Certain Series Related to the Triple Sine Function

Shin-ya Koyama and Nobushige Kurokawa

Running title. Triple Sine Function Abstract. We compute special values of Dirichlet series whose coefficients are given by the inverse of certain binomial coefficients via the triple sine function. 2000 Mathematics Subject Classification: 11M06

1 Introduction

In the famous proof of the irrationality of ζ(3) due to Apéry [1], he used the following expression:

ζ(3) = \frac{5}{2} ∑_{n=1}^{∞} \frac{(−1)^{n−1}}{n^{3}\binom{2n}{n}} .

We refer to van der Poorten [2] and Koecher [3] for explanations and backgrounds. Also Apéry gave a proof of the irrationality of ζ(2) by using

ζ(2) = 3 ∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2n}{n}}.

We can find in [2] and [3] that

∑_{n=1}^{∞} \frac{1}{\binom{2n}{n}} = \frac{1}{3} + \frac{2\sqrt{3}π}{27} ,

∑_{n=1}^{∞} \frac{1}{n\binom{2n}{n}} = \frac{\sqrt{3}π}{9} ,

∑_{n=1}^{∞} \frac{1}{n^{2}\binom{2n}{n}} = \frac{π^{2}}{18},

∑_{n=1}^{∞} \frac{1}{n^{4}\binom{2n}{n}} = \frac{π^{4}}{3240}.

Unfortunately, it remains open to find the expression for

∑_{n=1}^{∞} \frac{1}{n^{3}\binom{2n}{n}}.

The purpose of this paper is to investigate this problem via the triple sine function S_{3}(x) studied in the previous papers [4, 5, 6, 7, 8]. We refer to the excellent survey of Manin [9]. Here the triple sine function is defined as

S_{3}(x) = e^{\frac{x^{2}}{2}} ∏_{n=1}^{∞} ((1 − \frac{x^{2}}{n^{2}} )^{n^{2}} e^{x^{2}}) ,  (1.1)

which is an entire function of order 3. This reminds us of the usual sine function, and actually we define the first sine function S_{1}(x) as

S_{1}(x) = 2 sin(πx) = 2πx ∏_{n=1}^{∞} (1 − \frac{x^{2}}{n^{2}} ) .

Our result is

Theorem 1

∑_{n=1}^{∞} \frac{1}{n^{3}\binom{2n}{n}} = 4π^{2} log S_{3} (\frac{1}{6}) .

This is obtained as a special case of

Theorem 2

∑_{n=1}^{∞} \frac{(2 sin πx)^{2n}}{n^{3}\binom{2n}{n}} = 4π^{2} log S_{3}(x)

for −1/2 ≦ x ≦ 1/2.

From Theorem 1 we moreover show the following identity:

Theorem 3 It holds that

∑_{n=1}^{∞} \frac{1}{n^{3}\binom{2n}{n}} = −\frac{4}{3}ζ(3) + \frac{\sqrt{3}π}{3} (2L(2, χ_{6}) − L(2, χ_{3})) ,

where χ_{6} and χ_{3} are the nontrivial characters modulo 6 and 3, respectively, and L(s, χ) is the Dirichlet L-function.

2 The Triple Sine Function

In this section we first recall basic properties of the triple sine function (1.1). We find from the definition that

log S_{3}(x) = \frac{x^{2}}{2} + ∑_{n=1}^{∞} (n^{2} log (1 − \frac{x^{2}}{n^{2}} ) + x^{2})

and thus

\frac{S′_{3}}{S_{3}} (x) = πx^{2} cot(πx),  (2.1)

where we used the identity

cot(πx) = \frac{x}{π} ∑_{n=−∞}^{∞} \frac{1}{x^{2} − n^{2}} .

Hence we have

log S_{3}(x) = π ∫_{0}^{x} t^{2} cot(πt)dt  (2.2)

as the both sides vanish when x = 0. By

cot(πt) = i\frac{1 + e^{−2iπt}}{1 − e^{−2iπt}} = i (1 + 2 ∑_{m=1}^{∞} e^{−2πimt}) , (Im(t) < 0)

it holds in Im(z) < 0 that

log S_{3}(z) = i ∫_{0}^{z} πt^{2} (1 + 2 ∑_{m=1}^{∞} e^{−2πimt}) dt,

where the contour is taken in Im(t) < 0. By integrating by parts, we compute

∫_{0}^{z} t^{2}e^{αt}dt = \frac{z^{2}e^{αz}}{α} − \frac{2ze^{αz}}{α^{2}} + \frac{2(e^{αz} − 1)}{α^{3}}.

Therefore the following expression holds for Im(z) < 0.

log S_{3}(z) = −\frac{2}{(2πi)^{2}} ∑_{n=1}^{∞} (\frac{e^{−2πizn} − 1}{n^{3}} + 2πiz \frac{e^{−2πizn}}{n^{2}} + \frac{(2πiz)^{2}}{2} \frac{e^{−2πizn}}{n} ) + \frac{πi}{3}z^{3}.

By taking the real part and by continuity, we have for x ∈ ℝ (0 < x < 1)

log S_{3}(x) = \frac{2}{(2π)^{2}} ∑_{n=1}^{∞} (\frac{cos(2πnx) − 1}{n^{3}} + \frac{2πx sin(2πnx)}{n^{2}} − \frac{(2πx)^{2} cos(2πnx)}{2n}) .

We appeal to the formula

log(2 sin πx) = − ∑_{n=1}^{∞} \frac{cos(2πnx)}{n}

to get

log S_{3}(x) = x^{2} log(2 sin πx) + \frac{1}{2π^{2}} ∑_{n=1}^{∞} \frac{cos(2πnx)}{n^{3}} + \frac{x}{π} ∑_{n=1}^{∞} \frac{sin(2πnx)}{n^{2}} − \frac{1}{2π^{2}} ζ(3). (2.3)

(We can prove this formula also by showing that both sides are 0 at x = 1 and that differentiations of both sides are equal to πx^{2} cot(πx).)

Now, letting x = \frac{1}{6} we obtain,

log S_{3} (\frac{1}{6}) = \frac{1}{2π^{2}} (∑_{n=1}^{∞} \frac{cos \frac{nπ}{3}}{n^{3}} − ζ(3)) + \frac{1}{6π} ∑_{n=1}^{∞} \frac{sin \frac{nπ}{3}}{n^{2}} .

The Dirichlet series with coefficients sin \frac{nπ}{3} and cos \frac{nπ}{3} are calculated as follows:

∑_{n=1}^{∞} \frac{sin \frac{nπ}{3}}{n^{2}} = \frac{\sqrt{3}}{2} ( ∑_{n≡1,2 (mod 6)} \frac{1}{n^{2}} − ∑_{n≡4,5 (mod 6)} \frac{1}{n^{2}} )

= \frac{\sqrt{3}}{2} (2 ( ∑_{n≡1 (mod 6)} \frac{1}{n^{2}} − ∑_{n≡5 (mod 6)} \frac{1}{n^{2}} ) − ( ∑_{n≡1,4 (mod 6)} \frac{1}{n^{2}} − ∑_{n≡2,5 (mod 6)} \frac{1}{n^{2}} ))

= \frac{\sqrt{3}}{2} (2L(2, χ_{6}) − L(2, χ_{3})) ,

∑_{n=1}^{∞} \frac{cos \frac{nπ}{3}}{n^{3}} = \frac{1}{2} ( ∑_{n≡1,5 (mod 6)} \frac{1}{n^{3}} − ∑_{n≡2,4 (mod 6)} \frac{1}{n^{3}} ) + ∑_{n≡0 (mod 6)} \frac{1}{n^{3}} − ∑_{n≡3 (mod 6)} \frac{1}{n^{3}}

= ∑_{n≡1,5 (mod 6)} \frac{1}{n^{3}} − \frac{1}{2} (  ∑_{n≡1,2,4,5 (mod 6)} \frac{1}{n^{3}} ) + ∑_{n≡0 (mod 6)} \frac{1}{n^{3}} − ∑_{n≡3 (mod 6)} \frac{1}{n^{3}}

= L(3, 1_{6}) − \frac{1}{2}L(3, 1_{3}) + ∑_{n≡0 (mod 6)} \frac{1}{n^{3}} − ∑_{n≡3 (mod 6)} \frac{1}{n^{3}}

= (1 − 2^{−3})(1 − 3^{−3})ζ(3) − \frac{1}{2}(1 − 3^{−3})ζ(3) + 6^{−3}ζ(3) − 3^{−3}(1 − 2^{−3})ζ(3)
= \frac{1}{3}ζ(3),

where 1_{m} denotes the trivial Dirichlet character modulo m. Thus we have Theorem 4

log S_{3} (\frac{1}{6}) = \frac{1}{2π^{2}} (−\frac{2}{3}ζ(3) + \frac{\sqrt{3}π}{6} (2L(2, χ_{6}) − L(2, χ_{3}))) .

Remark 5 When x = \frac{1}{2} in (2.3), we have

log S_{3} (\frac{1}{2}) = \frac{1}{4} log 2 + \frac{1}{2π^{2}} ∑_{n=1}^{∞} \frac{(−1)^{n} − 1}{n^{3}}
= \frac{1}{4} log 2 − \frac{1}{π^{2}} ∑_{n:odd} \frac{1}{n^{3}}
= \frac{1}{4} log 2 − \frac{7}{8π^{2}} ζ(3).

Hence

ζ(3) = \frac{8π^{2}}{7} log (S_{3} (\frac{1}{2})^{−1} 2^{\frac{1}{4}} )

as in [4], [7], [8].

3 Proofs of Theorems

Theorem 1 is a special case of Theorem 2. Theorem 3 is obtained from Theorems 1 and 4. Hence it suffices to prove Theorem 2. It is known by Euler ([2, p.203], [3, p.62]) that

∑_{n=1}^{∞} \frac{(2 sin(πx))^{2n}}{n^{2}\binom{2n}{n}} = 2π^{2}x^{2}

for −1/2 ≦ x ≦ 1/2. Therefore it follows that

∑_{n=1}^{∞} \frac{1}{n^{2}} \binom{2n}{n}^{−1}2^{2n}(sin(πx))^{2n−1} cos(πx) = 2π^{2}x^{2} cot(πx).

Integrating both sides, we have

\frac{1}{π} ∑_{n=1}^{∞} \frac{1}{n^{3}} \binom{2n}{n}^{−1}2^{2n−1}(sin(πx))^{2n} = 2π^{2} ∫_{0}^{x} t^{2} cot(πt)dt
= 2π log S_{3}(x)

by the equation (2.2). This completes the proof of Theorem 2.

References

[1] R. Apéry: Irrationalité de ζ2 et ζ3. Astérisque 61 (1979) 11-13.

[2] A. van der Poorten: A proof that Euler missed...: Apéry’s proof of the irrationality of ζ(3). Math. Intell. 1 (1979) 195-203.

[3] M. Koecher: Letters, Math. Intell. 2 (1980) 62-64.

[4] N. Kurokawa: Multiple sine functions and Selberg zeta functions. Proc. Japan Acad. 67A (1991) 61-64.

[5] N. Kurokawa: Gamma factors and Plancherel measures. Proc. Japan Acad. 68A (1992) 256-260.

[6] N. Kurokawa: Multiple zeta functions: an example. In \underline{Zeta Functions in Geometry}, volume 21 of Advanced Studies in Pure Math., pages 219-226, Kinokuniya, Tokyo 1992.

[7] N. Kurokawa and S. Koyama: Multiple sine functions. Forum Math. (in press)

[8] N. Kurokawa and M. Wakayama: On ζ(3). J. Ramanujan Math. Soc. 16 (2001) 205-214.

[9] Yu. I. Manin: Lectures on zeta functions and motives (according to Deninger and

Kurokawa). Astérisque 228 (1995) 121-163. Department of Mathematics, Keio University, 223-8522, Japan Department of Mathematics, Tokyo Institute of Technology, 152-8551, Japan