On the Hurwitz Function for Rational Arguments

V. S. Adamchik

Department of Computer Science Carnegie Mellon University Pittsburgh, USA [email protected]

Abstract

Using functional properties of the Hurwitz zeta function and symbolic deriva- tives of the trigonometric functions, the function ζ(2n + 1, p/q) is expressed in several ways in terms of other mathematical functions and numbers, including in particular the Glaisher numbers.

2000 Mathematics Subject Classification. Primary 11M35, 33B99. Secondary 11B75, 33E20. Key Words and Phrases. Riemann zeta function, Hurwitz zeta function, multiple gamma function, Stirling numbers, Bernoulli numbers, Euler numbers, Glaisher’s numbers, derivatives of the cotangent.

1 Introduction

The Hurwitz zeta function, one of the fundamental transcendental functions, is traditionally defined (see [3, 4, 15, 17]) by the series

X∞ 1 ζ(s, a) = , <(s) > 1, a∈ / Z− (1) (k + a)s 0 k=0 The series converges absolutely for <(s) = σ > 1 and the convergence is uniform in every half-plane σ ≥ 1 + δ (δ > 0). Therefore, ζ(s, a) is analytic function of s in the half-plane <(s) = σ > 1. It satisfies the following integral representation (see [3], theorem 12.2) Z 1 ∞ xs−1e−ax ζ(s, a) = −x dx, <(s) > 1, <(a) > 0 (2) Γ(s) 0 1 − e x 1 1 Using the Taylor expansion 1/(1 − e ) = − x + 2 + O(x) and splitting (2) into two integrals Z ∞ µ ¶ 1 s−1 −ax 1 1 1 ζ(s, a + 1) = x e x − + dx Γ(s) 0 1 − e x 2 Z µ ¶ 1 ∞ 1 1 + xs−1e−ax − dx Γ(s) 0 x 2 we can analytically continue ζ(s, a) into the strip −1 < <(s) < 1. Evaluating the second integral, replacing a by a + 1 and making use of 1 ζ(s, a + 1) = ζ(s, a) − , <(a) > 0 (3) as we obtain 1−s −s Z ∞ µ ¶ a a 1 s−1 −ax 1 1 1 ζ(s, a) = + + x e x − + dx, s − 1 2 Γ(s) 0 1 − e x 2 (4) <(s) > −1, <(a) > 0

This could be further generalized to (see [17, p.93])

Xn Γ(k + s − 1) B ζ(s, a) = a−s + k a1−s−k+ Γ(s) k k=0 Ã ! Z ∞ Xn 1 1 Bk (5) xs−1e−ax − a1−k dx, Γ(s) 1 − ex k 0 k=0 n <(s) > −2b c − 1, <(a) > 0 2

where Bk are the Bernoulli numbers. We can also analytically continue the Hurwitz zeta function to the whole complex s-plane (except for a simple pole at s = 1) by means of the contour integral Z Γ(1 − s) xs−1eax ζ(s, a) = x dx (6) 2πi C 1 − e where the contour C is a loop that starts from −∞ along the lower side of the real axis, encircle the origin and then returns to −∞ along the upper side of the real axis. Among all integral representations for ζ(s, a), the Hermite integral has a definite place. Its derivation is based on the summation formula of Plana (see [4, p.22] or [17, p.90]) Z Z X∞ 1 ∞ ∞ f(ix) − f(−ix) f(k) = f(0) + f(x) dx + i dx (7) 2 e2πx − 1 k=0 0 0

2 Choosing f(k) in (7) as 1 f(k) = (k + a)s after some obvious evaluations, we readily find Z a−s a1−s ∞ (a + ix)−s − (a − ix)−s ζ(s, a) = + + i 2πx dx, 2 s − 1 0 e − 1 (8) <(a) > 0

Noting that 2 x (a + ix)−s − (a − ix)−s = sin(s arctan( )) (9) i (a2 + x2)s/2 a we rewrite (8) in an equivalent form ¡ ¡ ¢¢ −s 1−s Z ∞ x a a sin s arctan a ζ(s, a) = + + 2 s/2 dx, 2 s − 1 0 (x2 + a2) (e2πx − 1) (10) s 6= 1, <(a) > 0

This integral is known as Hermite’s integral of ζ(s, z). The Riemann zeta function ζ(s) is a special case of the Hurwitz zeta function

ζ(s, 1) = ζ(s)

Many series and integral representations for ζ(s) follows straightforwardly from correspon- dent representations for the Hurwitz zeta function. The same as the latter, the Riemann zeta function is an analytic function in the whole complex s-plane, except the pole at s = 1. Another special case of the Hurwitz function is the Bernoulli polynomials

B (a) ζ(−n, a) = − n+1 , n ∈ N (11) n + 1 0 This formula follows directly from integral (6) by means of the residue theorem Z µ ¶ Γ(n + 1) x−n−1eax 1 xeax ζ(−n, a) = x dx = −n! Res n+2 x (12) 2πi C 1 − e x=0 x e − 1 and the generating function for the Bernoulli numbers

xeax X∞ xn = B (a) (13) ex − 1 n n! k=0

3 The Hurwitz function is closely related to the multiple gamma function Γn(z) defined as a generalization of the classical Euler gamma function Γ(z), by the following recurrence- functional equation (for references and further reading see [1, 2, 5, 17]):

Γn+1(z) Γn+1(z + 1) = , z ∈ C, n ∈ N, Γn(z) Γ1(z) = Γ(z), (14)

Γn(1) = 1.

The simplest functional relationship between ζ(t, z) and Γn(z) is given by (we refer the reader to [1] for other relations of this type)

0 0 log Γ2(z + 1) + z log Γ(z) = ζ (−1, z) − ζ (−1), <(z) > 0, (15)

0 d where ζ (t, z) = dt ζ(t, z) is defined as an analytic continuation provided by the Hermite integral (10):

2 2 Z ∞ x 2 2 0 z z z 2 z arctan( z ) + x log(x + z ) ζ (−1, z) = log z − − log z + 2πx dx, 2 4 2 0 e − 1 (16) <(z) > 0.

The constant ζ0(−1) = ζ0(−1, 0) is known as the Glaisher-Kinkelin constant A, defined by 1 log A = − ζ0(−1). (17) 12 It satisfies the following integral representation Z 1 + log(2π) 1 ∞ x log x log A = − 2 x dx. (18) 12 2π 0 e − 1 The constant A originally appeared in papers by Kinkelin [11] and Glaisher [7, 8] on the asymptotic expansion when n → ∞ of the following product

11p 22p . . . nnp , p ∈ N.

For p = 1, we have µ ¶ 1 11 22 . . . nn = n!n Γ (n + 1) = A exp ζ0(−1, n + 1) − (19) 2 12

Generally, ³ ´ 11p 22p . . . nnp = exp ζ0(−p, n + 1) − ζ0(−p) (20)

4 From the integral (6) one can derive the Hurwitz series representation à ! Γ(1 − s) i X∞ e−2πina X∞ e2πina ζ(s, a) = e−πis/2 − eπis/2 (21) (2π)1−s n1−s n1−s n=1 n=1

valid in the half-plane <(s) < 0 and 0 < a < 1. If a 6= 1 this representation is also valid for <(s) < 1. The Dirichlet series in (21) can be rewritten in terms of the polylogarithm Lim(z):

Γ(1 − s) i ¡ ¢ ζ(s, a) = e−πis/2Li (e2πia) − eπis/2Li (e−2πia) (22) (2π)1−s 1−s 1−s

or, equivalently, (for <(s) > 1)

Γ(s) ¡ ¢ ζ(1 − s, a) = e−πis/2Li (e−2πia) + eπis/2Li (e2πia) (23) (2π)s s s

Setting a = 1, yields the functional equation for the Riemann zeta function

Γ(s) πs ζ(1 − s) = 2 cos( )ζ(s) (24) (2π)s 2

The result holds for all admissible s by analytic continuation. 2πia It is easy to see that the polylogarithm Lis(e ) is a linear combination of the Hurwitz zeta functions when parameter a is rational. Indeed, rearranging terms according to the residue classes mod q, we obtain

q q X X∞ e2πinp/q X X∞ 1 Li (e2πip/q) = = e2πinp/q s (qk + n)s (qk + n)s n=1 k=0 n=1 k=0 q 1 X n = e2πinp/qζ(s, ) qs q n=1 Now replacing the polylogarithms in (23) by the above finite sum, we obtain

q µ ¶ ¡ p¢ 2Γ(s) X πs 2πkp ¡ k ¢ ζ 1 − s, = cos − ζ s, , 1 ≤ p ≤ q (25) q (2πq)s 2 q q k=1 which holds true for all s 6= 0. The main object of this paper is to derive a family of closed form representations for p ζ(2n + 1, q ) by evaluating symbolic derivatives of certain trigonometric functions. We will also show how the results presented here relate to those that were obtained in other works.

5 2 Reflection formula

In this section we express symbolic derivatives of the cotangent function in a closed form. Simple observation d cot(z) = −(1 + cot2(z) dz

d2 cot(z) = 2 cot(z)(1 + cot2(z) dz2

dn suggests that dzn cot(z), n ∈ N is a polynomial in cot(z). Generally, some trigonometric and hyperbolic functions fall to the class of functions whose derivatives are polynomials in terms of the same function (see [10, 12, 13]).

Lemma 2.1 For any integer n > 1, µ ¶ d n Xn k! nno cot(z) = (2i)n(cot(z) − i) (i cot(z) − 1)k (26) dz 2k k j=1

© j ª where k are the Stirling subset numbers, defined by [16] ½ ¾ ½ ¾ nno n − 1 n − 1 nno = k + , = δ (27) k k k − 1 0 n

Proof. We start with the identity µ ¶ 2 cot(z) = i 1 + e2iz − 1

Setting eiz → y, we have

µ ¶n µ ¶n µ ¶ µ ¶n d eiz→y d 2 d 2 cot(z) = in+1 y 1 + = in+1 y dz dy y2 − 1 dy y2 − 1

1 Expanding y2−1 into a geometric series and then differentiating, we obtain

µ ¶n ∞ d eiz→y X cot(z) = −(2i)n+1 kny2k dz k=1 Next, we make a use of (see [12]) µ ¶ Xn nno 1 X∞ z k k! zk = kn , n ∈ N |z| < 1 (28) k 1 + z 1 + z k=1 k=1

6 which is proved by means of the generating function ½ ¾ Xn n kn = (−1)j(−k) (29) j j j=1

where (k)j is the Pochhammer symbol. Substituting (29) into the right hand side of (28), yields ½ ¾ µ ¶ 1 X∞ Xn n z k (−1)j(−k) z + 1 j j z + 1 k=1 j=1 Changing the order of summation and evaluating the inner sum µ ¶ X∞ m! y m (−k) yk = m 1 − y y − 1 k=1 proves (28). Therefore, in a view of (28), the derivative of the cotangent immediately reduces to

µ ¶n n+1 n n o µ 2 ¶k d eiz→y (2i) X n y cot(z) = k! dz y2 − 1 k 1 − y2 j=1 Finally, we set y = eiz into the right hand side to obtain formula (26). ¤

In the similar way we can derive a representation for the derivatives of the tangent µ ¶ d n Xn k! nno tan(z) = (−2i)n(tan(z) − i) (i tan(z) − 1)k (30) dz 2k k j=1 To complete the picture, by changing z → iz, we obtain representations for hyperbolic functions µ ¶ d n Xn (−1)kk! nno coth(z) = 2n(coth(z) − 1) (coth(z) + 1)k (31) dz 2k k j=1 µ ¶ d n Xn (−1)kk! nno tanh(z) = 2n(tanh(z) − 1) (tanh(z) + 1)k (32) dz 2k k j=1 Theorem 2.2 For any integer n > 1 and 0 < x < 1, ζ(n, 1 − x) + (−1)nζ(n, x) = ½ ¾ (2πi)n ¡ ¢ Xn−1 k! n − 1 ¡ ¢ (33) i cot(πx) + 1 i cot(πx) − 1 k 2(n − 1)! 2k k k=1 © j ª where k are the Stirling subset numbers.

7 Proof. The result follows from the reflection formula for the Hurwitz function

π dn−1 ζ(n, 1 − x) + (−1)nζ(n, x) = − cot(πx), n > 1, 0 < x < 1 (34) (n − 1)! dxn−1 and Lemma 2.1. ¤

Theorem 2.3 For n, p, q ∈ N and 1 ≤ p ≤ q, µ ¶ p p π2n+1 d 2n sin(2πp/q) ζ(2n + 1, 1 − ) − ζ(2n + 1, ) = − lim (35) q q (2n)! y→0 dy cos(2y) − cos(2πp/q)

Proof. We replace n by 2n + 1 and x by p/q in formula (33) and then use identity (28) to obtain p p i(−1)n(2π)2n+1 X∞ ζ(2n + 1, 1 − ) − ζ(2n + 1, ) = lim k2ny2k q q (2n)! y→exp( πip ) q k=1 where the series is understood as a meromorphic function in y ∈ C µ ¶ X∞ 1 d 2n 1 k2ny2k = − y 22n dy y2 − 1 k=1 Hence, µ ¶ p p i(−1)n(2π)2n+1 d 2n 1 ζ(2n + 1, 1 − ) − ζ(2n + 1, ) = − lim y (36) q q 22n(2n)! πip dy y2 − 1 y→exp( q ) It is easy to verify by direct evaluation that µ ¶ µ ¶ ¯ d 2n 1 d 2n 1 ¯ y = − z ¯ 2 2 ¯ dy y − 1 dz z − 1 1 z= y for n = 1, 2,... and, therefore, µ ¶ µ ¶ d 2n 1 d 2n 1 lim y = − lim y πip dy y2 − 1 πip dy y2 − 1 y→exp( q ) y→exp(− q ) Thus, formula (36) transforms to p p i(−1)nπ2n+1 ζ(2n + 1, 1 − ) − ζ(2n + 1, ) = − · q q (2n)! Ã µ ¶ µ ¶ ! d 2n 1 d 2n 1 lim y − lim y πip dy y2 − 1 πip dy y2 − 1 y→exp( q ) y→exp(− q )

8 Replacing y by y eπip/q and collecting terms, yields

µ ¶ 2πp p p (−1)nπ2n+1 d 2n 2 sin( )y2 ζ(2n + 1, 1 − ) − ζ(2n + 1, ) = − lim y q q q (2n)! y→1 dy 4 2πp 2 y − 2 cos( q )y + 1

which finally leads to (35) upon substitution y → exp(iy). ¤

3 Special values of ζ(2n + 1, p/q)

p p 1 5 1 3 1 2 1 In this section we consider special cases of ζ(2n + 1, q ), q = 6 , 6 , 4 , 4 , 3 , 3 , 2 when it is ex- pressible in terms of other transcendental functions and constants. Originally this problem (2n) p was solved by K.S. Kolbig [14], who expressed ψ ( q ) in finite sums of the Bernoulli num- p bers, using functional properties of the polylogarithm. In our approach, ζ(2n + 1, q ) are expressed in terms of generating functions of trigonometric functions. The case q = 2 is trivial and follows immediately from the multiplication formula (which, in turns, can be easily proved by rearranging terms in (1) according to the residue classes mod n)

Xn−1 k ζ(s, nz) = n−s ζ(s, z + ), n ∈ N (37) n k=0 Setting z = 1/2 and n = 1 in (37), we obtain 1 ζ(s, ) = (2s − 1) ζ(s) (38) 2 Next, we will derive other special cases of ζ(2n+1, p/q) based on the reflection formula (35).

Theorem 3.1 For n ∈ N,

1 ) ζ(2n + 1, ) n 2n+1 4 2n ¡ 2n+1 ¢ (−1) (2π) = 2 2 − 1 ζ(2n + 1) ± E2n (39) 3 4(2n)! ζ(2n + 1, ) 4

where En are the Euler numbers.

Proof. From the multiplication formula (37) with n = 4, z = 1/4 and Theorem 2.3 with p = 1, q = 4 we find ¯ µ ¶2n ¯ 1 ¡ ¢ π2n+1 d ¯ ζ(2n + 1, ) = 22n 22n+1 − 1 ζ(2n + 1) + sec(2y)¯ 4 2(2n)! dy ¯ y=0

9 Recalling the generating function for the Euler numbers

X∞ zn sech(z) = E n n! k=0 we conclude the proof. ¤

Formula (39) along with (33) suggests a functional relation between the Stirling and Euler numbers.

Lemma 3.2 For n ∈ N, µ ¶ ½ ¾ X2n i − 1 k 2n (1 − i) k! = E (40) 2 k 2n k=1

µ ¶ ½ ¾ X2n 1 + i k 2n (i + 1) − k! = E (41) 2 k 2n k=1 √ where i = −1.

Proof. We prove the first identity (40). By virtue of (28), we have µ ¶ ½ ¾ X2n i − 1 k 2n X∞ (1 − i) k! = −2i lim k2nzk 2 k z→i k=1 k=1 where the series is understood by the analytic continuation. Furthermore, since µ ¶ X∞ d n z knzk = z dz 1 − z k=1 we have µ ¶ ½ ¾ µ ¶ X2n i − 1 k 2n d 2n z (1 − i) k! = −2i lim z 2 k z→i dz 1 − z k=1 It is not difficult to verify the following chain of transformations µ ¶ µ ¶ µ ¶ µ ¶ d 2n z d 2n 1 1 d 2n 1 1 lim z = lim z = lim z − z→i dz 1 − z z→i dz 1 − z 2 z→i dz 1 − z 1 + z µ ¶ µ ¶ d 2n z d 2n z = lim z = i lim z z→i dz 1 − z2 z→1 dz 1 + z2

10 Therefore, µ ¶ ½ ¾ µ ¶ µ ¶ X2n i − 1 k 2n d 2n 2z (1 − i) k! = lim z (42) 2 k z→1 dz 1 + z2 k=1 On the other hand, µ ¶ µ ¶ d 2n d 2n 2y E2n = lim sech(z) = lim y (43) z→0 dz y→1 dy 1 + y2 where y = ez. Comparing (42) with (43), concludes the proof. ¤

Similarly to Theorem 3.1 we derive closed form representations for q = 3 and q = 6.

Theorem 3.3 For n ∈ N, 1 ζ(2n + 1, ) ) 6 1 = (32n+1 − 1)(22n+1 − 1)ζ(2n + 1) 5 2 ζ(2n + 1, ) 6 ¯ √ µ ¶2n ¯ π2n+1 3 d 1 ¯ ± ¯ (44) 2(2n)! dy 2 cos(2y) − 1¯ y=0

1 ζ(2n + 1, ) ) 3 1 = (32n+1 − 1)ζ(2n + 1) 2 2 ζ(2n + 1, ) 3 ¯ √ µ ¶2n ¯ π2n+1 3 d 1 ¯ ± ¯ (45) 2(2n)! dy 2 cos(2y) + 1¯ y=0

In particular, setting n = 1 and n = 3 in Theorems 3.1 and 3.3, we obtain

1 1 2π3 ζ(3, ) = π3 + 28ζ(3) ζ(3, ) = √ + 13ζ(3) 4 3 3 3 1 √ 1 5π5 ζ(3, ) = 2 3π3 + 91ζ(3) ζ(5, ) = + 496ζ(5) 6 4 3 1 2π5 1 22π5 ζ(5, ) = √ + 121ζ(5) ζ(5, ) = √ + 3751ζ(5) 3 3 3 6 3

11 In [9] (see also [6], p. 79), Glaisher defined special numbers Hn and Gn analogous to the Eulerian numbers 3 X∞ xn = H (46) 2(2 cos(x) − 1) n n! k=0 3 X∞ G xn = n (47) 2(2 cos(x) + 1) n + 1 n! k=0 Here are the first few values 3 H = ,H = 3,H = 33 0 2 2 4 H6 = 903,H8 = 46113,H2n+1 = 0 and 1 G = ,G = 1,G = 5 0 2 2 4 G6 = 49,G8 = 809,G2n+1 = 0 It is easy to see that representations (44) and (45) can be rewritten in terms of the Glaisher constants.

Corollary 3.4 For n ∈ N, 1 ) ζ(2n + 1, ) 2n+1 6 1 2n+1 2n+1 (2π) = (3 − 1)(2 − 1)ζ(2n + 1) ± √ H2n 5 2 2 3(2n)! ζ(2n + 1, ) 6 1 ) ζ(2n + 1, ) 2n+1 3 1 2n+1 (2π) = (3 − 1)ζ(2n + 1) ± √ G2n 2 2 2 3(2n + 1)! ζ(2n + 1, ) 3 On the other hand, inverting Corollary 3.4, Glaisher’s numbers are just a combination of the Hurwitz functions: √ µ ¶ 3 (2n)! 1 5 H = ζ(2n + 1, ) − ζ(2n + 1, ) , n ∈ N (48) 2n (2π)2n+1 6 6 √ µ ¶ 3 (2n + 1)! 1 2 G = ζ(2n + 1, ) − ζ(2n + 1, ) , n ∈ N (49) 2n (2π)2n+1 3 3

Acknowledgments

This research was supported by NSF grant CCR-0204003.

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