Multiple Sine Functions

Nobushige Kurokawa (Tokyo Institute of Technology) Shin-ya Koyama (Keio University)

Running title. Multiple sine functions Abstract. We construct a theory of multiple sine functions generalizing the usual sine function. As applications we have an expression for special values of zeta functions and we calculate gamma factors of Selberg zeta functions.

Mathematics Subject Classification 2000: Primary 11M36, 11M06

1 Introduction

Multiple sine functions are generalizations of the usual sine function

S_{1}(z) = 2 sin(πz) = 2πz ∏_{n=1}^{∞} (1 − \frac{z^{2}}{n^{2}} ) .  (1.1)

The double sine function S_{2}(z) was firstly studied by Hölder [H] in 1886 from

S_{2}(z) = e^{z} ∏_{n=1}^{∞} ((\frac{1 − \frac{z}{n}}{1 + \frac{z}{n}} )^{n} e^{2z}) .  (1.2)

Here we construct multiple sine functions S_{r}(z) for r ≥ 3 also, and we study their basic properties containing periodicity, special values, and algebraic differential equations. (Basic results were reported in [Ku1, Ku2, Ku3]; see also [Ma] for a survey.)

For example, the triple sine function is given by

S_{3}(z) = e^{\frac{z^{2}}{2}} ∏_{n=1}^{∞} ((1 − \frac{z^{2}}{n^{2}} )^{n^{2}} e^{z^{2}})  (1.3)

= exp (∫_{0}^{z} πt^{2} cot(πt)dt) .  (1.4)

Then we have the following expression for the famous mysterious value ζ(3) = ∑_{n=1}^{∞} \frac{1}{n^{3}} :

ζ(3) = \frac{8π^{2}}{7} log (S_{3} (\frac{1}{2})^{−1} 2^{\frac{1}{4}} ) .  (1.5)

We notice that this expression (1.5) originates from an attempt of Euler [E] in 1772.

After general study of multiple sine functions, this paper presents an application to the explicit calculation of the gamma factors of Selberg zeta functions in terms of multiple gamma functions Γ_{r}(s) of Barnes [B]. In this process, normalized multiple sine functions and the differential equations satisfied by the multiple sine functions and expressions like (1.5) are used crucially. The result is as follows.

Let M = Γ\G/K be a compact locally symmetric space of rank one. We denote by Z_{M}(s, σ) the Selberg zeta function with a unitary representation σ of Γ:

Z_{M}(s, σ) = ∏_{p∈Prim(M)} ∏_{λ≥0} det(1 − σ(p)N(p)^{−s−λ}),

where Prim(M) is the set of prime geodesics of M with their norm N(p) = exp(length(p)) and λ runs over a certain semi-lattice [G]. It is known that Z_{M}(s, σ) has an analytic continuation to all s ∈ ℂ as a meromorphic function of order dim M and has the following functional equation [Se, G, W]:

Z_{M}(2ρ_{0} − s, σ) = Z_{M}(s, σ) exp (vol(M) dim(σ) ∫_{0}^{s−ρ_{0}} μ_{M}(it)dt)  (1.6)

with ρ_{0} > 0 and μ_{M}(t) being the Plancherel measure.

We determine the gamma factor of Z_{M}(s, σ) and obtain the functional equation of symmetric type:

Theorem 1.1 Let M = Γ\G/K be an even dimensional compact locally symmetric space of rank one. Put

Γ_{M}(s, σ) = det (\sqrt{Δ_{M′} + ρ_{0}^{2}} + s − ρ_{0})^{vol(M) dim(σ)(−1)^{dim M/2}}

where det means the regularized determinant, and M′ = G′/K is the compact dual symmetric space with Δ_{M′} its Laplacian. Then

Γ_{M}(s, σ) = \{ \begin{matrix}(Γ_{2n}(s)Γ_{2n}(s + 1))^{vol(M) dim(σ)(−1)^{(dim M)/2−1}} & G = SO(1, 2n) \\ ( ∏_{k=0}^{n} Γ_{2n}(s + k)^{\binom{n}{k}}^{2})^{vol(M) dim(σ)(−1)^{(dim M)/2−1}} & G = SU(1, n) \\ ( ∏_{k=0}^{2n−1} Γ_{4n}(s + k)^{\frac{1}{2n}}^{\binom{2n}{k}}^{\binom{2n}{k+1}})^{−vol(M) dim(σ)} & G = Sp(1, n) \\ (Γ_{16}(s)Γ_{16}(s + 1)^{10}Γ_{16}(s + 2)^{28}Γ_{16}(s + 3)^{28} \\ ×Γ_{16}(s + 4)^{10}Γ_{16}(s + 5))^{−vol(M) dim(σ)} & G = F_{4} \end{matrix} (1.7)

The completed zeta function Ẑ_{M}(s, σ) = Γ_{M}(s, σ)Z_{M}(s, σ) satisfies the symmetric functional equation: Ẑ_{M}(s, σ) = Ẑ_{M}(2ρ_{0} − s, σ).

Remark 1.2 In the case of a Riemann surface (G = SO(1, 2) ≅ SU(1, 1)), this result was proved by Vigneras [Vi] and Cartier-Voros [CV]. If M is odd dimensional, the gamma factor is trivial (see Section 4). In the general case by using the Selberg trace formula we moreover have the following determinant expression similar to [Sa, Vo, Ko]:

Ẑ_{M}(s, σ) = e^{Q((s−ρ_{0})^{2})} det(Δ_{M} − s(2ρ_{0} − s)),

where Q is a polynomial with deg Q ≤ dim M/2.

Remark 1.3 This paper is an English version of a part of the following lecture note:

N. Kurokawa “Lectures on Multiple Sine Functions” (April-July, 1991, University of Tokyo, notes taken by S. Koyama, pp. 1-119).

Acknowledgment. The first author would like to express his hearty thanks to Christopher Deninger and Yu. I. Manin for their interests on the original lecture of 1991.

2 Multiple Sine Functions

In this section we introduce multiple sine functions, which will play the central role throughout this paper. We first introduce the multiple Hurwitz zeta function (see Barnes [B]). For ω_{1}, ..., ω_{r} > 0 and z ∈ ℂ, we put \underline{ω} = (ω_{1}, ..., ω_{r}) and

ζ_{r}(s, z, \underline{ω}) := ∑_{n≥0}(n · \underline{ω} + z)^{−s},  (2.1)

where n = (n_{1}, ..., n_{r}) ≥ 0 means n_{i} ≥ 0 and n_{i} ∈ ℤ for 1 ≤ i ≤ r, and n · \underline{ω} = n_{1}ω_{1} + ⋯ + n_{r}ω_{r}. The series (2.1) absolutely converges for Re(s) > r. It is analytically continued to s ∈ ℂ as a meromorphic function by the usual method (Barnes [B]) and holomorphic at s ∈ ℂ − {1, 2, ..., r}. We define the multiple gamma function by

Γ_{r}(z, \underline{ω}) := exp ζ′_{r}(0, z, \underline{ω}) = exp ( \frac{∂}{∂s}ζ_{r}(s, z, \underline{ω})|_{s=0}) ,

which was originally studied by Barnes [B]. We note that ζ_{1}(s, z, ω) = ω^{−s}ζ(s, \frac{z}{ω}) with ζ(s, z) the usual Hurwitz zeta function. Hence Γ_{1}(z, ω) = (2π)^{− \frac{1}{2}} Γ(\frac{z}{ω})ω^{\frac{z}{ω} − \frac{1}{2}} by Lerch’s formula. We define the r-ple sine functions S_{r}(z, \underline{ω}) and S_{r}(z) by

S_{r}(z, \underline{ω}) := Γ_{r}(z, \underline{ω})^{−1}Γ_{r}(|\underline{ω}| − z, \underline{ω})^{(−1)^{r}}  (2.2)

with |\underline{ω}| = ω_{1} + ⋯ + ω_{r} and for r ≥ 2

S_{r}(z) : = exp (\frac{z^{r−1}}{r − 1}) ∏_{n=1}^{∞} (P_{r} (\frac{z}{n}) P_{r} (−\frac{z}{n})^{(−1)^{r−1}})^{n^{r−1}}

= exp (\frac{z^{r−1}}{r − 1})\begin{matrix}∞ \\ ∏ & ′ \\ n=−∞ \end{matrix} P_{r} (\frac{z}{n})^{n^{r−1}}  (2.3)

with P_{r}(u) := (1 − u) exp(u + \frac{u^{2}}{2} + ⋯ + \frac{u^{r}}{r}). For example,

S_{2}(z) = e^{z} ∏_{n=1}^{∞} ((\frac{1 − \frac{z}{n}}{1 + \frac{z}{n}} )^{n} e^{2z}) ,

S_{3}(z) = e^{\frac{z^{2}}{2}} ∏_{n=1}^{∞} ((1 − \frac{z^{2}}{n^{2}} )^{n^{2}} e^{z^{2}}) .

We put

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

Taking r = 1 gives the usual sine function:

S_{1}(z, ω) = \frac{2π}{Γ(\frac{z}{ω})Γ(1 − \frac{z}{ω})} = 2 sin(\frac{πz}{ω}).

We set

S_{r}(z) := S_{r}(z; (1, ..., 1))

for simplicity. Thus

S_{1}(z) = S_{1}(z, 1) = S_{1}(z) = 2 sin(πz).

The double sine function S_{2}(z) was firstly studied by Hölder [H]. Later Shintani [Sh] used S_{2}(z, (ω_{1}, ω_{2})) to construct class fields over real quadratic fields. (Unfortunately they did not name the functions.) To distinguish multiple sine functions, we call S_{r}(z) the primitive multiple sine function and S_{r}(z, \underline{ω}) the normalized multiple sine function. The intimate relation between these two kinds of multiple sine functions is the main theme of this paper.

Theorem 2.1 The multiple sine function S_{r}(z, \underline{ω}) satisfies the following identities:

(a) For \underline{ω} = (ω_{1}, ..., ω_{r}) ∈ ℝ_{+}^{r} put \underline{ω}(i) = (ω_{1}, ..., ω_{i−1}, ω_{i+1}, ..., ω_{r}) ∈ ℝ_{+}^{r−1}, then we have

S_{r}(z + ω_{i}, \underline{ω}) = S_{r}(z, \underline{ω})S_{r−1}(z, \underline{ω}(i))^{−1},  (2.4)

where we put S_{0}(z, ·) ≡ −1.

(b) For a positive integer N, we have

S_{r}(Nz, \underline{ω}) = ∏_{0≤k_{i}≤N−1} S_{r} (z + \frac{k · \underline{ω}}{N} , \underline{ω}) ,  (2.5)

where the product is taken over the vectors k = (k_{1}, ..., k_{r}).

∏_{{0≤k_{i}≤N−1 \atop k≠0}} S_{r} (\frac{k · \underline{ω}}{N} , \underline{ω}) = N.

(d)

S_{r}(0, \underline{ω}) = 0.

(e) We have for any c > 0 the homogeneity

S_{r}(cz, c\underline{ω}) = S_{r}(z, \underline{ω}).

Proof. Since

ζ_{r}(s, z + ω_{i}, \underline{ω}) = ∑_{{n_{1},...,nr≥0 \atop n_{i}≥1}}(n_{1}ω_{1} + ⋯ + n_{r}ω_{r} + z)^{−s}
= ζ_{r}(s, z, \underline{ω}) − ζ_{r−1}(s, z, \underline{ω}(i)),

Γ_{r}(z + ω_{i}, \underline{ω}) = Γ_{r}(z, \underline{ω})Γ_{r−1}(z, \underline{ω}(i))^{−1}. Hence by |\underline{ω}| − (z + ω_{i}) = |\underline{ω}(i)| − z, we have

S_{r}(z + ω_{i}, \underline{ω}) = Γ_{r}(z + ω_{i}, \underline{ω})^{−1}Γ_{r}(|\underline{ω}| − (z + ω_{i}), \underline{ω})^{(−1)^{r}}
= (Γ_{r}(z, \underline{ω})Γ_{r−1}(z, \underline{ω}(i))^{−1})^{−1} (Γ_{r}(|\underline{ω}| − z, ω)Γ_{r−1}(|\underline{ω}(i)| − z, \underline{ω}(i)))^{(−1)^{r}}
= S_{r}(z, \underline{ω})S_{r−1}(z, \underline{ω}(i))^{−1},

which leads to (a).

Next we put

ξ_{r}(s, z, \underline{ω}) := −ζ_{r}(s, z, \underline{ω}) + (−1)^{r}ζ_{r}(s, |\underline{ω}| − z, \underline{ω}),  (2.6)

then

S_{r}(z, \underline{ω}) = exp(ξ′_{r}(0, z, \underline{ω})).  (2.7)

Since we need the details of the behavior of ξ_{r}(s, z, \underline{ω}) around s = 0, we describe the integral representation given by Riemann’s method:

ξ_{r}(s, z, \underline{ω}) = −\frac{Γ(1 − s)}{2πi} ∫_{C} φ(t, z, \underline{ω})(−t)^{s−1}dt,

where

φ(t, z, \underline{ω}) = \frac{−e^{−zt} + (−1)^{r}e^{(z−|\underline{ω}|)t}}{(1 − e^{−ω_{1}t}) ⋯ (1 − e^{−ω_{r}t})}

and C is the union of C_{1} : +∞ → +ε > 0, C_{2} : εe^{iθ} (0 ≤ θ ≤ 2π) and C_{3} : +ε → +∞. Thus ξ_{r}(s, z, \underline{ω}) is meromorphic in s ∈ ℂ. Put the coefficients a_{m}(z, \underline{ω}) ∈ ℂ to be

φ(t, z, \underline{ω}) = ∑_{m≥−r} a_{m}(z, \underline{ω})t^{m}

around t = 0. We compute ξ_{r}(−n, z, \underline{ω}) = (−1)^{n}n!a_{n}(z, \underline{ω}) and in particular ξ_{r}(0, z, \underline{ω}) = a_{0}(z, \underline{ω}).

To prove (b) we first compute that

ζ_{r}(s, Nz, \underline{ω}) = ∑_{n_{i}≥0}(n_{1}ω_{1} + ⋯ + n_{r}ω_{r} + Nz)^{−s}

= N^{−s} ∑_{n_{i}≥0} (\frac{n_{1}ω_{1} + ⋯ + n_{r}ω_{r}}{N} + z)^{−s}

= N^{−s} ∑_{0≤k_{i}≤N−1} ζ_{r} (s, z + \frac{k · \underline{ω}}{N} , \underline{ω}) .

Thus

ξ_{r}(s, Nz, \underline{ω}) = −ζ_{r}(s, Nz, \underline{ω}) + (−1)^{r}ζ_{r} (s, N (\frac{|\underline{ω}|}{N} − z) , \underline{ω})

= N^{−s} (− ∑_{0≤k_{i}≤N−1} ζ_{r} (s, z + \frac{k · \underline{ω}}{N} , \underline{ω})

+(−1)^{r} ∑_{0≤k_{i}≤N−1} ζ_{r} (s, \frac{|\underline{ω}|}{N} − z + \frac{k · \underline{ω}}{N} , \underline{ω}))

= N^{−s} ∑_{0≤k_{i}≤N−1} ξ_{r} (s, z + \frac{k · \underline{ω}}{N} , \underline{ω}) .

So we have

ξ′_{r}(0, Nz, \underline{ω}) = ∑_{0≤k_{i}≤N−1} ξ′_{r} (0, z + \frac{k · \underline{ω}}{N} , \underline{ω})

−(log N) ∑_{0≤k_{i}≤N−1} ξ_{r} (0, z + \frac{k · \underline{ω}}{N} , \underline{ω}) .

Therefore it suffices to show ξ_{r}(0, z, \underline{ω}) = 0. More generally we can show ξ_{r}(−n, z, \underline{ω}) = 0 for any even integer n ≥ 0. Indeed we see the function φ(t, z, \underline{ω}) is an odd function in t.

Then (c) is deduced from

\frac{S_{r}(Nz, \underline{ω})}{S_{r}(z, \underline{ω})} = ∏_{k≠0} S_{r} (z + \frac{k · \underline{ω}}{N} , \underline{ω})

with z = 0 substituted (see the proof of (d) below).

The assertion (d) follows from the following calculation:

ζ_{r}(s, z, \underline{ω}) = z^{−s} + ∑_{{n_{i}≥0 \atop n≠0}}(n_{1}ω_{1} + ⋯ + n_{r}ω_{r} + z)^{−s}

= z^{−s} − \frac{Γ(1 − s)}{2πi} ∫_{C} (\frac{1}{(1 − e^{−ω_{1}t}) ⋯ (1 − e^{−ω_{r}t})} − 1) e^{−zt}(−t)^{s−1}dt
= z^{−s} + O(1)

as z → 0. Thus ζ′_{r}(s, z, \underline{ω}) = −z^{−s} log z + O(1) and so ζ′_{r}(0, z, \underline{ω}) = − log z + O(1), which leads to

Γ_{r}(z, \underline{ω}) ∼ \frac{1}{c_{r}(\underline{ω})z}

as z → 0 with some constant c_{r}(\underline{ω}). We reach the result by substituting z = 0 to (2.2), since we have Γ_{r}(|\underline{ω}|, \underline{ω}) ≠ 0, ∞.

Lastly (e) follows from ξ_{r}(s, cz, c\underline{ω}) = c^{−s}ξ_{r}(s, z, \underline{ω}) and ξ_{r}(0, z, \underline{ω}) = 0.

Remark 2.2 The relation (c) indicates algebraicity of values at “division points”. For example let ϵ = \frac{5+\sqrt{21}}{2} be the fundamental unit of ℚ(\sqrt{21}) and take r = 2, \underline{ω} = (1, ϵ), N = 3. Then (c) is:

\begin{matrix}∏ & ′ \\ k_{i}=0,1,2 \end{matrix} S_{2} (\frac{k_{1} + k_{2}ϵ}{3}, (1, ϵ)) = 3.

In [Sh] Shintani proved a deep result on a similar product:

S_{2} (\frac{1}{3}, (1, ϵ)) S_{2} (1 + \frac{ϵ}{3}, (1, ϵ)) S_{2} (\frac{2 + 2ϵ}{3}, (1, ϵ)) = \sqrt{\frac{\frac{1+\sqrt{21}}{2} − \sqrt{\frac{3+\sqrt{21}}{2}}}{2}} .

We will deal with values at division points such as

S_{2} (\frac{ω_{1}}{2} , (ω_{1}, ω_{2})) = S_{2} (\frac{ω_{2}}{2} , (ω_{1}, ω_{2})) = \sqrt{2}

in a forthcoming paper [KK].

Remark 2.3 The above properties of the multiple sine functions S_{r}(z, \underline{ω}) generalize the well-known formulas of the usual sine function S_{1}(z, ω) = 2 sin \frac{πz}{ω}:

2 sin(Nθ) = ∏_{k=0}^{N−1} 2 sin (θ + \frac{kπ}{N} )  and  ∏_{k=1}^{N−1} 2 sin \frac{kπ}{N} = N.  (2.8)

Proposition 2.4 We have an expression:

S_{r}(z, \underline{ω}) = e^{Q\underline{ω}(z)}z(z − |\underline{ω}|)^{(−1)^{r−1}} \begin{matrix}∏ & ′ \\ n≥0 \end{matrix} P_{r} (−\frac{z}{n · \underline{ω}} ) P_{r} (\frac{z}{(n + 1) · \underline{ω}} )^{(−1)^{r−1}}

with Q\underline{ω}(z) a polynomial with deg Q\underline{ω} ≤ r and 1 := (1, ⋯, 1). Proof. We first compute

\frac{∂^{m}}{∂z^{m}} ζ_{r}(s, z, \underline{ω}) = (−1)^{m}s(s + 1) ⋯ (s + m − 1) ∑_{n≥0} \frac{1}{(z + n · \underline{ω})^{s+m}} .

It is absolutely convergent for Re(s) > r−m. In particular it converges at s = 0 if m ≥ r+1. We further compute that

\frac{∂^{m+1}}{∂z^{m}∂s}ζ_{r}(s, z, \underline{ω}) = (−1)^{m}(ms^{m−1} + ⋯ + (m − 1)!) ∑_{n≥0} \frac{1}{(z + n · \underline{ω})^{s+m}}

− (−1)^{m}s(s + 1) ⋯ (s + m − 1) ∑_{n≥0} \frac{log(z + n · \underline{ω})}{(z + n · \underline{ω})^{s+m}} .

Therefore if m ≥ r + 1, we have

\frac{∂^{m+1}}{∂z^{m}∂s}|_{s=0} ζ_{r}(s, z, \underline{ω}) = (−1)^{m}(m − 1)! ∑_{n≥0} \frac{1}{(z + n · \underline{ω})^{m}}

and

\frac{∂^{m+1}}{∂z^{m}∂s}|_{s=0} ζ_{r}(s, |\underline{ω}| − z, \underline{ω}) = (m − 1)! ∑_{n≥0} \frac{1}{(|\underline{ω}| − z + n · \underline{ω})^{m}}
= (−1)^{m}(m − 1)! ∑_{n≥0} \frac{1}{(z − (n + 1) · \underline{ω})^{m}} .

So we have for m ≥ r + 1,

\frac{d^{m}}{dz^{m}} log S_{r}(z, \underline{ω}) = \frac{∂^{m+1}}{∂z^{m}∂s}|_{s=0} ξ_{r}(s, z, \underline{ω})
= (−1)^{m+1}(m − 1)! ∑_{n≥0} (\frac{1}{(z + n · \underline{ω})^{m}} + \frac{(−1)^{r−1}}{(z − (n + 1) · \underline{ω})^{m}} ) ,

which is absolutely convergent for z ∉ \{−n · \underline{ω}, (n + 1) · \underline{ω} | n ≥ 0\}.

Next we deduce some properties of S_{r}(z). We recall that

S_{r}(z) = exp (\frac{z^{r−1}}{r − 1})\begin{matrix}∞ \\ ∏ & ′ \\ n=−∞ \end{matrix} P_{r} (\frac{z}{n})^{n^{r−1}} , (r ≥ 2)

where the product is taken over all nonzero integers n. We also defined

S_{1}(z) := 2πz\begin{matrix}∞ \\ ∏ & ′ \\ n=−∞ \end{matrix} P_{1} (\frac{z}{n}) = 2πz ∏_{n=1}^{∞} (1 − \frac{z^{2}}{n^{2}} ) = 2 sin πz.

Theorem 2.5 For r ≥ 2, we have S_{r}(0) = 1 and

\frac{S′_{r}}{S_{r}} (z) = πz^{r−1} cot(πz).

Consequently it holds that

S_{r}(z) = exp (∫_{0}^{z} πt^{r−1} cot(πt)dt) ,  (2.9)

where the contour lies in ℂ \ {±1, ±2, … }. Proof. We compute

\frac{S′_{r}}{S_{r}} (z) = z^{r−2} + ∑_{n=1}^{∞} n^{r−1} (\frac{1}{z − n} + \frac{(−1)^{r−1}}{z + n} + \frac{1}{n} ∑_{k=1}^{r} (\frac{z}{n})^{k−1} (1 + (−1)^{k+r−1}))
= z^{r−2} (\frac{1}{z} + ∑_{n=1}^{∞} \frac{2z}{z^{2} − n^{2}} )
= z^{r−1}π cot(πz).

Theorem 2.6 For r ≥ 2, the multiple sine function S_{r}(z) satisfies the following second order algebraic differential equation:

S′′_{r}(z) = (1 − z^{1−r})S′_{r}(z)^{2}S_{r}(z)^{−1} + (r − 1)z^{−1}S_{r}(z) − π^{2}z^{r−1}S_{r}(z)  (2.10)

with S_{r}(0) = 1 and S′_{r}(0) = \{ \begin{matrix}1 & (r = 2) \\ 0 & (r ≥ 3) \end{matrix} .

Proof. By the previous theorem we have

\frac{d}{dz} (\frac{1}{πz^{r−1}} \frac{S′_{r}}{S_{r}} (z)) = −\frac{π}{sin^{2} πz}
= −π(cot^{2}(πz) + 1)
= −π ((\frac{1}{πz^{r−1}} \frac{S′_{r}}{S_{r}} (z))^{2} + 1) .

Remark 2.7 We see (2.10) is analogous to Painlevé’s differential equation of type III. Moreover the multiple cosine function

C_{r}(z) = ∏_{{n=−∞ \atop n :odd}}^{∞} P_{r} (\frac{z}{(\frac{n}{2})})^{( \frac{n}{2} )^{r−1}} = S_{r}(2z)^{2^{1−r}}S_{r}(z)^{−1}

satisfies

\frac{C′_{r}}{C_{r}} (z) = −πz^{r−1} tan(πz)

and the algebraic differential equation (2.10).

The polylogarithm function Li_{k}(x) is defined by

Li_{k}(x) := ∑_{n=1}^{∞} \frac{x^{n}}{n^{k}} .

Theorem 2.8 For r ≥ 2, the following representations hold:

S_{r}(z) = exp (−\frac{(r − 1)!}{(2πi)^{r−1}} ∑_{k=0}^{r−1} \frac{(2πiz)^{k}}{k!} Li_{r−k}(e^{−2πiz}) + \frac{πi}{r}z^{r} + \frac{(r − 1)!}{(2πi)^{r−1}} ζ(r)) ,

(Im(z) < 0) (2.11)

S_{r}(z) = exp (−\frac{(r − 1)!}{(−2πi)^{r−1}} ∑_{k=0}^{r−1} \frac{(−2πiz)^{k}}{k!} Li_{r−k}(e^{2πiz}) − \frac{πi}{r}z^{r} + \frac{(r − 1)!}{(−2πi)^{r−1}} ζ(r)) ,

(Im(z) > 0) (2.12)

S_{r}(z) = (2 sin πz)^{z^{r−1}} exp ((−1)^{\frac{r}{2}} \frac{(r − 1)!}{(2π)^{r−1}} ∑_{{1≤k≤r−3 \atop k:odd}} \frac{(−1)^{\frac{k−1}{2}} (2πz)^{k}}{k!} ∑_{n=1}^{∞} \frac{cos(2πnz)}{n^{r−k}}

−(−1)^{\frac{r}{2}} \frac{(r − 1)!}{(2π)^{r−1}} ∑_{{0≤k≤r−2 \atop k:even}} \frac{(−1)^{\frac{k}{2}} (2πz)^{k}}{k!} ∑_{n=1}^{∞} \frac{sin(2πnz)}{n^{r−k}} ) ,

(2 ≤ r ∈ 2ℤ, 0 ≤ z < 1) (2.13)

S_{r}(z) = (2 sin πz)^{z^{r−1}} exp (−(−1)^{\frac{r−1}{2}} \frac{(r − 1)!}{(2π)^{r−1}} ∑_{{0≤k≤r−3 \atop k:even}} \frac{(−1)^{\frac{k}{2}} (2πz)^{k}}{k!} ∑_{n=1}^{∞} \frac{cos(2πnz)}{n^{r−k}}

−(−1)^{\frac{r−1}{2}} \frac{(r − 1)!}{(2π)^{r−1}} ∑_{{1≤k≤r−2 \atop k:odd}} \frac{(−1)^{\frac{k−1}{2}} (2πz)^{k}}{k!} ∑_{n=1}^{∞} \frac{sin(2πnz)}{n^{r−k}}

+(−1)^{\frac{r−1}{2}} \frac{(r − 1)!}{(2π)^{r−1}}ζ(r)) .

(3 ≤ r ∈ 1 + 2ℤ, 0 ≤ z < 1) (2.14)

Proof. When Im(z) < 0, by taking the contour t = uz (0 ≤ u ≤ 1) in (2.9) and taking into account that

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

we see

S_{r}(z) = exp (i ∫_{0}^{z} πt^{r−1} (1 + 2 ∑_{m=1}^{∞} e^{−2πimt}) dt) .

We reach the conclusion by calculating each term by integrating by parts:

∫_{0}^{1} t^{r−1}e^{αt}dt = (−1)^{r−1}(r − 1)!\frac{e^{α}}{α^{r}} (∑_{k=0}^{r−1} \frac{(−1)^{k}}{k!} α^{k} − e^{−α}) .

This completes the proof of (2.11). When Im(z) > 0, we deduce (2.12) similarly. For proving (2.13) and (2.14), it suffices to look at the logarithmic derivatives of the both sides since it is easy to see that the both sides equal 1 at z = 0. The direct calculation shows that the logarithmic derivatives of the right hand sides of (2.13) and (2.14) are equal to πz^{r−1} cot(πz) by trivial cancellations and the identity

log(2 sin πz) = − ∑_{n=1}^{∞} \frac{cos(2πnz)}{n}.

Alternatively we can show that

S_{r}(z) = (2 sin πz)^{z^{r−1}} exp (−\frac{(r − 1)!}{(2πi)^{r−1}} ∑_{k=0}^{r−2} \frac{(2πiz)^{k}}{k!} Li_{r−k}(e^{−2πiz})

+\frac{πi}{r}z^{r} − πiz^{r} + \frac{πi}{2}z^{r−1} + \frac{(r − 1)!}{(2πi)^{r−1}} ζ(r)) .

(0 ≤ z < 1) (2.15)

We will look at the logarithmic derivative of (2.15), since the both sides of (2.15) equal 1 at z = 0. By (2.9), the left hand side turns to

\frac{S′_{r}}{S_{r}} (z) = πz^{r−1} cot(πz).  (2.16)

We will show that the logarithmic derivative of the right hand side of (2.15) equals to (2.16). We have

\frac{d}{dz} log(right hand side of (2.15))

= \frac{d}{dz} (z^{r−1} log(2 sin(πz)) − \frac{(r − 1)!}{(2πi)^{r−1}} ∑_{k=0}^{r−2} \frac{(2πiz)^{k}}{k!} Li_{r−k}(e^{−2πiz})

+\frac{πi}{r}z^{r} − πiz^{r} + \frac{πi}{2}z^{r−1}) . (2.17)

The first term in the right hand side of (2.17) is equal to

(r − 1)z^{r−2} log(2 sin(πz)) + πz^{r−1} cot(πz)

whose second term agrees to (2.16). So it suffices to show

\frac{d}{dz} (−\frac{(r − 1)!}{(2πi)^{r−1}} ∑_{k=0}^{r−2} \frac{(2πiz)^{k}}{k!} Li_{r−k}(e^{−2πiz}) + \frac{πi}{r}z^{r} − πiz^{r} + \frac{πi}{2}z^{r−1})
= −(r − 1)z^{r−2} log(2 sin(πz)). (2.18)

By the formula

Li′_{k}(x) = \frac{1}{x}Li_{k−1}(x) (k ≥ 2),

the former part in the left hand side of (2.18) is equal to

−\frac{(r − 1)!}{(2πi)^{r−1}} ∑_{k=0}^{r−2} (L(k) − L(k + 1)) = \frac{(r − 1)!}{(2πi)^{r−1}} L(r − 1),  (2.19)

where we put

L(k) = \{ \begin{matrix}\frac{(2πi)^{k}}{(k − 1)!}z^{k−1}Li_{r−k}(e^{−2πiz}) & (1 ≤ k ≤ r − 1) \\ 0 & (k = 0). \end{matrix}

Then (2.19) is equal to

(r − 1)z^{r−2}Li_{1}(e^{−2πiz}) = −(r − 1)z^{r−2} log(1 − e^{−2πiz})
= −(r − 1)z^{r−2} log(e^{−πiz}(e^{πiz} − e^{−πiz}))
= −(r − 1)z^{r−2} (log(2 sin(πz)) + (\frac{1}{2} − z) πi) .

The term −(r − 1)z^{r−2}(\frac{1}{2} − z)πi cancells with the latter part of the left hand side of (2.18). This completes the proof of (2.15) and we obtain (2.13) and (2.14) by taking the absolute value.

Examples 2.9 (a) For r = 2

S_{2}(z) = (2 sin πz)^{z} exp (\frac{1}{2π} ∑_{n=1}^{∞} \frac{sin(2πnz)}{n^{2}}) .

In particular

S_{2} (\frac{1}{2}) = 2^{\frac{1}{2}} ,

S_{2} (\frac{1}{4}) = 2^{\frac{1}{8}} exp (\frac{1}{2π} ∑_{{n=1 \atop n:odd}}^{∞} \frac{(−1)^{\frac{n−1}{2}}}{n^{2}} )

= 2^{\frac{1}{8}} exp (\frac{1}{2π} L(2, χ_{−4})) ,

where χ_{−4} is the nontrivial Dirichlet character mod 4. Hence

L(2, χ_{−4}) = 2π log (S_{2} (\frac{1}{4}) 2^{− \frac{1}{8}} ) .

(b) For r = 3

S_{3}(z) = (2 sin πz)^{z^{2}} exp (\frac{1}{2π^{2}} ∑_{n=1}^{∞} \frac{cos(2πnz)}{n^{3}} + \frac{z}{π} ∑_{n=1}^{∞} \frac{sin(2πnz)}{n^{2}} − \frac{1}{2π^{2}} ζ(3)) .

In particular

S_{3} (\frac{1}{2}) = 2^{\frac{1}{4}} exp (\frac{1}{2π^{2}} ∑_{n=1}^{∞} \frac{(−1)^{n} − 1}{n^{3}})

= 2^{\frac{1}{4}} exp (−\frac{1}{π^{2}} ∑_{n:odd} \frac{1}{n^{3}} )

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

Hence

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

Theorem 2.10 The following identities hold: (a)

S_{r}(z + 1) = \frac{S′_{r}(1)}{2π} ∏_{k=1}^{r} S_{k}(z)^{\binom{r−1}{k−1}}.

(b)

S_{r}(Nz) = A_{r}(N) (S_{r}(z) ⋯ S_{r} (z + \frac{N − 1}{N}))^{N^{r−1}} ∏_{k=1}^{r−1} ∏_{a=1}^{N−1} S_{k} (z + \frac{a}{n})^{(−1)^{r−k}\binom{r−1}{k−1}a^{r−k}N^{k−1}}

with

A_{r}(N)^{−1} = (S_{r} (\frac{1}{N} ) ⋯ S_{r} (\frac{N − 1}{N}))^{N^{r−1}} ∏_{k=1}^{r−1} ∏_{a=1}^{N−1} S_{k} (\frac{a}{n})^{(−1)^{r−k}\binom{r−1}{k−1}a^{r−k}N^{k−1}} .

Proof. The logarithmic derivatives of the both sides of (a) coincide by Theorem 2.5, since

(z + 1)^{r−1} = ∑_{k=1}^{r−1} \binom{r − 1}{k − 1}z^{k−1}.

Calculating the both sides (divided by z) at z = 0 leads to (a).

For proving (b) we first appeal to (2.9) to obtain

\frac{d}{dz} log S_{r}(Nz) = Nπ(Nz)^{r−1} cot(πNz).

By the well-known formula (2.8), we see that

N cot(πNz) = \frac{1}{π} \frac{d}{dz} log S_{1}(Nz) = ∑_{a=0}^{N−1} \frac{1}{π} \frac{d}{dz} log S_{1} (z + \frac{a}{n}) = ∑_{a=0}^{N−1} cot π (z + \frac{a}{n}) .

Therefore we have

\frac{d}{dz} log S_{r}(Nz) = N^{r−1}πz^{r−1} ∑_{a=0}^{N−1} cot π (z + \frac{a}{N} ) .

Here we note that

z^{r−1} = (z + \frac{a}{N} )^{r−1} + ∑_{k=1}^{r−1} \binom{r − 1}{k − 1} (−\frac{a}{N} )^{r−k} (z + \frac{a}{N} )^{k−1} .

Thus

\frac{d}{dz} log S_{r}(Nz)

= N^{r−1}π ∑_{a=0}^{N−1} ((z + \frac{a}{N} )^{r−1} + ∑_{k=1}^{r−1} \binom{r − 1}{k − 1} (−\frac{a}{N} )^{r−k} (z + \frac{a}{N} )^{k−1}) cot π (z + \frac{a}{N} )
= N^{r−1} ∑_{a=0}^{N−1} \frac{d}{dz} (log S_{r} (z + \frac{a}{N} ) + ∑_{k=1}^{r−1}(−1)^{r−k}\binom{r − 1}{k − 1} (\frac{a}{N} )^{r−k} log S_{k} (z + \frac{a}{N} )) ,

which leads to the result. Remark 2.11 The constant S′_{r}(1) appearing in (a) is completely determined in Lemma 3.1. Examples 2.12

S′_{2}(1) = −2π ; S_{2}(z + 1) = −S_{2}(z)S_{1}(z),

S′_{3}(1) = −2π ; S_{3}(z + 1) = −S_{3}(z)S_{2}(z)^{2}S_{1}(z),
S′_{4}(1) = −2π exp(−6ζ′(−2)) ; S_{4}(z + 1) = − exp(−6ζ′(−2))S_{4}(z)S_{3}(z)^{3}S_{2}(z)^{3}S_{1}(z),

S′_{5}(1) = −2π exp(−12ζ′(−2)) ; S_{5}(z + 1) = − exp(−12ζ′(−2))S_{5}(z)S_{4}(z)^{4}S_{3}(z)^{6}S_{2}(z)^{4}S_{1}(z). Lemma 2.13 Let c(r, k) ∈ ℤ be defined by

c(r, k) = \frac{1}{k} ∑_{l=1}^{k}(−1)^{l−1}\binom{k}{l}l^{r}.

Then c(r, k) satisfies that

(−x)^{r−1} = ∑_{k=1}^{r} c(r, k)\binom{x + k − 1}{k − 1}

for an indeterminate x. In particular c(r, r) = (−1)^{r−1}(r − 1)!. Proof. Let S(n, k) be the Stirling number of the second kind [A] (13.3.16), which is the coefficient in the expansion

x^{n} = ∑_{k=0}^{n} S(n, k)(x)_{k},

where (x)_{k} = x(x − 1) ⋯ (x − k + 1). We will compute e^{xt} in two ways. First we deduce that

e^{xt} = (1 + (e^{t} − 1))^{x} = ∑_{k=0}^{∞} \frac{(x)_{k}}{k!} (e^{t} − 1)^{k}.

Secondly we calculate that

e^{xt} = ∑_{n=0}^{∞} (∑_{k=0}^{n} S(n, k)(x)_{k}) \frac{t^{n}}{n!} = ∑_{k=0}^{∞} (∑_{n≥k} \frac{S(n, k)}{n!}t^{n}) (x)_{k}.

Then we have

∑_{n≥k} \frac{S(n, k)}{n!}t^{n} = \frac{(e^{t} − 1)^{k}}{k!} = \frac{1}{k!} ∑_{l=0}^{k}(−1)^{k−l}e^{tl}\binom{k}{l} = \frac{1}{k!} ∑_{l=0}^{k}(−1)^{k−l}\binom{k}{l} ∑_{n=0}^{∞} \frac{(tl)^{n}}{n!} .

Therefore

S(n, k) = \frac{1}{k!} ∑_{l=0}^{k}(−1)^{k−l}\binom{k}{l}l^{n} = \frac{c(n, k)(−1)^{k−1}}{(k − 1)!} .

Thus

x^{n} = ∑_{k=0}^{n} \frac{c(n, k)(−1)^{k−1}}{(k − 1)!} (x)_{k},

hence

x^{n−1} = ∑_{k=0}^{n} c(n, k)\binom{−x + k − 1}{k − 1}.

Now, the fact c(n, n) = (−1)^{n−1}(n − 1)! is seen from comparing the coefficients of x^{n−1}. We recall that

S_{r}(z) = S_{r}(z, (1, ⋯, 1)) = Γ_{r}(z)^{−1}Γ_{r}(r − z)^{(−1)^{r}}.  (2.20)

Its relation to S_{r}(z) is given by the following theorem (the constant C_{r} will be determined in Theorem 3.5):

Theorem 2.14 For r = 1, 2, 3, ..., there exists a constant C_{r} such that

S_{r}(z) = C_{r} ∏_{k=1}^{r} S_{k}(z)^{c(r,k)}.  (2.21)

Proof. We remark that r = 1 case holds with C_{1} = 1 since

S_{1}(z) = S_{1}(z) = 2 sin(πz)

and c(1, 1) = 1. Hereafter we assume r ≥ 2. We first deduce that

S_{r}(z) = e^{P (z)} ∏_{k=1}^{r} S_{k}(z)^{c(r,k)}  (2.22)

for some polynomial P (z) such that deg P ≤ r. It suffices to show that

\frac{d^{r+1}}{dz^{r+1}} log S_{r}(z) = \frac{d^{r+1}}{dz^{r+1}} log (∏_{k=1}^{r} S_{k}(z)^{c(r,k)}) .  (2.23)

The left hand side is equal to

\frac{d^{r+1}}{dz^{r+1}} log (e^{\frac{z^{r−1}}{r−1}} \begin{matrix}∞ \\ ∏ & ′ \\ n=−∞ \end{matrix} P_{r} (\frac{z}{n})^{n^{r−1}})
= \frac{d^{r+1}}{dz^{r+1}} (\frac{z^{r−1}}{r − 1} + \begin{matrix}∞ \\ ∑ & ′ \\ n=−∞ \end{matrix} n^{r−1} (log (1 − \frac{z}{n}) + \frac{z}{n} + \frac{1}{2} (\frac{z}{n})^{2} + ⋯ + \frac{1}{r} (\frac{z}{n})^{r}))
= (−1)^{r}r!\begin{matrix}∞ \\ ∑ & ′ \\ n=−∞ \end{matrix} \frac{n^{r−1}}{(z − n)^{r+1}} .  (2.24)

Then as in the proof of Proposition 2.4, we have

\frac{d^{r+1}}{dz^{r+1}} log S_{r}(z) = (−1)^{r}r! ∑_{n=0}^{∞} _{r}H_{n} (\frac{1}{(z + n)^{r+1}} + \frac{(−1)^{r−1}}{(z − n − r)^{r+1}} ) ,

where _{r}H_{n} = \binom{n+r−1}{r−1}. Therefore

∑_{k=1}^{r} c(r, k)\frac{d^{r+1}}{dz^{r+1}} log S_{k}(z) = (−1)^{r}r! (∑_{n=0}^{∞} (∑_{k=1}^{r} c(r, k) _{k}H_{n}) \frac{1}{(z + n)^{r+1}}

+ ∑_{n=0}^{∞} (∑_{k=1}^{r} c(r, k) _{k}H_{n}(−1)^{k−1}) \frac{1}{(z − n − k)^{r+1}} ) .

The first sum over k is equal to (−n)^{r−1} = (−1)^{r−1}n^{r−1} by the previous lemma. In the second sum we replace n by n − k to get

∑_{n=0}^{∞} (∑_{k=1}^{r} c(r, k) _{k}H_{n−k}(−1)^{k−1}) \frac{1}{(z − n)^{r+1}} .

Here the sum over k is equal to n^{r−1}, because

(−1)^{k−1} _{k}H_{n−k} = (−1)^{k−1}\binom{n − 1}{k − 1} = (−1)^{k−1} \frac{(n − 1) ⋯ (n − k + 1)}{(k − 1)!}
= \frac{(−n + k − 1) ⋯ (−n + 1)}{(k − 1)!} = \binom{−n + k − 1}{k − 1} = _{k}H_{−n}.

Hence

∑_{k=1}^{r} c(r, k)\frac{d^{r+1}}{dz^{r+1}} log S_{k}(z) = (−1)^{r}r! (∑_{n=0}^{∞} \frac{(−1)^{r−1}n^{r−1}}{(z + n)^{r+1}} + ∑_{n=0}^{∞} \frac{n^{r−1}}{(z − n)^{r+1}} )

= \frac{d^{r+1}}{dz^{r+1}} log S_{r}(z)

by (2.24), and we reach (2.23). Thus we obtain (2.22).

Next we prove by induction on r that the polynomial P (z) is a constant. It holds by (2.22) that

S_{r}(z + 1) = e^{P (z+1)} ∏_{k=1}^{r} S_{k}(z + 1)^{c(r,k)}.  (2.25)

The left hand side is computed by Theorem 2.10 (a) as

c′S_{r}(z)S_{r−1}(z)^{r−1} ⋯ S_{k}(z)^{\binom{r−1}{k−1}} ⋯ S_{1}(z)

for some constant c′, which equals by (2.22) and by the assumption of induction for S_{1}(z), ..., S_{r−1}(z)

c′′ (e^{P (z)}S_{r}(z)^{c(r,r)} ⋯ S_{1}(z)^{c(r,1)}) (S_{r−1}(z)^{c(r−1,r−1)} ⋯ S_{1}(z)^{c(r−1,1)})^{r−1} ⋯ S_{1}(z)^{c(1,1)}

= c′′e^{P (z)}S_{r}(z)^{a(r)}S_{r−1}(z)^{a(r−1)} ⋯ S_{1}(z)^{a(1)}

with

a(k) = ∑_{l=k}^{r} \binom{r − 1}{l − 1}c(l, k).

The right hand side of (2.25) is by (2.4) equal to

e^{P (z+1)} ∏_{k=1}^{r} (S_{k}(z)S_{k−1}(z)^{−1})^{c(r,k)} = −e^{P (z+1)}S_{r}(z)^{c(r,r)}S_{r−1}(z)^{c(r,r−1)−c(r,r)} ⋯ S_{1}(z)^{c(r,1)−c(r,2)}

since c(r, r) = 1 and S_{0}(x, ·) = −1. Thus we have by comparing the both sides of (2.25)

c′′e^{P (z)}S_{r}(z)^{a(r)} ⋯ S_{1}(z)^{a(1)} = −e^{P (z+1)}S_{r}(z)^{c(r,r)}S_{r−1}(z)^{c(r,r−1)−c(r,r)} ⋯ S_{1}(z)^{c(r,1)−c(r,2)}.

So there exist b(k) ∈ ℤ such that

−c′′e^{P (z)−P (z+1)} = ∏_{k=1}^{r} S_{k}(z)^{b(k)}.  (2.26)

We can compare the order of zeros at z = −n (n = 1, 2, 3, ...) of both sides of (2.26) by using the identity

Γ_{k}(z)^{−1} = e^{Q_{k}(z)}z ∏_{n=1}^{∞} P_{k} (−\frac{z}{n})^{_{k}H_{n}}

with some Q_{k}(z) ∈ ℂ[z] such that deg Q_{k} ≤ k, which can be proved in exactly the same way as in the proof of Proposition 2.4 (see the proof of Theorem 3.7 below).

Thus we have for n = 1, 2, 3, ... that

∑_{k=1}^{r} b(k) _{k}H_{n} = ∑_{k=1}^{r} b(k)\binom{n + k − 1}{k − 1} = 0.  (2.27)

The left hand side of (2.27) is a polynomial in n whose degree is less than r. Therefore we have b(k) = 0 for k = 1, 2, ..., r. Thus c′′e^{P (z)−P (z+1)} = 1, and it is necessary that

P (z) = a + bz  (2.28)

for some constants a and b with b ∈ 2π\sqrt{−1}ℤ.

It remains to show that b = 0. We look at the identity

S_{r}(2z) = e^{P (2z)}S_{r}(2z)^{c(r,r)} ⋯ S_{1}(2z)^{c(r,1)}.  (2.29)

The left hand side of (2.29) is equal to by Theorem 2.10 (b) (N = 2)

S_{r}(2z) = c′′′ (S_{r}(z)S_{r} (z + \frac{1}{2}))^{2^{r−1}}

× S_{r−1} (z + \frac{1}{2})^{−\binom{r−1}{r−2}2^{r−2}} S_{r−2} (z + \frac{1}{2})^{\binom{r−1}{r−3}2^{r−3}} ⋯ S_{1} (z + \frac{1}{2})^{(−1)^{r−1}} .

So, by the assumption of the induction

S_{r}(2z) = c′′′′e^{2^{r−1}(P (z)+P(z+ \frac{1}{2}))} ∏_{k=1}^{r} S_{k}(z)^{c(k)}S_{k} (z + \frac{1}{2})^{d(k)}  (2.30)

for some c(k), d(k) ∈ ℤ. On the other hand the right hand side of (2.29) is equal to

e^{P (2z)} (S_{r}(z)S_{r} (z + \frac{1}{2})^{\binom{r}{1}} ⋯ S_{r} (z + \frac{r}{2})^{\binom{r}{r}})^{c(r,r)}

× (S_{r−1}(z) ⋯ S_{r−1} (z + \frac{r − 1}{2})^{\binom{r−1}{r−1}})^{c(r,r−1)} ⋯ (S_{1}(z)S_{1} (z + \frac{1}{2}))^{c(r,1)}

= e^{P (2z)} ∏_{k=1}^{r} S_{k}(z)^{c′(k)}S_{k} (z + \frac{1}{2})^{d′(k)} (2.31)

for some c′(k), d′(k) ∈ ℤ, where we used the formulas (2.4) and (2.5) for \underline{ω} = (1, ..., 1):

S_{r}(z + 1) = S_{r}(z)S_{r−1}(z)^{−1}

and

S_{r}(2z) = ∏_{k=0}^{r} S_{r} (z + \frac{k}{2})^{\binom{r}{k}} .

By comparing the order of zeros at z = −n and z = −n − \frac{1}{2} of (2.30) and (2.31) for n = 1, 2, 3, ..., we have c(k) = c′(k) and d(k) = d′(k) for k = 1, 2, ..., r. Hence

c′′′′e^{2^{r−1}(P (z)+P(z+ \frac{1}{2}))} = e^{P (2z)}.

Taking (2.28) into account, it follows that c′′′′e^{2^{r}bz+2^{r}a+2^{r−2}b} = e^{2bz+a} for all z ∈ ℂ. Hence b = 0 by r ≥ 2.

The following differential equation is crucial for later use.

Theorem 2.15

\frac{S′_{r}}{S_{r}} (z) = (−1)^{r−1}\binom{z − 1}{r − 1}π cot(πz).  (2.32)

Proof. The logarithmic derivative of (2.21) shows

\frac{S′_{r}}{S_{r}} (z) = ∑_{k=1}^{r} c(r, k)\frac{S′_{k}}{S_{k}} (z).

So by inverting it holds for some c′(r, k) ∈ ℚ that

\frac{S′_{r}}{S_{r}} (z) = ∑_{k=1}^{r} c′(r, k)\frac{S′_{k}}{S_{k}} (z).

Hence by Proposition 2.5 it follows that

\frac{S′_{r}}{S_{r}} (z) = (∑_{k=1}^{r} c′(r, k)z^{k−1}) π cot(πz).

Thus it suffices to prove that

∑_{k=1}^{r} c′(r, k)z^{k−1} = (−1)^{r−1}\binom{z − 1}{r − 1}.  (2.33)

By inverting (2.21) we have for some constant C′_{r} that

S_{r}(z) = C′_{r} ∏_{k=1}^{r} S_{k}(z)^{c′(r,k)}.

Let N be the least common multiple of the denominators of c′(r, k). We will compare the order of zeros of the both sides of

S_{r}(z)^{N} = C′_{r}^{N} ∏_{k=1}^{r} S_{k}(z)^{Nc′(r,k)}

at z = −m for m = 1, 2, 3, .... For the left hand side it is equal to the order of poles of Γ_{r}(z)^{N} at z = −m, which is N _{r}H_{m}. On the other hand for the right hand side it is equal

to N ∑_{k=1}^{r} c′(r, k)(−m)^{k−1}. Hence

∑_{k=1}^{r} c′(r, k)(−m)^{k−1} = _{r}H_{m} = \frac{(m + r − 1) ⋯ (m + 1)}{(r − 1)!}

for m = 1, 2, 3, .... Therefore as a polynomial in x, it holds that

∑_{k=1}^{r} c′(r, k)x^{k−1} = \frac{(−x + r − 1) ⋯ (−x + 1)}{(r − 1)!} = \frac{(−1)^{r−1}(x − 1) ⋯ (x − r + 1)}{(r − 1)!}

= (−1)^{r−1}\binom{x − 1}{r − 1}.

3 Calculations of Constants and Special Values

In this section we determine the constants C_{r} for r ≥ 2. As its application we obtain an expression of ζ(3) in terms of the triple sine function.

Lemma 3.1

S′_{r}(1) = −2π exp (−2 ∑_{{1<l<r \atop odd}} \binom{r − 1}{l − 1}ζ′(1 − l)) .

Proof. The case r = 1 is easily seen from S′_{1}(1) = −2π. Suppose that r ≥ 2. Then by the expressions (2.13) and (2.14) we have

S′_{r}(1) = \{ \begin{matrix}−2π exp (−\frac{(r − 1)!}{(2π)^{r−1}} ∑_{{1≤k≤r−3 \atop odd}} \frac{(2π)^{k}(−1)^{\frac{k−r+1}{2}}}{k!} ζ(r − k)) & (r ∈ 2ℤ) \\ −2π exp (−\frac{(r − 1)!}{(2π)^{r−1}} ∑_{{2≤k≤r−3 \atop even}} \frac{(2π)^{k}(−1)^{\frac{k−r+1}{2}}}{k!} ζ(r − k)) & (r ∈ 1 + 2ℤ, r ≥ 3) \end{matrix} ,

since the function f(z) = (2 sin πz)^{z^{r−1}} satisfies f(1) = 0 and

f′(1) = lim_{z→1}(2 sin πz)^{z^{r−1}−1} \frac{2 sin πz}{z − 1} = −2π.

Putting r − k = l, we see the both cases equal

S′_{r}(1) = −2π exp (−(r − 1)! ∑_{{1<l<r \atop odd}} \frac{(−1)^{\frac{l−1}{2}}}{(r − l)!(2π)^{l−1}} ζ(l))
= −2π exp (−(r − 1)! ∑_{{1<l<r \atop odd}} \frac{2}{(r − l)!(l − 1)!}ζ′(1 − l))
= −2π exp (−2 ∑_{{1<l<r \atop odd}} \binom{r − 1}{l − 1}ζ′(1 − l)) ,

where we used

ζ(l) = \frac{(2π)^{l−1}2(−1)^{\frac{l−1}{2}}}{(l − 1)!} ζ′(1 − l)

coming from the functional equation for ζ(s). Lemma 3.2 Let a(r, k) ∈ ℚ satisfy

\binom{X + r − 2}{r − 1} = ∑_{k=1}^{r−1} a(r, k)X^{k}.

Then we have

S_{r}(1) = exp (−2 ∑_{{2≤k≤r−1 \atop even}} a(r, k)ζ′(−k)) = exp (−2 ∑_{{3≤l≤r \atop odd}} a(r, l − 1)ζ′(1 − l)) . (3.1)

Proof. Since

ζ_{r}(s, z) = ζ_{r}(s, z, (1, ..., 1)) = ∑_{n_{1},...,n_{r}}(n_{1} + n_{2} + ⋯ + n_{r} + z)^{−s} = ∑_{n=0}^{∞} _{r}H_{n}(n + z)^{−s},

we see

ζ_{r}(s, 1) = ∑_{n=0}^{∞} _{r}H_{n}(n + 1)^{−s} = ∑_{n=1}^{∞} _{r}H_{n−1}n^{−s} = ∑_{n=1}^{∞} (∑_{k=1}^{r−1} a(r, k)n^{k}) n^{−s}
= ∑_{k=1}^{r−1} a(r, k)ζ(s − k).

Hence ζ′_{r}(s, 1) = ∑_{k=1}^{r−1} a(r, k)ζ′(s − k) and

Γ_{r}(1) = exp(ζ′_{r}(0, z))|_{z=1} = exp (∑_{k=1}^{r−1} a(r, k)ζ′(−k)) .  (3.2)

On the other hand

ζ_{r}(s, r − 1) = ∑_{n=0}^{∞} _{r}H_{n}(n + r − 1)^{−s} = ∑_{n=1}^{∞} _{r}H_{n−r+1}n^{−s}.

Since _{r}H_{n−r+1} = \frac{n(n−1)···(n−r+2)}{(r−1)!} = (−1)^{r−1} _{r}H_{−n−1}, we have

ζ_{r}(s, r − 1) = (−1)^{r−1} ∑_{n=1}^{∞} _{r}H_{−n−1}n^{−s} = (−1)^{r−1} ∑_{n=1}^{∞} (∑_{k=1}^{r−1} a(r, k)(−n)^{k}) n^{−s}
= (−1)^{r−1} ∑_{k=1}^{r−1} a(r, k)(−1)^{k}ζ(s − k).

Therefore

Γ_{r}(r − 1) = exp ((−1)^{r−1} ∑_{k=1}^{r−1} a(r, k)(−1)^{k}ζ′(−k)) .  (3.3)

The lemma follows from (3.2), (3.3) and S_{r}(1) = Γ_{r}(1)^{−1}Γ_{r}(r − 1)^{(−1)^{r}}. Remark 3.3 The number a(r, k) is a shifted version of the Stirling number of the first kind s(r, k) [A] (13.3.15):

(X)_{r} = ∑_{k=0}^{∞} s(r, k)X^{k}.

Lemma 3.4

∑_{k=l}^{r} c(r, k)a(r, l − 1) = (−1)^{l−1}\binom{r − 1}{l − 1}.

Proof. We have

(−1)^{r−1}(X − 1)^{r−1} = ∑_{k=1}^{r} c(r, k)\binom{X + k − 2}{k − 1}
= ∑_{k=1}^{r} c(r, k) (∑_{l=1}^{k} a(k, l − 1)X^{l−1})
= ∑_{l=1}^{r} (∑_{k=l}^{r} c(r, k)a(k, l − 1)) X^{l−1}.

Comparing the coefficients of X^{l−1} for the both sides leads to the result. Theorem 3.5 The constant C_{r} in Theorem 2.14 is given by

C_{r} = \{ \begin{matrix}1 & (r ∈ 2ℤ) \\ e^{2ζ′(1−r)} & (r ∈ 1 + 2ℤ, r ≥ 3) \end{matrix} .

Proof. Since S_{1}(1) = 0 and S′_{1}(1) = −2π, we have from (2.21) that

S′_{r}(1) = −2πC_{r} ∏_{k=2}^{r} S_{k}(1)^{c(r,k)}.

We will compute

C_{r} = −\frac{S′_{r}(1)}{2π} ∏_{k=2}^{r} S_{k}(1)^{−c(r,k)}.

It equals

exp (−2 ∑_{{1<l<r \atop odd}} \binom{r − 1}{l − 1}ζ′(1 − l) + 2 ∑_{{1<l≤r \atop odd}} (∑_{k=l}^{r} c(r, k)a(r, l − 1)) ζ′(1 − l)) ,  (3.4)

since Lemma 3.2 gives that

S_{k}(1) = exp (−2 ∑_{{1<l≤k \atop odd}} a(k, l − 1)ζ′(1 − l)) .

The sum over k in (3.4) is computed as \binom{r−1}{k−1} by Lemma 3.4 since l are odd. The theorem follows. Examples 3.6 We have

S_{1}(z) = S_{1}(z)
S_{2}(z) = S_{2}(z)^{−1}S_{1}(z)
S_{3}(z) = e^{2ζ′(−2)}S_{3}(z)^{2}S_{2}(z)^{−3}S_{1}(z)
S_{4}(z) = S_{4}(z)^{−6}S_{3}(z)^{12}S_{2}(z)^{−7}S_{1}(z)

and thus

S_{1}(z) = S_{1}(z)
S_{2}(z) = S_{2}(z)^{−1}S_{1}(z)
S_{3}(z) = e^{−ζ′(−2)}S_{3}(z)^{\frac{1}{2}} S_{2}(z)^{− \frac{3}{2}} S_{1}(z)
S_{4}(z) = e^{−2ζ′(−2)}S_{4}(z)^{− \frac{1}{6}} S_{3}(z)S_{2}(z)^{− \frac{11}{6}} S_{1}(z).

Theorem 3.7 It holds that

S_{r}(z) = C_{r} ∏_{k=1}^{r} Γ_{k}(z)^{−c(r,k)} (∏_{k=1}^{r} Γ_{k}(−z)^{−c(r,k)})^{(−1)^{r−1}} .

Proof. By substituting (2.20) to (2.21), we have

S_{r}(z) = C_{r} ∏_{k=1}^{r} Γ_{k}(z)^{−c(r,k)} ∏_{k=1}^{r} Γ_{k}(k − z)^{(−1)^{k}c(r,k)}.  (3.5)

The formula Γ_{k}(k − z) = Γ_{k}(k − 1 − z)Γ_{k−1}(k − 1 − z)^{−1} gives that

Γ_{k}(k − z) = ∏_{j=0}^{k} Γ_{j}(−z)^{a(k,j)}

with a(k, j) = (−1)^{k−j}\binom{k}{j} ∈ ℤ. We note a(k, 0) = (−1)^{k}. Thus we can put b(r, k) ∈ ℤ so that

∏_{k=1}^{r} Γ_{k}(k − z)^{(−1)^{k}c(r,k)} = (∏_{k=1}^{r} Γ_{k}(−z)^{−b(r,k)})^{(−1)^{r−1}} .

Therefore (3.5) becomes

S_{r}(z) = C_{r} ∏_{k=1}^{r} Γ_{k}(z)^{−c(r,k)} (∏_{k=1}^{r} Γ_{k}(−z)^{−b(r,k)})^{(−1)^{r−1}} .  (3.6)

To show that b(r, k) = c(r, k), we compute the order of zeros at z = n (n = 1, 2, 3, ...) for the both sides of (3.6). Some direct calculations show that

\frac{∂^{k+2}}{∂z^{k+1}∂s}ζ_{k}(s, z, \underline{ω})|_{s=0} = \frac{∂^{k+1}}{∂z^{k+1}} log (z \begin{matrix}∏ & ′ \\ n≥0 \end{matrix} P_{k} (−\frac{z}{n_{1}ω_{1} + ⋯ n_{k}ω_{k}} )) ,

where n ≥ 0 means the same as in (2.1). Thus we have

Γ_{k}(z, \underline{ω})^{−1} = e^{Q_{k}(z,\underline{ω})}z \begin{matrix}∏ & ′ \\ n≥0 \end{matrix} P_{k} (−\frac{z}{n · \underline{ω}} )

with some polynomial Q_{k} whose degree in z is not greater than k. When \underline{ω} = (1, ..., 1) it becomes

Γ_{k}(z)^{−1} = e^{Q_{k}(z)}z ∏_{n=1}^{∞} P_{k} (−\frac{z}{n})^{_{k}H_{n}}

with deg Q_{k} ≤ k. Hence the order of zeros at z = n (n = 1, 2, 3, ...) of (3.6) is

n^{r−1} = ∑_{k=1}^{r}(−1)^{r−1}b(r, k) _{k}H_{n}.

As this is valid for n = 1, 2, 3, ..., we deduce that b(r, k) = c(r, k).

Theorem 3.8 (a)
ζ(3) = \frac{8π^{2}}{7} log (S_{3} (\frac{1}{2})^{−1} 2^{\frac{1}{4}} ) .  (3.7)

(b)

ζ(3) = \frac{16π^{2}}{3} log (S_{3} (\frac{1}{2})^{−1} 2^{\frac{3}{8}} ) .  (3.8)

(c)

ζ(3) = 4π^{2} log(S_{3}(1)).  (3.9)

Proof. The assertion (a) is already proved in Example 2.9(b). For proving (b) we take r = 2 in (2.21) with Theorem 3.5 and have S_{2}(z) = S_{2}(z)^{−1}S_{1}(z), from which it follows that S_{2}(z) = S_{2}(z)^{−1}S_{1}(z). Putting r = 3 in (2.21) with Theorem 3.5 we have

S_{3}(z) = e^{2ζ′(−2)}S_{3}(z)^{2}S_{2}(z)^{−3}S_{1}(z),

or S_{3}(z) = e^{−ζ′(−2)}S_{3}(z)^{\frac{1}{2}} S_{2}(z)^{− \frac{3}{2}} S_{1}(z). Substituting z = \frac{1}{2} gives

S_{3} (\frac{1}{2}) = e^{−ζ′(−2)}S_{3} (\frac{1}{2})^{\frac{1}{2}} 2^{\frac{1}{4}} ,

where we used S_{2}(\frac{1}{2}) = \sqrt{2} in Example 2.9(a), which follows from

S_{2}(z) = (2 sin πz)^{z} exp (\frac{1}{2π} ∑_{n=1}^{∞} \frac{sin(2πnz)}{n^{2}})

for 0 ≤ z < 1. So using −ζ′(−2) = \frac{1}{4π^{2}} ζ(3) we have (3.8) from (3.7). Finally (c) follows from (3.1) for r = 3, which turns to S_{3}(1) = exp(−ζ′(−2)) = exp(\frac{1}{4π^{2}} ζ(3)).

Expectation 3.9 We expect S_{r}(ℚ) ⊂ \overline{ℚ} ∪ \{∞\} and S_{r}(ℚ) ⊂ \overline{ℚ} ∪ \{∞\}. These would

imply the transcendency of \frac{ζ(3)}{π^{2}} and ζ(3) by (3.7), (3.8) or (3.9) (cf. [KK] and [KW]).

Here we notice only that S_{2}(\frac{1}{2}) = \sqrt{2} as in Example 2.9(a) and S_{2}(\frac{1}{2}) = \sqrt{2} by S_{2}(\frac{1}{2}) = S_{2}(\frac{1}{2})^{−1}S_{1}(\frac{1}{2}) = \sqrt{2}. Similarly S_{2}(\frac{m}{2}) = (−1)^{[ \frac{m+1}{4} ]}2^{\frac{m}{2}} and S_{2}(\frac{m}{2}) = (−1)^{[ \frac{m}{4} ]}2^{1− \frac{m}{2}} for odd integers m: the former follows from S_{2}(\frac{1}{2}) = \sqrt{2} using S_{2}(z + 1) = −S_{2}(z)S_{1}(z) listed in Examples 2.12, and the latter is obtained by S_{2}(\frac{m}{2}) = S_{2}(\frac{m}{2})^{−1}S_{1}(\frac{m}{2}). (See Remark 2.2 also.)

4 Plancherel Measures

The Plancherel measure μ_{M}(t) and the constant ρ_{0} > 0 in the functional equation (1.6) are calculated by Miatello[Mi] as follows:

(0) G = SO(1, 2n − 1) (⇔ dim M : odd)

ρ_{0} = n − 1,
μ_{M}(it) : polynomial.

(1) G = SO(1, 2n)

dim M = 2n,

ρ_{0} = n − \frac{1}{2},

μ_{M}(it) = (−1)^{n}P_{M}(t)π tan(πt),

P_{M}(t) = \frac{2}{(2n − 1)!}t ∏_{k=1}^{n−1} (t^{2} − (k − \frac{1}{2})^{2}) .

(2) G = SU(1, 2n − 1)

dim M = 4n − 2,

ρ_{0} = n − \frac{1}{2},

μ_{M}(it) = −P_{M}(t)π tan(πt),

P_{M}(t) = \frac{2}{(2n − 1)!(2n − 2)!}t ∏_{k=1}^{n−1} (t^{2} − (k − \frac{1}{2})^{2}) .

(3) G = SU(1, 2n)

dim M = 4n,

ρ_{0} = n,

μ_{M}(it) = −P_{M}(t)π cot(πt),

P_{M}(t) = \frac{2}{(2n)!(2n − 1)!}t^{3} ∏_{k=1}^{n−1}(t^{2} − k^{2})^{2}.

(4) G = Sp(1, n)

ρ_{0} = n + \frac{1}{2},

dim M = 4n,

μ_{M}(it) = P_{M}(t)π tan(πt),

P_{M}(t) = \frac{2}{(2n + 1)!(2n − 1)!}t (t^{2} − (n − \frac{1}{2})^{2}) ∏_{k=1}^{n−1} (t^{2} − (k − \frac{1}{2})^{2})^{2} .

(5) G = F_{4}

ρ_{0} = \frac{11}{2},

dim M = 16,

μ_{M}(it) = P_{M}(t)π tan(πt),

P_{M}(t) = \frac{2}{11!4 · 5 · 6 · 7}t (t^{2} − \frac{1}{4})^{2} (t^{2} − \frac{9}{4})^{2} (t^{2} − \frac{25}{4}) (t^{2} − \frac{49}{4}) (t^{2} − \frac{81}{4}) .

In this section we will give a new expression of the Plancherel measures, which suggests the Betti type interpretation for the coefficients. In what follows we omit the case (0) since the gamma factor is “trivial” corresponding to the nonexistence of discrete series. We use the following combinatorial results:

Lemma 4.1 For integers n and m we have:

2nH_{m} +_{2n} H_{m−1} = \frac{(2m + 2n − 1)(m + 1) ⋯ (m + 2n − 2)}{(2n − 1)!}  (4.1)
= mult(m(m + n), Δ_{S^{2n}}),  (4.2)

∑_{k=0}^{n} \binom{n}{k}^{2} _{2n}H_{m−k} = \frac{(2m + n)(m + 1)^{2} ⋯ (m + n − 1)^{2}}{n!(n − 1)!}  (4.3)
= mult(m(m + n), Δ_{P_{ℂ}^{n}}),  (4.4)

∑_{k=0}^{2n−1} \frac{1}{2n} \binom{2n}{k}\binom{2n}{k + 1} _{4n}H_{m−k}

= \frac{(2m + 2n + 1)(m + 1)((m + 2) ⋯ (m + 2n − 1))^{2}(m + 2n)}{(2n + 1)!(2n − 1)!}  (4.5)
= mult(m(m + 2n + 1), Δ_{P_{H}^{n}}),  (4.6)

16H_{m} + 10 _{16}H_{m−1} + 28 _{16}H_{m−2} + 28 _{16}H_{m−3} + 10 _{16}H_{m−4} +_{16} H_{m−5}
= \frac{(2m+1)(m+1)(m+2)(m+3)(m+4)^{2}(m+5)^{2}(m+6)^{2}(m+7)^{2}(m+8)(m+9)(m+10)}{11!·4·5·6·7}  (4.7)
= mult(m(m + 11), Δ_{P_{O}^{2}}).  (4.8)

Proof. The identities (4.2), (4.4) are due to Cartan [C]. More generally the results of Cahn-Wolf [CW] give (4.2), (4.4), (4.6), (4.8). These are considered as real analytic analogs of the “Hirzebruch proportionality principle”.

It is easy to see (4.1). We compute (4.3) as follows:

∑_{k=0}^{n} \binom{n}{k}^{2} _{2n}H_{m−k} = ∑_{k=0}^{n} \binom{n}{k}^{2}\binom{m − 1 + 2n − k}{2n − 1}

= ∑_{k≥0} \binom{n}{k}^{2}\binom{m − 1 + n + k}{2n − 1}

= ∑_{k≥0} \binom{n}{k}^{2} ∑_{j≥0} \binom{k}{j}\binom{m − 1 + n}{2n − 1 − j},

where we used the Vandermond convolution ∑_{k≥0} \binom{m}{k}\binom{n}{l−k} = \binom{m+n}{l}. By changing the order of the sums it equals

∑_{j≥0} \binom{m − 1 + n}{2n − 1 − j} ∑_{k≥0} \binom{n}{k}^{2}\binom{k}{j}

= ∑_{j≥0} \binom{m − 1 + n}{2n − 1 − j}\binom{n}{j} ∑_{k≥0} \binom{n}{k}\binom{n − j}{n − k}

= ∑_{j≥0} \binom{m − 1 + n}{2n − 1 − j}\binom{n}{j}\binom{2n − j}{n}

= ∑_{j≥0} (\frac{2n}{m}\binom{m − 1 + n}{m − 1}\binom{m}{n − j}\binom{n}{j} − \frac{n}{m}\binom{m − 1 + n}{m − 1}\binom{m}{n − j}\binom{n − 1}{j − 1})

= \frac{2n}{m}\binom{m − 1 + n}{m − 1}\binom{m + n}{n} − \frac{n}{m}\binom{m − 1 + n}{m − 1}\binom{m + n − 1}{n − 1}
= \frac{2m + n}{n}\binom{m − 1 + n}{n − 1}^{2}.

Here we reached the right hand side of (4.3).

The identity (4.5) is proved as follows:

∑_{k=0}^{2n−1} \frac{1}{2n} \binom{2n}{k}\binom{2n}{k + 1} _{4n}H_{m−k}

= ∑_{k=0}^{2n−1} \frac{1}{2n}\binom{2n}{k}\binom{2n}{k + 1}\binom{4n + m − k − 1}{4n − 1}

= ∑_{k=0}^{2n−1} \frac{1}{2n}\binom{2n}{k + 1}\binom{2n}{k}\binom{2n + m + k}{4n − 1}  (k ↦ 2n − 1 − k)

= \frac{1}{2n} ∑_{k≥0} \binom{2n}{k + 1}\binom{2n}{k} ∑_{j≥0} \binom{k}{j}\binom{2n + m}{4n − 1 − j}

= \frac{1}{2n} ∑_{j≥0} \binom{2n + m}{4n − 1 − j} ∑_{k≥0} \binom{2n}{k + 1}\binom{2n}{k}\binom{k}{j}

= \frac{1}{2n} ∑_{j≥0} \binom{2n + m}{4n − 1 − j} ∑_{k≥0} \binom{2n}{k + 1}\binom{2n}{j}\binom{2n − j}{2n − k}

= \frac{1}{2n} ∑_{j≥0} \binom{2n + m}{4n − 1 − j}\binom{2n}{j}\binom{4n − j}{2n + 1}

= \frac{1}{2n} ∑_{j≥0} (\frac{4n}{m}\binom{2n + m}{2n + 1}\binom{m}{2n − 1 − j}\binom{2n}{j}

−\frac{2n}{m}\binom{2n + m}{2n + 1}\binom{m}{2n − 1 − j}\binom{2n − 1}{j − 1})

= \frac{2}{m}\binom{2n + m}{2n + 1}\binom{m + 2n}{2n − 1} − \frac{1}{m}\binom{2n + m}{2n + 1}\binom{m + 2n − 1}{2n − 2}.

This is equal to the right hand side of (4.5). We can verify (4.7) by direct calculations.

Theorem 4.2

P_{M}(t + ρ_{0}) = \{ \begin{matrix}_{2n}H_{t} + _{2n}H_{t} & G = SO(1, 2n) \\ ∑_{k=0}^{n} \binom{n}{k}^{2} _{2n}H_{t−k} & G = SU(1, n) \\ ∑_{k=0}^{2n−1} \frac{1}{2n}\binom{2n}{k}\binom{2n}{k + 1} _{4n}H_{t−k} & G = Sp(1, n) \\ _{16}H_{t} + 10 _{16}H_{t−1} + 28 _{16}H_{t−2} \\ +28 _{16}H_{t−3} + 10 _{16}H_{t−4} +_{16} H_{t−5} & G = F_{4}. \end{matrix}

Proof. Since we see P_{M}(m+ρ_{0}) = mult(m(m+2ρ_{0}), Δ_{M′}), the theorem holds as polynomials in t.

The following result represents that Plancherel measures are sums of logarithmic derivatives of multiple sine functions.

Theorem 4.3

exp (∫_{0}^{s−ρ_{0}} μ_{M}(it)dt)^{(−1)^{(dim M)/2}}

= \{ \begin{matrix}S_{2n}(s)S_{2n}(s + 1) & G = SO(1, 2n) \\ ∏_{k=0}^{n} S_{2n}(s + k)^{\binom{n}{k}}^{2} & G = SU(1, n) \\ ∏_{k=0}^{2n−1} S_{4n}(s + k)^{\frac{1}{2n}}^{\binom{2n}{k}}^{\binom{2n}{k+1}} & G = Sp(1, n) \\ S_{16}(s)S_{16}(s + 1)^{10}S_{16}(s + 2)^{28}S_{16}(s + 3)^{28}S_{16}(s + 4)^{10}S_{16}(s + 5) & G = F_{4}. \end{matrix}(4.9)

Proof. We first prove the case of SO(1, 2n). When s = ρ_{0} = n − \frac{1}{2}, the left hand side clearly equals to 1. The right hand side is computed as

S_{2n} (n − \frac{1}{2}) S_{2n} (n + \frac{1}{2}) = \frac{Γ_{2n} (n + \frac{1}{2})}{Γ_{2n} (n − \frac{1}{2})} \frac{Γ_{2n} (n − \frac{1}{2})}{Γ_{2n} (n + \frac{1}{2})} = 1.

So it suffices to compare the logarithmic derivative for the both sides of (4.9). For the right hand side we have

\frac{S′_{2n}}{S_{2n}} (s) + \frac{S′_{2n}}{S_{2n}} (s + 1) = − (\binom{s − 1}{2n − 1} + \binom{s}{2n − 1}) π cot(πs).

On the other hand the logarithmic derivative of the left hand side of (4.9) is equal to −P_{M}(s−n+\frac{1}{2})π cot(πs). Therefore all we have to prove is that P_{M}(s−n+\frac{1}{2}) = \binom{s−1}{2n−1}+\binom{s}{2n−1}. We compute

P_{M} (s − n + \frac{1}{2}) = \frac{2}{(2n − 1)!} (s − n + \frac{1}{2}) ∏_{k=1}^{n−1} ((s − n + \frac{1}{2})^{2} − (k − \frac{1}{2})^{2})

= (2s − 2n + 1)\frac{(s − 1)(s − 2) ⋯ (s − 2n + 2)}{(2n − 1)!}
= \frac{s(s − 1) ⋯ (s − 2n + 2)}{(2n − 1)!} + \frac{(s − 1)(s − 2) ⋯ (s − 2n + 1)}{(2n − 1)!}
= \binom{s}{2n − 1} + \binom{s − 1}{2n − 1}

G  K  G′  M′
SO(1, n) SO(n) SO(1 + n) S^{n}
SU(1, n) SU(n) SU(1 + n) P_{ℂ}^{n}
Sp(1, n) Sp(n) Sp(1 + n) P_{H}^{n}
F_{4}  Spin(9)  F′_{4}  P_{O}^{2}

Table 1: Compact Duals

as desired. The other cases are similarly proved by our using Lemma 4.1 and Theorem 4.2. Let M′ = G′/K be the compact dual symmetric spaces which are given in Table 1. We put

ζ (s, z, \sqrt{Δ_{M′} + ρ_{0}^{2}}) := ∑_{λ}(λ + z)^{−s}  (4.10)

where the sum is taken over all eigenvalues λ of \sqrt{Δ_{M′} + ρ_{0}^{2}} with Δ_{M′} being the Laplacian on M′.

Theorem 4.4

ζ (s, z − ρ_{0}, \sqrt{Δ_{M′} + ρ_{0}^{2}})

= \{ \begin{matrix}ζ_{2n}(s, z) + ζ_{2n}(s, z + 1) & G = SO(1, 2n) \\ ∑_{k=0}^{n} \binom{n}{k}^{2}ζ_{2n}(s, z + k) & G = SU(1, n) \\ ∑_{k=0}^{2n−1} \frac{1}{2n}\binom{2n}{k}\binom{2n}{k + 1}ζ_{4n}(s, z + k) & G = Sp(1, n) \\ ζ_{16}(s, z) + 10ζ_{16}(s, z + 1) + 28ζ_{16}(s, z + 2) \\ +28ζ_{16}(s, z + 3) + 10ζ_{16}(s, z + 4) + ζ_{16}(s, z + 5) & G = F_{4}. \end{matrix}

Proof. By expressing λ in terms of an eigenvalue μ of Δ_{M′}, we have

ζ (s, z − ρ_{0}, \sqrt{Δ_{M′} + ρ_{0}^{2}}) = ∑_{μ} (\sqrt{μ + ρ_{0}^{2}} + (z − ρ_{0}))^{−s} .

Now we carry out an explicit calculation for the case G = SU(1, n) by using Lemma 4.1. All other cases can be treated similarly. Since μ = m(m + 2ρ_{0}) for m = 0, 1, 2, ..., it holds that

ζ (s, z − ρ_{0}, \sqrt{Δ_{M′} + ρ_{0}^{2}}) = ∑_{m=0}^{∞} mult(m(m + 2ρ_{0}), Δ_{M′})(m + z)^{−s}.

Since we have ρ_{0} = \frac{n}{2} and M′ = P_{ℂ}^{n},

∑_{m=0}^{∞} mult(m(m + 2ρ_{0}), Δ_{P_{ℂ}^{n}})(m + z)^{−s} = ∑_{m=0}^{∞} ∑_{k=0}^{n} \binom{n}{k}^{2} _{2n}H_{m−k}(m + z)^{−s}

= ∑_{k=0}^{n} \binom{n}{k}^{2} ∑_{m=0}^{∞} _{2n}H_{m}(m + k + z)^{−s}

= ∑_{k=0}^{n} \binom{n}{k}^{2}ζ_{2n}(s, z + k).

Thus in particular we see that (4.10) is regular at s = 0.

Let A be an operator whose eigenvalues are 0 < a_{1} ≤ a_{2} ≤ a_{3} ≤ · · ·. We define the regularized determinant by

det(A) = ∐_{n=1}^{∞} a_{n} := exp (−ζ′_{A}(0)) ,

when the spectral zeta function ζ_{A}(s) := ∑_{n=1}^{∞} a_{n}^{−s} is regular at s = 0. (Cf. Deninger [D] and

Manin [Ma].) For example, the multiple gamma functions have the following determinant expressions:

Γ_{r}(z, \underline{ω}) = det(D\underline{ω} + z)^{−1}

and

Γ_{r}(z) = det(D_{r} + z)^{−1},

where

D\underline{ω} = ω_{1} \frac{∂}{∂t_{1}} + ⋯ + ω_{r} \frac{∂}{∂t_{r}} : ℂ[t_{1}, ⋯, t_{r}] ⟶ ℂ[t_{1}, ⋯, t_{r}]

and D_{r} := \frac{∂}{∂t_{1}} + ⋯ + \frac{∂}{∂t_{r}}. Consequently the multiple sine functions can also be expressed by some regularized determinants.

The regularity of (4.10) at s = 0 allows us to define

det (\sqrt{Δ_{M′} + ρ_{0}^{2}} + z) = exp (−ζ′ (0, z, \sqrt{Δ_{M′} + ρ_{0}^{2}})) .

Corollary 4.5

det (\sqrt{Δ_{M′} + ρ_{0}^{2}} + s − ρ_{0})^{−1}

= \{ \begin{matrix}Γ_{2n}(s)Γ_{2n}(s + 1) & G = SO(1, 2n) \\ ∏_{k=0}^{n} Γ_{2n}(s + k)^{\binom{n}{k}}^{2} & G = SU(1, n) \\ ∏_{k=0}^{2n−1} Γ_{4n}(s + k)^{\frac{1}{2n}}^{\binom{2n}{k}}^{\binom{2n}{k+1}} & G = Sp(1, n) \\ Γ_{16}(s)Γ_{16}(s + 1)^{10}Γ_{16}(s + 2)^{28}Γ_{16}(s + 3)^{28} \\ ×Γ_{16}(s + 4)^{10}Γ_{16}(s + 5) & G = F_{4} \end{matrix} (4.11)

Corollary 4.6

exp (∫_{0}^{s−ρ_{0}} μ_{M}(it)dt)^{(−1)^{(dim M)/2}} = (\begin{matrix}det & (\sqrt{Δ_{M′} + ρ_{0}^{2}} + (s − ρ_{0})) \\ det & (\sqrt{Δ_{M′} + ρ_{0}^{2}} − (s − ρ_{0})) \end{matrix})^{(−1)^{dim M/2}} . (4.12)

Proof. This is an immediate consequence from (4.9) and (4.11).

Proof of Theorem 1.1 The identities (1.7) are obtained from (4.11). The symmetric functional equation is deduced by (4.12).

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