Γ (z) Γ (1 − z) =

π , sin(πz)

(6.21)

also called Euler’s reflection formula. In particular, for z = 1/2, (6.21) yields Γ (1/2)2 = π. By (6.1), Γ (z) > 0 for z > 0. Therefore Γ (1/2) =

√

π.

(6.22)

From (6.6) and (6.22) we obtain, by induction on n,  (2n − 1)!! √ 1 = π Γ n+ 2 2n

(n = 1, 2, 3, . . . ).

(6.23)

We now prove the third functional equation for the gamma-function, i.e., the following

74

6 The Euler Gamma-Function

Theorem 6.2 (Gauss’ multiplication formula)  k n−1 1 = (2π) 2 n 2 −nz Γ (nz) Γ z+ n k=0

n−1

(n = 1, 2, 3, . . . ).

(6.24)

Proof We write (6.14) in the form m z (m − 1)! m lim m→∞ z(z + 1) · · · (z + m − 1) m→∞ z + m m z (m − 1)! . = lim m→∞ z(z + 1) · · · (z + m − 1)

Γ (z) = lim

It follows that n−1 n nz−1  k = Γ z+ Γ (nz) k=0 n

n nz−1 (nm)nz (nm − 1)! lim m→∞ nz(nz + 1) · · · (nz + nm − 1) n−1 m z+k/n (m − 1)! × lim m→∞ (z + k/n)(z + k/n + 1) · · · (z + k/n + m − 1) k=0 n nz−1 m nz+(n−1)/2 (m − 1)!n n nm m→∞ (nm)nz (nm − 1)! (n−1)/2 m (m − 1)!n n nm−1 . = lim m→∞ (nm − 1)! = lim

Thus the function

n−1 n nz−1  k Γ z+ Γ (nz) k=0 n

is independent of z, and hence equals identically its value at z = 1/n. Therefore n−1 n−1 n nz−1  k = Γ z+ Γ (k/n). Γ (nz) k=0 n k=1

(6.25)

By (6.21),

n−1 k=1

2 Γ (k/n)

=

n−1 k=1

Γ (k/n) Γ (1 − k/n) 

n−1 kπ −1 kπ −1 n−1 π sin = π sin . = n n k=1 k=1 n−1



6.2 Euler’s Limit Formula and Consequences

75

Hence, by (6.25),

n−1  n−1 n nz−1  k kπ −1/2 (n−1)/2 = π Γ z+ sin . Γ (nz) k=0 n n k=1

(6.26)

Since x n − 1 = (x − 1)(x n−1 + x n−2 + · · · + 1), the roots of the polynomial x n−1 + x n−2 + · · · + 1 are the nth roots of 1 different from 1. Therefore x n−1 + x n−2 + · · · + 1 =

n−1



x − e2kπi/n



k=1

whence, for x = 1, n =

n−1

n−1      − ekπi/n ekπi/n − e−kπi/n 1 − e2kπi/n =

k=1

=

k=1

n−1 

− 2i ekπi/n sin

k=1

= 2n−1

n−1 k=1

sin

kπ π = 2n−1 (−i)n−1 ei n n

n(n−1) 2

n−1

sin

k=1

kπ n

kπ . n

Consequently n−1

sin

k=1

n kπ = n−1 , n 2

(6.27) 

and (6.24) follows from (6.26) and (6.27). In the special case n = 2, (6.24) is the Legendre duplication formula: Γ (z) Γ (z + 1/2) =

√ 1−2z π2 Γ (2z),

(6.28)

which for z = 1/2 yields again (6.22). Theorem 6.3 (Weierstrass) The Weierstrass factorization formula ∞  1 z −z/n 1 = = eγz e 1+ , Γ (z + 1) z Γ (z) n n=1

where γ is Euler’s constant (5.18), holds for all z ∈ C.

(6.29)

76

6 The Euler Gamma-Function

Proof By (6.19) and (5.18),  1 = lim z(1 + z)(1 + z/2) · · · (1 + z/N )e−z log N N →∞ Γ (z)

= z lim (1 + z)e−z (1 + z/2)e−z/2 · · · (1 + z/N )e−z/N N →∞





N 1 − log N × exp z n n=1



N N  z −z/n 1 − log N lim e = z exp z lim 1+ N →∞ N →∞ n n n=1 n=1 = z eγz

∞  z −z/n e 1+ , n n=1



and (6.29) follows from (6.6).

The sequence z n = −n (n = 1, 2, 3, . . . ) has exponent of convergence β = 1, and the least integer satisfying (3.9) for |z n | = n is p = 1. Hence the infinite product ∞  n=1

1+

z −z/n e n

in (6.29) is indeed the Weierstrass canonical product (3.11) for the sequence z n = −n and, by Theorem 3.5, is an entire function of order 1. Thus, by (6.29), 1/Γ (z + 1) is an entire function with zeros z = −1, −2, −3, . . . , corresponding to the poles of Γ (z + 1) = zΓ (z). With the notation of Corollary 3.2, for the entire function (6.29) we have G(z) = γz, whence α = β = p = q = 1. Since (zΓ (z))−1 is entire, we get Γ (z) = 0 for all z ∈ C,

(6.30)

which is also an easy consequence of (6.21).

6.3 Stirling’s Formula Let R− = (−∞, 0 ] denote the set of real numbers ≤0, and let log Γ (z) = log |Γ (z)| + i arg Γ (z)

(6.31)

be the value of the logarithm of Γ (z) with arg Γ (z) = 0 for z > 0, and extended by continuity for z ∈ C \ R− . By (6.30), and since the poles of Γ (z) are in R− , in

6.3 Stirling’s Formula

77

the simply connected open set C \ R− the function (6.31) has no branch points and therefore is one-valued and regular. We prove the following Theorem 6.4 (Stirling’s formula) For any integer r ≥ 0 and for any ε > 0, log Γ (z) =



z−

√ 1 log z − z + log 2π (6.32) 2

 r

1 1 B2h + Or,ε + (2h − 1)2h z 2h−1 |z|2r +1 h=1

as z → ∞ in the sector −π + ε ≤ arg z ≤ π − ε. Here B2h is the (2h)th Bernoulli number, log Γ (z) is as in (6.31), i.e. continuous in C \ R− with arg Γ (z) = 0 for z > 0, and similarly log z is the principal logarithm, i.e., log z = log |z| + i arg z with −π < arg z < π. In particular we get the following improvement upon (5.6):  √ 1 log n − n + log 2π (6.33) log n! = n + 2 r  1

1 B2h (n → ∞). + O + r (2h − 1)2h n 2h−1 n 2r +1 h=1 Proof By (6.6) and (6.9), n! = Γ (n + 1) = n Γ (n) whence log n! = log Γ (n) + log n, so that (6.33) follows from (6.32) with z = n → ∞. Fix z ∈ C \ R− , and let N be a positive integer. We have 

Nz N! z−1 = z log N − log(z + N ) − . log 1 + z(z + 1) · · · (z + N ) k k=1 N

log

We apply to the last sum the Euler–MacLaurin summation formula (5.26) with M = 1, f (t) = log(1 + (z − 1)/t) = log(t + z − 1) − log t, and q = 2r + 1. A straightforward computation yields   Nz N! 1 z−1 = z− log z − N log 1 + z(z + 1) · · · (z + N ) 2 N    z−1 z 1 log 1 + + log 1 − − z− 2 N z+N 

r r

1 1 B2h B2h 1 − − 2h−1 + 2h−1 2h−1 (2h − 1)2h (z + N − 1) N (2h − 1)2h z h=1 h=1

log

−

r

h=1

1 B2h − (2h − 1)2h 2r + 1

N 1

B2r +1 ({t}) 1 dt + (z + t − 1)2r +1 2r + 1

N 1

B2r +1 ({t}) dt. t 2r +1

78

6 The Euler Gamma-Function

Making in this formula N → ∞, and assuming r ≥ 1, we get by (6.14)  1 log Γ (z) = z − log z − z + Cr (6.34) 2 ∞  r

1 B2h B2r +1 ({t}) 1 − dt, + 2h−1 (2h − 1)2h z 2r + 1 (z + t − 1)2r +1 h=1 1

where Cr = 1 −

r

h=1

1 B2h + (2h − 1)2h 2r + 1

∞

B2r +1 ({t}) dt t 2r +1

1

is independent of z, and the integrals are absolutely convergent since r ≥ 1 and B2r +1 ({t}) is bounded. Substituting t − 1 = τ and integrating by parts we obtain, using (4.25), 1 − 2r + 1

∞ 1

B2r +1 ({t}) 1 dt = − (z + t − 1)2r +1 2r + 1

∞ 0

B2r +1 ({τ }) dτ (z + τ )2r +1

B2r +2 1 1 = − 2r +1 (2r + 1)(2r + 2) z 2r + 2

∞ 0

(6.35)

B2r +2 ({τ }) dτ . (z + τ )2r +2

Let −π + ε ≤ arg z ≤ π − ε, and let ϑ = π − | arg z |, whence ε ≤ ϑ ≤ π. By applying Carnot’s theorem to the triangle of vertices 0, τ and z + τ we see that |z + τ |2 = |z|2 + τ 2 − 2|z|τ cos ϑ ≥ |z|2 + τ 2 − 2|z|τ cos ε. Therefore ∞ 0

B2r +2 ({τ }) dτ (z + τ )2r +2

∞ 0

dτ ≤ |z + τ |2r +2

∞ 0

(|z|2

+

τ2

dτ . − 2|z|τ cos ε)r +1

With the substitution τ = |z|(cos ε + u sin ε) the last integral becomes 1 (|z| sin ε)2r +1

+∞ − cot ε

du 1 < 2 r +1 (1 + u ) (|z| sin ε)2r +1

∞ −∞

du . (1 + u 2 )r +1

Hence, by (6.35), 1 − 2r + 1

∞ 1

B2r +1 ({t}) 1 dt r,ε 2r +1 for | arg z | ≤ π − ε, 2r +1 (z + t − 1) |z|

6.3 Stirling’s Formula

79

and (6.34) yields, for r ≥ 1 and | arg z | ≤ π − ε, log Γ (z) =

Since



z−

1 |z|2h−1

1 logz − z + Cr (6.36) 2

 r

1 1 B2h . + O + r,ε (2h − 1)2h z 2h−1 |z|2r +1 h=1



1 for h = 1, . . . , r, r + 1 and |z| → ∞, |z|

by (6.36) we get log Γ (z) =



 1 1 log z − z + Cr + Or,ε . z− 2 |z|

(6.37)

In particular, for z = n → ∞, 1  1 log n − n + Cr + Or . log n! = log Γ (n) + log n = n + 2 n Comparing this asymptotic formula with (5.6) we get, for all r ≥ 1, Cr = log

√ 2π,

(6.38)

and (6.32) follows from (6.36) and (6.38). Moreover, by (6.37) with a fixed r ≥ 1 and by (6.38), log Γ (z) =



z−

 √ 1 1 log z − z + log 2π + Oε , 2 |z|

which shows that (6.32) holds also for r = 0.

(6.39) 

We incidentally remark that (6.38) can also be obtained by taking the logarithm of the Legendre duplication formula (6.28), and then replacing log Γ (z), log Γ (z + 1/2) and log Γ (2z) with the values given by (6.37). With the same meaning as in (5.31), Stirling’s formula (6.32) can symbolically be written as log Γ (z) ∼



√ 1 1 B2h log z − z + log 2π + , 2h−1 2 (2h − 1)2h z h=1 ∞

z−

where, by (5.13), the series diverges for every z. Stirling’s formula (6.32) can be further generalized as follows.

80

6 The Euler Gamma-Function

Corollary 6.2 For any integer s ≥ 0, for any η > 0 and for any arbitrarily small ε > 0, √   log Γ (z + α) = z + B1 (α) log z − z + log 2π +

s

(−1)k+1

k=1



(6.40) 

1 Bk+1 (α) 1 + Os,η,ε k(k + 1) z k |z|s+1

uniformly for z, α ∈ C satisfying |α| ≤ η, |z| ≥ |α| + ε, | arg(z + α) | ≤ π − ε, where B1 (α), B2 (α), B3 (α), . . . are the Bernoulli polynomials. Proof Owing to (4.9), we can write (6.32) in the form log Γ (z) =



z−

√ 1 log z − z + log 2π (6.41) 2

 s

1 Bk+1 1 (−1)k+1 + Os,ε + k s+1 k(k + 1) z |z| k=1

for any integer s ≥ 0, as z → ∞ in the sector | arg z | ≤ π − ε. In (6.41) we change z to z + α. We get   √ α 1  log z + log 1 + − z − α + log 2π log Γ (z + α) = z + α − 2 z 

s 

α −k B 1 k+1 . z −k 1 + (−1)k+1 +O + k(k + 1) z |z|s+1 k=1 Since |α| ≤ η and |α/z| ≤ 1 − ε/(η + ε), the Taylor expansions of log(1 + α/z) and (1 + α/z)−k yield s

√ αm+1 1 log z − z + log 2π − α + (−1)m 2 (m + 1)z m m=0  s s s

 m

Bk+1 −k αm−k 1 m+1 α k+1 + α− (−1) + (−1) 2 m=1 mz m k(k + 1) m=k m − k z m k=1 

1 +O |z|s+1 s 

√ 1 αm+1 log z − z + log 2π + = z+α− (−1)m 2 (m + 1)z m m=1

 m s s m+1

−k Bk+1 αm−k − αm /2 1 m+1 α k+1 + (−1) + (−1) m − k k(k + 1) mz m z m k=1 m=1 m=1 

1 +O |z|s+1

log Γ (z + α) =



z+α−

6.3 Stirling’s Formula

=



81

m+1 s

√ − (m + 1)αm /2 1 m+1 α log z − z + log 2π + z+α− (−1) 2 m(m + 1) m=1

  

m

−k Bk+1 αm−k 1 1 . (−1)m−k + O + m − k k(k + 1) z m |z|s+1 k=1

Since (−1)m−k

−k m−k



=

m−1 m−k

 =

 m−1 , k−1

we get





 −k 1 m(m + 1) m − 1 1 m+1 (−1)m−k = = . k(k + 1) m − k m(m + 1) k(k + 1) k − 1 m(m + 1) k + 1 Therefore, using (4.20), √ 1 log z − z + log 2π z+α− 2   s m

m+1

1 (−1) m+1 m m+1 m+1 m−k Bk+1 α α α + − + m k + 1 m(m + 1) 2 z m=1 k=1 

1 +O |z|s+1 √   = z + B1 (α) log z − z + log 2π 

s

Bm+1 (α) 1 1 . (−1)m+1 + O + m(m + 1) z m |z|s+1 m=1 

log Γ (z + α) =

 Note that in the special case α = 0 (6.40) becomes (6.41), i.e., (6.32). Corollary 6.3 Let x1 , x2 ∈ R be fixed, x1 ≤ x2 . For x + i y in the vertical strip x1 ≤ x ≤ x2 , the following asymptotic formula holds: |Γ (x + i y)| =

√



 1 1 π 2π |y|x− 2 e− 2 |y| 1 + Ox1 ,x2 |y|

(y → ±∞).

(6.42)

Proof In (6.40) we take s = 0, η = max{|x1 |, |x2 |}, α = x, z = i y. For sufficiently large |y| we get

 √ 1 1 log(i y) − i y + log 2π + O 2 |y|

   √ π 1 1 , = x − + i y log |y| + i sgn y − i y + log 2π + O 2 2 |y|

log Γ (x + i y) =



x + iy −

82

6 The Euler Gamma-Function

whence   log |Γ (x + i y)| = Re log Γ (x + i y)

  √ π 1 1 log |y| − |y| + log 2π + O . = x− 2 2 |y| 

Taking exponentials we get (6.42).

6.4 The Psi-Function The logarithmic derivative of the gamma-function is traditionally denoted by ψ: ψ(z) : =

Γ  (z) , Γ (z)

(6.43)

and is called by some authors the digamma-function. Some formulae for Γ (z) turn into simpler formulae for ψ(z). By taking logarithms and then differentiating, the functional equations (6.6), (6.21) and (6.24) for the gamma-function yield, respectively,

and ψ(nz) =

1 ψ(1 + z) = ψ(z) + , z

(6.44)

ψ(1 − z) = ψ(z) + π cot(πz)

(6.45)

n−1

k 1  ψ z+ + log n n n k=0

(n = 1, 2, 3, . . . ).

(6.46)

By (6.30), and since the poles z = 0, −1, −2, . . . of Γ (z) are all simple, ψ(z) is regular in C \ {0, −1, −2, . . . }, with simple poles at z = 0, −1, −2, . . . , all with residue −1. Taking the logarithm of (6.29) we get log Γ (z) = − log z − γz +

 z − log 1 + , n n

∞ 

z n=1

(6.47)

6.4 The Psi-Function

83

where γ is Euler’s constant. Differentiating (6.47) we obtain 1 1 z 1 1 + − = −γ − + , z n=1 n n+z z n=1 n(n + z) ∞

ψ(z) = −γ −

∞

(6.48)

where term-by-term differentiation is justified by Theorems 2.1 and 2.2. By (6.44) and (6.48), ψ(1 + z) = −γ +

∞

n=1

z . n(n + z)

(6.49)

For any a, b ∈ C \ {0, −1, −2, . . . } we have, by (6.48), ∞ ∞ 

1 1  1 1 1 1 ψ(b) − ψ(a) = − + = , − − a b n=1 n + a n+b n+a n+b n=0 (6.50) whence, for a = b, ∞

ψ(b) − ψ(a) 1 = . b−a (n + a)(n + b) n=0 Thus, using the function ψ, one can sum a series of the type ∞

n=0

1 P(n)

(6.51)

where P is a polynomial of degree 2 with distinct roots ∈ / N. Since, by (6.48), ψ  (z) =

∞

∞

1 1 1 + = , 2 2 z (n + z) (n + z)2 n=1 n=0

(6.52)

for a = 0, −1, −2, . . . we get ψ  (a) =

∞

n=0

1 , (n + a)2

which yields the sum of (6.51) if P has a double root. Plainly this argument can be iterated to sum (6.51) if P is a polynomial of degree > 2. For instance, if deg P = 3 and the roots of P are distinct and ∈ / N, we get

84

6 The Euler Gamma-Function ∞

n=0

∞ 1  1 1 1 1 = − (n + a)(n + b)(n + c) b−a n+a n+b n+c n=0

 ∞

1 1 − (n + a)(n + c) (n + b)(n + c) n=0 n=0

 ψ(c) − ψ(a) ψ(c) − ψ(b) 1 − = b−a c−a c−b ψ(b) ψ(c) ψ(a) + + . = (b − a)(a − c) (c − b)(b − a) (a − c)(c − b) 1 = b−a

∞

Since Γ (1) = 1, from (6.43) and (6.48) we get ψ(1) = Γ  (1) = −γ − 1 +

∞ 

1 n=1

and clearly

∞ 

1

n

n=1

Therefore

−

n

−

1 , n+1

1 = 1. n+1

ψ(1) = Γ  (1) = −γ.

(6.53)

From (6.44) and (6.53) we obtain, by induction on n, ψ(n) = −γ +

n−1

1 k=1

k

(n = 1, 2, 3, . . . ).

(6.54)

We now apply (3.28) to the entire function (6.29). For F(z) = 1/Γ (z + 1) we get F  (z) Γ  (z + 1) = − = −ψ(z + 1). F(z) Γ (z + 1) Hence, by (3.28), ∞

n=1

(−n)−k = (−1)k ζ(k) =

1 ψ (k−1) (1) (k − 1)!

(k = 2, 3, 4, . . . )

(6.55)

where ζ is the Riemann zeta-function. By (6.53) and (6.55) we get the Taylor expansion ∞

(−1)k ζ(k)z k−1 (|z| < 1). (6.56) ψ(1 + z) = −γ + k=2

6.4 The Psi-Function

85

We remark that the Taylor series (6.56) easily yields another proof of Euler’s formulae (4.16) for ζ(2k). From (6.44) and (6.45) we have ψ(1 + z) − ψ(1 − z) =

1 − π cot(πz), z

whence, by (4.15), ψ(1 + z) − ψ(1 − z) =

∞

z 2k−1 (−1)k−1 (2π)2k B2k (2k)! k=1

(|z| < 1).

(6.57)

Changing z to −z in (6.56), ψ(1 − z) = −γ −

∞

ζ(k)z k−1 .

k=2

Subtracting this formula from (6.56) we get ψ(1 + z) − ψ(1 − z) =

∞ ∞

  1 + (−1)k ζ(k)z k−1 = 2 ζ(2k)z 2k−1 . k=2

(6.58)

k=1

Comparing (6.57) and (6.58) we obtain 2 ζ(2k) = (−1)k−1 (2π)2k B2k /(2k)!, i.e., (4.16). Lemma 6.1 For fixed q ∈ R and ε > 0, let f (z) be regular and satisfy f (z) q,ε

1 |z|q

(z → ∞)

(6.59)

in the sector −π + ε/2 ≤ arg z ≤ π − ε/2. Then f  (z) q,ε

1 |z|q+1

(z → ∞)

for −π + ε ≤ arg z ≤ π − ε. Proof Let z ∈ C satisfy −π + ε ≤ arg z ≤ π − ε, and let t be on the circumference γz,ε of centre z and radius |z| sin(ε/2). A picture shows that | arg t | ≤ π − ε/2 and |t| ≥ |z| − |z| sin

 ε ε = |z| 1 − sin . 2 2

From Cauchy’s integral formula we get f  (z) =

1 2πi

 γz,ε

f (t) dt, (t − z)2

(6.60)

86

6 The Euler Gamma-Function

whence, by (6.59) and (6.60), | f  (z)| ≤

1 2π

 γz,ε

| f (t)| 1 |dt| q,ε (|z| sin(ε/2))2 2π(|z| sin(ε/2))2

 γ,ε

|dt| |t|q

1 ε ≤ 2π|z| sin 2 q q 2π(|z| sin(ε/2)) |z| (1 − sin(ε/2)) 2 1 1 = q,ε . |z|q+1 sin(ε/2) (1 − sin(ε/2))q |z|q+1  Theorem 6.5 (Stirling’s formula for ψ) For any integer s ≥ 0, for any η > 0 and for any arbitrarily small ε > 0, ψ(z + α) =

Γ  (z + α) Γ (z + α)

(6.61)

B1 (α) 1 Bk+1 (α) 1 (−1)k + Os,η,ε + k+1 z k+1 z |z|s+2 k=1 s

= log z +



uniformly for z, α ∈ C satisfying |α| ≤ η, |z| ≥ |α| + ε, | arg(z + α)| ≤ π − ε. In the special case α = 0 we get, for any integer r ≥ 0, ψ(z) = log z −

 r

1 1 B2h 1 − + O r,ε 2z h=1 2h z 2h |z|2r +2

(6.62)

as z → ∞ in the sector −π + ε ≤ arg z ≤ π − ε. Proof Let 

r

√ 1 1 B2h f (z) : = log Γ (z) − z − log z + z − log 2π − 2h−1 2 (2h − 1)2h z h=1

for | arg z | ≤ π − ε/2. By (6.32) we have f (z) r,ε Since

1 |z|2r +1

(z → ∞).

B2h 1 1 + f (z) = ψ(z) − log z + , 2z h=1 2h z 2h r



(6.62) follows from Lemma 6.1.

6.4 The Psi-Function

87

Similarly, applying Lemma 6.1 (with z replaced by z + α) to the function s

√   Bk+1 (α) 1 log Γ (z + α) − z + B1 (α) log z + z − log 2π − (−1)k+1 , k(k + 1) z k k=1



we get (6.61) from (6.40). Theorem 6.6 (Gauss) For any z ∈ C such that Re z > 0, +∞

ψ(z) = 0

e−t e−t z − t 1 − e−t

 dt.

(6.63)

Proof For z in the half-plane Re z > 0 we have +∞ τ τ z 1 −t z −t z lim e dt = lim e dt = e−u du τ →+∞ z τ →+∞ 0

0

because

0

  1 1 = lim 1 − e−τ z = , τ →+∞ z z

 −τ z  e  = e−τ Re z → 0

(τ → +∞).

(6.64)

Changing z to z + m we get +∞ e−t (z+m) dt = 0

1 z+m

(m = 0, 1, 2, . . . ).

(6.65)

Hence, by (6.48), ∞ ψ(z) = −γ −

n 

∞

e−t z dt + lim

n→∞

0

= −γ + lim

∞ n

n→∞ 0 ∞

= −γ + lim

n→∞ 0

m=1

m=1

e−tm −

0 n

e−tm dt −

∞

e−t (z+m) dt

0

 e−t (z+m) − e−t z dt

m=1

e−t − e−t z − e−t (n+1) + e−t (z+n+1) dt. 1 − e−t

Combining this formula with (5.24) we obtain



88

6 The Euler Gamma-Function

∞

ψ(z) = 0

e−t e−t z − t 1 − e−t



Since lim

t→0

and, by (6.64),

∞ dt − lim

n→∞ 0

1 − e−t z −t (n+1) e dt. 1 − e−t

(6.66)

1 − e−t z = z 1 − e−t

1 − e−t z = 1, t→+∞ 1 − e−t lim

the function |(1 − e−t z )/(1 − e−t )| is bounded for t ∈ (0, +∞). Thus there exists a constant K = K (z) > 0, independent of t, such that   1 − e−t z   1 − e−t

   < K for all t > 0. 

Therefore, by (6.65),  ∞  ∞  1 − e−t z −t (n+1)    ≤ K e−t (n+1) dt = K → 0 (n → ∞), (6.67) e dt   1 − e−t n+1 0

0



and (6.63) follows from (6.66) and (6.67).

6.5 Binet’s Integral Formulae It is natural to ask whether, for suitable fixed z, in Stirling’s formula (6.39) the remainder term  √ 1 log z + z − log 2π log Γ (z) − z − 2 can be given a closed integral form independent of Bernoulli numbers and polynomials. Binet gave an affirmative answer to this question via the function ψ(z), with two different integral formulae valid for any z ∈ C in the half-plane Re z > 0 (so that in Binet’s formulae the ε appearing in (6.39) is immaterial). Theorem 6.7 (Binet’s first integral formula) For any z ∈ C such that Re z > 0, log Γ (z) =



√ 1 log z − z + log 2π + z− 2

+∞

0

1 1 1 − + t 2 t e −1



e−t z dt. t

6.5 Binet’s Integral Formulae

89

Proof Integrating along the segment of endpoints 1 and z we get by (6.65) z log z =

du = u

1

z

+∞ +∞  z +∞ −t e − e−t z −tu −tu dt, du e dt = dt e du = t

1

0

0

1

0

(6.68) where the interchange of integrations is justified by z 1

+∞ z +∞ z |du| |z − 1| −tu −t Re u |du| |e | dt = |du| e dt = ≤ < +∞. Re u min{1, Re z} 0

1

0

1

Thus, by (6.63), (6.65) and (6.68), ψ(z) =

Γ  (z) = Γ (z)

  ∞ −t ∞ −t  e e 1 e−t z − e−t z 1 + t dt = − e−t z − t dt t e −1 t e −1 0 ∞

= 0

0

 ∞ ∞

1 1 1 1 e−t − e−t z −t z dt − − + t e−t z dt e dt − t 2 2 t e −1 0

= log z −

1 − 2z

∞

0

0



1 1 1 − + t e−t z dt. 2 t e −1

We use this formula to integrate ψ(u) = Γ  (u)/Γ (u) along the segment of endpoints 1 and z. We get z log Γ (z) =

1 log u du − 2

1



= z−

z

du − u

1

1 2

∞

z du 1

∞

log z − z + 1 − 0

0

 1 1 1 − + t e−tu dt 2 t e −1

 z 1 1 1 − + t dt e−tu du, 2 t e −1 1

where the interchange of integrations is allowed as in (6.68) because    1 1  − + 1 , 2 t t e −1 by (4.8), tends to 0 as t → 0 and to 1/2 as t → +∞, and hence is bounded for t ∈ (0, +∞). Therefore

90

6 The Euler Gamma-Function

log Γ (z) =



1 log z − z + 1 − z− 2

∞

0

=



1 log z − z + C + z− 2

∞

0

where

∞

C = 1− 0

 −t 1 1 e − e−t z 1 − + t dt (6.69) 2 t e −1 t  −t z 1 1 1 e − + t dt, 2 t e −1 t

 −t 1 1 e 1 − + t dt 2 t e −1 t

(6.70)

is a constant. Note that the integrals in (6.69) and (6.70) are absolutely convergent because, again by (4.8), 

   1 1 1 1 1  = − + t . lim   t→0 2 t e −1 t 12 Moreover, denoting 

   1 1 1 1   , − + t M = max  t≥0 2 t e −1 t  we get, by (6.65),  ∞

 −t z  ∞   1 1 1 e M   − + t dt  ≤ M e−t Re z dt = .  2 t e −1 t Re z 0

0

Hence for z > 0, z → +∞, (6.69) yields log Γ (z) =



z−

1 1 log z − z + C + O . 2 z

Comparing this asymptotic formula with (6.39) we obtain C = log theorem follows from (6.69).

√

2π, and the 

Theorem 6.8 (Binet’s second integral formula) For any z ∈ C such that Re z > 0, log Γ (z) =



√ 1 log z − z + log 2π + 2 z− 2

+∞ 0

where the branch of arc tangent is defined by arctan(t/z) = a path from 0 to t/z in the half-plane Re u > 0.

arctan(t/z) dt, e2πt − 1  t/z 0

du/(1 + u 2 ) along

6.5 Binet’s Integral Formulae

91

Proof We apply Theorem 5.4 with the function f (v) : = (z + v)−2 and with r = 0. Then (5.34) yields, for any integer s > 0,

1 1 1 + + 2z 2 k=1 (z + k)2 2(z + s)2 s−1

1 1 +i = − z z+s

+∞

0

 1 dt 1 − 2 2 2πt (z + it) (z − it) e − 1

+∞

−i 0

Since

(6.71)

 1 dt 1 . − 2 2 2πt (z + s + it) (z + s − it) e − 1

1 1 4it z − = − 2 2 2 (z + it) (z − it) (z + t 2 )2   1 1   (z + s + it)2 − (z + s − it)2

we have

(6.72)

  4t|z + s|  = .  |(z + s)2 + t 2 |2

Plainly lim arg(z + s) = 0, so that for any s > s0 (z) we get | arg(z + s) | < π/4, s→+∞   whence Re (z + s)2 > 0, |(z + s)2 + t 2 | ≥ |z + s|2 . Thus for s > s0 (z) we have   +∞

   1 dt 1   −  (z + s + it)2 (z + s − it)2 e2πt − 1 

(6.73)

0

+∞ ≤ 4 |z + s|

≤

4 |z + s|3

t dt 2 2 2 2πt |(z + s) + t | e − 1

0 +∞

t dt → 0 −1

e2πt 0

(s → +∞).

Adding 1/(2z 2 ) to both sides of (6.71) and making s → +∞ we get, by (6.72) and (6.73), +∞ ∞

1 1 tz dt 1 . = + 2 +4 2 2 + t 2 )2 e2πt − 1 (z + k) z 2z (z k=0 0

Writing w in place of z and using (6.52) we obtain

92

6 The Euler Gamma-Function

1 1 + +4 ψ (w) = w 2w 2 

+∞ 0

(w 2

w t dt 2 2 2πt +t ) e −1

(Re w > 0).

(6.74)

For Re z > 0 and for any w on the segment [1, z] of endpoints 1 and z we have |w| ≤ max{1, |z|}. Also, let δ = π/2 − | arg w |, whence 0 < δ ≤ π/2. If δ ≥ π/4 then, for any t > 0, |w2 + t 2 | ≥ |w|2 ≥ (Re w)2 . If δ < π/4 then |w 2 + t 2 | ≥ |w|2 sin(2δ) > |w|2 sin δ = |w| Re w ≥ (Re w)2 . In either case |w 2 + t 2 | ≥ (Re w)2 ≥ (min{1, Re z})2 ,   w   (w 2 + t 2 )2

whence

   ≤ max{1, |z|} ,  (min{1, Re z})4

which shows that the integral on the right-hand side of (6.74) converges absolutely and uniformly for w ∈ [1, z]. Thus, after integrating (6.74) for w ∈ [1, z], we may interchange the integrations. We get 1 Γ  (z) = log z − + C1 − 2 ψ(z) = Γ (z) 2z

+∞ 0

t dt

(z 2

+

t 2 )(e2πt

− 1)

,

where C1 is a constant. Integrating again, log Γ (z) =



1 log z + (C1 − 1)z + C2 + 2 z− 2

+∞ arctan(t/z) dt, e2πt − 1

(6.75)

0

where C2 is a constant. For t, z > 0 we have 0 ≤ arctan(t/z) ≤ t/z, whence +∞ +∞ 1 1 arctan(t/z) t dt dt ≤ = O . 0 ≤ e2πt − 1 z e2πt − 1 z 0

0

Thus for z > 0, z → +∞, we get 1 1 log z + (C1 − 1)z + C2 + O . 2 z √ Comparing with (6.39) we obtain C1 = 0, C2 = log 2π. The theorem follows from (6.75).  log Γ (z) =



z−

Chapter 7

Linear Differential Equations

In this chapter we deal with some basic facts concerning ordinary linear differential equations in the analytic domain, culminating in Fuchs’ theory on regular singular points. Since several special functions of interest in various applications, including the hypergeometric functions 2 F1 , Φ = 1 F1 and Ψ discussed in the next chapter, are solutions of homogeneous linear differential equations of second order, we shall give a full proof of Fuchs’ theorem in the case of order n = 2. For a treatment of Fuchs’ theory in the general n case, the reader is referred to E. L. Ince’s book quoted in the bibliography.

7.1 Cauchy’s Method We expound in this section Cauchy’s original method, based on the so-called dominant functions, to prove the existence, in the analytic domain, of a unique solution of an ordinary homogeneous linear differential equation of order n with initial conditions, in a disc where the coefficients of the differential equation are regular functions. For simplicity, and in view of subsequent applications, we confine ourselves to treating linear differential equations, but such a restriction is not essential in this section, and Cauchy’s method can be adapted to more general differential equations of order n in the analytic domain. We begin with the following definition. Definition 7.1 Let the functions f (z) and g(z) be regular at z 0 ∈ C. If   (k)  f (z 0 ) ≤ g (k) (z 0 ) for all k = 0, 1, 2, . . . , we say that g(z) is dominant over f (z) at z 0 . © Springer International Publishing Switzerland 2016 C. Viola, An Introduction to Special Functions, UNITEXT - La Matematica per il 3+2 102, DOI 10.1007/978-3-319-41345-7_7

93

94

7 Linear Differential Equations

If g(z) is dominant over f (z) at z 0 , and if the Taylor series g(z) =

∞  g (k) (z 0 ) (z − z 0 )k k! k=0

(7.1)

converges absolutely at a point z, then also f (z) =

∞  f (k) (z 0 ) (z − z 0 )k k! k=0

(7.2)

converges absolutely at z. Hence the radius of convergence of the series (7.1) does not exceed the radius of convergence of (7.2). Theorem 7.1 (Cauchy) Let R > 0, z 0 ∈ C, and n ∈ N, n ≥ 1. Let a1 (z), . . . , an (z) be n functions regular in the closed disc |z − z 0 | ≤ R. Then, given n arbitrary constants w0 , w0 , . . . , w0(n−1) ∈ C, there exists a unique function w(z), regular in the open disc |z − z 0 | < R, satisfying the differential equation w (n) = a1 (z) w (n−1) + · · · + an−1 (z) w  + an (z) w

(7.3)

and the initial conditions w(z 0 ) = w0 , w  (z 0 ) = w0 , . . . , w (n−1) (z 0 ) = w0(n−1) . Proof Let aν (z) =

∞ 

aν (z − z 0 )

(ν = 1, . . . , n)

(7.4)

(7.5)

=0

be the Taylor expansions in the disc |z − z 0 | ≤ R of the coefficients of the differential equation (7.3). If the Cauchy problem (7.3), (7.4) has a solution w(z) =

∞ 

cr (z − z 0 )r

(7.6)

r =0

in a neighbourhood of z 0 then, by (7.4), c0 = w0 , c1 =

w0 w (n−1) , . . . , cn−1 = 0 . 1! (n − 1)!

(7.7)

7.1 Cauchy’s Method

95

Differentiating n times the series (7.6) and substituting in (7.3) the n + 1 series for w(z), w  (z), . . . , w (n) (z) thus obtained, as well as the n series (7.5), we obtain linear recurrence formulae yielding successively cn , cn+1 , . . . from the values (7.7). Such formulae can be written as cr =

r 

αr s cr −s

(r = n, n + 1, . . . ),

(7.8)

s=1

where αr s are certain linear combinations of aν with positive coefficients. Thus if a solution (7.6) of the Cauchy problem exists, it is unique. It remains to prove that the radius of convergence of the series (7.6) with the coefficients (7.7), (7.8) is ≥ R. Let M1 , . . . , Mn > 0 be any constants such that, in the disc |z − z 0 | ≤ R, |aν (z)| ≤ Mν

(ν = 1, . . . , n),

with the additional condition M1 >

n . R

(7.9)

From (7.5) we get, by Cauchy’s integral formula, aν =

1 aν() (z 0 ) = ! 2πi

Hence |aν | ≤

Mν 2π R +1

 |z−z 0 |=R

aν (z) dz. (z − z 0 )+1

 |dz| = |z−z 0 |=R

Mν . R

(7.10)

Define functions A1 (z), . . . , An (z) as follows: Aν (z) : =

∞  Mν Aν (z − z 0 ) = z − z0 =0 1− R

Then Aν (z) = Mν

∞   z − z 0  =0

R

(ν = 1, . . . , n).

(7.11)

(|z − z 0 | < R),

whence, by (7.10), |aν | ≤

Mν = Aν . R

Thus Aν (z) is dominant over aν (z) at z 0 (ν = 1, . . . , n).

(7.12)

96

7 Linear Differential Equations

Consider the auxiliary differential equation W (n) = A1 (z) W (n−1) + . . . + An−1 (z) W  + An (z) W. Let W (z) =

∞ 

Cr (z − z 0 )r

(7.13)

(7.14)

r =0

be its (formal) solution satisfying the initial conditions C0 = |c0 |, C1 = |c1 |, . . . , Cn−1 = |cn−1 |, and let Cr =

r 

βr s Cr −s

(r = n, n + 1, . . . )

(7.15)

(7.16)

s=1

be the recurrence formulae analogous with (7.8), where βr s are linear combinations of Aν with the same positive coefficients as in αr s . Thus, by (7.12), |αr s | ≤ βr s with βr s > 0.

(7.17)

From (7.8), (7.15), (7.16) and (7.17) we get, by induction on r , |cr | ≤ Cr In particular

∞ 

(r = 0, 1, 2, . . . ).

|cr (z − z 0 )r | ≤

r =0

∞ 

(7.18)

Cr |z − z 0 |r .

r =0

We will prove that, apart from the case c0 = c1 = · · · = cn−1 = 0 which yields W (z) ≡ 0 by (7.15) and (7.16), the radius of convergence of the series (7.14) is R. Then, by (7.18), W (z) is dominant over w(z) at z 0 , whence the radius of convergence of the series (7.6) is ≥ R, as desired. We make the change of variable z − z 0 = R Z . Then, for ν = 1, . . . , n, ν dν W ν d W = R dZ ν dz ν

and, by (7.11), Aν (z) =

Mν . 1− Z

7.1 Cauchy’s Method

97

Thus (7.13) becomes (1 − Z )

dn W dW dn−1 W + Mn R n W. = M R + . . . + Mn−1 R n−1 1 dZ n dZ n−1 dZ

The series (7.14) is W (Z ) =

∞ 

γr Z r

(7.19)

(7.20)

r =0

with γr = Cr R r ≥ 0.

(7.21)

Since we have excluded the trivial case c0 = c1 = · · · = cn−1 = 0, from (7.15), (7.16) and (7.17) we get γr = Cr R r > 0 for r ≥ n.

(7.22)

Substituting (7.20) into (7.19) we easily obtain the recursion r ! γr = (r − n)(r − 1)! γr −1 +

n 

(r − s)! Ms R s γr −s (r = n, n + 1, . . . ).

s=1

(7.23) By (7.21) and (7.23),  (r − s)! r − n + M1 R γr −1 + Ms R s γr −s r r ! s=2 n

γr =

≥

r − n + M1 R γr −1 (r ≥ n) r

(7.24)

whence, by (7.9) and (7.22), 0 < γn < γn+1 < γn+2 < . . . .

(7.25)

Hence from (7.24) we get, for r ≥ n + 1,  γr −s Ms R s r − n + M1 R γr + = . γr −1 r (r − s + 1) · · · (r − 1) r γ r −1 s=2 n

By (7.25), γr −s /γr −1 < 1 for r ≥ 2n and s = 2, . . . , n. Thus, by (7.26), lim

r →∞

γr = 1. γr −1

(7.26)

98

7 Linear Differential Equations

Therefore the series (7.20) has radius of convergence 1, whence (7.14) has radius of convergence R, as claimed. 

7.2 Singular Points of the Coefficients If the coefficients aν (z) of (7.3) are regular functions in a neighbourhood of a point z 0 ∈ C except possibly at z 0 , one generally expects the solutions w of the differential equation (7.3) to have a branch point at z 0 , as the following example shows. Let a(z) have a simple pole at z = 0 with residue r : a(z) =

r + b(z), z

where b(z) is regular in a disc |z| < R, and assume that the constant r is not an integer: r = Res a(z) ∈ C \ Z. z=0

Let z 1 ∈ C be any fixed point with 0 < |z 1 | < R. The solution z w(z) = exp

a(t) dt

(7.27)

z1

of the first order equation

w  = a(z)w

has a branch point at z = 0, because moving z, inside the punctured disc 0 < |z| < R, along a circuit beginning and ending at z 1 and enclosing 0 changes the value of (7.27) from 1 to e2πir = 1. We discuss in detail the case of a second order homogeneous linear differential equation. We write the equation in the form w  + p1 (z) w  + p2 (z) w = 0,

(7.28)

and assume that the functions p1 (z) and p2 (z) are one-valued and regular in a punctured disc  (7.29) P = {z ∈ C  0 < |z − z 0 | < R}, where z 0 is a singular point (i.e., either a pole or an isolated essential singularity) for at least one of p1 (z) and p2 (z).  Let D1 = {z ∈ C  |z − z 1 | < 1 } and D2 = {z ∈ C  |z − z 2 | < 2 } be two overlapping discs contained in P, and, by Theorem 7.1, let w1 (z) be any solution of (7.28) in D1 . We take a point z 3 ∈ D1 ∩ D2 , lying on the segment of endpoints z 1 , z 2

7.2 Singular Points of the Coefficients

99

and satisfying 1 − |z 1 − z 3 | < 2 − |z 2 − z 3 |, i.e., closer to ∂D1 than to ∂D2 . We can apply Theorem 7.1 to the differential equation (7.28) in the disc  D3 = {z ∈ C  |z − z 3 | < 3 },

3 = 2 − |z 2 − z 3 |,

with initial conditions w(z 3 ) = w1 (z 3 ) and w  (z 3 ) = w1 (z 3 ). Thus w1 (z) can be analytically continued to D1 ∪ D3 . Then we take a point z 4 ∈ D3 \ D1 lying on the segment of endpoints z 2 , z 3 , and apply Theorem 7.1 to (7.28) in the disc  D4 = {z ∈ C  |z − z 4 | < 4 },

4 = 2 − |z 2 − z 4 |,

with initial conditions w(z 4 ) = w1 (z 4 ) and w  (z 4 ) = w1 (z 4 ), so that w1 (z) is analytically continued to D1 ∪ D4 . Iterating this process finitely many times, we can analytically continue the solution w1 (z) of (7.28) to D1 ∪ D2 . This shows that for any simply connected open set Ω ⊂ P such that z 1 ∈ Ω, w1 (z) can be analytically continued to the whole Ω. Now, starting from any point z ∈ Ω, we move in the positive sense along a simple closed circuit Γ ⊂ P enclosing z 0 . By compactness of Γ , we can apply Theorem 7.1 to the equation (7.28) finitely many times, using successive overlapping discs contained in P with centres lying on Γ . Let W1 (z) be the value thus obtained by analytic continuation from w1 (z), after returning along Γ to the initial point z ∈ Ω. Since the differential equation (7.28) is identically satisfied along Γ , W1 (z) is a solution of (7.28) in Ω. Similarly, let w2 (z) be another solution of (7.28) in Ω, and let W2 (z) be the solution of (7.28) in Ω obtained from w2 (z) by analytic continuation along Γ . If w1 (z) and w2 (z) are linearly independent, there exist constants h 11 , h 12 , h 21 , h 22 , depending on w1 (z) and w2 (z) but not on Γ , such that for any z ∈ Ω 

Plainly

W1 (z) = h 11 w1 (z) + h 12 w2 (z) W2 (z) = h 21 w1 (z) + h 22 w2 (z).    h 11 h 12     h 21 h 22  = 0,

(7.30)

(7.31)

for otherwise W1 (z) and W2 (z) would be linearly dependent. Then there would exist constants μ1 , μ2 , not both zero, such that μ1 W1 (z) + μ2 W2 (z) = 0 identically in Ω, whence, moving along Γ in the negative sense, one would get by analytic continuation μ1 w1 (z) + μ2 w2 (z) = 0 identically in Ω, contradicting the linear independence of w1 (z) and w2 (z). Let w(z) be any not identically vanishing solution of (7.28) in Ω, and let k1 , k2 ∈ C be such that w(z) = k1 w1 (z) + k2 w2 (z).

100

7 Linear Differential Equations

If W (z) is the solution of (7.28) in Ω obtained from w(z) by analytic continuation after describing once the circuit Γ in the positive sense, we get W (z) = k1 W1 (z) + k2 W2 (z), whence, by (7.30), W (z) = (k1 h 11 + k2 h 21 )w1 (z) + (k1 h 12 + k2 h 22 )w2 (z). Thus the condition for w(z) to be invariant, up to a multiplicative constant λ, after describing the circuit Γ , i.e. W (z) = λw(z), is that λ satisfies k1 h 11 + k2 h 21 = λk1 and k1 h 12 + k2 h 22 = λk2 , i.e.,  (h 11 − λ)k1 + h 21 k2 = 0 (7.32) h 12 k1 + (h 22 − λ)k2 = 0. This linear system is satisfied by suitable (k1 , k2 ) = (0, 0) if and only if    h 11 − λ h 21     h 12 h 22 − λ  = 0,

(7.33)

i.e., if and only if λ is an eigenvalue of the matrix

h 11 h 12 . h 21 h 22

The equation (7.33) is called the characteristic equation of the singular point z 0 . Although h 11 , h 12 , h 21 , h 22 depend on the fundamental system w1 (z), w2 (z) of solutions of (7.28) in Ω, it is easily seen that the rootsλ1 , λ2 of the characteristic equation (7.33) depend on z 0 but are independent of w1 (z), w2 (z) . Indeed, the non-zero solutions w(z) of (7.28) which are invariant, up to a multiplicative constant λ, after describing once in the positive sense a circuit Γ enclosing z 0 are obviously independent of the fundamental system w1 (z), w2 (z) , and so are the corresponding quotients λ = W (z)/w(z). By (7.31), the roots λ1 , λ2 of (7.33) satisfy λ1 = 0 and λ2 = 0. Taking λ = λ1 in (7.32), we can find a solution (k1 , k2 ) = (0, 0) of (7.32), which is unique up to a constant factor unless the rank of (7.33) for λ = λ1 is zero, i.e., unless h 11 = h 22 = λ1 , h 12 = h 21 = 0. In this case we get λ1 = λ2 and, by (7.30), all the solutions of (7.28) in Ω, after describing Γ , are transformed by multiplication by λ1 . In either case, whether the rank of (7.33) for λ = λ1 is 1 or 0, let (k1 , k2 ) = (0, 0) be a solution of (7.32) for λ = λ1 . Let w1∗ (z) = k1 w1 (z) + k2 w2 (z)

7.2 Singular Points of the Coefficients

101

be the corresponding solution of (7.28) invariant up to multiplication by λ1 , and W1∗ (z) = k1 W1 (z) + k2 W2 (z) = λ1 w1∗ (z). Define r1 =

1 log λ1 . 2πi

(7.34)

Since the difference of any two logarithms of λ1 is 2lπi with l ∈ Z, r1 is defined modulo 1. Let z − z 0 = eiϑ ( > 0, ϑ ∈ R). Then the function (z − z 0 )r1 = r1 eir1 ϑ is multiplied by e2πir1 = λ1 after describing once in the positive sense the circuit Γ . Since the functions w1∗ (z) and (z − z 0 )r1 are both multiplied by λ1 , the quotient ϕ1 (z) : = w1∗ (z)/(z − z 0 )r1 is unchanged after describing Γ . Therefore the differential equation (7.28) has the solution w1∗ (z) = (z − z 0 )r1 ϕ1 (z),

(7.35)

where the function ϕ1 (z) is one-valued, regular and not identically zero in the punctured disc P defined by (7.29). We distinguish two cases. First case: λ1 = λ2 . Let 1 log λ2 . r2 = (7.36) 2πi We repeat the above discussion with λ2 in place of λ1 , and we conclude that (7.28) has the solutions w1∗ (z) = (z − z 0 )r1 ϕ1 (z) and w2∗ (z) = (z − z 0 )r2 ϕ2 (z)

(7.37)

/ Z, and with r1 and r2 , given by (7.34) and (7.36), defined modulo 1, r1 − r2 ∈ with ϕ1 (z) and ϕ2 (z) one-valued, regular and not identically zero in P. Plainly the two solutions (7.37) are linearly independent, for otherwise one would get (z − z 0 )r1 ϕ1 (z) = C(z − z 0 )r2 ϕ2 (z) with a constant C = 0, so that, after describing the circuit Γ , λ1 (z − z 0 )r1 ϕ1 (z) = λ1 C(z − z 0 )r2 ϕ2 (z) = Cλ2 (z − z 0 )r2 ϕ2 (z), whence λ1 = λ2 . Therefore, if λ1 = λ2 the two functions (7.37) form a fundamental system of solutions of (7.28), and all the solutions of (7.28) in P are given by w(z) = C1 (z − z 0 )r1 ϕ1 (z) + C2 (z − z 0 )r2 ϕ2 (z), with arbitrary constants C1 and C2 . Second case: λ1 = λ2 . We consider the fundamental system of solutions of (7.28) obtained by associating with the solution w1∗ (z) given by (7.35) any other solution

102

7 Linear Differential Equations

w

2 (z), regular in Ω, such that w1∗ (z) and w

2 (z) are linearly independent. If W1∗ (z)

2 (z) are the solutions of (7.28) obtained by analytic continuation from w1∗ (z) and W and w

2 (z) respectively after describing the circuit Γ , (7.30) becomes 

W1∗ (z) = λ1 w1∗ (z)

2 (z) = h 21 w1∗ (z) + h 22 w

2 (z). W

Thus the characteristic equation (7.33) is    λ1 − λ h 21     0 h 22 − λ  = 0,

(7.38)

i.e., (λ1 − λ)(h 22 − λ) = 0; and since λ1 = λ2 is a double root of (7.38), we get h 22 = λ1 = λ2 . Hence (7.30) is now 

W1∗ (z) = λ1 w1∗ (z)

2 (z) = h 21 w1∗ (z) + λ1 w

2 (z). W

Therefore

2 (z) w

2 (z) h 21 W = ∗ + . ∗ W1 (z) w1 (z) λ1

On describing the circuit Γ , arg(z − z 0 ) increases by 2π, whence the function h 21 h 21 h 21 log(z − z 0 ) = log |z − z 0 | + arg(z − z 0 ) 2πiλ1 2πiλ1 2πλ1 changes to h 21 h 21 log(z − z 0 ) + . 2πiλ1 λ1 Thus, defining the constant H :=

h 21 , 2πiλ1

(7.39)

we see that the function ψ2 (z) : =

w

2 (z) − H log(z − z 0 ) w1∗ (z)

is unchanged after describing Γ . We conclude that, in the present case λ1 = λ2 , (7.28) has the two linearly independent solutions

7.2 Singular Points of the Coefficients

103

w1∗ (z) = (z − z 0 )r1 ϕ1 (z) and

 w

2 (z) = (z − z 0 )r1 ϕ1 (z) H log(z − z 0 ) + ψ2 (z) , (7.40)

where r1 is given by (7.34), ϕ1 (z) is one-valued, regular and not identically zero in the punctured disc P, and ψ2 (z) is one-valued and meromorphic in P. Thus all the solutions of (7.28) in P are given by   w(z) = (z − z 0 )r1 ϕ1 (z) C1 + C2 H log(z − z 0 ) + ψ2 (z) with arbitrary constants C1 and C2 . We remark that ψ2 (z) can be defined up to an arbitrary additive constant C, because replacing in (7.40) ψ2 (z) by ψ2 (z) + C changes

2 (z) + Cw1∗ (z), with w1∗ (z) and w

2 (z) + Cw1∗ (z) linearly independent. w

2 (z) to w Moreover, if H = 0 the second solution w

2 (z) in (7.40) can be written as  ψ2 (z)  , (z − z 0 )r1 (H ϕ1 (z)) log(z − z 0 ) + H so that, by changing H ϕ1 (z) to ϕ1 (z) and ψ2 (z)/H to ψ2 (z), one may assume H = 1 in (7.40). This is the so-called logarithmic case, corresponding to (7.38) with h 22 = λ1 and with rank 1 for λ = λ1 since, by (7.39), h 21 = 2πiλ1 H = 0. If H = 0, i.e. if h 21 = 0, the characteristic equation (7.38) is    λ1 − λ 0     0 λ1 − λ  = 0. Then the linear independence of w1∗ (z) and w

2 (z) implies that ψ2 (z) is not constant, and we are in the case already discussed of (7.33) with rank zero for λ = λ1 . With the above discussion we have proved the following Theorem 7.2 Let the functions p1 (z) and p2 (z) be one-valued and regular in the punctured disc  P = {z ∈ C  0 < |z − z 0 | < R}, where z 0 is a singular point for at least one of p1 (z) and p2 (z). Then the differential equation (7.41) w  + p1 (z) w  + p2 (z) w = 0 has in P two linearly independent solutions w1 (z) and w2 (z) of the form w1 (z) = (z − z 0 )r1 ϕ1 (z) and w2 (z) = (z − z 0 )r2 ϕ2 (z)

(7.42)

/ Z, given by (7.34) and (7.36) and with ϕ1 (z) and ϕ2 (z) with r1 , r2 ∈ C, r1 − r2 ∈ one-valued and regular in P if the roots λ1 , λ2 of the characteristic equation (7.33) are distinct, or of the form

104

7 Linear Differential Equations

w1 (z) = (z − z 0 )r1 ϕ1 (z) and

 w2 (z) = (z − z 0 )r1 ϕ1 (z) H log(z − z 0 ) + ψ2 (z) (7.43)

with r1 ∈ C given by (7.34), H ∈ C given by (7.39), ϕ1 (z) one-valued and regular in P and ψ2 (z) one-valued and meromorphic in P if the characteristic equation (7.33) has the double root λ1 .

7.3 Fuchs’ Theorem If no additional conditions are assumed on the nature of the singularity z 0 for the coefficients p1 (z) and p2 (z) of the differential equation (7.41), one generally expects z 0 to be an isolated essential singularity for the functions ϕ1 (z) and ϕ2 (z) in (7.42) or (7.43). In particular, by Picard’s second theorem (Theorem 1.4), ϕ1 (z) has in general infinitely many zeros in any neighbourhood of z 0 . Thus if the characteristic equation (7.33) has the double root λ1 , by (7.43) ψ2 (z) = w2 (z)/w1 (z) − H log(z − z 0 ) is generally expected to have infinitely many poles in any neighbourhood of z 0 , and therefore not even to possess a Laurent series expansion at z 0 . However, in most cases of differential equations (7.41) arising from applications, the functions ϕ1 (z) and ϕ2 (z), or ϕ1 (z) and ψ2 (z), turn out to have a pole, or even to be regular, at z 0 . We incidentally remark that if ϕ1 (z) has a pole or a zero at z 0 , i.e., if ϕ1 (z) = (z − z 0 ) N Φ1 (z) with N ∈ Z and with Φ1 (z) regular and = 0 at z 0 , then (z − z 0 )r1 ϕ1 (z) = (z − z 0 )r1 +N Φ1 (z). Since r1 is defined modulo 1, one can replace r1 with r1 + N and hence ϕ1 (z) with Φ1 (z). Thus in this case one can assume ϕ1 (z) to be regular and = 0 at z 0 , and similarly for ϕ2 (z). Therefore, a natural question to ask is what conditions can be assumed on the behaviour of p1 (z) and p2 (z) at z 0 to ensure that z 0 is a pole or a regular point for ϕ1 (z) and ϕ2 (z), or for ϕ1 (z) and ψ2 (z). An answer to this question is given by the following Theorem 7.3 (Fuchs) Under the assumptions of Theorem 7.2, the functions ϕ1 (z), ϕ2 (z) and ψ2 (z) in (7.42) and (7.43) have at most a pole at z 0 , if and only if p1 (z) has at most a simple pole and p2 (z) at most a double pole at z 0 . If such conditions are satisfied, z 0 is said to be a regular singular point, or a fuchsian singular point, of the differential equation (7.41). Proof First we prove that Fuchs’ conditions on p1 (z) and p2 (z) are necessary. We assume that (7.41) has the fundamental system of solutions (7.42) or (7.43), where ϕ1 (z) and ϕ2 (z), or ϕ1 (z) and ψ2 (z), either are regular or have a pole (of any order) at z 0 . We can easily express the coefficients p1 (z) and p2 (z) of (7.41) through the solutions w1 (z) and w2 (z). From the identities w1 + p1 (z)w1 + p2 (z)w1 = 0 and w2 + p1 (z)w2 + p2 (z)w2 = 0,

(7.44)

7.3 Fuchs’ Theorem

105

multiplying the first by −w2 , the second by w1 and summing, we obtain w1 w2 − w2 w1 + p1 (z)(w1 w2 − w2 w1 ) = 0. Dividing by the wronskian w1 w2 − w2 w1 , which is = 0, we get p1 (z) = −

w1 w2 − w2 w1 d = − log(w1 w2 − w2 w1 ) w1 w2 − w2 w1 dz  

d 2 d w2 , = − log w1 dz dz w1

(7.45)

and, by the first of (7.44), p2 (z) = −

w1 w − p1 (z) 1 . w1 w1

(7.46)

Let F be the set of functions F(z) satisfying F(z) = (z − z 0 )α G(z), where α ∈ C is any constant and G(z) is any function meromorphic in the disc |z − z 0 | < R. Then the following properties hold: (i) Products, quotients and derivatives of functions in F belong to F, since  F  (z) = (z − z 0 )α−1 αG(z) + (z − z 0 )G  (z) . (ii) If F(z) ∈ F then F  (z)/F(z) has at most a simple pole at z 0 . For, let G(z) = (z − z 0 )n G 0 (z) with n ∈ Z and G 0 (z) regular and = 0 at z 0 . Then F(z) = (z − z 0 )α+n G 0 (z), whence F  (z) = (α + n)(z − z 0 )α+n−1 G 0 (z) + (z − z 0 )α+n G 0 (z),

(7.47)

and therefore α+n G  (z) F  (z) = + 0 . F(z) z − z0 G 0 (z) (iii) If F(z) ∈ F then F  (z)/F(z) has at most a double pole at z 0 . For, by (7.47), F  (z) = (α + n)(α + n − 1)(z − z 0 )α+n−2 G 0 (z) + 2(α + n)(z − z 0 )α+n−1 G 0 (z) + (z − z 0 )α+n G 0 (z),

106

7 Linear Differential Equations

whence F  (z) (α + n)(α + n − 1) 2(α + n) G 0 (z) G 0 (z) + = + . F(z) (z − z 0 )2 z − z 0 G 0 (z) G 0 (z) If (7.41) has the fundamental system of solutions (7.42), then w1 , w2 ∈ F whence, d  w2  ∈ F. By (7.45), − p1 (z) is the logarithmic by the above property (i), w12 dz w1 derivative of a function belonging to F and therefore, by (ii), has at most a simple pole at z 0 . Also, again by (ii), w1 /w1 has at most a simple pole at z 0 , whence p1 (z) w1 /w1 has at most a double pole. By (iii) w1 /w1 has at most a double pole, whence, by (7.46), p2 (z) has at most a double pole at z 0 . If (7.41) has the fundamental system of solutions (7.43), then w1 ∈ F and, by (7.45), − p1 (z) is the logarithmic derivative of w12

 d H log(z − z 0 ) + ψ2 (z) = w12 (z − z 0 )−1 H + (z − z 0 )ψ2 (z) dz

which belongs to F by property (i). By (ii), − p1 (z) has at most a simple pole at z 0 . Moreover, as in the previous case, by applying (7.46) we conclude that p2 (z) has at most a double pole at z 0 . Next we prove that Fuchs’ conditions on p1 (z) and p2 (z) are sufficient. We do this through a constructive method, due to Frobenius, which allows us to determine the exponents r1 and r2 in (7.42) or (7.43) so that, according to a remark above, ϕ1 (z) and ϕ2 (z) are regular and = 0 at z 0 , and also to find the Taylor expansions of ϕ1 (z) and ϕ2 (z) and the Laurent expansion of ψ2 (z) at z 0 . By Fuchs’ conditions we have p1 (z) =

A(z) B(z) and p2 (z) = , z − z0 (z − z 0 )2

(7.48)

where A(z) and B(z) are regular at z 0 and therefore are sums of Taylor series A(z) =

∞  n=0

an (z − z 0 )n ,

B(z) =

∞ 

bn (z − z 0 )n

(|z − z 0 | < R),

(7.49)

n=0

with R as in the statement of Theorem 7.2. Hence the differential equation (7.41) can be written as (z − z 0 )2 w  + A(z)(z − z 0 )w  + B(z)w = 0.

(7.50)

7.3 Fuchs’ Theorem

107

We seek solutions of (7.50) of the form w(z) = (z − z 0 )r

∞ 

cn (z − z 0 )n ,

(7.51)

n=0

with complex constants r and cn to be determined, where c0 is an arbitrary constant satisfying (7.52) c0 = 0. From (7.51) we get (z − z 0 )w  = (z − z 0 )r

∞  (r + n)cn (z − z 0 )n n=0

and (z − z 0 )2 w  = (z − z 0 )r

∞  (r + n)(r + n − 1)cn (z − z 0 )n . n=0

Substituting in (7.50) and dividing by the common factor (z − z 0 )r we obtain, by (7.49),



 ∞ ∞ ∞  n n n (r + n)(r + n − 1)cn (z − z 0 ) + (r + n)cn (z − z 0 ) an (z − z 0 ) n=0

n=0

+

 ∞ n=0

cn (z − z 0 )n

 ∞

n=0

bn (z − z 0 )n

= 0,

n=0

i.e.,

∞ n    (r + n)(r + n − 1)cn + (r + m)an−m + bn−m cm (z − z 0 )n = 0. n=0

m=0

Hence the condition for (7.51) to satisfy the differential equation (7.50) is  (r + n)(r + n − 1) + (r + n)a0 + b0 cn = −

n−1   (r + m)an−m + bn−m cm (n = 0, 1, 2, . . . ), (7.53) m=0

 provided the Taylor series cn (z − z 0 )n with the coefficients cn obtained from (7.53) has radius of convergence > 0. We can conveniently write the recurrence formula (7.53) by making use of auxiliary functions f ν (t) which we define as follows:

108

7 Linear Differential Equations

f 0 (t) = t (t − 1) + a0 t + b0 , f ν (t) = aν t + bν (ν = 1, 2, 3, . . . ).

(7.54)

Then (7.53) becomes f 0 (r + n)cn = −

n−1 

f n−m (r + m)cm

(n = 0, 1, 2, . . . ).

(7.55)

m=0

For n = 0 this yields f 0 (r )c0 = 0 whence, by (7.52), f 0 (r ) = r (r − 1) + a0 r + b0 = 0.

(7.56)

Since, by (7.49), a0 = A(z 0 ) and b0 = B(z 0 ), (7.56) can be written as  r 2 + A(z 0 ) − 1 r + B(z 0 ) = 0.

(7.57)

Thus the exponent r in (7.51) must be a root of (7.57), which is called the indicial equation of the regular singular point z 0 . We denote the roots of (7.57) by r1 and r2 with Re r1 ≥ Re r2 ,

(7.58)

s = r1 − r2 ,

(7.59)

and we set whence Re s ≥ 0. If s is not an integer, then λ1 = e2πir1 = λ2 = e2πir2 , and according to the discussion in Sect. 7.2 we expect to find through the recursion (7.55) the coefficients cn in (7.51) yielding two linearly independent solutions w1 (z) and w2 (z) of (7.50), corresponding to r = r1 and r = r2 respectively. If s is an integer ≥ 0 then λ1 = λ2 , and through (7.55) we shall find only one solution w1 (z) of the form (7.51), corresponding to r = r1 . Taking r = r1 in (7.55), we seek coefficients cn(1) such that f 0 (r1 + n)cn(1) = −

n−1 

f n−m (r1 + m)cm(1)

(n = 1, 2, 3, . . . ).

(7.60)

m=0

Since f 0 (t) = (t − r1 )(t − r2 ), from (7.59) we get f 0 (r1 + n) = n(n + s) = 0

(n = 1, 2, 3, . . . )

(7.61)

because Re s ≥ 0 yields s = −n. Hence the recurrence formula (7.60) yields successively c1(1) , c2(1) , c3(1) , . . . from c0(1) = 0 arbitrarily chosen. We claim that the radius of convergence of the Taylor series

7.3 Fuchs’ Theorem

109 ∞ 

cn(1) (z − z 0 )n

(7.62)

n=0

thus obtained is ≥ R, where R is as in (7.49). Take any  with 0 <  < R, and let M > 0 be a constant such that max |A(z)| ≤ M

and

|z−z 0 |≤

max |B(z)| ≤ M.

|z−z 0 |≤

By Cauchy’s integral formula 1 A(n) (z 0 ) = an = n! 2πi we get |an | ≤

M 2πn+1

 |z−z 0 |=

A(z) dz (z − z 0 )n+1

 |dz| = |z−z 0 |=

M , n

and similarly |bn | ≤

M . n

Therefore, by (7.54), | f ν (t)| ≤ |aν | |t| + |bν | ≤

M (|t| + 1) ν

(ν = 1, 2, 3, . . . ).

(7.63)

From (7.61) we get | f 0 (r1 + n)| = n |n + s| ≥ n(n + Re s) ≥ n 2 .

(7.64)

Thus, by (7.60), (7.63) and (7.64), n−1        n 2 cn(1)  ≤ | f 0 (r1 + n)| cn(1)  ≤ | f n−m (r1 + m)| cm(1)  m=0

≤ M whence, for n ≥ 1,

n−1  |r1 | + m + 1  (1)  cm , n−m m=0

110

7 Linear Differential Equations

 (1) 

 n−1 n−1  (1)  c    (1)  cm M |r |r | + m + 1 | M 1 1 m c  ≤ ≤ +1 n n−m n m=0 n  n n n−m m=0   n−1  n−1  K  cm(1)  M(|r1 | + 1)  cm(1)  ≤ , ≤ n n−m n m=0 n−m m=0

(7.65)

where K = max{1, M(|r1 | + 1)}.

(7.66)

Since K ≥ 1 we have 1 ≤ K ≤ K 2 ≤ . . . , whence n−1 

K m ≤ n K n−1 .

(7.67)

m=0

Then, by induction on n, n    (1)  c  ≤ c(1)  K n 0 

(n = 0, 1, 2, . . . ),

since, by (7.65) and (7.67),

n−1  (1)  n−1  (1)    (1)  cm K  c0  K m c  ≤ K ≤ n n m=0 n−m n m=0 n−m    n−1  

 (1)  K n K c0(1)   m K c0(1)  n−1   = K ≤ nK = c0 . nn m=0 nn      Hence cn(1)  |z − z 0 |n ≤ c0(1) (K |z − z 0 |/)n (n = 0, 1, 2, . . . ). Since the geometric series

∞  K |z − z 0 | n  n=0 converges for |z − z 0 | < /K , the radius of convergence of (7.62) is ≥ /K . Thus we have proved that the function w1 (z) = (z − z 0 )r1

∞ 

cn(1) (z − z 0 )n

n=0

is a solution of (7.50) in the punctured disc  P1 = {z ∈ C  0 < |z − z 0 | < /K }.

(7.68)

7.3 Fuchs’ Theorem

111

For any simply connected open set Ω ⊂ P such that Ω ∩ P1 = ∅, by applying finitely many times Theorem 7.1 to the differential equation (7.50) on successive overlapping discs as we did in Sect. 7.2, we can analytically continue the solution w1 (z) of (7.50) to the whole Ω. This shows that w1 (z) can be analytically continued from P1 to P. We conclude that the radius of convergence of (7.62) is ≥ R, as claimed. We now take r = r2 in (7.55). In place of (7.60), we seek cn(2) such that f 0 (r2 +

n)cn(2)

= −

n−1 

f n−m (r2 + m)cm(2)

(n = 1, 2, 3, . . . ),

(7.69)

m=0

where f 0 (r2 + n) = n(n − s). If s = r1 − r2 = ν for an integer ν ≥ 1 then f 0 (r2 + ν) = ν(ν − s) = 0, so that the (2) , . . . . If s = r1 − r2 = 0 then (7.69) recursion (7.69) does not determine cν(2) , cν+1 coincides with (7.60), and one finds again the series (7.62) previously obtained. Thus if s ∈ N, through (7.55) we get only the solution (7.68) of (7.50), as expected. If s ∈ / N then f 0 (r2 + n) = n(n − s) = 0, and (7.69) yields c1(2) , c2(2) , c3(2) , . . . from an arbitrarily chosen c0(2) = 0. We now define S = sup n≥1

n , |n − s|

whence S < +∞ since n/|n − s| → 1 for n → ∞. Then in place of (7.64) we get | f 0 (r2 + n)| = n |n − s| =

n2 n2 ≥ n/|n − s| S

and we can argue as in the previous case r = r1 , with (7.66) replaced by K  = max{1, M S(|r2 | + 1)}. Thus

 n    (2)  c  ≤ c(2)  K n 0 

(n = 0, 1, 2, . . . ),

and we conclude as above that the series ∞  n=0

cn(2) (z − z 0 )n

112

7 Linear Differential Equations

has radius of convergence ≥ R. The solutions w1 (z) = (z − z 0 )r1

∞ 

cn(1) (z − z 0 )n and w2 (z) = (z − z 0 )r2

n=0

∞ 

cn(2) (z − z 0 )n

n=0

of the differential equation (7.50) thus obtained are linearly independent by the argument we used to prove the linear independence of (7.37), or by remarking that the wronskian    w1 (z) w2 (z)  (1) (2) r1 +r2 −1    + ...  w (z) w  (z)  = (r2 − r1 )c0 c0 (z − z 0 ) 1 2 cannot be identically zero. It remains to find the second solution w2 (z) of (7.50) when s = r1 − r2 ∈ N. In this case λ1 = e2πir1 = λ2 = e2πir2 , and in accordance with (7.43) we seek the quotient w2 (z)/w1 (z), where w1 (z) is given by (7.68). From (7.45), (7.48) and (7.49) we get

∞  d  w2  A(z) d a0 log w12 = − = − − an (z − z 0 )n−1 . dz dz w1 z − z0 z − z 0 n=1 Integrating and taking exponentials we have w12

 ∞ d  w2  an −a0 n (z − z 0 ) , = C(z − z 0 ) exp − dz w1 n n=1

where C = 0 is a constant. Thus, by (7.68), 

−2 ∞ d  w2  −a0 −2r1 (1) n = C(z − z 0 ) cn (z − z 0 ) dz w1 n=0

 ∞ an (z − z 0 )n . (7.70) × exp − n n=1 Since c0(1) = 0, the function  ∞ n=0

cn(1) (z − z 0 )n

−2

 ∞ an (z − z 0 )n exp − n n=1

7.3 Fuchs’ Theorem

113

is regular at z 0 , and therefore in a disc |z − z 0 | <  ≤ R has a Taylor expansion  ∞

cn(1) (z − z 0 )n

n=0

with

−2

 ∞ ∞  an (z − z 0 )n = exp − αn (z − z 0 )n , n n=1 n=0  −2 α0 = c0(1) = 0,

(7.71)

where the coefficients αn are determined by the an in (7.49) and the cn(1) given by the recurrence formula (7.60). Moreover, by (7.57), r1 + r2 = 1 − a0 , whence, by (7.59), −a0 − 2r1 = r1 + r2 − 1 − 2r1 = −(s + 1). Since in the present case s is an integer ≥ 0, (7.70) becomes, for 0 < |z − z 0 | <  , ∞  d  w2  = C(z − z 0 )−(s+1) αn (z − z 0 )n dz w1 n=0

α0 α1 αs = C + + . . . + + α + α (z − z ) + . . . . s+1 s+2 0 (z − z 0 )s+1 (z − z 0 )s z − z0

Integrating again we get, for constants C = 0 and C  , w2 α0 αs−1 = C − − ... − + αs log(z − z 0 ) s w1 s(z − z 0 ) z − z0

αs+2 2  (z − z 0 ) + . . . + C + αs+1 (z − z 0 ) + 2 = H log(z − z 0 ) + ψ2 (z), with H = Cαs , where ψ2 (z) is given in 0 < |z − z 0 | <  by the Laurent expansion ψ2 (z) = C −

α0 αs−1 − ... − + C s s(z − z 0 ) z − z0

αs+2 2 (z − z 0 ) + . . . . (7.72) + αs+1 (z − z 0 ) + 2

Therefore, as predicted in (7.43),  w2 (z) = w1 (z) H log(z − z 0 ) + ψ2 (z) ,

(7.73)

where, by (7.71) and (7.72), ψ2 (z) has a pole of order s at z 0 if s > 0, or is regular  at z 0 if s = 0. This completes the proof of Fuchs’ theorem.

Chapter 8

Hypergeometric Functions

8.1 Totally Fuchsian Differential Equations Let the functions p1 (z) and p2 (z) be one-valued and regular for any sufficiently large |z|, say for |z| > R > 0. Then the differential equation w  + p1 (z) w  + p2 (z) w = 0

(8.1)

is said to satisfy Fuchs’ conditions at the point z = ∞, whenever Z = 0 is a regular singular point of the differential equation obtained from (8.1) with the substitution z = 1/Z . Since dw dw = −Z 2 , dz dZ   d2 w d2 w dw 2 d 2 dw − Z = Z4 , = −Z + 2Z 3 2 dz dZ dZ dZ 2 dZ

(8.2)

the substitution z = 1/Z changes (8.1) to d2 w + dZ 2



 p1 (1/Z ) dw p2 (1/Z ) 2 − + w = 0. Z Z2 dZ Z4

Hence Fuchs’ conditions for (8.1) hold at z = ∞ if and only if p1 (1/Z ) has at least a simple zero and p2 (1/Z ) at least a double zero at Z = 0, i.e., if and only if p1 (z) has at least a simple zero and p2 (z) at least a double zero at z = ∞. Definition 8.1 The differential equation (8.1) is said to be totally fuchsian when p1 (z) and p2 (z) are one-valued functions meromorphic in C and satisfying Fuchs’ conditions at all the poles in C and at z = ∞.

© Springer International Publishing Switzerland 2016 C. Viola, An Introduction to Special Functions, UNITEXT - La Matematica per il 3+2 102, DOI 10.1007/978-3-319-41345-7_8

115

116

8 Hypergeometric Functions

If (8.1) is totally fuchsian, p1 (z) and p2 (z) vanish at z = ∞ and hence are regular in C\{z 1 , . . . , z N }, where z n (n = 1, . . . , N ) is a pole for at least one of p1 (z) and p2 (z). Thus p1 (z) and p2 (z) can be decomposed as in (2.1), where the entire function G(z) vanishes at z = ∞ and therefore is identically zero. Thus Fuchs’ conditions at z 1 , . . . , z N yield p1 (z) =

N  n=1

 kn z + ln hn and p2 (z) = z − zn (z − z n )2 n=1

with constants h n , kn , ln ∈ C, where

N

N 

kn = 0 since p2 (z) vanishes at z = ∞ with

n=1

order ≥ 2. We conclude that (8.1) is totally fuchsian if and only if p1 (z) and p2 (z) are rational functions of the form p1 (z) =

PN −1 (z) Q 2N −2 (z) and p2 (z) = , (z − z 1 ) · · · (z − z N ) (z − z 1 )2 · · · (z − z N )2

where the numerators are any polynomials satisfying deg PN −1 ≤ N − 1 and deg Q 2N −2 ≤ 2N − 2.

8.2 The Hypergeometric Differential Equation The prototype of totally fuchsian equation is the hypergeometric differential equation: (8.3) z(1 − z)w  + (γ − (α + β + 1)z)w − αβw = 0, where α, β, γ are any complex parameters. The hypergeometric equation (8.3) was introduced by Euler, and later studied by Gauss, Kummer, Riemann and many other authors. The coefficients of (8.3) are p1 (z) =

γ − (α + β + 1)z , z(1 − z)

p2 (z) = −

αβ , z(1 − z)

and Fuchs’ conditions are plainly satisfied at z = 0, 1, ∞. Using Frobenius’ method described in the proof of Theorem 7.3, we seek solutions of (8.3) of the form (7.42) in the punctured disc 0 < |z| < 1. With the notation (7.48), for z 0 = 0 we have A(z) =

γ − (α + β + 1)z = (γ − (α + β + 1)z)(1 + z + z 2 + · · · ) 1−z = γ + (γ − α − β − 1)(z + z 2 + z 3 + · · · ),

8.2 The Hypergeometric Differential Equation

and B(z) = −

117

αβz = −αβ(z + z 2 + z 3 + · · · ). 1−z

Thus at z 0 = 0 the auxiliary functions (7.54) are f 0 (t) = t (t + γ − 1), f ν (t) = (γ − α − β − 1)t − αβ

(ν = 1, 2, 3, . . .),

(8.4)

and the indicial equation (7.57) is r (r + γ − 1) = 0.

(8.5)

According to (7.58) and the subsequent notation, we take r1 = 0 and r2 = 1 − γ, if Re γ ≥ 1, or r1 = 1 − γ and r2 = 0, if Re γ < 1. / Z, i.e., if γ ∈ / Z, there are two linearly independent If s = r1 − r2 = ±(γ − 1) ∈ solutions of (8.3) of the form (7.42), which we can write as ∞ 

cn z n with c0 = 1

(8.6)

n=0

and z 1−γ

∞ 

 cn z n with  c0 = 1

(8.7)

n=0

for 0 < |z| < 1. In the case γ ∈ Z, we have the solution (8.6) if γ = 1, 2, 3, . . . , or the solution (8.7) if γ = 0, −1, −2, . . . , while the second solution of (8.3), linearly independent of (8.6) or (8.7), in general contains a logarithmic term in accordance with (7.43). Assuming γ ∈ C\{0, −1, −2, . . . }, we compute the Taylor coefficients cn of the solution (8.6). By (8.4), the recurrence formula (7.55) with r = 0 is n(n + γ − 1)cn = −

n−1 

(γ − α − β − 1)m − αβ cm . m=0

From our choice c0 = 1 we get γc1 = αβ, whence c1 =

αβ , γ

(8.8)

118

8 Hypergeometric Functions

and, for n ≥ 2, n(n + γ − 1)cn = αβ +

n−1 

(α + β − γ + 1)m + αβ cm

(8.9)

m=1

= αβ +

n−1 

(α + m)(β + m) − m(m + γ − 1) cm . m=1

This yields cn =

(α)n (β)n (γ)n n!

(n = 0, 1, 2, . . . ),

(8.10)

where the Pochhammer symbols are defined by (α)0 = 1, (α)n = α(α + 1) · · · (α + n − 1)

(n = 1, 2, 3, . . . ),

(8.11)

and similarly for (β)n and (γ)n . We have seen that (8.10) holds for n = 0, 1. Assuming (8.10) for n = 1, . . . , q, we get from (8.9) (q + 1)(q + γ)cq+1 = αβ + = αβ + =

q 

(α)m (β)m (α + m)(β + m) − m(m + γ − 1) (γ)m m! m=1 q q   (α)m+1 (β)m+1 (α)m (β)m − (γ)m m! (γ)m−1 (m − 1)! m=1 m=1

(α)q+1 (β)q+1 , (γ)q q!

whence cq+1 =

(α)q+1 (β)q+1 , (γ)q+1 (q + 1)!

which proves (8.10) by induction on n. Thus (8.6) is the Euler–Gauss hypergeometric function 2 F1 (α, β; γ; z), defined by 2 F1 (α, β; γ; z) =

∞  (α)n (β)n z n (γ)n n! n=0

(γ = 0, −1, −2, . . . ),

(8.12)

where the subscripts 2 and 1 of F indicate that the coefficients in the Taylor series (8.12) have two Pochhammer symbols in the numerator and one in the denominator. If α = −N or β = −N , where N ∈ N, the coefficients in the series (8.12) vanish for n > N , so that (8.12) is a polynomial in z of degree N . If α, β = 0, −1, −2, . . . we get

8.2 The Hypergeometric Differential Equation

119

      (α)n (β)n (γ)n+1 (n + 1)!    = lim  (γ + n)(n + 1)  = 1, · lim    n→∞ n→∞ (α + n)(β + n)  (γ)n n! (α)n+1 (β)n+1 and the Taylor series (8.12) has radius of convergence 1. We point out that several elementary functions are special cases of 2 F1 . For example, (1 + z)α = 2 F1 (−α, β; β; −z), log(1 + z) = 2 F1 (1, 1; 2; −z), z arctan z = 2 F1 (1/2, 1; 3/2; −z 2 ). z We also remark that, by (8.12), ∞

∞

 (α)n (β)n z n−1  (α)n+1 (β)n+1 z n d = 2 F1 (α, β; γ; z) = dz (γ)n (n − 1)! (γ)n+1 n! n=1 n=0 =

∞ αβ  (α + 1)n (β + 1)n z n γ n=0 (γ + 1)n n!

=

αβ 2 F1 (α + 1, β + 1; γ + 1; z), γ

whence, by induction on m, we get the differentiation formula dm 2 F1 (α, β; γ; z) dz m (α)m (β)m = 2 F1 (α + m, β + m; γ + m; z) (γ)m

(8.13) (m = 1, 2, 3, . . . ).

If γ ∈ C\{1, 2, 3, . . . }, we compute the coefficients  cn in (8.7). By (8.4) and the recurrence formula (7.55) with r = 1 − γ, we have n(n − γ + 1) cn = −

n−1 

(γ − α − β − 1)(m − γ + 1) − αβ  cm . m=0

If in (8.14) we make the involutory substitution ⎧ ⎪ α − γ+1 ⎨α =   β = β − γ+1 ⎪ ⎩ γ = 2 − γ,

(8.14)

120

8 Hypergeometric Functions

 γ , we get (8.8) with α, β, γ, c replaced by  α, β, c, respectively. Hence, by (8.10),  cn =

n ( α)n (β) (α − γ + 1)n (β − γ + 1)n = ( γ )n n! (2 − γ)n n!

(n = 0, 1, 2, . . . ).

Thus the solution (8.7) is z 1−γ 2 F1 (α − γ + 1, β − γ + 1; 2 − γ; z)

(γ = 1, 2, 3, . . . ).

(8.15)

8.3 Euler’s Integral Representation of 2 F1 As we proved in Sect. 8.2, for α, β, γ = 0, −1, −2, . . . the hypergeometric series (8.12) has radius of convergence 1. However, for any arbitrarily large R > 1 and for any simply connected open set Ω such that   Ω ⊂ {z ∈ C  |z| < R, z = 1}, Ω ∩ {z ∈ C  |z| < 1} = ∅, if we apply finitely many times Theorem 7.1 to the hypergeometric differential equation (8.3) on successive overlapping discs as we did in Sect. 7.2, we can analytically continue the hypergeometric function 2 F1 (α, β; γ; z) to the whole Ω. In particular, the hypergeometric function can be analytically continued to the cut plane C\[1, +∞), and this analytic continuation is again denoted by 2 F1 (α, β; γ; z). If Re γ > Re α > 0, the analytic continuation of 2 F1 (α, β; γ; z) to C\[1, +∞) can also be obtained through one of the most important formulae concerning 2 F1 , namely the Euler integral representation, stated in the following Theorem 8.1 (Euler) If Re γ > Re α > 0 and z ∈ C\[1, +∞), then Γ (γ) 2 F1 (α, β; γ; z) = Γ (α) Γ (γ − α) =

1 B(α, γ − α)

1 0

1 0

t α−1 (1 − t)γ−α−1 dt (1 − t z)β

(8.16)

t α−1 (1 − t)γ−α−1 dt, (1 − t z)β

where B and Γ are the Euler beta- and gamma-functions. Proof The second equality in (8.16) is a consequence of (6.13). By (6.7), Γ (α + n) = α(α + 1) · · · (α + n − 1) Γ (α) = (α)n Γ (α),

(8.17)

8.3 Euler’s Integral Representation of 2 F1

121

and by (6.10) and (6.13), Γ (α + n) B(α + n, γ − α) 1 = = Γ (γ + n) Γ (γ − α) Γ (γ − α)

1

t α−1+n (1 − t)γ−α−1 dt.

0

Hence (8.17) yields (α)n Γ (α + n) Γ (γ) = (γ)n Γ (α) Γ (γ + n) 1 Γ (γ) = t α−1+n (1 − t)γ−α−1 dt Γ (α) Γ (γ − α)

(8.18) (n = 0, 1, 2, . . . ).

0

Substituting this into (8.12) we get, for |z| < 1, 1 ∞  zn Γ (γ) (β)n t α−1+n (1 − t)γ−α−1 dt (8.19) 2 F1 (α, β; γ; z) = Γ (α) Γ (γ − α) n=0 n! 0

=

Γ (γ) Γ (α) Γ (γ − α)

1

t α−1 (1 − t)γ−α−1

0

∞  (β)n n=0

n!

(t z)n dt,

where the interchange of sum and integral is justified by absolute convergence, because 1 ∞    n  α−1+n  (β)n  |z| t (1 − t)γ−α−1  dt n! n=0 0

∞ 

|z|n ≤ (|β|)n n! n=0

1

t Re α−1+n (1 − t)Re γ−Re α−1 dt

0

Γ (Re α) Γ (Re γ − Re α) = 2 F1 (Re α, |β|; Re γ; |z|). Γ (Re γ) For 0 ≤ t ≤ 1, |z| < 1, the binomial series expansion yields −β

(1 − t z)

=

  −β (t z)n = (−1) n n=0

∞  n=0

∞ 

(−1)n

n

∞  (−β)(−β − 1) · · · (−β − n + 1) (β)n (t z)n = (t z)n . n! n! n=0

122

8 Hypergeometric Functions

Therefore, by (8.19), Γ (γ) 2 F1 (α, β; γ; z) = Γ (α) Γ (γ − α)

1 0

t α−1 (1 − t)γ−α−1 dt. (1 − t z)β

If z varies in a neighbourhood of a point z 0 such that | arg(1 − z 0 ) | < π, i.e., z 0 ∈ C\[1, +∞), and t varies in [0, 1], then  

(1 − t z)β  = |1 − t z|Re β exp − (Im β) arg(1 − t z) is bounded from below by a positive constant. Thus, for Re γ > Re α > 0, the integral in (8.16) is a regular function of z in C\[1, +∞). Hence (8.16) yields the analytic  continuation of 2 F1 (α, β; γ; z) from |z| < 1 to z ∈ C\[1, +∞). By (8.12), 2 F1 (α, β; γ; z)

= 2 F1 (β, α; γ; z).

(8.20)

Therefore, if Re γ > Re β > 0 and z ∈ C\[1, +∞), (8.16) yields 1

t β−1 (1 − t)γ−β−1 dt. (1 − t z)α

(8.21)

Re γ > max{Re α, Re β} and min{Re α, Re β} > 0,

(8.22)

Γ (γ) 2 F1 (α, β; γ; z) = Γ (β) Γ (γ − β)

0

Moreover, if

then, by (8.16) and (8.21), Γ (γ) 2 F1 (α, β; γ; z) = Γ (α) Γ (γ − α) =

Γ (γ) Γ (β) Γ (γ − β)

1 0

1 0

t α−1 (1 − t)γ−α−1 dt (1 − t z)β t β−1 (1 − t)γ−β−1 dt, (1 − t z)α

whence the integral transformation formula 1 0

t α−1 (1 − t)γ−α−1 Γ (α) Γ (γ − α) dt = β (1 − t z) Γ (β) Γ (γ − β)

for any z ∈ C\[1, +∞).

1 0

t β−1 (1 − t)γ−β−1 dt (1 − t z)α

(8.23)

8.3 Euler’s Integral Representation of 2 F1

123

Under the assumption (8.22), we give another proof of (8.23) independent of the hypergeometric function 2 F1 . By (6.10) and (6.13) we have Γ (β) Γ (γ − β) Γ (γ)

1 0

1 = B(β, γ − β) 0

11 = 0 0

0 0

(8.24)

t α−1 (1 − t)γ−α−1 dt (1 − t z)β

t α−1 (1 − t)γ−α−1 u β−1 (1 − u)γ−β−1 dt du (1 − t z)β

11  =

t α−1 (1 − t)γ−α−1 dt (1 − t z)β

α 

t 1−t

u (1 − u)(1 − t z)

β



γ (1 − t)(1 − u)

dt du . t (1 − t)u(1 − u)

The last double integral suggests the change of variables u = U, (1 − u)(1 − t z)

t = T, 1−t which easily yields (1 − t)(1 − u) =

1 , 1 + T + U + T U (1 − z)

dt du dT dU = . t (1 − t)u(1 − u) TU

Thus, if z ∈ R, z < 1, from (8.24) we get Γ (β) Γ (γ − β) Γ (γ)

1 0

t α−1 (1 − t)γ−α−1 dt (1 − t z)β +∞ +∞ = 0

0

T α−1 U β−1

γ dT dU, 1 + T + U + T U (1 − z)

which is symmetric in α and β. This proves (8.23) for z ∈ R, z < 1, and hence for any z ∈ C\[1, +∞) by analytic continuation. If neither Re γ > Re α > 0 nor Re γ > Re β > 0, the analytic continuation of to C\[1, +∞) can be reduced to (8.16) or (8.21) as follows. A straightforward computation yields, for any α, β ∈ C, γ ∈ C\{0, −1, −2, . . . }, and k ∈ N, k ≥ 1,

2 F1 (α, β; γ; z)

124

8 Hypergeometric Functions

γ(γ − β + 1)

(α + 1)k (β + 1)k (α + 1)k (β)k + βγ (γ + 2)k k! (γ + 2)k k! (α)k (β)k (α + 1)k−1 (β + 1)k−1 = γ(γ + 1) . − β(γ − α) (γ + 2)k−1 (k − 1)! (γ)k k!

Multiplying by z k with |z| < 1 and summing for k ≥ 1, we get

γ(γ − β + 1) 2 F1 (α + 1, β; γ + 2; z) − 1 +

βγ 2 F1 (α + 1, β + 1; γ + 2; z) − 1 − β(γ − α)z 2 F1 (α + 1, β + 1; γ + 2; z)

= γ(γ + 1) 2 F1 (α, β; γ; z) − 1 , whence 2 F1 (α, β; γ; z)

=

γ−β+1 (8.25) 2 F1 (α + 1, β; γ + 2; z) γ+1 β(γ − (γ − α)z) + 2 F1 (α + 1, β + 1; γ + 2; z). γ(γ + 1)

By applying (8.25) successively n times, we get 2 F1 (α, β; γ; z)

=

n 

cmn (α, β; γ; z) 2 F1 (α + n, β + m; γ + 2n; z),

(8.26)

m=0

where, as is easy to prove by induction on n, for every m, n with 0 ≤ m ≤ n (γ)2n cmn (α, β; γ; z) is a polynomial in α, β, γ, z.

(8.27)

n > max{− Re α, Re(α − γ)},

(8.28)

If we choose then Re(γ + 2n) > Re(α + n) > 0, whence, by (8.16) and (8.26), 2 F1 (α, β; γ; z) n 

=

Γ (γ + 2n) cmn (α, β; γ; z) Γ (α + n) Γ (γ − α + n) m=0

1 0

t α+n−1 (1 − t)γ−α+n−1 dt. (1 − t z)β+m

Here each term in the sum on the right-hand side is a regular function of z in C\[1, +∞).

8.3 Euler’s Integral Representation of 2 F1

125

Theorem 8.2 Let γ = 0, −1, −2, . . . , and let Re(γ − α − β) > 0.

(8.29)

Then the hypergeometric series (8.12) is totally convergent in the closed disc |z| ≤ 1. Moreover ∞  (α)n (β)n n=0

(γ)n n!

= lim 2 F1 (α, β; γ; z) = z→1−

Γ (γ) Γ (γ − α − β) . Γ (γ − α) Γ (γ − β)

(8.30)

Proof If α = −N or β = −N with N ∈ N, (8.12) is a polynomial, so the total convergence is trivial. Otherwise, for |z| ≤ 1 we have            (α)n (β)n z n   ≤  Γ (γ)  ·  Γ (α + n) Γ (β + n)  ,       (γ) n! Γ (α) Γ (β) Γ (γ + n) Γ (n + 1)  n

(8.31)

since n! = Γ (n + 1) by (6.9) and (α)n = Γ (α + n)/Γ (α) by (8.17), and similarly for (β)n and (γ)n . By (6.40) with s = 0 and z = n we get, for n → ∞, 1  √ 1 log n − n + log 2π + O . log Γ (α + n) = n + α − 2 n Using the same asymptotic formula with α replaced by β, γ or 1 we obtain log

1 Γ (α + n) Γ (β + n) = (α + β − γ − 1) log n + O . Γ (γ + n) Γ (n + 1) n

Taking exponentials and then absolute values, we get by (6.3)       1   α+β−γ−1   Γ (α + n) Γ (β + n)      1+O 1 = n Re(α+β−γ)−1 1 + O .  Γ (γ + n) Γ (n + 1)  = n n n By (8.29),



n Re(α+β−γ)−1 < +∞.

n

Hence, by (8.31), the series (8.12) is totally convergent for |z| ≤ 1. In particular, (8.12) is uniformly convergent for 0 ≤ z ≤ 1, whence lim 2 F1 (α, β; γ; z) =

z→1−

∞  (α)n (β)n n=0

(γ)n n!

.

(8.32)

If Re γ > Re α > 0,

(8.33)

126

8 Hypergeometric Functions

under the assumption (8.29) we have, for 0 < t < 1, 0 ≤ z ≤ 1,   α−1  t (1 − t)γ−α−1   ≤ t Re α−1 (1 − t)δ−1    (1 − t z)β 

with δ =

(8.34)

Re(γ − α), if Re β ≤ 0 Re(γ − α − β), if Re β > 0,

whence δ > 0. Therefore 1

t Re α−1 (1 − t)δ−1 dt < +∞.

(8.35)

0

By (8.34) and (8.35), the integral in (8.16) is uniformly convergent for 0 ≤ z ≤ 1. Hence, making z → 1− in (8.16), we can interchange limit and integral. By (6.10) and (6.13) we get Γ (γ) lim 2 F1 (α, β; γ; z) = z→1− Γ (α) Γ (γ − α)

1

t α−1 (1 − t)γ−α−β−1 dt

(8.36)

0

Γ (γ) Γ (γ) Γ (γ − α − β) = B(α, γ − α − β) = . Γ (α) Γ (γ − α) Γ (γ − α) Γ (γ − β) If instead of (8.33) we assume the weaker inequalities Re α > −1 and Re(γ − α) > −1, then Re(α + 1) > 0, Re((γ + 2) − (α + 1)) = Re(γ − α + 1) > 0, and, by (8.29), Re((γ + 2) − (α + 1) − β) = Re(γ − α − β) + 1 > 0, so that the two hypergeometric functions on the right-hand side of (8.25) satisfy (8.29) and (8.33). Applying (8.36) to such hypergeometric functions, by (6.6) we obtain γ − β + 1 Γ (γ + 2) Γ (γ − α − β + 1) γ + 1 Γ (γ − α + 1) Γ (γ − β + 2) αβ Γ (γ + 2) Γ (γ − α − β) + γ(γ + 1) Γ (γ − α + 1) Γ (γ − β + 1) Γ (γ) Γ (γ − α − β) = . Γ (γ − α) Γ (γ − β)

lim 2 F1 (α, β; γ; z) =

z→1−

Iterating this process n times, under the assumptions (8.29) and Re α > −n and Re(γ − α) > −n

8.3 Euler’s Integral Representation of 2 F1

127

we get lim 2 F1 (α, β; γ; z) =

z→1−

Γ (γ) Γ (γ − α − β) . Γ (γ − α) Γ (γ − β)

(8.37)

Since n is arbitrary, we can choose n satisfying (8.28). Hence (8.37) holds under the assumption (8.29) only. From (8.32) and (8.37) we get (8.30). 

8.4

2 F1

as a Function of the Parameters

Theorem 8.3 For any fixed z ∈ C\[1, +∞), 2 F1 (α, β; γ; z)

Γ (γ) is an entire function of α, β and γ. Proof If Re γ > Re α > 0,

(8.38)

from (8.16) we get 2 F1 (α, β; γ; z)

Γ (γ)

1 1 = Γ (α) Γ (γ − α)

1 0

t α−1 (1 − t)γ−α−1 dt. (1 − t z)β

(8.39)

For any fixed z ∈ C\[1, +∞), the integral on the right-hand side of (8.39) is uniformly convergent for ε ≤ Re α ≤ A, |β| ≤ B and ε ≤ Re(γ − α) ≤ C, where A, B and C can be taken arbitrarily large and ε > 0 arbitrarily small. Moreover, for any 0 < t < 1 the integrand is plainly an entire function of α, β and γ, and 1/Γ (α) and 1/Γ (γ − α) are entire functions of α and γ by Theorem 6.3. Hence (8.39) is an entire function of β, and a regular function of α and γ for Re α > 0 and Re(γ − α) > 0, i.e., in the region (8.38). If (8.38) does not hold, by (8.26) and (8.17) we get 2 F1 (α, β; γ; z)

Γ (γ)

=

n  m=0

(γ)2n cmn (α, β; γ; z)

2 F1 (α

+ n, β + m; γ + 2n; z) . Γ (γ + 2n) (8.40)

128

8 Hypergeometric Functions

By the previous argument and by (8.27), each term in the sum on the right-hand side of (8.40) is an entire function of β, and a regular function of α and γ for Re α > −n and Re(γ − α) > −n. Since n can be taken arbitrarily large, choosing n as in (8.28) we conclude that (8.40) is an entire function of α, β and γ.  The six functions 2 F1 (α

± 1, β; γ; z),

2 F1 (α, β

± 1; γ; z),

2 F1 (α, β; γ

± 1; z)

are called contiguous to 2 F1 (α, β; γ; z). The function 2 F1 (α, β; γ; z) and any two functions contiguous to it are related by homogeneous three-terms linear identities, whose coefficients are polynomials in α, β, γ, z of partial degrees 0 or 1 in each of

α, β, z, and of partial degrees 0, 1 or 2 in γ. Thus there are 26 = 15 such identities, due to Gauss, which can be proved by direct substitution of the hypergeometric series (8.12). The full list of the fifteen Gauss contiguity formulae can be found in vol. 1, pp. 103–104, of the treatise by A. Erdélyi et al. quoted in the bibliography. To simplify the notation, it is customary to abbreviate 2 F1 (α, β; γ; z) 2 F1 (α, β

= F,

2 F1 (α

± 1; γ; z) = F(β ± 1),

± 1, β; γ; z) = F(α ± 1),

2 F1 (α, β; γ

± 1; z) = F(γ ± 1).

As an example, we prove the contiguity formula relating F, F(α + 1), F(β + 1), namely (α − β)F − α F(α + 1) + β F(β + 1) = 0. Multiplying the identity (α − β) − (α + n) + (β + n) = 0 by (α)n (β)n z n (γ)n n! with |z| < 1, we get   (α − β)(α)n (β)n − α(α + 1)n (β)n + (α)n β(β + 1)n Summing for n ≥ 0 we obtain (8.41).

zn = 0. (γ)n n!

(8.41)

8.5 Linear Transformations

129

8.5 Linear Transformations The group of fractional linear transformations Z =

az + b cz + d

(a, b, c, d ∈ C, ad − bc = 0)

carrying the points z = 0, 1, ∞ onto the points Z = 0, 1, ∞ in any order is obviously isomorphic to the symmetric group S3 of order 3! = 6. Its elements are easily seen to be 1 Z = 1 − z, Z = , z z z−1 1 Z= ,Z= ,Z= . z−1 z 1−z

Z = z,

(8.42)

If we apply any one of the transformations (8.42) to the differential equation (8.3), we change (8.3) to a differential equation with regular singular points at Z = 0, 1, ∞, which is again of type (8.3) where the parameters α, β, γ are suitably transformed. We begin with the simplest case, i.e., with the transformation Z = 1 − z. Since dw dZ dw dw = = − dz dZ dz dZ and d2 w d2 w dZ d2 w d  dw  − = − = = , dz 2 dz dZ dZ 2 dz dZ 2 (8.3) becomes Z (1 − Z )

dw d2 w − αβw = 0, + α + β − γ + 1 − (α + β + 1)Z dZ 2 dZ

(8.43)

which is the hypergeometric differential equation with the change of parameters (α, β, γ) −→ (α, β, α + β − γ + 1) in (8.3). Applying the solutions (8.12) and (8.15) of (8.3) to the differential equation (8.43), and then substituting Z = 1 − z, we get two further solutions of (8.3) in the cut plane C\(−∞, 0 ], namely 2 F1 (α,

β; α + β − γ + 1; 1 − z)

and, combining with (8.20),

(γ − α − β = 1, 2, 3, . . . )

(8.44)

130

8 Hypergeometric Functions

(1 − z)γ−α−β 2 F1 (γ − α, γ − β; γ − α − β + 1; 1 − z)

(8.45)

(γ − α − β = 0, −1, −2, . . . ), which are linearly independent if γ − α − β ∈ / Z. By (8.2), the transformation Z = 1/z changes (8.3) to Z (1 − Z )

dw αβ d2 w + w = 0. + 1 − α − β + (γ − 2)Z dZ 2 dZ Z

Here we apply the change of function w = Z α W . We easily obtain the equation

dW d2 W − α(α − γ + 1)W = 0, + α − β + 1 − (2α − γ + 2)Z 2 dZ dZ (8.46) which is of type (8.3) with the change of parameters Z (1 − Z )

(α, β, γ) −→ (α, α − γ + 1, α − β + 1). Applying (8.12) and (8.15) to (8.46) and then substituting Z = 1/z, W = Z −α w = z α w, we get two new solutions of (8.3) in the region C\[ 0, 1 ], namely z −α 2 F1 (α, α − γ + 1; α − β + 1; 1/z)

(β − α = 1, 2, 3, . . . )

(8.47)

(β − α = 0, −1, −2, . . . ),

(8.48)

and z −β 2 F1 (β, β − γ + 1; β − α + 1; 1/z)

which are linearly independent if β − α ∈ / Z. The remaining three transformations in the second row of (8.42) yield no further solutions of (8.3). This is a consequence of the involutory transformation formula (8.49) below. Theorem 8.4 (Euler) For any α, β ∈ C, γ ∈ C\{0, −1, −2, . . . } and z ∈ C \[1, +∞) we have 2 F1 (α, β; γ; z)

= (1 − z)−α 2 F1 (α, γ − β; γ; z/(z − 1))

(8.49)

2 F1 (α, β; γ; z)

= (1 − z)γ−α−β 2 F1 (γ − α, γ − β; γ; z).

(8.50)

and

Proof Let B = γ − β,

Z =

z . z−1

(8.51)

Clearly Re γ > Re β > 0 if and only if Re γ > Re B > 0. We temporarily assume this. For any z ∈ C\[1, +∞), we substitute t = 1 − T in the integral (8.21). We get

8.5 Linear Transformations

131

Γ (γ) 2 F1 (α, β; γ; z) = Γ (β) Γ (γ − β) = (1 − z)

−α

= (1 − z)

−α

1 0

T γ−β−1 (1 − T )β−1 dT (1 − z + T z)α

Γ (γ) Γ (B) Γ (γ − B) 2 F1 (α,

1 0

T B−1 (1 − T )γ−B−1 dT (1 − T Z )α

B; γ; Z ),

whence (8.49). Since, by Theorem 8.3, for any z ∈ C\[1, +∞) both sides of (8.49) divided by Γ (γ) are entire functions of β and γ, by analytic continuation on β and γ the restrictive assumption Re γ > Re β > 0 can be dropped. Thus (8.49) holds for any α, β ∈ C, for any γ ∈ C\{0, −1, −2, . . . } and for any z ∈ C\[1, +∞). By the symmetry property (8.20) and by (8.49) we get 2 F1 (α, β; γ; z)

= (1 − z)−β 2 F1 (γ − α, β; γ; z/(z − 1)).

(8.52)

With the notation (8.51) we have Z ∈ C\[1, +∞) if and only if z ∈ C\[1, +∞). If we apply first (8.49) and then (8.52) we obtain 2 F1 (α, β; γ; z)

= (1 − z)−α 2 F1 (α, B; γ; Z ) = (1 − z)−α (1 − Z )−B 2 F1 (γ − α, B; γ; Z /(Z − 1)), 

whence (8.50).

We remark that the transformation formula (8.50) is noteworthy because, unlike (8.49) or other identities, it relates the values of two hypergeometric functions at the same point z. We now combine (8.49) with the action of the group (8.42). The transformation Z = z/(z − 1), together with the change of function w = (1 − Z )α W , changes (8.3) to Z (1 − Z )

dW d2 W − α(γ − β)W = 0, + γ − (α − β + γ + 1)Z dZ 2 dZ

(8.53)

corresponding to the change of parameters in (8.3) (α, β, γ) −→ (α, γ − β, γ). We apply (8.12) and (8.15) to (8.53), and then substitute Z = z/(z − 1), W = (1 − Z )−α w = (1 − z)α w. Since, as we remarked, Z ∈ C\[1, +∞) if and only if z ∈ C\[1, +∞), we get the following solutions of (8.3) in the cut plane C\[1, +∞): (1 − z)−α 2 F1 (α, γ − β; γ; z/(z − 1))

(γ = 0, −1, −2, . . . )

(8.54)

132

8 Hypergeometric Functions

and (8.55) (−z)1−γ (1 − z)γ−α−1 2 F1 (α − γ + 1, 1 − β; 2 − γ; z/(z − 1)) (γ = 1, 2, 3, . . . ). By (8.49), the functions (8.12) and (8.54) coincide. Moreover (8.49) yields, for γ = 2, 3, 4, . . . , z 1−γ 2 F1 (α − γ + 1, β − γ + 1; 2 − γ; z) = z

1−γ

(1 − z)

γ−α−1

2 F1 (α

(8.56)

− γ + 1, 1 − β; 2 − γ; z/(z − 1))

(note that for γ = 1 the identities (8.49) and (8.56) are the same). By (8.56), we see that (8.15) and (8.55) coincide up to the constant factor (−1)1−γ . Hence the transformation Z = z/(z − 1) yields no new solutions of (8.3). We remark that, by applying (8.49), (8.45) becomes z α−γ (1 − z)γ−α−β 2 F1 (γ − α, 1 − α; γ − α − β + 1; (z − 1)/z).

(8.57)

The transformation Z = (z − 1)/z, together with the change of function w = (1 − Z )α W , changes (8.3) to Z (1 − Z )

dW d2 W − α(α − γ + 1)W = 0, + α + β − γ + 1 − (2α − γ + 2)Z 2 dZ dZ

corresponding to the change of parameters (α, β, γ) −→ (α, α − γ + 1, α + β − γ + 1). This equation yields the following solutions of (8.3) in the cut plane C\(−∞, 0 ]: z −α 2 F1 (α, α − γ + 1; α + β − γ + 1; (z − 1)/z) (γ − α − β = 1, 2, 3, . . . ) (8.58) and (8.59) z β−γ (z − 1)γ−α−β 2 F1 (γ − β, 1 − β; γ − α − β + 1; (z − 1)/z) (γ − α − β = 0, −1, −2, . . . ). By (8.49), the solution (8.58) coincides with (8.44), and (8.59) coincides with (8.45) up to the constant factor (−1)γ−α−β . Moreover, since (8.45) equals (8.57), (8.59) can also be written as z α−γ (z − 1)γ−α−β 2 F1 (γ − α, 1 − α; γ − α − β + 1; (z − 1)/z)

(8.60)

(γ − α − β = 0, −1, −2, . . . ).

8.5 Linear Transformations

133

Finally, the transformation Z = 1/(1 − z), together with the change of function w = Z α W , changes (8.3) to Z (1 − Z )

dW d2 W + α − β + 1 − (α − β + γ + 1)Z − α(γ − β)W = 0, 2 dZ dZ

corresponding to the change of parameters (α, β, γ) −→ (α, γ − β, α − β + 1). This equation yields the solutions of (8.3) in the region C\[ 0, 1 ]: (1 − z)−α 2 F1 (α, γ − β; α − β + 1; 1/(1 − z))

(β − α = 1, 2, 3, . . . ) (8.61)

and (1 − z)−β 2 F1 (β, γ − α; β − α + 1; 1/(1 − z)) (β − α = 0, −1, −2, . . . ). (8.62) By (8.49), the solution (8.61) coincides with (8.47) up to the factor (−1)α , and (8.62) coincides with (8.48) up to the factor (−1)β . Altogether, through the three transformations in the first row of (8.42) we have obtained the following three pairs of solutions of (8.3): 

2 F1 (α, β; γ; z) z 1−γ 2 F1 (α − γ

(γ = 0, −1, −2, . . . ) + 1, β − γ + 1; 2 − γ; z)

(γ = 1, 2, 3, . . . )

(8.63)

for z ∈ C\[1, +∞), with (8.63) linearly independent if γ ∈ / Z; ⎧ ⎪ (γ − α − β = 1, 2, 3, . . . ) ⎨2 F1 (α, β; α + β − γ + 1; 1 − z) (1 − z)γ−α−β 2 F1 (γ − α, γ − β; γ − α − β + 1; 1 − z) ⎪ ⎩ (γ − α − β = 0, −1, −2, . . . )

(8.64)

for z ∈ C\(−∞, 0 ], with (8.64) linearly independent if γ − α − β ∈ / Z; 

z −α 2 F1 (α, α − γ + 1; α − β + 1; 1/z) z −β 2 F1 (β, β − γ + 1; β − α + 1; 1/z)

(β − α = 1, 2, 3, . . . ) (β − α = 0, −1, −2, . . . )

(8.65)

for z ∈ C\[ 0, 1 ], with (8.65) linearly independent if β − α ∈ / Z. If γ ∈ Z, γ − α − β ∈ Z and β − α ∈ Z, the general solution of (8.3) cannot be expressed as a linear combination of (8.63), (8.64) or (8.65). Thus, if γ, α + β, α − β ∈ Z, a function of logarithmic type (7.73) cannot be avoided in the general solution of (8.3).

134

8 Hypergeometric Functions

We conclude this discussion by giving an application of (8.50). Let α, β, γ = 0, −1, −2, . . . and Re(γ − α − β) < 0. Then

(8.66)

  lim (1 − z)γ−α−β  = lim (1 − z)Re(γ−α−β) = +∞.

z→1−

z→1−

Moreover Re(γ − (γ − α) − (γ − β)) = Re(α + β − γ) > 0 whence, by (8.30), lim 2 F1 (γ − α, γ − β; γ; z) =

z→1−

Γ (γ) Γ (α + β − γ) = 0. Γ (α) Γ (β)

Therefore, as a complement to Theorem 8.2, from the identity (8.50) we get lim 2 F1 (α, β; γ; z) =

z→1−

Γ (γ) Γ (α + β − γ) lim (1 − z)γ−α−β = ∞, z→1− Γ (α) Γ (β)

under the assumptions (8.66). With the next theorem we express 2 F1 (α, β; γ; z) as a linear combination of the functions (8.64), or of (8.58) and (8.60), if γ − α − β ∈ / Z; of (8.65), or of (8.61) and (8.62), if β − α ∈ / Z. Theorem 8.5 / Z then, for Let α, β, γ ∈ C with

γ = 0, −1, −2, . . . . If γ − α − β ∈ any z ∈ C\ (−∞, 0 ] ∪ [ 1, +∞) , 2 F1 (α, β; γ; z)

=

Γ (γ) Γ (γ − α − β) 2 F1 (α, β; α + β − γ + 1; 1 − z) Γ (γ − α) Γ (γ − β)

Γ (γ) Γ (α + β − γ) (1 − z)γ−α−β 2 F1 (γ − α, γ − β; γ − α − β + 1; 1 − z) Γ (α) Γ (β) Γ (γ) Γ (γ − α − β) −α z 2 F1 (α, α − γ + 1; α + β − γ + 1; (z − 1)/z) = Γ (γ − α) Γ (γ − β) Γ (γ) Γ (α + β − γ) α−γ z (1 − z)γ−α−β + Γ (α) Γ (β) × 2 F1 (γ − α, 1 − α; γ − α − β + 1; (z − 1)/z). (8.67) +

If β − α ∈ / Z then, for any z ∈ C\[ 0, +∞), Γ (γ) Γ (β − α) (−z)−α 2 F1 (α, α − γ + 1; α − β + 1; 1/z) Γ (γ − α) Γ (β) (8.68) Γ (γ) Γ (α − β) (−z)−β 2 F1 (β, β − γ + 1; β − α + 1; 1/z) + Γ (α) Γ (γ − β) Γ (γ) Γ (β − α) = (1 − z)−α 2 F1 (α, γ − β; α − β + 1; 1/(1 − z)) Γ (γ − α) Γ (β)

2 F1 (α,β; γ; z)

=

8.5 Linear Transformations

+

135

Γ (γ) Γ (α − β) (1 − z)−β 2 F1 (β, γ − α; β − α + 1; 1/(1 − z)). Γ (α) Γ (γ − β)

Proof If γ − α − β ∈ / Z, the general solution of the hypergeometric differential equation (8.3) is a linear combination of the functions (8.64) with constant coefficients (i.e., independent of z). In particular, since γ = 0, −1, −2, . . . , there exist coefficients C1 = C1 (α, β, γ) and C2 = C2 (α, β, γ) such that 2 F1 (α, β; γ; z)

= C1 2 F1 (α, β; α + β − γ + 1; 1 − z)

+ C2 (1 − z) for all

γ−α−β

2 F1 (γ

(8.69)

− α, γ − β; γ − α − β + 1; 1 − z)



z ∈ C\ (−∞, 0 ] ∪ [ 1, +∞) .

(8.70)

In order to find C1 and C2 , we temporarily assume Re(α + β) < Re γ < 1. We take the limits of (8.69) for z → 1− and for z → 0+ . Since   lim (1 − z)γ−α−β  = lim (1 − z)Re(γ−α−β) = 0,

z→1−

z→1−

by (8.69) and by Theorem 8.2 we obtain C1 = lim

z→1−

2 F1 (α, β; γ; z)

=

Γ (γ) Γ (γ − α − β) Γ (γ − α) Γ (γ − β)

(8.71)

and C1

Γ (α + β − γ + 1) Γ (1 − γ) Γ (γ − α − β + 1) Γ (1 − γ) + C2 Γ (β − γ + 1) Γ (α − γ + 1) Γ (1 − α) Γ (1 − β) = lim 2 F1 (α, β; γ; z) = 1. z→0+

Hence, by (8.71), C2 =

Γ (1 − α) Γ (1 − β) (8.72) Γ (γ − α − β + 1) Γ (1 − γ)   Γ (γ) Γ (γ − α − β) Γ (α + β − γ + 1) Γ (1 − γ) × 1− · . Γ (γ − α) Γ (γ − β) Γ (α − γ + 1) Γ (β − γ + 1)

With the values (8.71) and (8.72) for C1 and C2 , we can apply Theorem 8.3 to (8.69). By analytic continuation on α, β and γ, we can drop the restrictive assumption Re(α + β) < Re γ < 1. Therefore, for any fixed z satisfying (8.70), the identity (8.69) holds for all α, β and γ such that γ = 0, −1, −2, . . . and γ − α − β ∈ / Z.

136

8 Hypergeometric Functions

In order to express C2 by a formula more satisfactory than (8.72), we apply (8.50) to the left-hand side of (8.69). Dividing by (1 − z)γ−α−β we obtain 2 F1 (γ

− α, γ − β; γ; z) = C1 (1 − z)α+β−γ 2 F1 (α, β; α + β − γ + 1; 1 − z) + C2 2 F1 (γ − α, γ − β; γ − α − β + 1; 1 − z).

Assuming now Re(α + β − γ) > 0 and making z → 1−, by Theorem 8.2 we get C2 = lim

z→1−

2 F1 (γ

− α, γ − β; γ; z) =

Γ (γ) Γ (α + β − γ) . Γ (α) Γ (β)

(8.73)

As before, by analytic continuation on α, β and γ, we can drop the restriction Re(α + β − γ) > 0. Then (8.67) follows from (8.69), (8.71) and (8.73), and by recalling that (8.58) coincides with (8.44), and (8.60) coincides with (8.45) up to the factor (−1)γ−α−β . If β − α ∈ / Z and z ∈ C\[ 0, +∞), we apply first (8.49) and then (8.67). We get 2 F1 (α, β; γ; z) =

Γ (γ) Γ (β − α) (1 − z)−α 2 F1 (α, γ − β; α − β + 1; 1/(1 − z)) Γ (γ − α) Γ (β)

(8.74) +

Γ (γ) Γ (α − β) (1 − z)−β 2 F1 (γ − α, β; β − α + 1; 1/(1 − z)). Γ (α) Γ (γ − β)

But, as we have shown, (8.61) coincides with (8.47) up to the factor (−1)α , and (8.62) coincides with (8.48) up to the factor (−1)β . Therefore (8.68) follows from (8.74). 

8.6 The Confluent Hypergeometric Function 1 F1 A natural generalization of the hypergeometric function (8.12) is p Fs (α1 , . . . , α p ;

γ1 , . . . , γs ; z) =

∞  (α1 )n · · · (α p )n z n (γ1 )n · · · (γs )n n! n=0

(8.75)

(γk = 0, −1, −2, . . . ; k = 1, . . . , s) for any p, s ∈ N, where α1 , . . . , α p ; γ1 , . . . , γs are complex parameters and (αh )n , (γk )n are Pochhammer symbols defined in (8.11). If −αh ∈ N for at least one h (1 ≤ h ≤ p), (8.75) is plainly a polynomial in z. If αh = 0, −1, −2, . . . for h = 1, . . . , p, (8.75) is a Taylor series with radius of convergence such that

8.6 The Confluent Hypergeometric Function 1 F1

137

⎧ ⎪ ⎨ +∞, if p ≤ s = 1, if p = s + 1 ⎪ ⎩ 0, if p > s + 1

(8.76)

because    (α1 )n · · · (α p )n (γ1 )n+1 · · · (γs )n+1 (n + 1)!   · = lim   n→∞ (γ1 )n · · · (γs )n n! (α1 )n+1 · · · (α p )n+1    (γ1 + n) · · · (γs + n)(n + 1)   = lim n s− p+1 . = lim   n→∞ n→∞ (α1 + n) · · · (α p + n) Beside the hypergeometric function 2 F1 , of special interest is the confluent hypergeometric function 1 F1 (α; γ; z), first introduced by Kummer and also commonly denoted by Φ(α; γ; z). Thus Φ(α; γ; z) = 1 F1 (α; γ; z) =

∞  (α)n z n (γ)n n! n=0

(γ = 0, −1, −2, . . . ).

(8.77)

If α = −N with N ∈ N, Φ(α; γ; z) is a polynomial in z of degree N . If α = 0, −1, −2, . . . , by (8.76) Φ(α; γ; z) is the sum of a Taylor series with radius of convergence +∞. Thus in either case Φ(α; γ; z) = 1 F1 (α; γ; z) is an entire function of z. Similarly to Theorem 8.3, for any fixed z ∈ C the function Φ ∗ (α; γ; z) : =

Φ(α; γ; z) = Γ (γ)

1 F1 (α; γ; z)

Γ (γ)

is an entire function of α and γ. For, by (8.17), Φ ∗ (α; γ; z) =

∞

 (α)n zn Φ(α; γ; z) = . Γ (γ) Γ (γ + n) n! n=0

(8.78)

By (6.29), 1/Γ (γ + n) is an entire function of γ. Hence each term in the series (8.78) is a polynomial in α and an entire function of γ, and the series is totally convergent, and hence uniformly convergent, for |α| ≤ A and |γ| ≤ C, where the constants A and C can be taken arbitrarily large, because, for any sufficiently large n,     α+n z   (α)n+1 z n+1 Γ (γ + n) n!     =   Γ (γ + n + 1) (n + 1)! · (α)n zn  γ+n n+1 A+n 1 ≤ |z| < , (n − C)(n + 1) 2

138

8 Hypergeometric Functions

whence      (α)n z n   (1/2)n < +∞.  Γ (γ + n) n!   n n Since Φ ∗ (α; γ; z) in (8.78) is an entire function of γ, it is defined, unlike Φ(α; γ; z), also for γ = 0, −1, −2, . . . . Since 1/Γ (γ + n) vanishes at γ = −n, −(n + 1), −(n + 2), . . . , (8.78) yields, for s = 1, 2, 3, . . . , Φ ∗ (α; 1 − s; z) =

∞  n=s

=

∞  n=s

zn (α)n Γ (n − s + 1) n! ∞

 (α)s (α + s)ν (α)n z n = z s+ν , (n − s)! n! ν! (s + ν)! ν=0

whence Φ ∗ (α; 1 − s; z) = (α)s z s Φ ∗ (α + s; s + 1; z)

(s = 1, 2, 3, . . . ).

(8.79)

Other elementary properties of Φ are similar to the corresponding properties of For instance, as in (8.13), we plainly have the differentiation formula

2 F1 .

dm (α)m Φ(α; γ; z) = Φ(α + m; γ + m; z) m dz (γ)m

(m = 1, 2, 3, . . . ).

(8.80)

Also, the analogues of Gauss’ contiguity formulae for 2 F1 hold for Φ. We abbreviate Φ(α; γ; z) = Φ,

Φ(α ± 1; γ; z) = Φ(α ± 1),

Φ(α; γ ± 1; z) = Φ(γ ± 1).

By direct substitution of the series (8.77) one easily proves the following contiguity formulae:

4

2

=6

(2α − γ + z)Φ − α Φ(α + 1) + (γ − α)Φ(α − 1) = 0, γ(α + z)Φ − αγ Φ(α + 1) − (γ − α)z Φ(γ + 1) = 0, (α − γ + 1)Φ − α Φ(α + 1) + (γ − 1)Φ(γ − 1) = 0, γ Φ − γ Φ(α − 1) − z Φ(γ + 1) = 0, (α + z − 1)Φ + (γ − α)Φ(α − 1) − (γ − 1)Φ(γ − 1) = 0, γ(γ + z − 1)Φ − (γ − α)z Φ(γ + 1) − γ(γ − 1)Φ(γ − 1) = 0.

(8.81)

8.6 The Confluent Hypergeometric Function 1 F1

139

The analogue of Euler’s integral representation (8.16) of 2 F1 is given by the following Theorem 8.6 If Re γ > Re α > 0, for any z ∈ C we have Γ (γ) Φ(α; γ; z) = Γ (α)Γ (γ − α)

1

e zt t α−1 (1 − t)γ−α−1 dt.

(8.82)

0

Proof We substitute (8.18) in (8.77). As in the proof of (8.19) we get 1 ∞  Γ (γ) zn Φ(α; γ; z) = t α−1+n (1 − t)γ−α−1 dt Γ (α)Γ (γ − α) n=0 n! 0

=

=

Γ (γ) Γ (α)Γ (γ − α) Γ (γ) Γ (α)Γ (γ − α)

1  ∞ 0

1

n=0

(zt)n α−1 t (1 − t)γ−α−1 dt n!

e zt t α−1 (1 − t)γ−α−1 dt.



0

As a consequence of (8.82) we obtain Kummer’s transformation formula (8.83) below. Theorem 8.7 (Kummer) For any α, γ, z ∈ C with γ = 0, −1, −2, . . . , Φ(α; γ; z) = e z Φ(γ − α; γ; −z).

(8.83)

Proof If Re γ > Re α > 0 we substitute t = 1 − T in the integral (8.82). We get Γ (γ) Φ(α; γ; z) = e Γ (α)Γ (γ − α)

1

z

e−zT T γ−α−1 (1 − T )α−1 dT

0

= e Φ(γ − α; γ; −z). z

Since both sides of (8.83) divided by Γ (γ) are entire functions of α and γ, by analytic continuation on α and γ we can drop the restrictive assumption Re γ > Re α > 0. Hence (8.83) holds for any α, γ, z ∈ C with γ = 0, −1, −2, . . . . 

8.7 The Confluent Hypergeometric Equation In this section we discuss the confluent hypergeometric function Φ(α; γ; z) = 1 F1 (α; γ; z) as a solution of the confluent hypergeometric differential equation.

140

8 Hypergeometric Functions

Since     ∞  1 n − 1 zn (α)n 1+ ··· 1 + , 2 F1 (α, β; γ; z/β) = 1 + (γ)n β β n! n=1 we get Φ(α; γ; z) = lim 2 F1 (α, β; γ; z/β). β→∞

(8.84)

The hypergeometric function 2 F1 (α, β; γ; z/β) satisfies the differential equation (8.3) with z replaced by z/β, i.e.,     dw α+β+1 z d2 w z − αw = 0, + γ − z 1− β dz 2 β dz

(8.85)

with regular singular points at z = 0, β, ∞. Making β → ∞ in (8.85), we get the confluent hypergeometric differential equation: z w + (γ − z) w  − αw = 0.

(8.86)

Hence (8.86) can be viewed as the limit case of (8.85) at the ‘confluence’ of the fuchsian points β and ∞ (whence the denomination). Writing (8.86) in the form (8.1), the coefficients p1 (z) and p2 (z) are p1 (z) =

γ−z , z

α p2 (z) = − . z

Thus z = 0 is a regular singular point of (8.86), while Fuchs’ conditions are not satisfied at the confluence point z = ∞. As we did in Sect. 8.2 for the hypergeometric differential equation (8.3), we seek solutions of (8.86) of the form (7.42) with z 0 = 0, i.e., in the present case, for 0 < |z| < +∞. With the notation (7.48), (7.49) we have A(z) = γ − z,

B(z) = −αz,

whence a0 = γ, a1 = −1, a2 = a3 = . . . = 0, b0 = 0, b1 = −α, b2 = b3 = . . . = 0. Thus the auxiliary functions (7.54) are f 0 (t) = t (t + γ − 1), f 1 (t) = −(t + α), f ν (t) = 0 (ν = 2, 3, . . . ),

(8.87)

8.7 The Confluent Hypergeometric Equation

141

and the indicial equation is again (8.5). Therefore the roots of the indicial equation are r1 = 0 and r2 = 1 − γ, if Re γ ≥ 1, r1 = 1 − γ and r2 = 0, if Re γ < 1. Hence if γ ∈ / Z, (8.86) has two linearly independent solutions of the form ∞ 

cn z n with c0 = 1

(8.88)

n=0

and z

1−γ

∞ 

 cn z n with  c0 = 1,

(8.89)

n=0

with radii of convergence +∞. If γ ∈ Z, we have the solution (8.88) if γ = 1, 2, 3, . . . , or the solution (8.89) if γ = 0, −1, −2, . . . . Let γ ∈ C\{0, −1, −2, . . . }. By (8.87), the recursion (7.55) with r = 0 is now cn =

n+α−1 cn−1 n(n + γ − 1)

(n = 1, 2, 3, . . . ).

(8.90)

From c0 = 1 we get, by induction on n, cn =

(α)n (γ)n n!

(n = 0, 1, 2, . . . ),

(8.91)

with Pochhammer symbols (α)n and (γ)n . Hence the solution (8.88) of the differential equation (8.86) is the confluent hypergeometric function Φ(α; γ; z) in (8.77), as expected by (8.84). If γ ∈ C\{1, 2, 3, . . . }, by (8.87) the recursion (7.55) with r = 1 − γ is  cn =

n+α−γ  cn−1 n(n − γ + 1)

(n = 1, 2, 3, . . . ).

With the involutory substitution 

α= α − γ+1 γ = 2 − γ

(8.92)

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8 Hypergeometric Functions

(8.92) becomes (8.90), with α, γ, c replaced by  α, γ , c respectively. Hence (8.91) yields (α − γ + 1)n ( α )n = (n = 0, 1, 2, . . . ).  cn = ( γ )n n! (2 − γ)n n! Thus the solution (8.89) is z 1−γ Φ(α − γ + 1; 2 − γ; z)

(γ = 1, 2, 3, . . . ).

(8.93)

Therefore, if γ ∈ / Z the general solution of (8.86) for z ∈ C\{0} is C1 Φ(α; γ; z) + C2 z 1−γ Φ(α − γ + 1; 2 − γ; z),

(8.94)

with arbitrary constants C1 and C2 . We have seen that the function Φ ∗ (α; γ; z) =

Φ(α; γ; z) Γ (γ)

in (8.78) is an entire function of γ. Hence, by analytic continuation on γ, Φ ∗ (α; γ; z) is a solution of (8.86) also for γ = 0, −1, −2, . . . , and hence for all α, γ ∈ C. Similarly, z 1−γ Φ ∗ (α − γ + 1; 2 − γ; z) = z 1−γ

Φ(α − γ + 1; 2 − γ; z) Γ (2 − γ)

is a solution of (8.86) also for γ = 1, 2, 3, . . . , hence for all α, γ ∈ C. However, as we shall now prove, the general solution of (8.86) can be written in the form C1 Φ ∗ (α; γ; z) + C2 z 1−γ Φ ∗ (α − γ + 1; 2 − γ; z) / Z. with arbitrary constants C1 and C2 , if and only if γ ∈ Using the differentiation formula (8.80) for m = 1, it is easy to compute the wronskian of the solutions Φ ∗ (α; γ; z) and z 1−γ Φ ∗ (α − γ + 1; 2 − γ; z) of (8.86). We recall that if w1 and w2 are any two solutions of the differential equation (8.1), the wronskian of w1 and w2 is given by Liouville’s formula    w1 w2  

 W = W (w1 , w2 ) =     = C exp − p1 (z) dz , w1 w2 where C is a constant (i.e., depending on w1 and w2 , but independent of z). Liouville’s formula is easily proved by remarking that    w1 w2  dW  =     = − p1 (z)W. w1 w2 dz

8.7 The Confluent Hypergeometric Equation

143

In particular, the wronskian of any two solutions of (8.86) is C exp

  γ dz = Ce z z −γ . 1− z

(8.95)

By (8.80), d ∗ d 1 Φ (α; γ; z) = Φ(α; γ; z) (8.96) dz Γ (γ) dz α = Φ(α + 1; γ + 1; z) = α Φ ∗ (α + 1; γ + 1; z). γΓ (γ) This easily yields

W Φ ∗ (α; γ; z), z 1−γ Φ ∗ (α − γ + 1; 2 − γ; z) = (1 − γ)z −γ Φ ∗ (α; γ; z) Φ ∗ (α − γ + 1; 2 − γ; z) + z 1−γ (α − γ + 1) Φ ∗ (α; γ; z) Φ ∗ (α − γ + 2; 3 − γ; z)



− α Φ ∗ (α + 1; γ + 1; z) Φ ∗ (α − γ + 1; 2 − γ; z) = Ce z z −γ by (8.95). Multiplying by z γ and then putting z = 0 we obtain

C = (1 − γ) Φ ∗ (α; γ; 0) Φ ∗ (α − γ + 1; 2 − γ; 0) 1 1 1 sin(πγ) = (1 − γ) = = , Γ (γ) Γ (2 − γ) Γ (γ)Γ (1 − γ) π where we have used Euler’s reflection formula (6.21). It follows that

sin(πγ) z −γ W Φ ∗ (α; γ; z), z 1−γ Φ ∗ (α − γ + 1; 2 − γ; z) = e z , π which shows that Φ ∗ (α; γ; z) and z 1−γ Φ ∗ (α − γ + 1; 2 − γ; z) are linearly independent if and only if γ ∈ / Z. In order to treat the case γ ∈ Z, it is convenient to introduce a new function Ψ (α; γ; z) which is a solution of (8.86) for all α, γ ∈ C. Also, as we shall prove, the general solution of (8.86) is C1 Φ ∗ (α; γ; z) + C2 Ψ (α; γ; z) with arbitrary constants C1 and C2 , if and only if α = 0, −1, −2, . . . .

144

8 Hypergeometric Functions

If γ ∈ / Z we define, for any α ∈ C and any z ∈ C\{0}, Γ (1 − γ) Φ(α; γ; z) (8.97) Γ (α − γ + 1) Γ (γ − 1) 1−γ z + Φ(α − γ + 1; 2 − γ; z) Γ (α) Γ (γ)Γ (1 − γ) ∗ Φ (α; γ; z) = Γ (α − γ + 1) Γ (γ − 1)Γ (2 − γ) 1−γ ∗ Φ (α − γ + 1; 2 − γ; z). + z Γ (α)

Ψ (α; γ; z) =

Ψ (α; γ; z) is called the confluent hypergeometric function of the second kind. Since Γ (γ − 1)/Γ (α) = 0 if and only if α = 0, −1, −2, . . . , and z 1−γ is a one-valued function of z if and only if γ ∈ Z, from (8.97) we see that, for γ ∈ / Z, Ψ (α; γ; z) is a multivalued function of z ∈ C\{0} if and only if α = 0, −1, −2, . . . . In this

case, its principal value is defined by taking as usual z 1−γ = exp (1 − γ) log z with log z = log |z| + i arg z, −π < arg z ≤ π, so that Ψ (α; γ; z) is one-valued and regular for z in the cut plane C\R− . By (8.94), Ψ (α; γ; z) is a solution of the differential equation (8.86) when γ ∈ / Z. Moreover, we shall show that Ψ (α; γ; z) takes finite values for γ ∈ Z, so that, by analytic continuation, Ψ (α; γ; z) is a solution of (8.86) for all α, γ ∈ C. From (8.97) we get, for any α ∈ C and any γ ∈ C\Z, Ψ (α; γ; z) =

Φ ∗ (α; γ; z) Φ ∗ (α − γ + 1; 2 − γ; z) π π

z 1−γ + sin(πγ) Γ (α − γ + 1) sin π(γ − 1) Γ (α) 

=

Φ ∗ (α; γ; z)

π sin(πγ) Γ (α − γ + 1)

 Φ ∗ (α − γ + 1; 2 − γ; z) − z 1−γ .

(8.98)

Γ (α)

First we extend Ψ (α; γ; z) for γ = s + 1 (s = 0, 1, 2, . . . ). If α = −m (m = 0, 1, 2, . . . ), since 1/Γ (−m) = 0 and Φ ∗ is an entire function of the parameters, (8.98) yields Ψ (−m; γ; z) =

π Φ ∗ (−m; γ; z) . sin(πγ) Γ (−m − γ + 1)

(8.99)

From the Taylor expansion of sin(πγ) and the Laurent expansion of Γ (−m − γ + 1) around γ = s + 1 (s = 0, 1, 2, . . . ) we easily obtain, by (6.8), lim sin(πγ) Γ (−m − γ + 1) =

γ→s+1

(−1)m π . (m + s)!

8.7 The Confluent Hypergeometric Equation

145

Thus (8.99) yields, for m = 0, 1, 2, . . . and s = 0, 1, 2, . . . , Ψ (−m; s + 1; z) = (−1)m (m + s)! Φ ∗ (−m; s + 1; z) (m + s)! = (−1)m Φ(−m; s + 1; z). s!

(8.100)

We now assume α = 0, −1, −2, . . . . For γ ∈ / Z, we replace in (8.98) Φ ∗ (α; γ; z) ∗ and Φ (α − γ + 1; 2 − γ; z) with the corresponding series (8.78). We get Ψ (α; γ; z) = π

f (α; γ; z) − g(α; γ; z) , sin(πγ)

(8.101)

where f (α; γ; z) : =

∞  1 zn (α)n Γ (α − γ + 1) n=0 Γ (γ + n) n!

(8.102)

∞ 1  (α − γ + 1)n z 1−γ+n . Γ (α) n=0 Γ (2 − γ + n) n!

(8.103)

and g(α; γ; z) : = By (8.17), (α − s)s+n (α)n = . Γ (α) Γ (α − s)

(8.104)

1 = 0 for n = 0, 1, . . . , s − 1. Γ (n − s + 1)

(8.105)

Plainly

By (8.102), (8.103), (8.104) and (8.105) we get g(α; s + 1; z) = =

∞ ∞ 1  (α − s)n z n−s 1  (α − s)s+n z n = Γ (α) n=s Γ (n − s + 1) n! Γ (α) n=0 n! (s + n)! ∞  (α)n z n 1 = f (α; s + 1; z). Γ (α − s) n=0 n! (s + n)!

146

8 Hypergeometric Functions

Thus (8.101) yields, by L’Hôpital’s rule, f (α; γ; z) − g(α; γ; z) sin(πγ)   ∂f ∂g − = (−1)s+1 . ∂γ ∂γ γ=s+1

Ψ (α; s + 1; z) = lim π γ→s+1

(8.106)

As in Sect. 6.4, we denote by ψ(t) =

Γ  (t) Γ (t)

the logarithmic derivative of the gamma-function. Then Γ  (t) d 1 ψ(t) = − . = − 2 dt Γ (t) Γ (t) Γ (t)

(8.107)

As we remarked, (8.78) is a series of entire functions of γ, uniformly convergent for γ in any bounded subset of C. Hence term-by-term differentiation with respect to γ is allowed. By (8.102) and (8.107) we get ∞

∞

 ψ(γ + n) (α)n z n ∂f ψ(α − γ + 1)  (α)n zn 1 = − ∂γ Γ (α − γ + 1) n=0 Γ (γ + n) n! Γ (α − γ + 1) n=0 Γ (γ + n) n! =

∞ 

zn (α)n 1 ψ(α − γ + 1) − ψ(γ + n) . Γ (α − γ + 1) n=0 Γ (γ + n) n!

Therefore 

∂f ∂γ

 γ=s+1

=

∞ 

(α)n z n 1 ψ(α − s) − ψ(s + n + 1) . (8.108) Γ (α − s) n=0 n! (s + n)!

Similarly, by (8.103) and (8.107), ∞ ∂g 1  z n+1 ∂ (α − γ + 1)n z −γ = ∂γ Γ (α) n=0 n! ∂γ Γ (2 − γ + n)  ∞ 1  z n+1−γ 1 ∂ = (α − γ + 1)n Γ (α) n=0 n! Γ (2 − γ + n) ∂γ

(8.109)

 ψ(2 − γ + n) (α − γ + 1)n log z + (α − γ + 1)n . − Γ (2 − γ + n) Γ (2 − γ + n)

8.7 The Confluent Hypergeometric Equation

147

In view of (8.105), we split the sum in (8.109) into the sums over the ranges 0 ≤ n ≤ s − 1 and n ≥ s (the first of these two sums being 0 if s = 0), and we put γ = s + 1. We get 

Since

∂g ∂γ

 γ=s+1

=

   s−1 1 1  1 ∂ (α − γ + 1)n Γ (α) n=0 n! z s−n Γ (n − s + 1) ∂γ γ=s+1  ψ(n − s + 1) (α − s)n − log z + (α − s)n Γ (n − s + 1) Γ (n − s + 1)   ∞ n−s  1 ∂ z + (α − γ + 1)n Γ (α) n=s n! (n − s)! ∂γ γ=s+1  − (α − s)n log z + (α − s)n ψ(n − s + 1) . (8.110)

∂ (α − γ + 1)n is a polynomial in γ, the quantity ∂γ 

∂ (α − γ + 1)n ∂γ

 γ=s+1

is finite. By (8.105), the only contribution to the first sum in (8.110) is given by the term ψ(n − s + 1) (α − s)n . Γ (n − s + 1) By (6.8), for n = 0, . . . , s − 1 we have 1 ψ(n − s + 1) = − = (−1)s−n (s − n − 1)!. Γ (n − s + 1) Res Γ (t) t=n−s+1

Hence the first sum in (8.110) equals s−1  (α − s)n (s − n − 1)!  n=0

n!

−

1 s−n . z

To compute the second sum in (8.110) we remark that, by (6.50),

148

8 Hypergeometric Functions

d d (γ)n = γ(γ + 1) · · · (γ + n − 1) dγ dγ = (γ + 1) · · · (γ + n − 1) + γ(γ + 2) · · · (γ + n − 1) + . . . + γ(γ + 1) · · · (γ + n − 2)   1 1 1 = (γ)n + + ... + γ γ+1 γ+n−1

= (γ)n ψ(γ + n) − ψ(γ) . Therefore 

∂ (α − γ + 1)n ∂γ

 γ=s+1



= (α − s)n ψ(α − s) − ψ(α + n − s) .

Thus the second sum in (8.110) is ∞ 

(α − s)n z n−s ψ(n − s + 1) + ψ(α − s) − ψ(α + n − s) − log z n! (n − s)! n=s

=

∞ 

(α − s)s+n z n ψ(n + 1) + ψ(α − s) − ψ(α + n) − log z . n! (s + n)! n=0

Hence (8.110) yields 

∂g ∂γ +

 γ=s+1

s−1 1  (α − s)n (s − n − 1)!  1 s−n − Γ (α) n=0 n! z

=

(8.111)

∞

1  (α − s)s+n z n ψ(n + 1) + ψ(α − s) − ψ(α + n) − log z . Γ (α) n=0 n! (s + n)!

From (8.106), (8.108) and (8.111) we obtain Ψ (α; s +1; z) =

∞

(−1)s+1  (α)n z n ψ(α − s) − ψ(s + n + 1) (8.112) Γ (α − s) n=0 n! (s + n)!

+

s−1 (−1)s  (α − s)n (s − n − 1)!  1 s−n − Γ (α) n=0 n! z ∞

+

(−1)s  (α − s)s+n z n Γ (α) n=0 n! (s + n)!

× ψ(n + 1) + ψ(α − s) − ψ(α + n) − log z .

8.7 The Confluent Hypergeometric Equation

149

Using again (8.104), from (8.112) we get the series expansion 1  (−1)n (α − s)n (s − n − 1)! Ψ (α; s +1; z) = Γ (α) n=0 n! z s−n s−1

(8.113)

∞

+

(−1)s+1  (α)n z n Γ (α − s) n=0 n! (s + n)!

× ψ(α + n) − ψ(n + 1) − ψ(s + n + 1) + log z (s = 0, 1, 2, . . . ; α = 0, −1, −2, . . . ).

Since Φ(α; s + 1; z) =

∞  n=0

∞

 (α)n z n (α)n z n = s! , (s + 1)n n! n! (s + n)! n=0

(8.113) can be written in the form

Ψ (α; s + 1; z) = Φ(α; s + 1; z) H log z + ψ2 (z) with H=

(−1)s+1 s! Γ (α − s)

and 1 ψ2 (z) = Φ(α; s + 1; z) +



s−1 1  (−1)n (α − s)n (s − n − 1)! Γ (α) n=0 n! z s−n

 ∞

(−1)s+1  (α)n z n ψ(α + n) − ψ(n + 1) − ψ(s + n + 1) , Γ (α − s) n=0 n! (s + n)!

in accordance with (7.72) and (7.73). In fact, here s = γ − 1 ∈ N has the same meaning as in (7.59), since in the present case the roots of the indicial equation are r1 = 0 and r2 = 1 − γ. From (8.97) we get, for γ ∈ / Z, Ψ (α; γ; z) = z 1−γ Ψ (α − γ + 1; 2 − γ; z),

−π < arg z ≤ π,

(8.114)

with the principal values of Ψ (α; γ; z), Ψ (α − γ + 1; 2 − γ; z) and z 1−γ . Using (8.114), we can define Ψ (α; γ; z) for γ = 0, −1, −2, . . . . We get Ψ (α; 1 − s; z) = z s Ψ (α + s; s + 1; z),

(s = 1, 2, 3, . . . ).

(8.115)

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8 Hypergeometric Functions

By (8.97), (8.100), (8.113) and (8.115) we see that Ψ (α; γ; z) is defined for all α, γ ∈ C, as claimed. As an analogue to Theorem 8.6, the next theorem gives an integral representation of Ψ (α; γ; z), showing that, for Re α > 0, Ψ (α; γ; z) can be expressed by a (possibly oblique) Laplace transform. Theorem 8.8 Let Re α > 0, and let z ∈ C\{0}. Let arg z denote the principal argument of z, i.e., −π < arg z ≤ π. For any ϕ satisfying − arg z −

π π < ϕ < − arg z + and − π < ϕ < π, 2 2

(8.116)

the following integral representation holds: eiϕ ∞

1 Ψ (α; γ; z) = Γ (α)

e−zt t α−1 (1 + t)γ−α−1 dt,

(8.117)

0

where the path of integration is the half-line from zero to infinity through eiϕ . Accordingly,

t α−1 = exp (α − 1)(log |t| + i arg t) with arg t = ϕ, and similarly

(1 + t)γ−α−1 = exp (γ − α − 1)(log |1 + t| + i arg(1 + t)) with arg 1 = 0 and arg(1 + t) → ϕ as t → ∞. If Re z > 0, i.e., if −π/2 < arg z < π/2, one can choose ϕ = 0 in (8.116), so that the integral in (8.117) can be taken to be the ordinary Laplace transform of t α−1 (1 + t)γ−α−1 . Proof By (8.116), cos(arg z + ϕ) > 0. Since  −zt  e  = e−|zt| cos(arg z+ϕ) and

 α−1  t  = e−ϕ Im α |t|Re α−1 ,

by the assumption Re α > 0 the integral in (8.117) is absolutely and uniformly convergent for z in a neighbourhood of any point z 0 satisfying −π/2 < arg z 0 + ϕ < π/2. Therefore, denoting the right-hand side of (8.117) by 1 w := Γ (α)

eiϕ ∞

e−zt t α−1 (1 + t)γ−α−1 dt,

0

(8.118)

8.7 The Confluent Hypergeometric Equation

151

we get 1 dw = − dz Γ (α) and d2 w 1 = dz 2 Γ (α)

eiϕ ∞

e−zt t α (1 + t)γ−α−1 dt

(8.119)

0

eiϕ ∞

e−zt t α+1 (1 + t)γ−α−1 dt,

0

whence z

dw d2 w − αw + (γ − z) dz 2 dz 1 = Γ (α)

eiϕ ∞



e−zt t α−1 (1 + t)γ−α−1 zt 2 − (γ − z)t − α dt

0

1 = − Γ (α)

eiϕ ∞

d −zt α e t (1 + t)γ−α dt dt

0



= −

1 e−zt t α (1 + t)γ−α Γ (α)

t=eiϕ ∞

= 0.

t=0

Thus the function w defined in (8.118) is a solution of the confluent hypergeometric equation (8.86). By (8.94), if γ ∈ / Z we get w = C1 Φ(α; γ; z) + C2 z 1−γ Φ(α − γ + 1; 2 − γ; z),

(8.120)

where C1 = C1 (α, γ) and C2 = C2 (α, γ) are independent of z. If Re γ < 1 we have  1−γ  z  = e(arg z) Im γ |z|1−Re γ ≤ eπ Im γ |z|1−Re γ , whence   lim z 1−γ  = 0.

z→0

We temporarily assume 0 < Re γ < 1. From (8.118) and (8.120) we obtain 1 C1 = lim w = z→0 Γ (α)

eiϕ ∞

t α−1 (1 + t)γ−α−1 dt,

0

(8.121)

152

8 Hypergeometric Functions

where the interchange of limit and integral  in (8.118) is allowed by uniform convergence of (8.118) near z = 0, because e−zt  = e−|zt| cos(arg z+ϕ) < 1, and  α−1   t (1 + t)γ−α−1  



| t |Re α−1 for t → 0 | t |Re γ−2 for t → ∞.

For any real number r > 0, let λr denote the arc  λr : = {t ∈ C  | t | = r, arg t from 0 to ϕ}. Since          t α−1 (1 + t)γ−α−1 dt  ≤ t α−1  (1 + t)γ−α−1  |dt|   λr

λr

 r

Re α−1



(1 + r )

Re γ−Re α−1

2πr 

r Re α for r → 0 r Re γ−1 for r → +∞,

our assumptions Re α > 0 and Re γ < 1 yield  lim

ε→0

t α−1 (1 + t)γ−α−1 dt = lim



t α−1 (1 + t)γ−α−1 dt = 0.

R→+∞

λε

(8.122)

λR

By Cauchy’s theorem,  R

 +

ε

λR

Reiϕ   t α−1 (1 + t)γ−α−1 dt = 0. − − εeiϕ

λε

Hence, by (8.122), eiϕ ∞

t

α−1

(1 + t)

0

γ−α−1

+∞ dt = t α−1 (1 + t)γ−α−1 dt.

(8.123)

0

From (6.11) and (6.13) we get +∞ Γ (α)Γ (1 − γ) . t α−1 (1 + t)γ−α−1 dt = Γ (α − γ + 1) 0

(8.124)

8.7 The Confluent Hypergeometric Equation

153

Therefore, by (8.121), (8.123) and (8.124), C1 =

Γ (1 − γ) . Γ (α − γ + 1)

(8.125)

From (8.120) and (8.80) we have zγ

dw α = C1 z γ Φ(α + 1; γ + 1; z) + C2 (1 − γ) Φ(α − γ + 1; 2 − γ; z) dz γ α−γ+1 z Φ(α − γ + 2; 3 − γ; z). + C2 2−γ

By the assumption Re γ > 0 and by (8.119) we get 1 lim z γ C2 (1 − γ) = − Γ (α) z→0

eiϕ ∞

e−zt t α (1 + t)γ−α−1 dt.

0

With the substitution zt = τ we obtain eiϕ ∞

zγ

e−zt t α (1 + t)γ−α−1 dt =

0

ei(arg z+ϕ) ∞

e−τ τ α (z + τ )γ−α−1 dτ .

0

we can plainly rotate the integration path for τ from the half-line As above, 0, ei(arg z+ϕ) ∞ to (0, +∞). Hence, by uniform convergence near z = 0, 1 lim C2 = (γ − 1)Γ (α) z→0 =

1 (γ − 1)Γ (α)

+∞ e−τ τ α (z + τ )γ−α−1 dτ

(8.126)

0

+∞ e−τ τ γ−1 dτ = 0

Γ (γ − 1) Γ (γ) = . (γ − 1)Γ (α) Γ (α)

Thus (8.97), (8.118), (8.120), (8.125) and (8.126) yield, for Re α > 0 and 0 < Re γ < 1, 1 Γ (α)

eiϕ ∞

e−zt t α−1 (1 + t)γ−α−1 dt

0

Γ (γ − 1) 1−γ Γ (1 − γ) Φ(α; γ; z) + z Φ(α − γ + 1; 2 − γ; z) Γ (α − γ + 1) Γ (α) = Ψ (α; γ; z),

=

154

8 Hypergeometric Functions

i.e., (8.117). Clearly, by analytic continuation on γ, the temporary assumption 0 < Re γ < 1 can be dropped. This completes the proof of the theorem.  With the next corollary we prove for the function Ψ (α; γ; z) an analogue of the differentiation formula (8.80) for Φ(α; γ; z). Corollary 8.1 dm Ψ (α; γ; z) = (−1)m (α)m Ψ (α + m; γ + m; z) (m = 1, 2, 3, . . . ). dz m (8.127) Proof If Re α > 0, by (8.117), (8.118) and (8.119) we get d 1 Ψ (α; γ; z) = − dz Γ (α)

eiϕ ∞

e−zt t α (1 + t)γ−α−1 dt

0

α = − Γ (α + 1)

eiϕ ∞

e−zt t α (1 + t)γ−α−1 dt

0

= −α Ψ (α + 1; γ + 1; z). By analytic continuation on α, the differentiation formula d Ψ (α; γ; z) = −α Ψ (α + 1; γ + 1; z) dz

(8.128)

holds for all α, γ ∈ C. Then (8.127) follows from (8.128) by induction on m.



Using (8.96), (8.98) and (8.128) we can compute the wronskian of Φ ∗ (α; γ; z) and Ψ (α; γ; z). A straightforward computation yields

W Φ ∗ (α; γ; z), Ψ (α; γ; z) ∗ πz −γ Φ (α; γ; z) Φ ∗ (α − γ + 1; 1 − γ; z) = − Γ (α) sin(πγ)

− αz Φ ∗ (α + 1; γ + 1; z) Φ ∗ (α − γ + 1; 2 − γ; z) = Ce z z −γ by (8.95). Multiplying by z γ and then putting z = 0 we get π Φ ∗ (α; γ; 0) Φ ∗ (α − γ + 1; 1 − γ; 0) Γ (α) sin(πγ) 1 1 1 π = − . = − Γ (α) sin(πγ) Γ (γ) Γ (1 − γ) Γ (α)

C = −

8.7 The Confluent Hypergeometric Equation

155

Therefore

1 W Φ ∗ (α; γ; z), Ψ (α; γ; z) = − e z z −γ . Γ (α) This shows that Φ ∗ (α; γ; z) and Ψ (α; γ; z) are linearly independent if and only if α = 0, −1, −2, . . . , as claimed.

8.8 Mellin–Barnes’ Integral Representations Let f : R+ → C and σ ∈ R satisfy +∞ | f (x)| x σ−1 dx < +∞,

(8.129)

0

and let  (σ) = {s ∈ C  s = σ + it, −∞ < t < +∞} denote the vertical line in C of real part σ. The Mellin transform of f is the function F : (σ) → C defined by +∞ F(s) =

f (x) x s−1 dx,

s = σ + it.

(8.130)

0

By (8.129), the integral (8.130) is absolutely convergent. With the substitution x = e−2πu the integral (8.130) becomes +∞ F(σ + it) = 2π

f e−2πu e−2π(σ+it)u du = 2π ϕ σ (t),

−∞

where +∞ ϕ σ (t) = ϕσ (u)e−2πitu du −∞

is the Fourier transform of the function ϕσ defined by

ϕσ (u) = f e−2πu e−2πσu for − ∞ < u < +∞.

(8.131)

156

8 Hypergeometric Functions

Therefore, under mild assumptions for f besides (8.129), applying Fourier’s integral σ (u) = ϕσ (−u) we get theorem ϕ

σ (−u) f e−2πu e−2πσu = ϕσ (u) = ϕ +∞  +∞ 1 2πiut ϕ σ (t)e dt = F(σ + it)e2πiut dt, = 2π −∞

−∞

whence

f e−2πu =

+∞

1 2πi

F(σ + it)e2πu(σ+it) d(σ + it), t=−∞

i.e., the Mellin inversion formula: 1 f (x) = 2πi



F(s) x −s ds,

x > 0.

(8.132)

(σ)

A standard example of Mellin’s transform is the gamma-function. Choosing f (x) = e−x in (8.130), for σ = Re s > 0 we get F(s) = Γ (s) from (6.1), so that Γ (s) is the Mellin transform of e−x . Thus the Mellin inversion formula (8.132) yields

e

−x

1 = 2πi

σ+i∞ 

Γ (s) x −s ds

(x > 0, σ > 0).

σ−i∞

By analytic continuation we obtain

e

−z

1 = 2πi

σ+i∞ 

Γ (s) z −s ds

(z ∈ C, Re z > 0, σ > 0).

(8.133)

σ−i∞

Some formulae, similar to (8.133), representing hypergeometric functions as Mellin integrals over vertical lines in C with integrands containing gamma-factors were first proved by Barnes through a method, independent of Fourier analysis, based on the residue theorem and the exponential decay of |Γ (s)| along vertical strips expressed by (6.42). We follow Barnes’ method, beginning with the Kummer confluent hypergeometric function Φ(α; γ; z) = 1 F1 (α; γ; z).

8.8 Mellin–Barnes’ Integral Representations

157

Theorem 8.9 Let k ∈ R, and let α, γ, z ∈ C with α, γ = 0, −1, −2, . . . and Re z < 0. Then  Γ (s + α) Γ (γ) 1 Φ(α; γ; z) = 1 F1 (α; γ; z) = Γ (−s)(−z)s ds, (8.134) Γ (α) 2πi Γ (s + γ) (k)

where (k) denotes the vertical line (k − i∞, k + i∞) deformed (if necessary) inside a bounded region to separate the poles s = 0, 1, 2, . . . of Γ (−s) from the poles s = −α, −α − 1, −α − 2, . . . of Γ (s + α), so that s = 0, 1, 2, . . . lie on the right and s = −α, −α − 1, −α − 2, . . . lie on the left of the integration path (this is s obviously possible

for any k ∈ R, since α = 0, −1, −2, . . . ). In (8.134), (−z) = exp s log(−z) with log(−z) ∈ R for z ∈ R, z < 0. Proof Let −z = eiϑ , α = α1 + iα2 , γ = γ1 + iγ2 and s = k + it for sufficiently large |t|. Then



|(−z)s | = exp Re s log(−z) = exp Re (k + it)(log + iϑ) = k e−ϑt . (8.135) Since k is fixed, for bounded = |z| we get by (6.42), for any ε > 0 and for t → ±∞,     Γ (s + α) s   Γ (s + γ) Γ (−s)(−z) 



π 1  |t|α1 −γ1 −k− 2 exp − |t + α2 | − |t + γ2 | + |t| − ϑt    2 π α1 −γ1 −k− 21 exp |ϑ| − + ε |t| .  |t| 2

(8.136) 

Since Re z < 0 we have |ϑ| < π/2. We take ε such that 0 < ε < π/2 − |ϑ|. Thus, by (8.136), the integral (8.134) is absolutely and uniformly convergent for z in any bounded region contained in the half-plane Re z < 0, and hence is a regular function of z in Re z < 0. For an integer N > max{k, |α1 |} and for a real T > |α2 | we consider the integral, with the same integrand as in (8.134), taken along the border of the region R in the s-plane obtained from the rectangle with vertices at k ± i T and N + 1/2 ± i T by deforming (if necessary) the left side (k − i T, k + i T ) to exclude from R the poles of Γ (s + α) and to include in R the poles s = 0, 1, . . . , N of Γ (−s), as in the statement of the theorem. If we keep N fixed and let T → +∞, from (8.136) we see that the integrals along the upper and lower sides (k + i T, N + 1/2 + i T ) and (k − i T, N + 1/2 − i T ) of R tend to zero. Therefore, by the residue theorem, the difference 1 2πi



N + 21 +i∞

 



− N + 21 −i∞

(k)

Γ (s + α) Γ (−s)(−z)s ds Γ (s + γ)

(8.137)

158

8 Hypergeometric Functions

equals the sum of residues of the integrand at the poles s = 0, 1, . . . , N of Γ (−s). By (6.8), the residue of Γ (−s) at s = n (0 ≤ n ≤ N ) is (−1)n+1 /n!. Thus (8.137) equals, by (8.17), N  (−1)n+1 Γ (α + n) (−z)n n! Γ (γ + n) n=0

= −

(8.138) N N  Γ (α)  (α)n z n (α)n Γ (α) z n = − . (γ)n Γ (γ) n! Γ (γ) n=0 (γ)n n! n=0

We now show that the integral over (N + 1/2 − i∞, N + 1/2 + i∞) in (8.137) tends to zero as N → +∞. Let 1 IN = 2πi

N + 21 +i∞



N + 21 −i∞

Γ (s + α) Γ (−s)(−z)s ds. Γ (s + γ)

With the substitution s = N + 1/2 + it, using (8.135), we get  +∞  N + 1 −ϑt  Γ (α + N + 1/2 + it)      2e |I N |  dt.  Γ (γ + N + 1/2 + it)  Γ (−N − 1/2 − it) |z| −∞

By Euler’s reflection formula (6.21), since | sin(π(N + 1/2 + it))| = cosh(πt), we obtain |Γ (−N − 1/2 − it)| =

π . cosh(πt) |Γ (N + 3/2 + it)|

Hence (8.139) |I N |  |z| N + 2 +∞     Γ (α + N + 1/2 + it)  1 e−ϑt   ×  Γ (γ + N + 1/2 + it)  |Γ (N + 3/2 + it)| cosh(πt) dt. 1

−∞

By repeated application of the functional equation (6.6) we get  1  3    1   1 3 + it + it · · · N + + it Γ + it , Γ N + + it = 2 2 2 2 2

8.8 Mellin–Barnes’ Integral Representations

159

whence, by (6.23),         Γ N + 3 + it  ≥ 1 · 3 · · · 2N + 1 Γ 1 + it  (8.140)   2   2 2 2 2        (2N + 1)!!  1 1 3    1   = Γ 2 + it  = √π Γ N + 2 Γ 2 + it . 2 N +1 Again by the reflection formula (6.21) we get    2     1 π π Γ  = Γ 1 + it Γ 1 − it = + it = .  2  2 2 sin(π/2 + iπt) cosh(πt) Thus (8.140) yields     Γ N + 3 + it  ≥ Γ√(N + 3/2) .   2 cosh(πt)

(8.141)

By Stirling’s formula (6.40) with z, α replaced by N + it and α + 1/2 respectively we obtain log Γ (α + N + 1/2 + it) = (α + N + it) log(N + it) − N − it + log

√

 2π + O

 1 , |N + it|

whence Γ (α + N + 1/2 + it) =

  √ 2π (N + it)α+N +it e−N −it 1 + O

1 |N + it|

 , (8.142)

and similarly for Γ (γ + N + 1/2 + it). Therefore         Γ (α + N + 1/2 + it)   1  = (N + it)α−γ  1 + O .    Γ (γ + N + 1/2 + it)  |N + it|  As with (8.135) we have α−γ        (N + it)α−γ  =  N α−γ   1 + it  N α1 −γ1      it  it  = N α1 −γ1 1 +  (α2 − γ2 ) exp − arg 1 + N N α1 −γ1      it π |α2 − γ2 | . exp ≤ N α1 −γ1 1 +  N 2

160

8 Hypergeometric Functions

Thus, for N → +∞,  α1 −γ1     Γ (α + N + 1/2 + it)     N α1 −γ1 1 + it    Γ (γ + N + 1/2 + it)   N α1 −γ1 ≤ N λα1 −γ1 (t),

(8.143)

where  λν (t) =

1 + t2 1

ν/2

for ν > 0 for ν ≤ 0.

(8.144)

+∞ e−ϑt dt. λα1 −γ1 (t) √ cosh(πt)

(8.145)

From (8.139), (8.141) and (8.143) we get |z| N + 2 N α1 −γ1 |I N |  Γ (N + 3/2) 1

−∞

√ √ π Since cosh(πt) ∼ e 2 |t| 2 as t → ±∞, the last integral converges for |ϑ| = | arg(−z)| < π/2, i.e., for Re z < 0, which we have assumed. Moreover, again by Stirling’s formula (6.40), Γ (N + 3/2)  N N +1 e−N . Thus (8.145) yields lim I N = 0

N →+∞

as claimed. Then, making N → +∞ in (8.137), (8.138), we get −

1 2πi

 (k)

∞

Γ (s + α) Γ (α)  (α)n z n Γ (−s)(−z)s ds = − , Γ (s + γ) Γ (γ) n=0 (γ)n n! 

i.e., (8.134).

We remark that, by substituting s → −s, z → −z and choosing α = γ in (8.134), we obtain (8.133). Next we prove Barnes’ integral representation of 2 F1 (α, β; γ; z). The argument is similar to the proof of Theorem 8.9 except that for the proof of Theorem 8.10 we shall require the temporary assumption |z| < 1, which ensures convergence of the hypergeometric series (8.12). Theorem 8.10 Let k ∈ R, let α, β, γ ∈ C\{0, −1, −2, . . . } and z ∈ C\[0, +∞). Then  1 Γ (s + α)Γ (s + β) Γ (γ) Γ (−s)(−z)s ds, 2 F1 (α, β; γ; z) = Γ (α)Γ (β) 2πi Γ (s + γ) (k)

(8.146)

8.8 Mellin–Barnes’ Integral Representations

161

where (k) denotes the vertical line (k − i∞, k + i∞) deformed (if necessary) inside a bounded region to separate the poles of Γ (−s) from the poles of Γ (s + α)Γ (s + β). Proof Let −z = eiϑ , α = α1 + iα2 , β = β1 + iβ2 , γ = γ1 + iγ2 and s = k + it for sufficiently large |t|. By (6.42) and (8.135) we get, for bounded = |z|, for any ε > 0 and for t → ±∞,     Γ (s + α)Γ (s + β) s  (8.147) Γ (−s)(−z)   Γ (s + γ)  

π α1 +β1 −γ1 −1  |t| exp − |t + α2 | + |t + β2 | − |t + γ2 | + |t| − ϑt 2

α1 +β1 −γ1 −1  |t| exp (|ϑ| − π + ε)|t| . Since z ∈ C\[0, +∞) we have |ϑ| = | arg(−z) | < π. We take ε such that 0 < ε < π − |ϑ|. By (8.147) the integral on the right-hand side of (8.146) is a regular function of z in C\[0, +∞). For an integer N > max{k, |α1 |, |β1 |} and for a real T > max{|α2 |, |β2 |}, let R be the region in the s-plane obtained from the rectangle with vertices at k ± i T and N + 1/2 ± i T by deforming (if necessary) the left side (k − i T, k + i T ) to exclude from R the poles of Γ (s + α)Γ (s + β) and to include in R the poles s = 0, 1, . . . , N of Γ (−s). For fixed N and for T → +∞, by (8.147) the integrals along the upper and lower sides of R tend to zero. Thus the difference 1 2πi



N + 21 +i∞

 



− N + 21 −i∞

(k)

Γ (s + α)Γ (s + β) Γ (−s)(−z)s ds Γ (s + γ)

(8.148)

equals the sum of residues of the integrand at the poles s = 0, 1, . . . , N of Γ (−s). Hence (8.148) equals N  (−1)n+1 Γ (α + n)Γ (β + n) (−z)n n! Γ (γ + n) n=0

(8.149)

Γ (α)Γ (β)  (α)n (β)n z n . = − Γ (γ) n=0 (γ)n n! N

Let 1 JN = 2πi

N + 21 +i∞



N + 21 −i∞

Γ (s + α)Γ (s + β) Γ (−s)(−z)s ds. Γ (s + γ)

162

8 Hypergeometric Functions

We substitute s = N + 1/2 + it. Similarly to (8.139) we obtain |JN |  |z| N + 2 (8.150)  +∞  Γ (α + N + 1/2 + it) Γ (β + N + 1/2 + it)  e−ϑt   ×  Γ (γ + N + 1/2 + it) Γ (N + 3/2 + it)  cosh(πt) dt. 1

−∞

By Stirling’s formula (6.40) with z replaced by N + it and α replaced by 3/2 we get Γ (N + 3/2 + it) =

√

2π (N + it)

N +1+it −N −it

e

  1+O

1 |N + it|

 .

Combining this with (8.142) and the analogous asymptotic formulae with β and γ in place of α we obtain     Γ (α + N + 1/2 + it) Γ (β + N + 1/2 + it)   α+β−γ−1      Γ (γ + N + 1/2 + it) Γ (N + 3/2 + it)   (N + it)      it α+β−γ−1  =  N α+β−γ−1   1 +  N      it α1 +β1 −γ1 −1 it   = N α1 +β1 −γ1 −1 1 +  (α2 + β2 − γ2 ) exp − arg 1 + N N  π  it α1 +β1 −γ1 −1 α1 +β1 −γ1 −1  |α2 + β2 − γ2 | ≤ N exp 1 +  N 2  N α1 +β1 −γ1 −1 λα1 +β1 −γ1 −1 (t), where λν (t) is defined by (8.144). Hence (8.150) yields |JN |  |z|

N + 21

N

α1 +β1 −γ1 −1

+∞ λα1 +β1 −γ1 −1 (t)

−∞

e−ϑt dt. cosh(πt)

(8.151)

Since |ϑ| < π, the last integral converges. If |z| < 1, by (8.151) we get lim JN = 0,

N →+∞

whence, making N → +∞ in (8.148), (8.149), −

1 2πi

 (k)

∞

Γ (s + α)Γ (s + β) Γ (α)Γ (β)  (α)n (β)n z n Γ (−s)(−z)s ds = − . Γ (s + γ) Γ (γ) n=0 (γ)n n!

8.8 Mellin–Barnes’ Integral Representations

163

Thus we have proved (8.146) under the additional assumption |z| < 1. Since 2 F1 (α, β; γ; z) is a regular function of z in C\[1, +∞) and the right-hand side of (8.146) is a regular function of z in C\[0, +∞), we conclude that (8.146) holds for all z ∈ C\[0, +∞) by analytic continuation. 

Bibliography

This short list is a selection of few among the numerous classical treatises dealing with the theory of functions of one complex variable and with special functions. These items have been selected with particular regard to the subjects treated in the present lecture notes. G. E. Andrews, R. Askey, R. Roy. Special Functions, Encyclopedia of Mathematics and its Applications, vol. 71, Cambridge University Press, New York, 1999. A. Erdélyi et al. Higher Transcendental Functions, 3 volumes, Bateman Manuscript Project, McGraw-Hill, New York, 1953. H. Hochstadt. The Functions of Mathematical Physics, Dover Publications, New York, 1986. E. L. Ince. Ordinary Differential Equations, Dover Publications, New York, 1956. K. Knopp. Theory of Functions, parts I and II, translated by F. Bagemihl, Dover Publications, New York, 1996. N. N. Lebedev. Special Functions and their Applications, translated by R. A. Silverman, Dover Publications, New York, 1972. E. C. Titchmarsh. The Theory of Functions, 2nd edition, Oxford University Press, London, 1968. E. T. Whittaker, G. N. Watson. A Course of Modern Analysis, 4th edition, Cambridge University Press, London, 1927.

© Springer International Publishing Switzerland 2016 C. Viola, An Introduction to Special Functions, UNITEXT - La Matematica per il 3+2 102, DOI 10.1007/978-3-319-41345-7

165

Index

A Asymptotic series, 61 Auxiliary functions, 107

B Barnes integral for 1 F1 , 157 Barnes integral for 2 F1 , 160 Bernoulli numbers, 41 Bernoulli polynomials, 44 Binet first formula, 88 Binet second formula, 90 Blaschke factor, 29 Borel–Carathéodory theorem, 1

C Canonical product, 31 Characteristic equation, 100 Confluent hypergeometric differential equation, 140 Confluent hypergeometric function, 137 Confluent hypergeometric function of the second kind, 144 Contiguity formulae for 1 F1 , 138 Contiguous hypergeometric functions, 128

D Digamma-function, 82 Dominant function, 93

E Entire function, 10 Euler beta-function B(x, y), 69 Euler constant γ , 55

Euler formulae for ζ (2k), 43 Euler gamma-function Γ (z), 67 Euler integral representation of 2 F1 , 120 Euler limit formula for Γ (z), 70 Euler product for Γ (z), 72 Euler product for sinz, 39 Euler reflection formula, 73 Euler transformation formulae for 2 F1 , 130 Euler–MacLaurin summation formula, 58 Eulerian integral of the first kind, 69 Eulerian integral of the second kind, 67 Exponent of convergence, 30 F Fourier integral theorem, 156 Fourier transform, 155 Fractional linear transformations, 129 Frobenius method, 106 Fuchsian singular point, 104 Functional equations for Γ (z), 68, 73 G Gauss contiguity formulae, 128 Gauss formula for ψ(z), 87 Gauss multiplication formula, 74 Genus of a function, 37 H Hadamard theorem, 31 Hypergeometric differential equation, 116 Hypergeometric function 2 F1 , 118 I Indicial equation, 108

© Springer International Publishing Switzerland 2016 C. Viola, An Introduction to Special Functions, UNITEXT - La Matematica per il 3+2 102, DOI 10.1007/978-3-319-41345-7

167

168

Index Pochhammer symbol (α)n , 118 Principal logarithm, 5 Psi-function ψ(z), 82

Isolated essential singularity, 11

K Kronecker summation formula, 62 Kummer transformation formula for 139

L Landau symbol O, 27 Laplace transform, 150 Legendre duplication formula, 75 Liouville formula, 142

1 F1 ,

R Regular function, 1 Regular singular point, 104 Riemann zeta-function ζ (s), 40 S Schottky theorem, 5 Semifactorial N !!, 50 Stirling formula, 51, 77, 80, 86 Sum of consecutive nth powers, 46

M Mellin inversion formula, 156 Mellin transform, 155 Meromorphic function, 15 Mittag-Leffler theorem, 16

T Totally convergent series, 16 Totally fuchsian differential equation, 115

O Order of a function, 27

V Vinogradov symbols  and , 27

P Partial summation, 53 Picard first theorem, 10 Picard second theorem, 12

W Wallis formula, 50 Weierstrass factorization for Γ (z), 75 Weierstrass factorization theorem, 22