Division Values of Multiple Sine Functions

Shin-ya Koyama

Abstract. We refine a formula on values of multiple sine functions at division points. As applications we prove a formula on a sum of reciprocal trigonometric values, and obtain multiple modularity of a three variable modular function, which concerns a generalization of the Dedekind η function.

Key words: Multiple sine function; division values; rationality.

1 Introduction

Throughout this paper we put r ≥ 1 to be an integer and let

D_{r} = {ω = (ω_{1}, ⋯, ω_{r}) ∈ C^{r} |ω_{1}, ⋯, ω_{r} and 1 belong to one side with respect to a line crossing 0} .

For ω ∈ D_{r} and x ∈ C \ {−m_{1}ω_{1} − ⋯ − m_{r}ω_{r} | m_{j} ∈ Z_{≥0}}, the multiple gamma function is defined by

Γ_{r}(x; ω) = exp (ζ′(0, x; ω)) ,

where ζ_{r}(s, x; ω) is the multiple Hurwitz zeta function defined by

ζ_{r}(s, x; ω) = ∑_{m_{1}=0}^{∞} ⋯ ∑_{m_{r}=0}^{∞}(m_{1}ω_{1} + ⋯ + m_{r}ω_{r} + x)^{−s}

which is absolutely convergent in Re(s) > r and has a meromorphic continuation to the entire plane. The multiple sine function

S_{r}(x; ω) = Γ_{r}(x; ω)^{−1}Γ_{r}(ω_{1} + ⋯ + ω_{r} − x; ω)^{(−1)^{r}}

was first introduced by Kurokawa [K1], which was a generalization of Shintani’s function F (x; (ω_{1}, ω_{2})) := S_{2}(x; (ω_{1}, ω_{2})). In pursueing Kronecker’s Jugendtraum for a real quadratic field, Shintani [S] introduced F (x; (ω_{1}, ω_{2})) as a generalization of the sine function S_{1}(x) := S_{1}(x; 1) = 2 sin(πx). More precisely, for an L-function L_{K}(s, χ) = ∑_{𝔞} χ(𝔞)N(𝔞)^{−s} for a real quadratic field K, he expressed the value

exp(−L′_{K}(0, χ))

in terms of a product of division values of S_{2}(x; (1, ε)) with ε the fundamental unit of K. This is regarded as a generalization of a well-known formula over Q:

exp(−L′(0, χ)) = ∏_{k=1}^{N−1} S_{1} (\frac{k}{N} )^{χ(k)/2} ,  (1.1)

where L(s, χ) = ∑_{n=1}^{∞} χ(n)n^{−s} is the Dirichle L-function for a non-trivial primitive even character χ modulo N. The formula (1.1) gives a reason for the appearance of the sine function in Kronecker’s Jugendtraum over Q in the sense that the division values of S_{1}(x) generate the maximal abelian extension of Q. Discovering a function with an analogous property for totally real number fields is an open unsolved problem. Unfortunately very few sequent works are seen after Shintani’s pioneering work [S] in 1977.

The fundamental importance of division values of multiple sine functions is also seen from another aspect concerning special values of zeta functions. Indeed there is another direction for the extension of the formula (1.1). In a previous paper [KK2] we gave a generalization of (1.1) to s ≠ 0. In this generalization, we fix the period to be trivial as we consider over Q, and simply denote S_{r}(x) := S_{r}(x, (1, ⋯, 1)). In the case of N = 1, we proved for n = 1, 2, 3, ⋯ that

exp(−ζ′(−2n)) = ∏_{k=1}^{n} S_{2n+1}(k)^{a(2n+1,k)}

with explicitly determined positive integers a(2n + 1, k), and we also proved for χ being primitive odd modulo N that

exp(−L′(−1, χ)) = ∏_{k=1}^{N−1} (S_{2} (\frac{k}{N} )^{N} S_{1} (\frac{k}{N} )^{k})^{χ(k)/2} ,

and for χ non-trivial primitive even modulo N that

exp(−L′(−2, χ)) = ∏_{k=1}^{N−1} (S_{3} (\frac{k}{N} )^{2N^{2}} S_{2} (\frac{k}{N} )^{2Nk−3N^{2}} S_{1} (\frac{k}{N} )^{k^{2}})^{χ(k)/2} .

These enable to express unknown special values such as

ζ(3) = 4π^{2} log S_{3}(1),
ζ(5) = −\frac{4π^{4}}{3} log(S_{5}(1)S_{5}(2)^{11}),
ζ(7) = \frac{8π^{6}}{45} log(S_{7}(1)S_{7}(2)^{57}S_{7}(3)^{302}),
L(2, (\frac{−4}{∗})) = −\frac{π}{4} log (S_{2} (\frac{1}{4})^{4} S_{1} (\frac{1}{4}) S_{2} (\frac{3}{4})^{−4} S_{1} (\frac{3}{4})^{−3})

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

L(3, (\frac{12}{∗})) = \frac{\sqrt{3}π^{2}}{432} log (S_{3}(\frac{1}{12})^{288}S_{2}(\frac{1}{12})^{−408}S_{1}(\frac{1}{12})S_{3}(\frac{5}{12})^{−288}S_{2}(\frac{5}{12})^{312}S_{1}(\frac{5}{12})^{−25}

S_{3}(\frac{7}{12})^{−288}S_{2}(\frac{7}{12})^{264}S_{1}(\frac{7}{12})^{−49}S_{3}(\frac{11}{12})^{288}S_{2}(\frac{11}{12})^{−164}S_{1}(\frac{11}{12})^{121}) .

As shown in these two aspects concerning generalizations of the Dirichlet class number formula (1.1), the study of division values of multiple sine functions seems to be of fundamental importance. Above all, rationality at division points is of central interest.

Fix once and for all an integer N ≥ 2. A basic relevant fact concerning the rationality of division values is the following formula proved in [KK1]:

∏’_{0≤k_{1},...,k_{r}≤N−1} S_{r} (\frac{k_{1}ω_{1} + ⋯ + k_{r}ω_{r}}{N}; ω) = N.  (1.2)

(We remark that in [KK1] everything is done for ω_{j} > 0, but the formula (1.2) holds for ω ∈ D_{r} by the same proof.) It would be significant to make (1.2) more precise to get more detailed information by determining values of various partial products of the left hand side of (1.2). Kurokawa [K2] recently discovered a beautiful refinement of it for r = 2. He proved that the contributions from terms with k_{j} = 0 (j = 1, 2) are equal. Namely, he showed that

∏_{k_{1}=1}^{N−1} S_{2} (\frac{k_{1}ω_{1}}{N}; ω) = ∏_{k_{2}=1}^{N−1} S_{2} (\frac{k_{2}ω_{2}}{N}; ω) = \sqrt{N}  (1.3)

and that

∏_{1≤k_{1},k_{2}≤N−1} S_{2} (\frac{k_{1}ω_{1} + k_{2}ω_{2}}{N}; ω) = 1.  (1.4)

He also obtained some applications which will be explained later.

In this paper we first generalize (1.3) and (1.4) to the case of r = 4. A generalization of (1.4) is easily described. We show the following theorem (Theorem 4.4 below):

Theorem 1.1

∏_{1≤k_{1},k_{2},k_{3},k_{4}≤N−1} S_{4} (\frac{k_{1}ω_{1} + ⋯ + k_{4}ω_{4}}{N}; ω) = 1.  (1.5)

One may wonder what happens when r = 3. Since the situation essentially depends on the parity of r, the formula like (1.4) and (1.5) is not true for r being odd. Actually in Theorem 3.3 below, we obtain an alternative expression for r = 3 in terms of the division values of the triple gamma function.

For describing a generalization of (1.3) to r = 4, we start by defining the contribution of k_{j} = 0 to the product (1.2). Let A_{j} be the partial product of (1.2) over terms with k_{j} = 0 and k_{l} ≠ 0 for l ≠ j. We similarly denote by A_{j,l} and A_{j,l,m} the partial products of (1.2) over terms with k_{j} = k_{l} = 0 and k_{j} = k_{l} = k_{m} = 0, respectively, where all other coefficients are nonzero. Then by (1.2) and (1.5) we see for r = 4 that

(∏_{j=1}^{4} A_{j}) ( ∏_{1≤j<l≤3} A_{j,l}) ( ∏_{1≤j<l<m≤3} A_{j,l,m}) = N.

One of our goals is to determine the contribution from k_{j} = 0 for each j = 1, 2, 3, 4 to (1.2). The key idea is that the contribution of k_{j} = 0 from A_{j,l} to (1.2) should be regarded as (A_{j,l})^{\frac{1}{2}} , since the product A_{j,l} is shared by k_{j} = 0 and k_{l} = 0. This idea is also applicable for A_{j,l,m}, which is identified as the product of the three contributions from k_{j} = 0, k_{l} = 0 and k_{m} = 0. In this way we reach the following definition:

Ã_{4} := A_{4} (∏_{j=1}^{3} A_{j,4})^{\frac{1}{2}} ( ∏_{1≤i<j≤3} A_{i,j,4})^{\frac{1}{3}} ,  (1.6)

which we call the contribution of terms with k_{4} = 0 to the product (1.2). We similarly define Ã_{j} for j = 1, 2, 3, 4. Taking (1.5) into account, the formula (1.2) is equivalently written as

∏_{j=1}^{4} Ã_{j} = N.  (1.7)

One of the purposes of this paper is to determine each contribution Ã_{j} in (1.7). Our main theorem is as follows. Without loss of generality we only consider the case j = 4.

Theorem 1.2 The contribution from terms with k_{4} = 0 to the product (1.2) is given by

Ã_{4} = N^{\frac{1}{4} + \frac{1}{72} (N ∑_{i≠j}( \frac{ω_{4}}{ω_{j}} − \frac{ω_{i}}{ω_{j}} )+ \frac{1}{N} ∑_{i≠j}( \frac{ω_{j}}{ω_{4}} − \frac{ω_{i}}{ω_{j}} ))} ∏_{i≠j} ∏’_{0≤k_{1},k_{2}≤N−1} (\frac{Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; (ω_{1}, ω_{2}, ω_{3}))}{Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4}))})^{\frac{1}{2}}

× ∏_{k=1}^{N−1} (\frac{Γ_{3} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4}))}{Γ_{3} (\frac{kω_{j}}{N}; (ω_{1}, ω_{2}, ω_{3}))})^{\frac{2}{3}} (\frac{Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}))}{Γ_{2} (\frac{kω_{j}}{N}; (ω_{j}, ω_{4}))})^{\frac{1}{6}} ,

where the sums and the product over i ≠ j are taken over all ordered pairs (i, j) ∈ {1, 2, 3}^{2} with i ≠ j.

Corollary 1 If ω_{1} = ω_{2} = ω_{3} = ω_{4}, then it holds that

Ã_{j} = N^{\frac{1}{4}} (j = 1, 2, 3, 4).

This shows that Theorem 1.1 is a generalization of (1.3).

Example. When N = 2 and ω = (1, 1, 1, 1), we can compute each factor of Ã_{4} by using the results proved in later sections as follows:

A_{4} = S_{3} (\frac{3}{2})^{\frac{1}{2}} = 2^{− \frac{1}{16}} e^{\frac{3}{8} ζ′(−2)},

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

( ∏_{1≤i<j≤3} A_{i,j,4})^{\frac{1}{3}} = 2^{− \frac{1}{4}} S_{3} (\frac{1}{2})^{\frac{3}{2}} = 2^{\frac{5}{16}} e^{\frac{9}{8} ζ′(−2)}.

By multiplying the both sides we have

Ã_{4} = 2^{\frac{1}{4}} .

Division values of multiple sine functions have modular interpretations as indicated in [K2]. Let Im(τ) > 0 and q = e^{2πiτ}. Kurokawa showed that by putting

F (τ) = q^{− \frac{1}{48}} ∏_{n=0}^{∞} (1 + q^{n+ \frac{1}{2}} )

= \frac{η(τ)^{2}}{η(\frac{τ}{2})η(2τ)}

with the Dedekind η function

η(τ) = q^{\frac{1}{24}} ∏_{n=1}^{∞} (1 − q^{n}) ,

it holds that

F (−\frac{1}{τ}) = F (τ)  (1.8)

and that

∑_{n=1}^{∞} \frac{(−1)^{n}}{n sin(πnτ)} + ∑_{n=1}^{∞} \frac{(−1)^{n}}{n sin(πnτ^{−1})} = −\frac{π}{12} (τ + \frac{1}{τ}) .  (1.9)

Indeed he showed that both (1.8) and (1.9) are equivalent to (1.4) for N = 2 and (ω_{1}, ω_{2}) = (1, τ). In Section 7 we generalize such modular interpretations to r = 4.

The first result is a generalization of (1.9).

Theorem 1.3 Let 0 < arg(τ_{3}) < arg(τ_{2}) < arg(τ_{1}) < π and 0 < Re(τ_{1} + τ_{2} + τ_{3} + 1). Then it holds that

∑_{n=1}^{∞} \frac{(−1)^{n}}{n sin(πnτ_{1}) sin(πnτ_{2}) sin(πnτ_{3})}

= 8πζ_{4} (0, \frac{τ_{1} + τ_{2} + τ_{3} + 1}{2}, (τ_{1}, τ_{2}, τ_{3}, 1)) − ∑_{n=1}^{∞} \frac{(−1)^{n}}{n sin(π \frac{n}{τ_{1}} ) sin(π \frac{nτ_{2}}{τ_{1}} ) sin(π \frac{nτ_{3}}{τ_{1}} )}

− ∑_{n=1}^{∞} \frac{(−1)^{n}}{n sin(π \frac{nτ_{1}}{τ_{2}} ) sin(π \frac{n}{τ_{2}} ) sin(π \frac{nτ_{3}}{τ_{2}} )} − ∑_{n=1}^{∞} \frac{(−1)^{n}}{n sin(π \frac{nτ_{1}}{τ_{3}} ) sin(π \frac{nτ_{2}}{τ_{3}} ) sin(π \frac{n}{τ_{3}} )},

where ζ_{4} (0, \frac{τ_{1}+τ_{2}+τ_{3}+1}{2}, (τ_{1}, τ_{2}, τ_{3}, 1)), the value of the multiple Hurwitz zeta function, is explicitly given in terms of τ_{j} (j = 1, 2, 3) in Lemma 7.1.

In the following theorem we obtain multiple modularity of the three variable function

F (τ_{1}, τ_{2}, τ_{3}) = ∏_{n_{1}=0}^{∞} ∏_{n_{2}=0}^{∞} ∏_{n_{3}=0}^{∞} (1 + q_{1}^{n_{1}+ \frac{1}{2}} q_{2}^{n_{2}+ \frac{1}{2}} q_{3}^{n_{3}+ \frac{1}{2}} )

for 0 < arg(τ_{3}) < arg(τ_{2}) < arg(τ_{1}) < π, 0 < Re(τ_{1} + τ_{2} + τ_{3} + 1) and q_{j} = e^{2πiτ_{j}} with j = 1, 2, 3.

Theorem 1.4 Put x = (τ_{1}+τ_{2}+τ_{3}+1)/2 and τ = (τ_{1}, τ_{2}, τ_{3}, 1). Then we have the following multiple modularity:

F (τ_{1}, τ_{2}, τ_{3}) = e^{πiζ_{4}(0,x;τ)} \frac{F (−\frac{1}{τ_{1}} , −\frac{τ_{2}}{τ_{1}} , −\frac{τ_{3}}{τ_{1}} ) F (\frac{τ_{1}}{τ_{3}} , \frac{τ_{2}}{τ_{3}} , −\frac{1}{τ_{3}} )}{F (\frac{τ_{1}}{τ_{2}} , −\frac{1}{τ_{2}} , −\frac{τ_{3}}{τ_{2}} )},

where ζ_{4}(0, x; τ), the value of the multiple Hurwitz zeta function, is explicitly given in terms of τ_{j} (j = 1, 2, 3) in Lemma 7.1.

Finally in Section 8 we investigate the behavior of the multiple sine function for real positive periods ω ∈ R_{>0}^{r}, and locate some two-division points. In the previous paper [KK3] we draw a graph of S_{2}(x; (ω_{1}, ω_{2})), and more recently Kurokawa [K3] did it for S_{3}(x; (ω_{1}, ω_{2}, ω_{3})). We present a generalization of these works to r = 4. We will show in Theorem 8.1 that S_{4}(x, ω) has four extremal values in the fundamental period (0, |ω|) with |ω| = ω_{1} + ω_{2} + ω_{3} + ω_{4}, and that each interval of (0, |ω|/2) and (|ω|/2, |ω|) has both a maximal and a minimal values. The behavior of S_{4}(x, ω) is roughly shown in Figure 1.

O y

x |ω| \frac{|ω|}{2}

1

α β

Figure 1. The graph of S(x) = S_{4}(x; ω).

We will see in Theorem 8.2 that there exist 2-division points both in (α, β) and (β, |ω|/2).

Thoughout the proof we use the following notation. Notation. For r = 3 and ω = (ω_{1}, ω_{2}, ω_{3}), we put for any integer N

ω_{1}^{N} = (ω_{1}, Nω_{2}, Nω_{3}),  ω_{2}^{N} = (Nω_{1}, ω_{2}, Nω_{3}),  ω_{3}^{N} = (Nω_{1}, Nω_{2}, ω_{3}),
ω_{1}^{(N)} = (Nω_{1}, ω_{2}, ω_{3}),  ω_{2}^{(N)} = (ω_{1}, Nω_{2}, ω_{3}),  ω_{3}^{(N)} = (ω_{1}, ω_{2}, Nω_{3}),

Acknowledgement. The author thanks Professor Nobushige Kurokawa for his showing the ideas in the preprint [K2] and his encouragement for this work.

2 Stirling Modular Forms

The Stirling modular form is defined as

ρ_{r}(ω)^{−1} = lim_{x→0} xΓ(x; ω).

We introduce an auxiliary zeta function as

ζ_{r}(s; ω) = ∑_{m_{1}=1}^{∞} ⋯ ∑_{m_{r}=1}^{∞}(m_{1}ω_{1} + ⋯ m_{r}ω_{r})^{−s}.

It is related to ζ_{r}(s, x; ω) as

lim_{x→0} (ζ_{r}(s, x; ω) − x^{−s})

= ζ_{r}(s; ω) + ∑_{j=1}^{r} ζ_{r−1}(s; ω(j)) + ∑_{{1≤j,k≤r \atop j<k}} ζ_{r−2}(s; ω(j, k)) + ⋯ + (ω_{1}^{−s} + ⋯ + ω_{r}^{−s}) ζ(s). (2.1)

Proposition 2.1

ζ_{2}(0; (ω_{1}, ω_{2})) = \frac{ω_{1}^{2} + ω_{2}^{2} + 3ω_{1}ω_{2}}{12ω_{1}ω_{2}} = \frac{1}{12} (\frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{1}} + 3) ,

ζ_{3}(0; ω) = −\frac{1}{24} (\frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{3}} + \frac{ω_{3}}{ω_{1}} + \frac{ω_{2}}{ω_{1}} + \frac{ω_{1}}{ω_{3}} + \frac{ω_{3}}{ω_{2}} + 3) = −\frac{1}{24} ∑_{i, j} \frac{ω_{i}}{ω_{j}} ,

where the sum is taken over all ordered pairs (i, j) with i, j ∈ {1, 2, 3}. Proof. We first show the following two identities in turn.

lim_{x→0} ζ_{2}(0, x; (ω_{1}, ω_{2})) = \frac{ω_{1}^{2} + ω_{2}^{2} + 3ω_{1}ω_{2}}{12ω_{1}ω_{2}} = \frac{1}{12} (\frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{1}} + 3) ,

lim_{x→0} ζ_{3}(0, x; ω) = \frac{1}{24} (\frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{3}} + \frac{ω_{3}}{ω_{1}} + \frac{ω_{2}}{ω_{1}} + \frac{ω_{1}}{ω_{3}} + \frac{ω_{3}}{ω_{2}} + 3) .

We appeal to an integral representation

lim_{x→0} ζ_{2}(s, x; ω) = −\frac{Γ(1 − s)}{2πi} ∫_{C} \frac{(−t)^{s−1}}{(1 − e^{−ω_{1}t})(1 − e^{−ω_{2}t})}dt,

where C is the standard coutour consisting of +∞ → ε > 0, εe^{iθ} (0 ≤ θ ≤ 2π), ε → +∞. By the expansion given by [B], we have

lim_{x→0} ζ_{2}(0, x; ω) = \frac{1}{2πi} ∫_{C} \frac{t^{−1}}{(1 − e^{−ω_{1}t})(1 − e^{−ω_{2}t})}dt
= \frac{1}{2πi} ∫_{C} ( \frac{_{2}S_{1}^{(3)}(0)}{t} − _{2}S_{1}^{(2)}(0) + _{2}S′_{1}(0)t + ⋯) \frac{dt}{t^{2}}
= _{2}S′_{1}(0)
= \frac{ω_{1}^{2} + ω_{2}^{2} + 3ω_{1}ω_{2}}{12ω_{1}ω_{2}} .

For the second identity by putting

f(t) = \frac{t}{(1 − e^{−ω_{1}t})(1 − e^{−ω_{2}t})(1 − e^{−ω_{3}t})}
= \frac{a}{t^{2}} + \frac{b}{t} + c + dt + et^{2} + ⋯,

we compute

lim_{x→0} ζ_{3}(0, x; ω) = \frac{1}{2πi} ∫_{C} \frac{t^{−1}}{(1 − e^{−ω_{1}t})(1 − e^{−ω_{2}t})(1 − e^{−ω_{3}t})}dt
= \frac{1}{2πi} ∫_{C} \frac{f(t)}{t^{2}}dt = d.

We calculate each coefficient in order as follows:

a = \frac{1}{ω_{1}ω_{2}ω_{3}} ,
b = \frac{ω_{1} + ω_{2} + ω_{3}}{2ω_{1}ω_{2}ω_{3}} ,
c = \frac{ω_{1}^{2} + ω_{2}^{2} + ω_{3}^{2} + 3(ω_{1}ω_{2} + ω_{2}ω_{3} + ω_{3}ω_{1})}{12ω_{1}ω_{2}ω_{3}},
d = \frac{1}{24} (\frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{3}} + \frac{ω_{3}}{ω_{1}} + \frac{ω_{2}}{ω_{1}} + \frac{ω_{1}}{ω_{3}} + \frac{ω_{3}}{ω_{2}} + 3) .

(Under the notation in [B], we write

a = _{3}S_{1}^{(4)}(0), b = − _{3}S_{1}^{(3)}(0), c = _{3}S_{1}^{(2)}(0), d = − _{3}S′_{1}(0), e = \frac{_{3}S′_{2}(0)}{2}.)

Hence we obtain the two values lim_{x→0} ζ_{2}(0, x; ω) and lim_{x→0} ζ_{3}(0, x; ω).

From (2.1) it follows that

lim_{x→0} ζ_{2}(0, x; ω) = 1 + ζ_{2}(0; ω) + ζ_{1}(0, ω_{1}) + ζ_{1}(0, ω_{2})
= 1 + ζ_{2}(0; ω) + (ω_{1}^{−s} + ω_{2}^{−s})_{s=0} ζ(0)
= ζ_{2}(0; ω),

and

lim_{x→0} ζ_{3}(0, x; ω) = 1 + ζ_{3}(0; ω) + ∑_{{1≤i,j≤3 \atop i≠j}} ζ_{2}(0; (ω_{i}, ω_{j})) + ∑_{j=1}^{3} ζ_{1}(0, ω_{j})
= 1 + ζ_{3}(0; ω) + \frac{1}{12} ( ∑_{{1≤i,j≤3 \atop i≠j}} \frac{ω_{i}}{ω_{j}} + 9) − \frac{3}{2}
= 1 + ζ_{3}(0; ω) + \frac{1}{12} ( ∑_{1≤i,j≤3} \frac{ω_{i}}{ω_{j}} + 6) − \frac{3}{2}
= ζ_{3}(0; ω) + \frac{1}{12} ∑_{1≤i,j≤3} \frac{ω_{i}}{ω_{j}} .

Hence

ζ_{3}(0; ω) = \frac{1}{12} ∑_{1≤i,j≤3} \frac{ω_{i}}{ω_{j}} − lim_{x→0} ζ_{3}(0, x; ω)
= −\frac{1}{24} ∑_{1≤i,j≤3} \frac{ω_{i}}{ω_{j}}

as desired. Proposition 2.2

ζ′_{2}(0; (ω_{1}, ω_{2})) = log \frac{2π}{ρ_{2}(ω_{1}, ω_{2})\sqrt{ω_{1}ω_{2}}} ,

ζ′_{3}(0; ω) = log \frac{ρ_{2}(ω_{1}, ω_{2})ρ_{2}(ω_{2}, ω_{3})ρ_{2}(ω_{3}, ω_{1})\sqrt{ω_{1}ω_{2}ω_{3}}}{(2π)^{\frac{3}{2}} ρ_{3}(ω)} .

Proof. By definition we have

ζ_{2}(s, x; (ω_{1}, ω_{2}))

= ∑_{m_{1}=0}^{∞} ∑_{m_{2}=0}^{∞}(m_{1}ω_{1} + m_{2}ω_{2} + x)^{−s}
= ∑_{m_{1}=1}^{∞} ∑_{m_{2}=1}^{∞}(m_{1}ω_{1} + m_{2}ω_{2} + x)^{−s} + ∑_{m_{1}=1}^{∞}(m_{1}ω_{1} + x)^{−s} + ∑_{m_{2}=1}^{∞}(m_{2}ω_{2} + x)^{−s} + x^{−s}

and thus

lim_{x→0} \frac{d}{ds} (ζ_{2}(s, x; (ω_{1}, ω_{2})) − x^{−s})|_{s=0} = ζ′_{2}(0; (ω_{1}, ω_{2})) + 2ζ′(0) − (log ω_{1}ω_{2})ζ(0).

As ζ′(0) = −\frac{log 2π}{2} and ζ(0) = −\frac{1}{2}, we have

ζ′_{2}(0; (ω_{1}, ω_{2})) = log 2π − \frac{log ω_{1}ω_{2}}{2} + lim_{x→0} (ζ′_{2}(0, x; (ω_{1}, ω_{2})) + log x)
= log 2π − \frac{log ω_{1}ω_{2}}{2} + lim_{x→0} (log Γ_{2}(x; (ω_{1}, ω_{2})) + log x)
= log 2π − \frac{log ω_{1}ω_{2}}{2} + lim_{x→0} log(xΓ_{2}(x; (ω_{1}, ω_{2})))
= log 2π − \frac{log ω_{1}ω_{2}}{2} − log ρ_{2}(ω_{1}, ω_{2})
= log \frac{2π}{ρ_{2}(ω_{1}, ω_{2})\sqrt{ω_{1}ω_{2}}} .

For the second identity we have

ζ_{3}(s, x; ω)
= ∑_{m_{1}=0}^{∞} ∑_{m_{2}=0}^{∞} ∑_{m_{3}=0}^{∞}(m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3} + x)^{−s}

= ∑_{m_{1}=1}^{∞} ∑_{m_{2}=1}^{∞} ∑_{m_{3}=1}^{∞}(m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3} + x)^{−s} + ∑_{m_{1}=1}^{∞} ∑_{m_{2}=1}^{∞}(m_{1}ω_{1} + m_{2}ω_{2} + x)^{−s}

+ ∑_{m_{2}=1}^{∞} ∑_{m_{3}=1}^{∞}(m_{2}ω_{2} + m_{3}ω_{3} + x)^{−s} + ∑_{m_{1}=1}^{∞} ∑_{m_{3}=1}^{∞}(m_{1}ω_{1} + m_{3}ω_{3} + x)^{−s}

+ ∑_{m_{1}=1}^{∞}(m_{1}ω_{1} + x)^{−s} + ∑_{m_{2}=1}^{∞}(m_{2}ω_{2} + x)^{−s} + ∑_{m_{3}=1}^{∞}(m_{3}ω_{3} + x)^{−s} + x^{−s},

and thus

lim_{x→0} \frac{d}{ds} (ζ_{3}(s, x; ω) − x^{−s})|_{s=0}
= ζ′_{3}(0; ω) + ζ′_{2}(0; (ω_{1}, ω_{2})) + ζ′_{2}(0; (ω_{2}, ω_{3})) + ζ′_{2}(0; (ω_{1}, ω_{3})) + 3ζ′(0) − (log ω_{1}ω_{2}ω_{3})ζ(0)
= ζ′_{3}(0; ω) + log \frac{(2π)^{3}}{ρ_{2}(ω_{1}, ω_{2})ρ_{2}(ω_{2}, ω_{3})ρ_{2}(ω_{1}, ω_{3})ω_{1}ω_{2}ω_{3}} − \frac{3}{2} log 2π + \frac{log ω_{1}ω_{2}ω_{3}}{2}
= ζ′_{3}(0; ω) + log \frac{(2π)^{\frac{3}{2}}}{ρ_{2}(ω_{1}, ω_{2})ρ_{2}(ω_{2}, ω_{3})ρ_{2}(ω_{1}, ω_{3})\sqrt{ω_{1}ω_{2}ω_{3}}} .

Hence we have

ζ′_{3}(0; ω) = log \frac{ρ_{2}(ω_{1}, ω_{2})ρ_{2}(ω_{2}, ω_{3})ρ_{2}(ω_{1}, ω_{3})\sqrt{ω_{1}ω_{2}ω_{3}}}{(2π)^{\frac{3}{2}}} + lim_{x→0} (ζ′_{3}(0, x; ω) + log x)
= log \frac{ρ_{2}(ω_{1}, ω_{2})ρ_{2}(ω_{2}, ω_{3})ρ_{2}(ω_{1}, ω_{3})\sqrt{ω_{1}ω_{2}ω_{3}}}{(2π)^{\frac{3}{2}}} + lim_{x→0} log(xΓ_{3}(x; ω))
= log \frac{ρ_{2}(ω_{1}, ω_{2})ρ_{2}(ω_{2}, ω_{3})ρ_{2}(ω_{1}, ω_{3})\sqrt{ω_{1}ω_{2}ω_{3}}}{(2π)^{\frac{3}{2}} ρ_{3}(ω)} .

Lemma 2.1

ρ_{r}(Nω)^{−1} = ρ_{r}(ω)^{−1}N^{1−lim_{x→0} ζ_{r}(0,x;ω)}
= ρ_{r}(ω)^{−1} × \begin{cases}N^{1− \frac{1}{12} ( \frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{1}} +3)} & (r = 2) \\ N^{1− \frac{1}{24} ∑_{1≤i,j≤3} \frac{ω_{i}}{ω_{j}}} . & (r = 3) \end{cases}

Proof. We compute that

ρ_{r}(Nω)^{−1} = lim_{x→0} xΓ_{r}(x, Nω)
= lim_{x→0} x exp (ζ′_{r}(0, x, Nω))
= lim_{x→0} x exp ( \frac{∂}{∂s}|_{s=0} ∑_{m≥0}(m · (Nω) + x)^{−s})
= lim_{x→0} x exp ( \frac{∂}{∂s}|_{s=0} N^{−s} ∑_{m≥0}(m · ω + \frac{x}{N})^{−s})
= lim_{x→0} x exp ( \frac{∂}{∂s}|_{s=0} N^{−s}ζ_{r}(s, \frac{x}{N}; ω))
= lim_{x→0} x exp (N^{−s}(− log N)ζ_{r}(s, \frac{x}{N}; ω) + N^{−s}ζ′_{r}(s, \frac{x}{N}; ω))_{s=0}
= lim_{x→0} x exp ((− log N)ζ_{r}(0, \frac{x}{N}; ω) + ζ′_{r}(0, \frac{x}{N}; ω))
= lim_{x→0} \frac{x}{N} Γ_{r} (\frac{x}{N}; ω) N^{1−ζ_{r}(0, \frac{x}{N};ω)}
= ρ_{r}(ω)^{−1}N^{1−lim_{x→0} ζ_{r}(0,x;ω)}.

The values lim_{x→0} ζ_{r}(0, x; ω) for r = 2, 3 are calculated in the proof of Proposition 2.1. . Lemma 2.2

\frac{ρ_{r}(ω)}{ρ_{r}(\frac{ω_{1}}{N_{1}} , ..., \frac{ω_{r}}{N_{r}})} = ∏_{k_{1}=1}^{N_{1}−1} ⋯ ∏_{k_{r}=1}^{N_{r}−1} Γ_{r} (\frac{k_{1}ω_{1}}{N_{1}} + ⋯ + \frac{k_{r}ω_{r}}{N_{r}}; ω) .

Proof.

\frac{ρ_{r}(ω)}{ρ_{r}(\frac{ω_{1}}{N_{1}} , ..., \frac{ω_{r}}{N_{r}})} = lim_{x→0} \frac{Γ_{2}(x; (\frac{ω_{1}}{N_{1}} , ..., \frac{ω_{r}}{N_{r}}))}{Γ_{2}(x; ω)}
= lim_{x→0} exp (ζ′_{r} (0, x, (\frac{ω_{1}}{N_{1}} , ..., \frac{ω_{r}}{N_{r}} )) − ζ′_{r}(0, x; ω)) .

Here we compute

ζ_{r} (s, x, (\frac{ω_{1}}{N_{1}} , ..., \frac{ω_{r}}{N_{r}} )) − ζ_{r}(s, x; ω)

= ∑_{m≥0} ((\frac{m_{1}ω_{1}}{N_{1}} + ⋯ + \frac{m_{r}ω_{r}}{N_{r}} + x)^{−s} − (m · ω + x)^{−s})

= ∑_{m≥0} ( ∑_{k_{1}=0}^{N_{1}−1} ⋯ ∑_{k_{r}=0}^{N_{r}−1} (\frac{(m_{1}N_{1} + k_{1})ω_{1}}{N_{1}} + ⋯ + \frac{(m_{r}N_{r} + k_{r})ω_{r}}{N_{r}} + x)^{−s}) − ∑_{m≥0}(m · ω + x)^{−s}

= ∑_{m≥0} ∑_{k_{1}=1}^{N_{1}−1} ⋯ ∑_{k_{r}=1}^{N_{r}−1} (\frac{(m_{1}N_{1} + k_{1})ω_{1}}{N_{1}} + ⋯ + \frac{(m_{r}N_{r} + k_{r})ω_{r}}{N_{r}} + x)^{−s}

= ∑_{m≥0} ∑_{k_{1}=1}^{N_{1}−1} ⋯ ∑_{k_{r}=1}^{N_{r}−1} (m · ω + \frac{k_{1}ω_{1}}{N_{1}} + ⋯ + \frac{k_{r}ω_{r}}{N_{r}} + x)^{−s}

= ∑_{k_{1}=1}^{N_{1}−1} ⋯ ∑_{k_{r}=1}^{N_{r}−1} ζ_{r} (s, \frac{k_{1}ω_{1}}{N_{1}} + ⋯ + \frac{k_{r}ω_{r}}{N_{r}} + x; ω) .

Therefore

\frac{ρ_{r}(ω)}{ρ_{r}(\frac{ω_{1}}{N_{1}} , ..., \frac{ω_{r}}{N_{r}})} = lim_{x→0} ∏_{k_{1}=1}^{N_{1}−1} ⋯ ∏_{k_{r}=1}^{N_{r}−1} exp (ζ′_{r} (0, \frac{k_{1}ω_{1}}{N_{1}} + ⋯ + \frac{k_{r}ω_{r}}{N_{r}} + x; ω))

= ∏_{k_{1}=1}^{N_{1}−1} ⋯ ∏_{k_{r}=1}^{N_{r}−1} Γ_{r} (\frac{k_{1}ω_{1}}{N_{1}} + ⋯ + \frac{k_{r}ω_{r}}{N_{r}}; ω) .

Lemma 2.3

\frac{ρ_{2}(ω_{1}, ω_{2})}{ρ_{2}(Nω_{1}, ω_{2})} = N^{^{\frac{1}{12}} (9−( \frac{ω_{1}N}{ω_{2}} + \frac{ω_{2}}{ω_{1}N} ))} ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{2}}{N}; (ω_{1}, ω_{2})) .

Proof. We apply Lemma 2.2 to the case r = 2, N_{1} = 1, N_{2} = N. Then

\frac{ρ_{2}(ω_{1}, ω_{2})}{ρ_{2}(ω_{1}, \frac{ω_{2}}{N})} = ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{2}}{N}; (ω_{1}, ω_{2})) .

By Lemma 2.1

ρ_{2}(ω_{1}, \frac{ω_{2}}{N}) = ρ_{2}(Nω_{1}, ω_{2})N^{1− \frac{1}{12} ( \frac{ω_{1}N}{ω_{2}} + \frac{ω_{2}}{ω_{1}N} +3)}

= ρ_{2}(Nω_{1}, ω_{2})N^{^{\frac{1}{12}} (9−( \frac{ω_{1}N}{ω_{2}} + \frac{ω_{2}}{ω_{1}N} ))}.

Hence we have the conclusion.

Lemma 2.4 (a) For a sequence of positive numbers ω_{j} (j = 1, 2, 3), we have

∏_{{1≤i,j≤3 \atop i≠j}} \frac{ρ_{2}(ω_{i}, ω_{j})}{ρ_{2}(Nω_{i}, ω_{j})} = N^{\frac{9}{2} − \frac{1}{12}(N+ \frac{1}{N}) ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{{1≤i,j≤3 \atop i≠j}} ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j})) .

(b) For a sequence of positive numbers ω_{j} (j = 1, 2, 3, 4), we have

∏_{{1≤i,j≤4 \atop i≠j}} \frac{ρ_{2}(ω_{i}, ω_{j})}{ρ_{2}(Nω_{i}, ω_{j})} = N^{9− \frac{1}{12}(N+ \frac{1}{N}) ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{{1≤i,j≤4 \atop i≠j}} ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j})) ,

where in the right hand side the sum over i ≠ j is taken over all ordered pairs (i, j) ∈ {1, 2, 3, 4}^{2} with i ≠ j.

Proof. We take a product of the result of the previous lemma over all ordered pairs (i, j) with i ≠ j. We see the coeffieients of N and 1/N in the exponent of N are both given by

−\frac{1}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}.

Lemma 2.5

\frac{ρ_{3}(ω)}{ρ_{3}(Nω_{1}, Nω_{2}, ω_{3})} = N^{^{\frac{1}{24}} (21−( \frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{1}} )−N( \frac{ω_{1}}{ω_{3}} + \frac{ω_{2}}{ω_{3}} )− \frac{1}{N} ( \frac{ω_{3}}{ω_{1}} + \frac{ω_{3}}{ω_{2}} ))} ∏_{k=1}^{N−1} Γ_{3} (\frac{kω_{3}}{N}; ω) .

Proof. We again apply Lemma 2.2 to the case r = 3, N_{1} = N_{2} = 1, N_{3} = N. Then

\frac{ρ_{3}(ω_{1}, ω_{2}, ω_{3})}{ρ_{3}(ω_{1}, ω_{2}, \frac{ω_{3}}{N})} = ∏_{k=1}^{N−1} Γ_{3} (\frac{kω_{3}}{N}; ω) .

By Lemma 2.1

ρ_{3}(ω_{1}, ω_{2}, \frac{ω_{3}}{N}) = ρ_{3}(Nω_{1}, Nω_{2}, ω_{3})N^{1− \frac{1}{24} (( \frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{1}} )+N( \frac{ω_{1}}{ω_{3}} + \frac{ω_{2}}{ω_{3}} )+ \frac{1}{N} ( \frac{ω_{3}}{ω_{1}} + \frac{ω_{3}}{ω_{2}} )+3)}
= ρ_{3}(Nω_{1}, Nω_{2}, ω_{3})N^{^{\frac{1}{24}} (21−( \frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{1}} )−N( \frac{ω_{1}}{ω_{3}} + \frac{ω_{2}}{ω_{3}} )− \frac{1}{N} ( \frac{ω_{3}}{ω_{1}} + \frac{ω_{3}}{ω_{2}} ))}.

Hence we have the conclusion.

Lemma 2.6 (a) For a sequence of positive numbers ω_{j} (j = 1, 2, 3), we have

∏_{l=1}^{3} \frac{ρ_{3}(ω_{i}, ω_{j}, ω_{l})}{ρ_{3}(Nω_{i}, Nω_{j}, ω_{l})} = N^{\frac{21}{8} − \frac{1}{24}(1+N+ \frac{1}{N}) ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{k=1}^{N−1} ∏_{l=1}^{3} Γ_{3} (\frac{kω_{l}}{N}; (ω_{1}, ω_{2}, ω_{3})) .

(b) For a sequence of positive numbers ω_{j} (j = 1, 2, 3, 4), we have

∏_{l=1}^{4} ∏_{{i,j≠l \atop i<j}} \frac{ρ_{3}(ω_{i}, ω_{j}, ω_{l})}{ρ_{3}(Nω_{i}, Nω_{j}, ω_{l})} = N^{\frac{21}{2} − \frac{1}{12}(1+N+ \frac{1}{N}) ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{k=1}^{N−1} ∏_{l=1}^{4} ∏_{{i,j≠l \atop i<j}} Γ_{3} (\frac{kω_{l}}{N}; (ω_{i}, ω_{j}, ω_{l}))

Proof. We take a product of the result of the previous lemma over distinct i, j, k ∈ {1, 2, 3, 4} with i < j. The product consists of twelve terms. The coeffients of 1, N and 1/N in the exponent are all ∑_{i≠j} \frac{ω_{i}}{ω_{j}}.

Lemma 2.7 For any j ∈ 1, 2, ..., r, it holds that

Γ_{r}(x + ω_{j}; ω) = Γ_{r}(x; ω)Γ_{r−1}(x; ω(j))^{−1}.

Proof. We prove for j = 1. The general cases are similarly shown. By definition we have Γ_{r}(x + ω_{1}) = e^{ζ′_{r}(0,x+ω_{1};ω)} Here we compute

ζ′_{r}(0, x + ω_{1}; ω) = (∑_{k_{1}≥0} ∑_{k_{2}≥0}(k_{1}ω_{1} + k_{2}ω_{2} + ⋯ + k_{r}ω_{r} + x + ω_{1})^{−s})′_{s=0}

= (∑_{k_{1}≥1} ∑_{k_{2}≥0} ⋯ ∑_{k_{r}≥0}(k_{1}ω_{1} + k_{2}ω_{2} + ⋯ + k_{r}ω_{r} + x)^{−s})′_{s=0}

= ζ′_{2}(0, x; ω) − (∑_{k_{2}≥0} ⋯ ∑_{k_{r}≥0}(k_{2}ω_{2} + ⋯ + k_{r}ω_{r} + x)^{−s})′_{s=0}
= ζ′_{r}(0, x; ω) − ζ′_{r−1}(0, x; ω(1)).

Hence Γ_{r}(x + ω_{1}; ω) = e^{ζ′_{r}(0,x;ω)−ζ′_{r−1}(0,x;ω(1))} = Γ_{r}(x; ω)Γ_{r−1}(x; ω(1))^{−1}.

Lemma 2.8

\frac{ρ_{3}(ω_{1}, ω_{2}, ω_{3})}{ρ_{2}(ω_{1}, ω_{2})ρ_{2}(ω_{2}, ω_{3})ρ_{2}(ω_{1}, ω_{3})} = \frac{\sqrt{ω_{1}ω_{2}ω_{3}}}{(2π)^{\frac{3}{2}} Γ_{3}(|ω|; ω)}.

Proof. We compute

\frac{ρ_{3}(ω_{1}, ω_{2}, ω_{3})}{ρ_{2}(ω_{1}, ω_{2})ρ_{2}(ω_{2}, ω_{3})ρ_{2}(ω_{1}, ω_{3})}

= lim_{x→0} \frac{x^{3}Γ_{2}(x; (ω_{1}, ω_{2}))Γ_{2}(x; (ω_{2}, ω_{3}))Γ_{2}(x; (ω_{1}, ω_{3}))}{xΓ_{3}(x; ω)}
= lim_{x→0} x^{2} exp (ζ′_{2}(0, x; (ω_{1}, ω_{2})) + ζ′_{2}(0, x; (ω_{2}, ω_{3})) + ζ′_{2}(0, x; (ω_{1}, ω_{3})) − ζ′_{3}(0, x; ω)) .

Here

ζ_{2}(s, x; (ω_{1}, ω_{2})) + ζ_{2}(s, x; (ω_{2}, ω_{3})) + ζ_{2}(s, x; (ω_{1}, ω_{3})) − ζ_{3}(s, x; ω)

= ∑_{m≥0} ((m_{1}ω_{1} + m_{2}ω_{2} + x)^{−s} + (m_{2}ω_{2} + m_{3}ω_{3} + x)^{−s} + (m_{1}ω_{1} + m_{3}ω_{3} + x)^{−s}
−(m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3} + x)^{−s})
= − ∑_{m≥1}(m · ω + x)^{−s} + ∑_{m_{1}=0}^{∞}(m_{1}ω_{1} + x)^{−s} + ∑_{m_{2}=0}^{∞}(m_{2}ω_{2} + x)^{−s} + ∑_{m_{3}=0}^{∞}(m_{3}ω_{3} + x)^{−s}
= −ζ_{3}(s, x + |ω|; ω) + ∑_{j=1}^{3} ζ_{1}(s, x, ω_{j}) − x^{−s}.

Thus

\frac{ρ_{3}(ω_{1}, ω_{2}, ω_{3})}{ρ_{2}(ω_{1}, ω_{2})ρ_{2}(ω_{2}, ω_{3})ρ_{2}(ω_{1}, ω_{3})}

= lim_{x→0} x^{2} exp (−ζ′_{3}(0, x + |ω|; ω) + ∑_{j=1}^{3} ζ′_{1}(0, x, ω_{j}) + log x)
= lim_{x→0} x^{3} exp (−ζ′_{3}(0, x + |ω|; ω) + ∑_{j=1}^{3} ζ′_{1}(0, x, ω_{j}))
= (Γ_{3}(|ω|; ω) ∏_{j=1}^{3} ρ_{1}(ω_{j}))^{−1} .

As we have

ρ_{1}(ω_{j})^{−1} = lim_{x→0} xΓ_{1}(x, ω_{j})
= lim_{x→0} xe^{ζ′_{1}(0,x,ω_{j})}
= lim_{x→0} xe^{(∑_{m=0}^{∞}(mω_{j}+x)^{−s}){′ \atop s=0}}
= lim_{x→0} xe^{(x^{−s}+∑_{m=1}^{∞}(mω_{j}+x)^{−s}){′ \atop s=0}}
= (lim_{x→0} xe^{− log x}) e^{(ω_{j}^{−s}ζ(s))′_{s=0}}
= e^{(− log ω_{j})ζ(0)+ζ′(0)}
= e^{\frac{1}{2} log ω_{j}− \frac{1}{2} log(2π)}
= \sqrt{\frac{ω_{j}}{2π}} ,

we reach the conclusion. Lemma 2.9

\frac{ρ_{3}(ω_{1}, ω_{2}, ω_{3})}{ρ_{3}(ω_{1}, ω_{2}, Nω_{3})}

= N^{^{\frac{1}{24}} (21−( \frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{1}} + \frac{1}{N} ( \frac{ω_{1}}{ω_{3}} + \frac{ω_{2}}{ω_{3}} )+N( \frac{ω_{3}}{ω_{1}} + \frac{ω_{3}}{ω_{2}} )))} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{1} + k_{2}ω_{2}}{N}; (ω_{1}, ω_{2}, ω_{3}))

Proof. When r = 3, N_{1} = N_{2} = N and N_{3} = 1, Lemma 2.2 shows that

\frac{ρ_{3}(ω_{1}, ω_{2}, ω_{3})}{ρ_{3}(\frac{ω_{1}}{N} , \frac{ω_{2}}{N} , ω_{3})} = ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{1} + k_{2}ω_{2}}{N}; (ω_{1}, ω_{2}, ω_{3})) .

Here by Lemma 2.1

ρ_{3} (\frac{ω_{1}}{N} , \frac{ω_{1}}{N} , ω_{3}) = ρ_{3}(ω_{1}, ω_{2}, Nω_{3})N^{1− \frac{1}{24} ( \frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{1}} +N( \frac{ω_{1}}{ω_{3}} + \frac{ω_{2}}{ω_{3}} )+ \frac{1}{N} ( \frac{ω_{3}}{ω_{1}} + \frac{ω_{3}}{ω_{2}} ))}
= ρ_{3}(ω_{1}, ω_{2}, Nω_{3})N^{^{\frac{1}{24}} (21− \frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{1}} −N( \frac{ω_{1}}{ω_{3}} + \frac{ω_{2}}{ω_{3}} )− \frac{1}{N} ( \frac{ω_{3}}{ω_{1}} + \frac{ω_{3}}{ω_{2}} ))}.

Lemma 2.10 For a sequence of positive numbers ω_{j} (j = 1, 2, 3), we have

∏_{l=1}^{3} \frac{ρ_{3}(ω_{i}, ω_{j}, ω_{l})}{ρ_{3}(ω_{i}, ω_{j}, Nω_{l})} = N^{\frac{21}{8} − \frac{1}{24}(1+N+ \frac{1}{N}) ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{i<j} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i} + k_{2}ω_{j}}{N}; (ω_{1}, ω_{2}, ω_{3})) .

For a sequence of positive numbers ω_{j} (j = 1, 2, 3, 4), we have

∏_{l=1}^{4} ∏_{{i,j≠l \atop i<j}} \frac{ρ_{3}(ω_{i}, ω_{j}, ω_{l})}{ρ_{3}(ω_{i}, ω_{j}, Nω_{l})} = N^{\frac{21}{2} − \frac{1}{12}(1+N+ \frac{1}{N}) ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{l=1}^{4} ∏_{{i,j≠l \atop i<j}} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i} + k_{2}ω_{j}}{N}; (ω_{i}, ω_{j}, ω_{l})) .

Proof. By the previous lemma, we have an expression for \frac{ρ_{3}(ω_{i},ω_{j},ω_{l})}{ρ_{3}(ω_{i},ω_{j},Nω_{l})}. Taking a product over l = 1, 2, 3 leads to the conclusion for the first identity.

For the second identity, all we have to do is to take a product over possible combinations i, j, l where l = 1, 2, 3, 4 and (i, j) runs through combinations of elements not equal to l.

3 The case r = 3

Theorem 3.1 In the product (1.2) for r = 3, the contribution from terms with k_{2} = k_{3} = 0 is expressed as

∏_{k=1}^{N−1} S_{3} (\frac{kω_{1}}{N}; ω)

= exp(−2ζ′_{3}(0; (ω_{1}, Nω_{2}, Nω_{3})) + 2ζ′_{3}(0; ω)
−ζ′_{2}(0; (ω_{1}, Nω_{2})) − ζ′_{2}(0; (ω_{1}, Nω_{3})) + ζ′_{2}(0; (ω_{1}, ω_{2})) + ζ′_{2}(0; (ω_{1}, ω_{3}))

−(log N)(2ζ_{3}(0; (ω_{1}, Nω_{2}, Nω_{3})) + ζ_{2}(0; (ω_{1}, Nω_{2})) + ζ_{2}(0; (ω_{1}, Nω_{3})) + ζ(0))).

The total contribution from terms with two of the coefficients k_{j} being zero is expressed as follows:

∏_{k=1}^{N−1} S_{3} (\frac{kω_{1}}{N}; ω) S_{3} (\frac{kω_{2}}{N}; ω) S_{3} (\frac{kω_{3}}{N}; ω) = N^{\frac{3}{2}} ∏_{k=1}^{N−1} \frac{∏_{{1≤i,j≤3 \atop i≠j}} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}))}{∏_{j=1}^{3} Γ_{3} (\frac{kω_{j}}{N} )^{2}}.

Proof.

∏_{k_{1}=1}^{N−1} S_{3} (\frac{k_{1}ω_{1}}{N}; ω)

= ∏_{k_{1}=1}^{N−1} \frac{1}{Γ_{3} (|ω| − \frac{k_{1}ω_{1}}{N}; ω) Γ_{3} (\frac{k_{1}ω_{1}}{N}; ω)}

= exp (− \frac{∂}{∂s}|_{s=0} ∑_{k_{1}=1}^{N−1} (ζ_{3} (s, |ω| − \frac{k_{1}ω_{1}}{N}; ω) + ζ_{3} (s, \frac{k_{1}ω_{1}}{N}; ω))) .

Here

∑_{k_{1}=1}^{N−1} ζ_{3} (s, |ω| − \frac{k_{1}ω_{1}}{N}; ω)

= ∑_{k_{1}=1}^{N−1} ∑_{m_{1}=0}^{∞} ∑_{m_{2}=0}^{∞} ∑_{m_{3}=0}^{∞} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3} + |ω| − \frac{k_{1}ω_{1}}{N} )^{−s}

= ∑_{k_{1}=1}^{N−1} ∑_{m_{1}=1}^{∞} ∑_{m_{2}=1}^{∞} ∑_{m_{3}=1}^{∞} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3} − \frac{k_{1}ω_{1}}{N} )^{−s}

= N^{s} ∑_{k_{1}=1}^{N−1} ∑_{m_{1}=1}^{∞} ∑_{m_{2}=1}^{∞} ∑_{m_{3}=1}^{∞} ((m_{1}N − k_{1})ω_{1} + m_{2}Nω_{2} + m_{3}Nω_{3})^{−s}

= N^{s} (ζ_{3}(s; (ω_{1}, Nω_{2}, Nω_{3})) − ∑_{m_{1}=1}^{∞} ∑_{m_{2}=1}^{∞} ∑_{m_{3}=1}^{∞}(m_{1}Nω_{1} + m_{2}Nω_{2} + m_{3}Nω_{3})^{−s})
= N^{s} (ζ_{3}(s; (ω_{1}, Nω_{2}, Nω_{3})) − N^{−s}ζ_{3}(s; ω))
= N^{s}ζ_{3}(s; (ω_{1}, Nω_{2}, Nω_{3})) − ζ_{3}(s; ω),

and

∑_{k_{1}=1}^{N−1} ζ_{3} (s, \frac{k_{1}ω_{1}}{N}; ω)

= ∑_{k_{1}=1}^{N−1} ∑_{m_{1}=0}^{∞} ∑_{m_{2}=0}^{∞} ∑_{m_{3}=0}^{∞} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3} + \frac{k_{1}ω_{1}}{N} )^{−s}

= N^{s} ∑_{k_{1}=1}^{N−1} ∑_{m_{1}=0}^{∞} ∑_{m_{2}=0}^{∞} ∑_{m_{3}=0}^{∞} ((m_{1}N + k_{1})ω_{1} + m_{2}Nω_{2} + m_{3}Nω_{3})^{−s}
= N^{s}(ζ_{3}(s; (ω_{1}, Nω_{2}, Nω_{3})) − N^{−s}ζ_{3}(s; ω))

+N^{s} ∑_{k_{1}=1}^{N−1} ∑_{m_{1}=0}^{∞} ∑_{m_{2}=1}^{∞} ((m_{1}N + k_{1})ω_{1} + m_{2}Nω_{2})^{−s}

+N^{s} ∑_{k_{1}=1}^{N−1} ∑_{m_{1}=0}^{∞} ∑_{m_{3}=1}^{∞} ((m_{1}N + k_{1})ω_{1} + m_{3}Nω_{3})^{−s}

+N^{s} ∑_{k_{1}=1}^{N−1} ∑_{m_{1}=0}^{∞} ((m_{1}N + k_{1})ω_{1})^{−s}

= N^{s}ζ_{3}(s; (ω_{1}, Nω_{2}, Nω_{3})) − ζ_{3}(s; ω)

+N^{s}(ζ_{2}(s; (ω_{1}, Nω_{2})) − N^{−s}ζ_{2}(s; (ω_{1}, ω_{2})))

+N^{s}(ζ_{2}(s; (ω_{1}, Nω_{3})) − N^{−s}ζ_{2}(s; (ω_{1}, ω_{3}))) +N^{s}ω_{1}^{−s}(1 − N^{−s})ζ(s)

= N^{s}(ζ_{3}(s; (ω_{1}, Nω_{2}, Nω_{3})) + ζ_{2}(s; (ω_{1}, Nω_{2})) + ζ_{2}(s; (ω_{1}, Nω_{3})) + ω_{1}^{−s}ζ(s))

−ζ_{3}(s; ω) − ζ_{2}(s; (ω_{1}, ω_{2})) − ζ_{2}(s; (ω_{1}, ω_{3})) − ω_{1}^{−s}ζ(s).

Thus we obtain the first identity in Theorem. Therefore we calculate

log ∏_{k=1}^{N−1} S_{3} (\frac{kω_{1}}{N}; ω) S_{3} (\frac{kω_{2}}{N}; ω) S_{3} (\frac{kω_{2}}{N}; ω)

= −2 ∑_{l=1}^{3} (ζ′_{3}(0; ω_{l}^{N}) − ζ′_{3}(0; ω)) − ∑_{i≠j} (ζ′_{2}(0; (ω_{i}, Nω_{j})) − ζ′_{2}(0; (ω_{i}, ω_{j})))

−(log N) (2 ∑_{l=1}^{3} ζ_{3}(0; ω_{l}^{N}) + ∑_{i,j} ζ_{2}(0; (ω_{i}, Nω_{j})) − \frac{3}{2})

= ∑_{l=1}^{3} log (\frac{ρ_{3}(ω_{l}^{N})}{ρ_{3}(ω)} \frac{ρ_{2}(ω_{l}, ω_{i})ρ_{2}(ω_{l}, ω_{j})ρ_{2}(ω_{i}, ω_{j})}{Nρ_{2}(ω_{l}, Nω_{i})ρ_{2}(ω_{l}, Nω_{j})ρ_{2}(Nω_{i}, Nω_{j})})^{2} + ∑_{i≠j} log \frac{ρ_{2}(ω_{i}, Nω_{j})\sqrt{N}}{ρ_{2}(ω_{i}, ω_{j})}

−(log N) (−\frac{1}{12} (1 + N + \frac{1}{N} ) ∑_{i≠j} \frac{ω_{i}}{ω_{j}} − \frac{9}{12} + \frac{1}{12} (N + \frac{1}{N} ) ∑_{i≠j} \frac{ω_{i}}{ω_{j}} + \frac{18}{12} − \frac{3}{2})

= ∑_{l=1}^{3} log ((\frac{ρ_{3}(ω_{l}^{N})}{ρ_{3}(ω)})^{2} \frac{ρ_{2}(ω_{l}, ω_{i})ρ_{2}(ω_{l}, ω_{j})}{ρ_{2}(ω_{l}, Nω_{i})ρ_{2}(ω_{l}, Nω_{j})}) + log N^{\frac{3}{2} − \frac{1}{6} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}}

+(log N) (\frac{1}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}} + \frac{9}{12})

= log (N^{\frac{9}{4} − \frac{1}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{l=1}^{3} (\frac{ρ_{3}(ω_{l}^{N})}{ρ_{3}(ω)})^{2} \frac{ρ_{2}(ω_{l}, ω_{i})ρ_{2}(ω_{l}, ω_{j})}{ρ_{2}(ω_{l}, Nω_{i})ρ_{2}(ω_{l}, Nω_{j})}) ,

where i and j denote the elements not being equal to l, when they are in a sum or a product over l.

Hence by Lemmas in the previous section,

∏_{k=1}^{N−1} S_{3} (\frac{kω_{1}}{N}; ω) S_{3} (\frac{kω_{2}}{N}; ω) S_{3} (\frac{kω_{2}}{N}; ω)

= N^{\frac{9}{4} − \frac{1}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{l=1}^{3} (\frac{ρ_{3}(ω_{l}^{N})}{ρ_{3}(ω)})^{2} ∏_{{1≤i,j≤3 \atop i≠j}} \frac{ρ_{2}(ω_{i}, ω_{j})}{ρ_{2}(ω_{i}, Nω_{j})}

= N^{( \frac{9}{4} − \frac{1}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}} )+(− \frac{21}{4} + \frac{1}{12}(1+N+ \frac{1}{N}) ∑_{i≠j} \frac{ω_{i}}{ω_{j}} )+( \frac{9}{2} − \frac{1}{12}(N+ \frac{1}{N}) ∑_{i≠j} \frac{ω_{i}}{ω_{j}} )} \frac{∏_{k=1}^{N−1} ∏_{{1≤i,j≤3 \atop i≠j}} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}))}{∏_{k=1}^{N−1} ∏_{j=1}^{3} Γ_{3} (\frac{kω_{j}}{N} )^{2}}

= N^{\frac{3}{2}} \frac{∏_{k=1}^{N−1} ∏_{{1≤i,j≤3 \atop i≠j}} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}))}{∏_{k=1}^{N−1} ∏_{j=1}^{3} Γ_{3} (\frac{kω_{j}}{N} )^{2}}.

Theorem 3.2 In the product (1.2) for r = 3, the contribution from terms with k_{3} = 0 is expressed as

∏_{k_{1}=1}^{N−1} ∏_{k_{2}=1}^{N−1} S_{3} (\frac{k_{1}ω_{1} + k_{2}ω_{2}}{N}; ω) = exp(−2 (ζ′_{3}(0; ω_{3}^{(N)}) − ζ′_{3}(0; ω_{1}^{N}) − ζ′_{3}(0; ω_{2}^{N}) + ζ′_{3}(0; ω))
− (2ζ′_{2}(0; (ω_{1}, ω_{2})) − ζ′_{2}(0; (Nω_{1}, ω_{2})) − ζ′_{2}(0; (ω_{1}, Nω_{2})))
− (log N)(2 (ζ_{3}(0; ω_{3}^{(N)}) − ζ_{3}(0; ω_{1}^{N}) − ζ_{3}(0; ω_{2}^{N}))

+ (ζ_{2}(0; (ω_{1}, ω_{2})) − ζ_{2}(0; (Nω_{1}, ω_{2})) − ζ_{2}(0; (ω_{1}, Nω_{2})))))

The total contribution from terms with only one of the coefficients k_{j} being zero is expressed as follows:

∏_{k_{1}=1}^{N−1} ∏_{k_{2}=1}^{N−1} S_{3} (\frac{k_{1}ω_{1} + k_{2}ω_{2}}{N}; ω) S_{3} (\frac{k_{1}ω_{2} + k_{2}ω_{3}}{N}; ω) S_{3} (\frac{k_{1}ω_{3} + k_{2}ω_{1}}{N}; ω)

= \frac{N^{− \frac{9}{4} + \frac{1}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{j=1}^{3} ∏_{k=1}^{N−1} Γ_{3} (\frac{kω_{j}}{N}; ω)^{4}}{( ∏_{{1≤i,j≤3 \atop i≠j}} ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}))) (∏_{i<j} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; ω)^{2})}.

Proof.

∏_{k_{1}=1}^{N−1} ∏_{k_{2}=1}^{N−1} S_{3} (\frac{k_{1}ω_{1} + k_{2}ω_{2}}{N}; ω)

= ∏_{k_{1}=1}^{N−1} ∏_{k_{2}=1}^{N−1} \frac{1}{Γ_{3} (|ω| − \frac{k_{1}ω_{1}+k_{2}ω_{2}}{N}; ω) Γ_{3} (\frac{k_{1}ω_{1}+k_{2}ω_{2}}{N}; ω)}
= exp (− \frac{∂}{∂s}|_{s=0} ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} (ζ_{3} (s, |ω| − \frac{k_{1}ω_{1} + k_{2}ω_{2}}{N}; ω) + ζ_{3} (s, \frac{k_{1}ω_{1} + k_{2}ω_{2}}{N}; ω))) .

Here

∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ζ_{3} (s, |ω| − \frac{k_{1}ω_{1} + k_{2}ω_{2}}{N}; ω)

= ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{m_{1}=0}^{∞} ∑_{m_{2}=0}^{∞} ∑_{m_{3}=0}^{∞} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3} + |ω| − \frac{k_{1}ω_{1} + k_{2}ω_{2}}{N} )^{−s}
= ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{m_{1}=1}^{∞} ∑_{m_{2}=1}^{∞} ∑_{m_{3}=1}^{∞} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3} − \frac{k_{1}ω_{1} + k_{2}ω_{2}}{N} )^{−s}
= N^{s} ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{m_{1}=1}^{∞} ∑_{m_{2}=1}^{∞} ∑_{m_{3}=1}^{∞} ((m_{1}N − k_{1})ω_{1} + (m_{2}N − k_{2})ω_{2} + m_{3}Nω_{3})^{−s}
= N^{s} ∑_{{m_{1}≥1 \atop N∤m_{1}}} ∑_{{m_{2}≥1 \atop N∤m_{2}}} ∑_{m_{3}≥1} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}Nω_{3})^{−s}
= N^{s} (ζ_{3}(s; (ω_{1}, ω_{2}, Nω_{3})) − ζ_{3}(s; (Nω_{1}, ω_{2}, Nω_{3})) − ζ_{3}(s; (ω_{1}, Nω_{2}, Nω_{3})) + ζ_{3}(s, N(ω_{1}, ω_{2}, ω_{3})))
= N^{s} (ζ_{3}(s; ω_{3}^{(N)}) − ζ_{3}(s; ω_{1}^{N}) − ζ_{3}(s; ω_{2}^{N}) + ζ_{3}(s, Nω))
= N^{s} (ζ_{3}(s; ω_{3}^{(N)}) − ζ_{3}(s; ω_{1}^{N}) − ζ_{3}(s; ω_{2}^{N})) + ζ_{3}(s; ω),

and

∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ζ_{3} (s, \frac{k_{1}ω_{1} + k_{2}ω_{2}}{N}; ω)

= ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{m_{1}=0}^{∞} ∑_{m_{2}=0}^{∞} ∑_{m_{3}=0}^{∞} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3} + \frac{k_{1}ω_{1} + k_{2}ω_{2}}{N} )^{−s}
= N^{s} ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{m_{1}=0}^{∞} ∑_{m_{2}=0}^{∞} ∑_{m_{3}=0}^{∞} ((m_{1}N + k_{1})ω_{1} + (m_{2}N + k_{2})ω_{2} + m_{3}Nω_{3})^{−s}

= N^{s} ( ∑_{{m_{1}≥1 \atop N∤m_{1}}} ∑_{{m_{2}≥1 \atop N∤m_{2}}} ∑_{m_{3}≥1} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}Nω_{3})^{−s} + ∑_{{m_{1}≥1 \atop N∤m_{1}}} ∑_{{m_{2}≥1 \atop N∤m_{2}}} (m_{1}ω_{1} + m_{2}ω_{2})^{−s})

= N^{s} (ζ_{3}(s; ω_{3}^{(N)}) − ζ_{3}(s; ω_{1}^{N}) − ζ_{3}(s; ω_{2}^{N}) + ζ_{3}(s, Nω)

+ζ_{2}(s; (ω_{1}, ω_{2}) − ζ_{2}(s; (Nω_{1}, ω_{2}) − ζ_{2}(s; (ω_{1}, Nω_{2})) + ζ_{2}(s; (Nω_{1}, Nω_{2}))

= N^{s} (ζ_{3}(s; ω_{3}^{(N)}) − ζ_{3}(s; ω_{1}^{N}) − ζ_{3}(s; ω_{2}^{N})) + ζ_{3}(s; ω)

+N^{s}(ζ_{2}(s; (ω_{1}, ω_{2})) − ζ_{2}(s; (Nω_{1}, ω_{2})) − ζ_{2}(s; (ω_{1}, Nω_{2}))) + ζ_{2}(s; (ω_{1}, ω_{2}))

Thus we have the first identity in Theorem. For the second identity we calculate

\frac{∂}{∂s}|_{s=0} ∑_{i<j} ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} (ζ_{3} (s, |ω| − \frac{k_{1}ω_{i} + k_{2}ω_{j}}{N}; ω) + ζ_{3} (s, \frac{k_{1}ω_{i} + k_{2}ω_{j}}{N}; ω))

= 2 ∑_{l=1}^{3} (ζ′_{3}(0; ω_{l}^{(N)}) − 2ζ′_{3}(0; ω_{l}^{N}) + ζ′_{3}(0; ω)) + ∑_{i≠j}(ζ′_{2}(0; (ω_{i}, ω_{j})) − ζ′_{2}(0; (Nω_{i}, ω_{j})))

+ (log N) (2 ∑_{l=1}^{3} (ζ_{3}(0; ω_{l}^{(N)}) − 2ζ_{3}(0; ω_{l}^{N})) + ∑_{i≠j}(\frac{ζ_{2}(0; (ω_{i}, ω_{j}))}{2} − ζ_{2}(0; (Nω_{i}, ω_{j}))))

= 2 log (∏_{i≠j} \frac{ρ_{2}(ω_{i}, ω_{j})}{ρ_{2}(ω_{i}, Nω_{j})}) (N^{3− \frac{1}{12} ( ∑_{i≠j} \frac{ω_{i}}{ω_{j}} +9)})^{2} ∏_{l=1}^{3} ((\frac{ρ_{3}(ω_{l}^{N})^{2}}{ρ_{3}(ω)} )^{2} \frac{ρ_{3}(ω)}{ρ_{3}(ω_{l}^{(N)})} \begin{matrix}√ & \overline{N} \\ (\sqrt{N^{2}}) & 2 \end{matrix} )

+ log ∏_{i≠j} (\frac{ρ_{2}(Nω_{i}, ω_{j})}{ρ_{2}(ω_{i}, ω_{j})}\sqrt{N})

+(log N) (\frac{1 + N + \frac{1}{N}}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}} + \frac{9}{12} + \frac{1}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}} + \frac{9}{12} − \frac{N + \frac{1}{N}}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}} − \frac{18}{12})

= log (∏_{i≠j} \frac{ρ_{2}(ω_{i}, ω_{j})}{ρ_{2}(ω_{i}, Nω_{j})}) (N^{3− \frac{1}{3} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ) ∏_{l=1}^{3} ((\frac{ρ_{3}(ω_{l}^{N})^{2}}{ρ_{3}(ω)} )^{2} \frac{ρ_{3}(ω)}{ρ_{3}(ω_{l}^{(N)})})^{2}

+(log N) (\frac{1}{6} ∑_{i≠j} \frac{ω_{i}}{ω_{j}} )

= log (∏_{i≠j} \frac{ρ_{2}(ω_{i}, ω_{j})}{ρ_{2}(ω_{i}, Nω_{j})}) (N^{3− \frac{1}{6} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ) ∏_{l=1}^{3} ((\frac{ρ_{3}(ω_{l}^{N})^{2}}{ρ_{3}(ω)} )^{2} \frac{ρ_{3}(ω)}{ρ_{3}(ω_{l}^{(N)})})^{2}

= log (N^{3− \frac{1}{6} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ) (N^{\frac{9}{2} − \frac{1}{12} (N+ \frac{1}{N}) ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{{1≤i,j≤3 \atop i≠j}} ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j})))

+ log (N^{− \frac{21}{8} + \frac{1}{24} (1+N+ \frac{1}{N}) ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} \frac{∏_{i<j} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N} )}{∏_{j=1}^{3} ∏_{k=1}^{N−1} Γ_{3} (\frac{kω_{j}}{N} )^{2}} )^{2}

= log (N^{\frac{9}{4} − \frac{1}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{i≠j}^{3} ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j})) (\frac{∏_{i<j} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N} )}{∏_{j=1}^{3} ∏_{k=1}^{N−1} Γ_{3} (\frac{kω_{j}}{N} )^{2}} )^{2}) .

Example. When N = 2, ω = (1, 1, 1), we have

S_{3} (1)^{3} = 2^{− \frac{7}{4}} (\frac{Γ_{3}(\frac{1}{2})^{6}}{Γ_{2}(\frac{1}{2})^{3}Γ_{3}(\frac{1}{2})^{6}Γ_{3}(1)^{3}} )^{2} = 2^{− \frac{7}{4}} (Γ_{2} (\frac{1}{2}) Γ_{3}(1))^{−6} .

As we will show later in the final section, it agrees to the results on the special values S_{3}(1) = e^{−ζ′(−2)}, Γ_{2}(\frac{1}{2}) = 2^{− \frac{7}{24}} e^{− \frac{ζ′(−1)}{2}} , and Γ_{3}(1) = e^{\frac{ζ′(−2)+ζ′(−1)}{2}} .

Theorem 3.3 In the product (1.2) for r = 3, the contribution from terms with none of the coefficients k_{j} being zero is expressed as follows:

∏_{k_{1}=1}^{N−1} ∏_{k_{2}=1}^{N−1} ∏_{k_{3}=1}^{N−1} S_{3} (\frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω)

= exp (2 (∑_{j=1}^{3} (ζ′_{3}(0; ω_{j}^{(N)}) − ζ′_{3}(0; ω_{j}^{N})) − (log N)ζ_{3}(0; ω)))

= N^{\frac{7}{4} − \frac{1}{12} ∑_{i<j} \frac{ω_{i}}{ω_{j}}} (\frac{∏’_{0≤k_{1},k_{2}≤N−1} ∏_{{i<j \atop 1≤i,j≤3}} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; ω)}{∏_{k=1}^{N−1} ∏_{l=1}^{3} Γ_{3} (\frac{kω_{l}}{N}; ω)})^{2} .

Proof.

∏_{k_{1}=1}^{N−1} ∏_{k_{2}=1}^{N−1} ∏_{k_{3}=1}^{N−1} S_{3} (\frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω)

= ∏_{k_{1}=1}^{N−1} ∏_{k_{2}=1}^{N−1} ∏_{k_{3}=1}^{N−1} \frac{1}{Γ_{3} (|ω| − \frac{k_{1}ω_{1}+k_{2}ω_{2}+k_{3}ω_{3}}{N}; ω) Γ_{3} (\frac{k_{1}ω_{1}+k_{2}ω_{2}+k_{3}ω_{3}}{N}; ω)}
= exp (− \frac{∂}{∂s}|_{s=0} ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{k_{3}=1}^{N−1} (ζ_{3} (s, |ω| − \frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω)

+ζ_{3} (s, \frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω))) .

Here

∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{k_{3}=1}^{N−1} ζ_{3} (s, |ω| − \frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω)

= ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{k_{3}=1}^{N−1} ∑_{m_{1}=0}^{∞} ∑_{m_{2}=0}^{∞} ∑_{m_{3}=0}^{∞} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3} + |ω| − \frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N} )^{−s}
= ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{k_{3}=1}^{N−1} ∑_{m_{1}=1}^{∞} ∑_{m_{2}=1}^{∞} ∑_{m_{3}=1}^{∞} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3} − \frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N} )^{−s}
= N^{s} ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{k_{3}=1}^{N−1} ∑_{m_{1}=1}^{∞} ∑_{m_{2}=1}^{∞} ∑_{m_{3}=1}^{∞} ((m_{1}N − k_{1})ω_{1} + (m_{2}N − k_{2})ω_{2} + (m_{3}N − k_{3})ω_{3})^{−s}
= N^{s} ∑_{{m_{1}≥1 \atop N∤m_{1}}} ∑_{{m_{2}≥1 \atop N∤m_{2}}} ∑_{{m_{3}≥1 \atop N∤m_{3}}} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3})^{−s}

and

∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{k_{3}=1}^{N−1} ζ_{3} (s, \frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω)

= ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{k_{3}=1}^{N−1} ∑_{m_{1}=0}^{∞} ∑_{m_{2}=0}^{∞} ∑_{m_{3}=0}^{∞} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3} + \frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N} )^{−s}
= N^{s} ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{k_{3}=1}^{N−1} ∑_{m_{1}=0}^{∞} ∑_{m_{2}=0}^{∞} ∑_{m_{3}=0}^{∞} ((m_{1}N + k_{1})ω_{1} + (m_{2}N + k_{2})ω_{2} + (m_{3}N + k_{3})ω_{3})^{−s}
= N^{s} ∑_{{m_{1}≥1 \atop N∤m_{1}}} ∑_{{m_{2}≥1 \atop N∤m_{2}}} ∑_{{m_{3}≥1 \atop N∤m_{3}}} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3})^{−s}

Hence

\frac{∂}{∂s}|_{s=0} ∑_{k_{1}=1}^{N−1} ∑_{k_{2}=1}^{N−1} ∑_{k_{3}=1}^{N−1} (ζ_{3} (s, |ω| − \frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω) + ζ_{3} (s, \frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω))

= \frac{∂}{∂s}|_{s=0} 2N^{s} ∑_{{m_{1}≥1 \atop N∤m_{1}}} ∑_{{m_{2}≥1 \atop N∤m_{2}}} ∑_{{m_{3}≥1 \atop N∤m_{3}}} (m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3})^{−s}

= \frac{∂}{∂s}|_{s=0} 2N^{s} ((1 − N^{−s})ζ_{3}(s; ω) − ∑_{j=1}^{3} (ζ_{3}(s; ω_{j}^{(N)}) − ζ_{3}(s; ω_{j}^{N})))

= \frac{∂}{∂s}|_{s=0} 2 ((N^{s} − 1)ζ_{3}(s; ω) − N^{s} ∑_{j=1}^{3} (ζ_{3}(s; ω_{j}^{(N)}) − ζ_{3}(s; ω_{j}^{N})))

= 2 (− ∑_{j=1}^{3} (ζ′_{3}(0; ω_{j}^{(N)}) − ζ′_{3}(0; ω_{j}^{N})) + (log N) (ζ_{3}(0; ω) − ∑_{j=1}^{3} (ζ_{3}(0; ω_{j}^{(N)}) − ζ_{3}(0; ω_{j}^{N}))))

= 2 (− ∑_{j=1}^{3} (ζ′_{3}(0; ω_{j}^{(N)}) − ζ′_{3}(0; ω_{j}^{N})) + (log N)ζ_{3}(0; ω)) ,

since

∑_{j=1}^{3} ζ′_{3}(0; ω_{j}^{(N)}) = ∑_{j=1}^{3} ζ′_{3}(0; ω_{j}^{N}) = \frac{1}{24} (∑_{i≠j} \frac{ω_{i}}{ω_{j}} (1 + N + \frac{1}{N} ) + 9) .

Thus we have the first identity. For the second identity we compute

2 (ζ′_{3}(0; ω_{j}^{(N)}) − ζ′_{3}(0; ω_{j}^{N})) = log (\frac{1}{N^{\frac{3}{2}}} ∏_{i<j} \frac{ρ_{2}(ω_{i}, ω_{j})}{ρ_{2}(Nω_{i}, Nω_{j})} ∏_{j=1}^{3} \frac{ρ_{3}(ω_{j}^{N})}{ρ_{3}(ω_{j}^{(N)})})^{2}

= log N^{−3} (N^{3− \frac{1}{12} ( ∑_{i≠j} \frac{ω_{i}}{ω_{j}} +9)})^{2} (∏_{j=1}^{3} \frac{ρ_{3}(ω_{j}^{N})}{ρ_{3}(ω_{j}^{(N)})})^{2}

= log N^{\frac{3}{2} − \frac{1}{6} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} (∏_{j=1}^{3} \frac{ρ_{3}(ω_{j}^{N})}{ρ_{3}(ω_{j}^{(N)})})^{2} .

We also compute by Lemmas in the previous section that

\frac{ρ_{3}(ω_{j}^{N})}{ρ_{3}(ω_{j}^{(N)})} = \frac{ρ_{3}(ω_{j}^{N})}{ρ_{3}(ω)} \frac{ρ_{3}(ω)}{ρ_{3}(ω_{j}^{(N)})}
= \frac{∏_{i<j} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; ω)}{∏_{l=1}^{3} ∏_{k=0}^{N−1} Γ_{3} (\frac{kω_{l}}{N}; ω)}.

Thus

∏_{k_{1}=1}^{N−1} ∏_{k_{2}=1}^{N−1} ∏_{k_{3}=1}^{N−1} S_{3} (\frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω)

= N^{\frac{1}{12} ( ∑_{i≠j} \frac{ω_{i}}{ω_{j}} +3)+ \frac{3}{2} − \frac{1}{6} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} (\frac{∏_{i<j} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; ω)}{∏_{l=1}^{3} ∏_{k=1}^{N−1} Γ_{3} (\frac{kω_{l}}{N}; ω)})^{2}
= N^{\frac{7}{4} − \frac{1}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} (\frac{∏_{i<j} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; ω)}{∏_{l=1}^{3} ∏_{k=1}^{N−1} Γ_{3} (\frac{kω_{l}}{N}; ω)})^{2} .

Example. When N = 2, ω = (1, 1, 1), we have

S_{3} (\frac{3}{2}) = 2^{\frac{5}{4}} (\frac{Γ_{3}(\frac{1}{2})^{6}Γ_{3}(1)^{3}}{Γ_{3}(\frac{1}{2})^{3}} )^{2} = 2^{\frac{5}{4}} Γ_{3} (\frac{1}{2})^{6} Γ_{3}(1)^{6}.

This agrees to the results we will show in the final section, since S_{3} (\frac{3}{2}) = 2^{− \frac{1}{8}} e^{\frac{3}{4} ζ′(−2)}.

4 The case r = 4

Let

A = ∏_{1≤k_{1},...,k_{4}≤N−1} S_{4} (\frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω) .

Then

∏’_{0≤k_{1},...,k_{4}≤N−1} S_{4} (\frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω) = A ∏_{j=1}^{4} A_{j} ∏_{{j,l=1 \atop j<l}}^{4} A_{j,l} ∏_{{j,l,m=1 \atop j<l<m}}^{4} A_{j,l,m}, (4.1)

where A_{j}, A_{j,l}, and A_{j,l,m} are the partial products defined in Section 1.

Theorem 4.1 The partial product A_{j,l,m} is expressed in terms of the triple sine functions as

A_{j,l,m} = N^{− \frac{1}{4}} ∏_{k_{n}=1}^{N−1} (S_{3} (\frac{k_{n}ω_{n}}{N}; ω(j)) S_{3} (\frac{k_{n}ω_{n}}{N}; ω(l)) S_{3} (\frac{k_{n}ω_{n}}{N}; ω(m)))^{\frac{1}{2}} . (4.2)

In the product (1.2) for r = 4, the total contribution from terms with three of the coefficients k_{j} being zero is

∏_{j<l<m} A_{j,l,m} = N^{2} ∏_{j=1}^{4} ∏_{k=1}^{N−1} \frac{∏_{i≠l} Γ_{2} (\frac{kω_{l}}{N}; (ω_{i}, ω_{l}))^{\frac{1}{2}}}{∏_{{1≤l≤4 \atop l≠j}} Γ_{3} (\frac{kω_{l}}{N}; ω(j))^{\frac{1}{2}}} .

Proof. We calculate A_{j,l,m} for j, l, m = 1, 2, 3, 4 with j < l < m. Let n ∈ {1, 2, 3, 4} be the number which is neither j, l nor m. Then we have

A_{j,l,m}^{2} = ∏_{k_{n}=1}^{N−1} S_{4} (\frac{k_{n}ω_{n}}{N}; ω) S_{4} (\frac{(N − k_{n})ω_{n}}{N}; ω)

= ∏_{k_{n}=1}^{N−1} S_{4} (\frac{k_{n}ω_{n}}{N}; ω) S_{4} (ω_{n} − \frac{k_{n}ω_{n}}{N}; ω)

= ∏_{k_{n}=1}^{N−1} S_{4} (\frac{k_{n}ω_{n}}{N}; ω) S_{4} (ω_{j} + ω_{l} + ω_{m} + \frac{k_{n}ω_{n}}{N}; ω)^{−1}

= ∏_{k_{n}=1}^{N−1} S_{4} (\frac{k_{n}ω_{n}}{N}; ω) S_{4} (ω_{j} + ω_{l} + \frac{k_{n}ω_{n}}{N}; ω)^{−1} S_{3} (ω_{j} + ω_{l} + \frac{k_{n}ω_{n}}{N}; ω(m))

= ∏_{k_{n}=1}^{N−1} S_{4} (\frac{k_{n}ω_{n}}{N}; ω) S_{4} (ω_{j} + \frac{k_{n}ω_{n}}{N}; ω)^{−1}

S_{3} (ω_{j} + \frac{k_{n}ω_{n}}{N}; ω(l)) S_{3} (ω_{j} + ω_{l} + \frac{k_{n}ω_{n}}{N}; ω(m))

= ∏_{k_{n}=1}^{N−1} S_{3} (\frac{k_{n}ω_{n}}{N}; ω(j)) S_{3} (ω_{j} + \frac{k_{n}ω_{n}}{N}; ω(l)) S_{3} (ω_{j} + ω_{l} + \frac{k_{n}ω_{n}}{N}; ω(m)) .

= ∏_{k_{n}=1}^{N−1} S_{3} (\frac{k_{n}ω_{n}}{N}; ω(j)) S_{3} (\frac{k_{n}ω_{n}}{N}; ω(l)) S_{2} (\frac{k_{n}ω_{n}}{N}; ω(l, j))^{−1}

S_{3} (ω_{j} + \frac{k_{n}ω_{n}}{N}; ω(m)) S_{2} (ω_{j} + \frac{k_{n}ω_{n}}{N}; ω(m, l))^{−1} .

= ∏_{k_{n}=1}^{N−1} S_{3} (\frac{k_{n}ω_{n}}{N}; ω(j)) S_{3} (\frac{k_{n}ω_{n}}{N}; ω(l)) S_{2} (\frac{k_{n}ω_{n}}{N}; ω(l, j))^{−1}

S_{3} (\frac{k_{n}ω_{n}}{N}; ω(m)) S_{2} (\frac{k_{n}ω_{n}}{N}; ω(m, j))^{−1} S_{2} (\frac{k_{n}ω_{n}}{N}; ω(m, l))^{−1} S_{1} (\frac{k_{n}ω_{n}}{N}; ω(m, l, j)) .

Hence we obtain

A_{j,l,m} = ∏_{k_{n}=1}^{N−1} (S_{3} (\frac{k_{n}ω_{n}}{N}; ω(j)) S_{3} (\frac{k_{n}ω_{n}}{N}; ω(l)) S_{3} (\frac{k_{n}ω_{n}}{N}; ω(m)))^{\frac{1}{2}}
(S_{2} (\frac{k_{n}ω_{n}}{N}; ω(l, j)) S_{2} (\frac{k_{n}ω_{n}}{N}; ω(m, j)) S_{2} (\frac{k_{n}ω_{n}}{N}; ω(m, l)))^{− \frac{1}{2}}

S_{1} (\frac{k_{n}ω_{n}}{N}; ω(m, l, j))^{\frac{1}{2}}

= N^{− \frac{1}{4}} ∏_{k_{n}=1}^{N−1} (S_{3} (\frac{k_{n}ω_{n}}{N}; ω(j)) S_{3} (\frac{k_{n}ω_{n}}{N}; ω(l)) S_{3} (\frac{k_{n}ω_{n}}{N}; ω(m)))^{\frac{1}{2}}

by (1.2) for r = 1 and 2. This proves (4.2). Theorem 3.1 leads to the second identity. Theorem 4.2 The partial product A_{i,j} is expressed in terms of the triple sine functions as

A_{j,l} = ∏_{1≤k_{m},k_{n}≤N−1} S_{3} (\frac{k_{m}ω_{m} + k_{n}ω_{n}}{N}; ω(j))^{\frac{1}{2}} S_{3} (\frac{k_{m}ω_{m} + k_{n}ω_{n}}{N}; ω(l))^{\frac{1}{2}} . (4.3)

In the product (1.2) for r = 4, the total contribution from terms with exactly two of the coefficients k_{j} being zero is

∏_{{1≤i,j≤4 \atop i≠j}} A_{i,j} = N^{− \frac{9}{2} + \frac{1}{12} ∑_{j≠l} \frac{ω_{j}}{ω_{l}}} \frac{∏_{l=1}^{4} ∏_{j≠l} ∏_{k=1}^{N−1} Γ_{3} (\frac{kω_{j}}{N}; ω(l))^{2}}{∏_{l=1}^{4} ( ∏_{{i<j \atop i,j≠l}} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; ω(l))) ( ∏_{i,j≠l} ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j})))}.

Proof. We first calculate A_{1,2}. It follows that

A_{1,2}^{2} = ∏_{1≤k_{3},k_{4}≤N−1} S_{4} (\frac{k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω)^{2}

= ∏_{1≤k_{3},k_{4}≤N−1} S_{4} (\frac{k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω) S_{4} (\frac{(N − k_{3})ω_{3} + (N − k_{4})ω_{4}}{N}; ω)

= ∏_{1≤k_{3},k_{4}≤N−1} S_{4} (\frac{k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω) S_{4} (ω_{3} + ω_{4} − \frac{k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω)

= ∏_{1≤k_{3},k_{4}≤N−1} S_{4} (\frac{k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω) S_{4} (ω_{1} + ω_{2} + \frac{k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω)^{−1}

= ∏_{1≤k_{3},k_{4}≤N−1} S_{4} (\frac{k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω) S_{4} (ω_{2} + \frac{k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω)^{−1}

×S_{3} (ω_{2} + \frac{k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω(1))

= ∏_{1≤k_{3},k_{4}≤N−1} S_{3} (\frac{k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω(2)) S_{3} (\frac{k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω(1))

×S_{2} (\frac{k_{3}ω_{3} + k_{4}ω_{4}}{N}; ω(1, 2))^{−1} .

Therefore denoting by m, n (m < n) the numbers in {1, 2, 3, 4} different from j and l, we showed for j, l = 1, 2, 3, 4 with j < l

A_{j,l} = ∏_{1≤k_{m},k_{n}≤N−1} S_{3} (\frac{k_{m}ω_{m} + k_{n}ω_{n}}{N}; ω(j))^{\frac{1}{2}} S_{3} (\frac{k_{m}ω_{m} + k_{n}ω_{n}}{N}; ω(l))^{\frac{1}{2}}

×S_{2} (\frac{k_{m}ω_{m} + k_{n}ω_{n}}{N}; ω(j, l))^{− \frac{1}{2}}

= ∏_{1≤k_{m},k_{n}≤N−1} S_{3} (\frac{k_{m}ω_{m} + k_{n}ω_{n}}{N}; ω(j))^{\frac{1}{2}} S_{3} (\frac{k_{m}ω_{m} + k_{n}ω_{n}}{N}; ω(l))^{\frac{1}{2}} .

This proves (4.3).

In the product ∏_{j,l} A_{j,l}, which is the product over terms with just two k_{j}’s are zero, the

contribution from S_{3}(·; ω(4)) is calculated by Theorem 3.2 as

∏_{{1≤i,j≤3 \atop i≠j}} ∏_{k_{1}=1}^{N−1} ∏_{k_{2}=1}^{N−1} S_{3} (\frac{k_{1}ω_{i} + k_{2}ω_{j}}{N}; ω(4))^{\frac{3}{2}}

= ( \frac{N^{− \frac{9}{4} + \frac{1}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{j=1}^{3} ∏_{k=1}^{N−1} Γ_{3} (\frac{kω_{j}}{N}; ω(4))^{4}}{∏_{{1≤i,j≤3 \atop i≠j}} ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j})) ∏_{i<j} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; ω(4))^{2}} )^{\frac{3}{2}}

= \frac{N^{− \frac{27}{8} + \frac{1}{8} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{j=1}^{3} ∏_{k=1}^{N−1} Γ_{3} (\frac{kω_{j}}{N}; ω(4))^{6}}{∏_{{1≤i,j≤3 \atop i≠j}} ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}))^{\frac{3}{2}} ∏_{1≤i<j≤3} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; ω(4))^{3}}

Taking a product of such expression leads to the second identity. Theorem 4.3 The partial product A_{j} is expressed in terms of the triple sine functions as

A_{j} = ∏_{1≤k_{1},k_{2},k_{3}≤N−1} S_{3} (\frac{k_{1}ω_{l} + k_{2}ω_{m} + k_{3}ω_{n}}{N}; ω(j))^{\frac{1}{2}} ,  (4.4)

where l < m < n are the elements not equal to j in {1, 2, 3, 4}. In the product (1.2) for r = 4, the contribution from terms with only one of the coefficients k_{j} being zero is

∏_{j=1}^{4} A_{j} = N^{\frac{7}{2} − \frac{1}{12} ∑_{{1≤i,j≤4 \atop i≠j}} \frac{ω_{i}}{ω_{j}}} \frac{∏’_{0≤k_{1},k_{2}≤N−1} ∏_{l=1}^{4} ∏_{{i<j \atop i,j≠l}} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; ω(l))}{∏_{k=0}^{N−1} ∏_{l=1}^{4} ∏_{{1≤j,l≤4 \atop j≠l}} Γ_{3} (\frac{kω_{j}}{N}; ω(l))}.

Proof. We first calculate A_{4}.

A_{4}^{2} = ∏_{1≤k_{1},k_{2},k_{3}≤N−1} S_{4} (\frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω)^{2}
= ∏_{1≤k_{1},k_{2},k_{3}≤N−1} S_{4} (\frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω) S_{4} (\frac{(N − k_{1})ω_{1} + (N − k_{2})ω_{2} + (N − k_{3})ω_{3}}{N}; ω)
= ∏_{1≤k_{1},k_{2},k_{3}≤N−1} S_{4} (\frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω) S_{4} (ω_{1} + ω_{2} + ω_{3} − \frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω)
= ∏_{1≤k_{1},k_{2},k_{3}≤N−1} S_{3} (\frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω(4)) .

where we used the formula [KK1, Theorem 2.1] S_{r}(z + ω_{i}; ω) = S_{r}(z; ω)S_{r−1}(z; ω(i))^{−1} and the functional equation

S_{r}(z; ω) = S_{r}(|ω| − z; ω)^{(−1)^{r−1}}.

Hence we proved.

A_{4} = ∏_{1≤k_{1},k_{2},k_{3}≤N−1} S_{3} (\frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{N}; ω(4))^{\frac{1}{2}} .

This proves (4.4) for j = 4. Theorem 3.3 gives

A_{4} = N^{\frac{7}{8} − \frac{1}{24} ∑_{{i<j \atop 1≤i,j≤3}} \frac{ω_{i}}{ω_{j}}} (\frac{∏’_{0≤k_{1},k_{2}≤N−1} ∏_{{i<j \atop 1≤i,j≤3}} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; ω(4))}{∏_{k=1}^{N−1} ∏_{l=1}^{3} Γ_{3} (\frac{kω_{l}}{N}; ω(4))}) .  (4.5)

Similarly we have A_{j} for j = 1, 2, 3, 4. Taking a product leads to the conclusion.

Theorem 4.4 In the product (1.2) for r = 4, the contribution from terms with none of the coefficients k_{j} being zero is given by

A = 1.

Proof. By considering the product of (4.4), (4.3), (4.2) over all possible combinations j < l < m, we get all possible values S_{1}(x, ω), S_{2}(x; ω′) and S_{3}(x; ω′′) for any possible one, two and three dimensional vectors ω, ω′ and ω′′ and for any possible linear combinations x = ∑ k_{j}ω_{j}/N with 0 ≤ k_{j} ≤ N − 1. Thus by using the formula (1.2) for r = 1, 2, 3

∏_{j=1}^{4} A_{j} ∏_{{j,l=1 \atop j<l}}^{4} A_{j,l} ∏_{{j,l,m=1 \atop j<l<m}}^{4} A_{j,l,m} = (∏_{j=1}^{4} ∏_{{0≤k_{l}≤N−1 \atop l≠j}} S_{3} (∑_{l≠j} \frac{k_{l}ω_{l}}{N}; ω(j)))^{\frac{1}{2}}

( ∏_{{j,l=1 \atop j<l}}^{4} ∏_{{0≤km≤N−1 \atop m≠j,l}} S_{2} ( ∑_{m≠j,l} \frac{k_{m}ω_{m}}{N}; ω(j, l)))^{− \frac{1}{2}}

( ∏_{{j,l,m=1 \atop j<l<m}}^{4} ∏_{{0≤kn≤N−1 \atop n≠j,l,m}} S_{1} (\frac{k_{n}ω_{n}}{N}; ω(j, l, m)))^{\frac{1}{2}}

= (N^{\binom{4}{3}}N^{−\binom{4}{2}}N^{\binom{4}{1}})^{\frac{1}{2}}
= (N^{4−6+4})^{\frac{1}{2}}
= N.

Then by (4.1) and (1.2) we have

N = AN,

which proves A = 1.

5 Proof of Main Theorem

Proposition 5.1

∏_{1≤k_{1},k_{2}≤N−1} ∏_{1≤i<j≤3} S_{3} (\frac{k_{1}ω_{i} + k_{2}ω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4}))

= \frac{N^{− \frac{9}{4} + \frac{1}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{i≠j} ∏_{k=1}^{N−1} Γ_{3} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4}))^{2}}{( ∏_{{1≤i,j≤3 \atop i≠j}} ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}))) (∏_{i<j} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4}))^{2})}.

Proof. By Theorem 3.2,

∏_{1≤k_{1},k_{2}≤N−1} ∏_{1≤i<j≤3} S_{3} (\frac{k_{1}ω_{i} + k_{2}ω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4}))

= ∏_{i≠j} exp(2ζ′_{3}(0; ω_{i}^{N}) − ζ′_{3}(0; ω_{4}^{(N)}) − ζ′_{3}(0, ω) + ζ′_{2}(0; (Nω_{i}, ω_{j})) − ζ′_{2}(0; (ω_{i}, ω_{j}))

+(log N) (2ζ_{3}(0; ω_{i}^{N}) − ζ_{3}(0; ω_{4}^{(N)}) + ζ_{2}(0; (Nω_{i}, ω_{j})) − \frac{ζ_{2}(0; (ω_{i}, ω_{j}))}{2})).

In the last line the coefficient of (log N) is calculated by Theorem 2.1 as

∑_{i≠j}(−\frac{2}{24} (\frac{ω_{i}}{Nω_{j}} + \frac{ω_{j}}{ω_{4}} + \frac{Nω_{4}}{ω_{i}} + \frac{Nω_{j}}{Nω_{i}} + \frac{ω_{4}}{ω_{j}} + \frac{ω_{i}}{Nω_{4}} + 3)

+\frac{1}{24} (\frac{ω_{i}}{ω_{j}} + \frac{ω_{j}}{Nω_{4}} + \frac{Nω_{4}}{ω_{i}} + \frac{ω_{j}}{Nω_{i}} + \frac{Nω_{4}}{ω_{j}} + \frac{ω_{i}}{Nω_{4}} + 3)

+\frac{1}{12} (N \frac{ω_{i}}{ω_{j}} + \frac{ω_{j}}{Nω_{i}} + 3) − \frac{1}{24} (\frac{ω_{i}}{ω_{j}} + \frac{ω_{j}}{ω_{i}} + 3))

= −\frac{4}{24} ∑_{j=1}^{3} (\frac{ω_{j}}{ω_{4}} + \frac{ω_{4}}{ω_{j}} ) .

Other parts are treated by Theorem 2.2 and Lemmas in Section 2 as

∏_{i≠j} \frac{ρ_{3}(ω_{i}, ω_{j}, Nω_{4})}{ρ_{3}(ω_{i}, ω_{j}, ω_{4})} (\frac{ρ_{3}(ω_{i}, ω_{j}, ω_{4})}{ρ_{3}(ω_{i}, Nω_{j}, Nω_{4})})^{2} \frac{ρ_{2}(ω_{i}, Nω_{j})}{ρ_{2}(ω_{i}, ω_{j})} (\frac{ρ_{2}(Nω_{j}, Nω_{4})}{ρ_{2}(ω_{j}, ω_{4})})^{2} N

= N^{− \frac{21}{4} + \frac{2}{24} ( ∑_{i≠j}( \frac{ω_{i}}{ω_{j}} + \frac{ω_{j}}{Nω_{4}} + \frac{Nω_{4}}{ω_{j}} ))} ∏_{i≠j} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i} + k_{2}ω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4}))^{−1}

× (N^{\frac{21}{4} − \frac{1}{24} ( ∑_{i≠j}( \frac{ω_{j}}{ω_{4}} + \frac{Nω_{j}}{ω_{i}} + \frac{Nω_{4}}{ω_{i}} + \frac{ω_{i}}{Nω_{j}} + \frac{ω_{i}}{ω_{4}} ))})^{2} ∏_{k=1}^{N−1} ∏_{i≠j} Γ_{3} (\frac{kω_{i}}{N} , (ω_{i}, ω_{j}, ω_{4}))^{2}

×N^{− \frac{9}{2} + \frac{1}{12}(N+ \frac{1}{N}) ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{k=1}^{N−1} ∏_{i≠j} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}))^{−1}

× (N^{−6+ \frac{2}{12} ∑_{j=1}^{3}( \frac{ω_{j}}{ω_{4}} + \frac{ω_{4}}{ω_{j}} +3)})^{2} × N^{6}

= N^{− \frac{9}{4} + \frac{1}{24} (2 ∑_{i≠j} \frac{ω_{i}}{ω_{j}} +4( ∑_{j=1}^{3} \frac{ω_{j}}{ω_{4}} + \frac{ω_{4}}{ω_{j}} ))}

×\frac{∏_{i≠j} ∏_{k=1}^{N−1} Γ_{3} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4}))^{2}}{( ∏_{{1≤i,j≤3 \atop i≠j}} ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}))) (∏_{i<j} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i}+k_{2}ω_{j}}{N}; ω)^{2})}

as desired. Proposition 5.2

∏_{i≠j} S_{3} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4})) = N^{3+ \frac{1}{12} ∑_{i≠j}( \frac{Nω_{4}}{ω_{j}} − \frac{Nω_{i}}{ω_{j}} + \frac{ω_{j}}{Nω_{4}} − \frac{ω_{j}}{Nω_{i}} )}

× ∏_{k=1}^{N−1} ∏_{i≠j} \frac{Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j})) Γ_{2} (\frac{kω_{j}}{N}; (ω_{j}, ω_{4}))}{Γ_{3} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4}))^{2}}.

Proof. By Theorem 3.1 and Lemmas in Section 2, we have

∏_{i≠j} S_{3} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4}))

= ∏_{i≠j} exp(2(ζ′_{3}(0; (ω_{i}, ω_{j}, ω_{4})) − ζ′_{3}(0; (ω_{i}, Nω_{j}, Nω_{4})))

+ζ′_{2}(0; (ω_{i}, ω_{j})) + ζ′_{2}(0; (ω_{i}, ω_{4})) − ζ′_{2}(0; (ω_{i}, Nω_{j})) − ζ′_{2}(0; (ω_{i}, Nω_{4})) −(log N)(2ζ_{3}(0; (ω_{i}, Nω_{j}, Nω_{4})) + 2ζ_{2}(0; (ω_{i}, Nω_{j})) + ζ(0)))

= ∏_{i≠j} (\frac{ρ_{3}(ω_{i}, Nω_{j}, Nω_{4})}{ρ_{3}(ω_{i}, ω_{j}, ω_{4})})^{2} \frac{ρ_{2}(ω_{i}, ω_{j})}{ρ_{2}(ω_{i}, Nω_{j})} \frac{ρ_{2}(ω_{i}, ω_{4})}{ρ_{2}(ω_{i}, Nω_{4})} (\frac{ρ_{2}(ω_{j}, ω_{4})}{ρ_{2}(Nω_{j}, Nω_{4})})^{2} N^{−1}

×N^{\frac{2}{24} ( \frac{ω_{i}}{Nω_{j}} + \frac{ω_{j}}{ω_{4}} + \frac{Nω_{4}}{ω_{i}} + \frac{Nω_{j}}{ω_{i}} + \frac{ω_{4}}{ω_{j}} + \frac{ω_{i}}{Nω_{4}} +3)− \frac{2}{12} ( \frac{ω_{i}}{Nω_{j}} + \frac{Nω_{j}}{ω_{i}} +3)+ \frac{1}{2}}

= N^{− \frac{21}{2} + \frac{2}{24} ( ∑_{i≠j} \frac{ω_{j}}{ω_{4}} + \frac{ω_{4}}{ω_{j}} +N( \frac{ω_{j}}{ω_{i}} + \frac{ω_{4}}{ω_{i}} )+ \frac{1}{N} ( \frac{ω_{i}}{ω_{j}} + \frac{ω_{i}}{ω_{4}} ))} ∏_{i≠j} ∏_{k=1}^{N−1} Γ_{3} (\frac{kω_{i}}{N}; (ω_{i}, ω_{j}, ω_{4}))^{−2}

×N^{9− \frac{1}{12} ∑_{i≠j}(N+ \frac{1}{N})( \frac{ω_{i}}{ω_{j}} + \frac{ω_{4}}{ω_{j}} )} ∏_{i≠j} ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{i}}{N}; (ω_{i}, ω_{j})) Γ_{2} (\frac{kω_{j}}{N}; (ω_{j}, ω_{4}))

× (N^{6− \frac{2}{12} ∑_{j=1}^{3}( \frac{ω_{j}}{ω_{4}} + \frac{ω_{4}}{ω_{j}} +3)})^{2} N^{−6}

×N^{^{\frac{1}{12}} ∑_{i≠j}( \frac{1}{N} ( \frac{ω_{i}}{ω_{4}} − \frac{ω_{i}}{ω_{j}} )+N( \frac{ω_{4}}{ω_{i}} − \frac{ω_{j}}{ω_{i}} )+ \frac{ω_{j}}{ω_{4}} + \frac{ω_{4}}{ω_{j}} +3)}

= N^{3+ \frac{1}{12} ∑_{i≠j}(N( \frac{ω_{4}}{ω_{j}} − \frac{ω_{i}}{ω_{j}} )+ \frac{1}{N} ( \frac{ω_{j}}{ω_{4}} − \frac{ω_{j}}{ω_{i}} ))} ∏_{i≠j} ∏_{k=1}^{N−1} \frac{Γ_{2} (\frac{kω_{i}}{N}; (ω_{i}, ω_{j})) Γ_{2} (\frac{kω_{j}}{N}; (ω_{j}, ω_{4}))}{Γ_{3} (\frac{kω_{i}}{N}; (ω_{i}, ω_{j}, ω_{4}))^{2}}

as desired. Proposition 5.3

∏_{j=1}^{3} A_{j,4} = N^{− \frac{9}{4} + \frac{1}{12} ∑_{i≠j} \frac{ω_{i}}{ω_{j}}} ∏_{i≠j} ∏_{k=1}^{N−1} (Γ_{3} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4})) Γ_{3} (\frac{kω_{j}}{N}; (ω_{1}, ω_{2}, ω_{3})))^{2}
( ∏_{{1≤i,j≤3 \atop i≠j}} ∏_{k=1}^{N−1} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j})))^{−1}

(∏_{i<j} ∏’_{0≤k_{1},k_{2}≤N−1} Γ_{3} (\frac{k_{1}ω_{i} + k_{2}ω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4})) Γ_{3} (\frac{k_{1}ω_{i} + k_{2}ω_{j}}{N}; (ω_{1}, ω_{2}, ω_{3})))^{−1} .

Proof. By (4.3) we have

∏_{j=1}^{3} A_{j,4}^{2} = ∏_{1≤k_{1},k_{2}≤N−1} ∏_{1≤i<j≤3} S_{3} (\frac{k_{1}ω_{i} + k_{2}ω_{j}}{N}; (ω_{1}, ω_{2}, ω_{3})) S_{3} (\frac{k_{1}ω_{i} + k_{2}ω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4})) .

(5.1)

The product of the first and the second factors in (5.1) is given by Theorem 3.2 and Proposition 5.1. Proposition follows.

Proposition 5.4

∏_{i<j} A_{i,j,4} = N^{\frac{3}{2} + \frac{1}{24} (N ∑_{i≠j}( \frac{ω_{4}}{ω_{j}} − \frac{ω_{j}}{ω_{4}} )+ \frac{1}{N} ∑_{i≠j}( \frac{ω_{i}}{ω_{4}} − \frac{ω_{i}}{ω_{j}} ))}

× ∏_{k=1}^{N−1} \frac{∏_{i≠j} Γ_{2} (\frac{kω_{j}}{N}; (ω_{j}, ω_{4}))^{\frac{1}{2}} Γ_{2} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}))}{(∏_{i≠j} Γ_{3} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4}))) (∏_{i≠j} Γ_{3} (\frac{kω_{j}}{N}; (ω_{1}, ω_{2}, ω_{3})))}

Proof. By (4.2) we deduce

∏_{i<j} A_{i,j,4} = ∏_{j=1}^{3} ∏_{k=1}^{N−1} S_{4} (\frac{kω_{j}}{N} )

= N^{− \frac{3}{4}} ∏_{k=1}^{N−1} (∏_{i≠j} S_{3} (\frac{kω_{j}}{N}; (ω_{i}, ω_{j}, ω_{4})) ∏_{j=1}^{3} S_{3} (\frac{kω_{j}}{N}; (ω_{1}, ω_{2}, ω_{3})))^{\frac{1}{2}} ,

where the last products are calculated by Proposition 5.2 and Theorem 3.1. The proof is complete.

Proof of Theorem 1.2. Each factor consisting Ã_{4} is calculated in (4.5) and Propositions 5.3-5.4. Taking a product leads to the conclusion.

6 Special Values

In this section we present some numerical datum which are necessary for the examples given in the preceding sections.

Notation. When ω = 1 = (1, ..., 1), we simply write Γ_{r}(x, 1) = Γ_{r}(x). Lemma 6.1 We have Γ_{2}(1) = e^{ζ′(−1)}, Γ_{2}(2) = e^{ζ′(−1)+ \frac{1}{2}} , Γ_{3}(1) = e^{\frac{ζ′(−2)+ζ′(−1)}{2}} , and Γ_{3}(2) = e^{\frac{ζ′(−2)−ζ′(−1)}{2}} . For n ≥ 3, it holds that

Γ_{2}(n) = exp (ζ′(−1) + \frac{n − 1}{2}) ∏_{k=1}^{n−2} k^{k+1−n},

and that

Γ_{3}(n) = exp (\frac{ζ′(−2) − (2n − 3)ζ′(−1)}{2} − \frac{(n − 1)(n − 2)}{4}) ∏_{k=1}^{n−1} k^{\frac{(n−k+1)(n−k+2)}{2}} .

Proof. For an integer n ≥ 1, we compute

ζ_{r}(s, n, 1) = ∑_{m≥0}(m · 1 + n)^{−s} = ∑_{k=n}^{∞} \frac{f_{r}(k − n)}{k^{s}},

where

f_{r}(n) = #{(k_{1}, ..., k_{r}) | k_{1} + ⋯ + k_{r} = n, k_{1}, ..., k_{r} ≥ 0} = _{n+r−1}C_{n}.

When r = 2 or 3, since

f_{r}(n) = \begin{cases}n + 1 & (r = 2) \\ \frac{(n+1)(n+2)}{2} & (r = 3) \end{cases} ,

we have

ζ_{2}(s, n, 1) = ∑_{k=n}^{∞} \frac{k − n + 1}{k^{s}}
= ζ(s − 1) − (n − 1)ζ(s) − ∑_{k=1}^{n−1} \frac{k − n + 1}{k^{s}},

and

ζ_{3}(s, n, 1) = ∑_{k=n}^{∞} \frac{(k − n + 1)(k − n + 2)}{2k^{s}}
= ∑_{k=n}^{∞} \frac{k^{2} − (2n − 3)k + (n − 1)(n − 2)}{2k^{s}}
= \frac{ζ(s − 2)}{2} − \frac{2n − 3}{2}ζ(s − 1) + \frac{(n − 1)(n − 2)}{2}ζ(s) − ∑_{k=1}^{n−1} \frac{(k − n + 1)(k − n + 2)}{2k^{s}}.

ζ′_{r}(0, n, 1) = \begin{cases}ζ′(−1) + \frac{n−1}{2} + ∑_{k=1}^{n−2}(k + 1 − n) log k & (r = 2) \\ \frac{1}{2}ζ′(−2) − \frac{2n−3}{2}ζ′(−1) − \frac{(n−1)(n−2)}{4} + ∑_{k=1}^{n−1} \frac{(n+1−k)(n+2−k)}{2} log k. & (r = 3) \end{cases}

Thus we have the desired result as Γ_{r}(n) = e^{ζ′_{r}(0,n,1)}. Lemma 6.2

Γ_{2} (\frac{1}{2}) = 2^{− \frac{7}{24}} e^{− \frac{ζ′(−1)}{2}} .

Proof. Putting r = 2, ω = 1 and N_{1} = N_{2} = 2 in Lemma 2.2, we have

\frac{ρ_{2}(1, 1)}{ρ_{2}(\frac{1}{2}, \frac{1}{2})} = Γ_{2} (\frac{1}{2})^{2} Γ_{2}(1) = Γ_{2} (\frac{1}{2})^{2} e^{ζ′(−1)}

by Lemma 6.1. Due to Lemma 2.1 the left hand side is equal to

(\frac{1}{2})^{1− \frac{5}{12}} = 2^{− \frac{7}{12}} .

Hence we have the conclusion. Lemma 6.3

Γ_{3} (\frac{1}{2}) = 2^{− \frac{11}{48}} exp (−\frac{3}{8}ζ′(−2) − \frac{1}{2}ζ′(−1)) .

Proof. Putting r = 3, ω = 1 and N_{1} = N_{2} = N_{3} = 2 in Lemma 2.1, we have

\frac{ρ_{3}(1, 1, 1)}{ρ_{3}(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})} = Γ_{3} (\frac{1}{2})^{3} Γ_{3}(1)^{3}Γ_{3} (\frac{3}{2}) .

From Lemma 2.1 the left hand side is equal to

(\frac{1}{2})^{1− \frac{3}{8}} = 2^{− \frac{5}{8}} .

By Lemma 6.1 Γ_{3}(1) = e^{\frac{ζ′(−2)+ζ′(−1)}{2}} , and by Lemma 2.7 we have

Γ_{3} (\frac{3}{2}) = Γ_{3} (\frac{1}{2}) Γ_{2} (\frac{1}{2})^{−1} = Γ_{3} (\frac{1}{2}) 2^{\frac{7}{24}} e^{\frac{ζ′(−1)}{2}} ,

appealing to the previous lemma. Then we have

2^{− \frac{5}{8}} = Γ_{3} (\frac{1}{2})^{4} exp (3\frac{ζ′(−2) + ζ′(−1)}{2} + \frac{ζ′(−1)}{2}) 2^{\frac{7}{24}}
= Γ_{3} (\frac{1}{2})^{4} exp (\frac{3ζ′(−2)}{2} + 2ζ′(−1)) 2^{\frac{7}{24}} .

Thus

Γ_{3} (\frac{1}{2}) = 2^{− \frac{11}{48}} exp (−\frac{3}{8}ζ′(−2) − \frac{1}{2}ζ′(−1)) .

7 Modular Interpretations

Lemma 7.1

ζ_{4} (0, \frac{ω_{1} + ω_{2} + ω_{3} + ω_{4}}{2} ; ω) = \frac{7(ω_{1}^{4} + ω_{2}^{4} + ω_{3}^{4} + ω_{4}^{4}) + 10 ∑_{{i<j \atop 1≤i,j≤4}} ω_{i}^{2}ω_{j}^{2} − 360ω_{1}ω_{2}ω_{3}ω_{4}}{5760ω_{1}ω_{2}ω_{3}ω_{4}} .

Proof. We appeal to an integral representation

ζ_{4}(s, x; ω) = −\frac{Γ(1 − s)}{2πi} ∫_{C} \frac{e^{−xt}(−t)^{s−1}}{(1 − e^{−ω_{1}t})(1 − e^{−ω_{2}t})(1 − e^{−ω_{3}t})(1 − e^{−ω_{4}t})}dt,

where C is the standard coutour consisting of +∞ → ε > 0, εe^{iθ} (0 ≤ θ ≤ 2π), ε → +∞. By putting

\frac{1}{(1 − e^{−ω_{1}t})(1 − e^{−ω_{2}t})(1 − e^{−ω_{3}t})(1 − e^{−ω_{4}t})} = \frac{a_{−4}}{t^{4}} + \frac{a_{−3}}{t^{3}} + \frac{a_{−2}}{t^{2}} + \frac{a_{−1}}{t} + a_{0} + O(t)

as t → 0, we compute

ζ_{4}(0, x; ω) = \frac{1}{2πi} ∫_{C} \frac{e^{−xt}t^{−1}}{(1 − e^{−ω_{1}t})(1 − e^{−ω_{2}t})(1 − e^{−ω_{3}t})(1 − e^{−ω_{4}t})}dt
= a_{0} − xa_{−1} + \frac{x^{2}a_{−2}}{2} − \frac{x^{3}a_{−3}}{3!} + \frac{x^{4}a_{−4}}{4!} .

We calculate each coefficient in order as follows:

a_{−4} = \frac{1}{ω_{1}ω_{2}ω_{3}ω_{4}} ,
a_{−3} = \frac{ω_{1} + ω_{2} + ω_{3} + ω_{4}}{2ω_{1}ω_{2}ω_{3}ω_{4}} ,
a_{−2} = \frac{1}{12ω_{1}ω_{2}ω_{3}ω_{4}} (ω_{1}^{2} + ω_{2}^{2} + ω_{3}^{2} + ω_{4}^{2} + 3 ∑_{{i<j \atop 1≤i,j≤4}} ω_{i}ω_{j}) ,
a_{−1} = \frac{1}{24ω_{1}ω_{2}ω_{3}ω_{4}} ( ∑_{1≤i,j≤4} ω_{i}ω_{j}^{2} + 3 ∑_{1≤i<j<k≤4} ω_{i}ω_{j}ω_{k}) ,
a_{0} = −\frac{1}{720ω_{1}ω_{2}ω_{3}ω_{4}} (ω_{1}^{4} + ω_{2}^{4} + ω_{3}^{4} + ω_{4}^{4} − 5 ∑_{{i<j \atop 1≤i,j≤4}} ω_{i}^{2}ω_{j}^{2} − 15 ∑_{i=1}^{4} ∑_{{1≤j,k≤4 \atop j,k≠i}} ω_{i}^{2}ω_{j}ω_{k}) .

Taking x = \frac{ω_{1}+ω_{2}+ω_{3}+ω_{4}}{2} and carrying out straightforward calculations lead to the result. Proof of Theorem 1.3. By Akatsuka’s product expression [A] of multiple sine functions we have

S_{4}(x; ω) = exp (πiζ_{4}(0, x; ω) + ∑_{j=1}^{4} ∑_{n=1}^{∞} \frac{1}{n} \frac{e(\frac{nx}{ω_{j}})}{∏_{l≠j}(1 − e(\frac{nω_{l}}{ω_{j}}))}) .  (7.1)

When x = (ω_{1} + ω_{2} + ω_{3} + ω_{4})/2, we compute

e (\frac{nx}{ω_{j}}) = ∏_{l=1}^{4} e (\frac{nω_{l}}{2ω_{j}} ) = (−1)^{n} ∏_{l≠j} e (\frac{nω_{l}}{2ω_{j}} ) .

Hence

S_{4}(x; ω) = exp (πiζ_{4}(0, x; ω) + ∑_{j=1}^{4} ∑_{n=1}^{∞} \frac{1}{n} ∏_{l≠j} \frac{(−1)^{n}e (\frac{nω_{l}}{2ω_{j}} )}{1 − e(\frac{nω_{l}}{ω_{j}})})
= exp (πiζ_{4}(0, x; ω) + ∑_{j=1}^{4} ∑_{n=1}^{∞} \frac{1}{n} ∏_{l≠j} \frac{(−1)^{n}}{e(−\frac{nω_{l}}{2ω_{j}}) − e(\frac{nω_{l}}{2ω_{j}})})
= exp (πiζ_{4}(0, x; ω) + ∑_{j=1}^{4} ∑_{n=1}^{∞} \frac{1}{n} ∏_{l≠j} \frac{(−1)^{n}}{−2i sin(π \frac{nω_{l}}{ω_{j}})})

= exp (πiζ_{4}(0, x; ω) + ∑_{j=1}^{4} ∑_{n=1}^{∞} \frac{1}{n} \frac{(−1)^{n}}{8i} ∏_{l≠j} \frac{1}{sin(π \frac{nω_{l}}{ω_{j}})}) .

When ω = (τ_{1}, τ_{2}, τ_{3}, 1) and x = (τ_{1} + τ_{2} + τ_{3} + 1)/2, Theorem 4.4 implies that S_{4}(x; ω) = 1. Thus

−8πζ_{4}(0, x; ω) + ∑_{j=1}^{4} ∑_{n=1}^{∞} \frac{(−1)^{n}}{n} ∏_{l≠j} \frac{1}{sin(π \frac{nω_{l}}{ω_{j}})} = 0.

This leads to

∑_{n=1}^{∞} \frac{(−1)^{n}}{n sin(πnτ_{1}) sin(πnτ_{2}) sin(πnτ_{3})}

= 8πζ_{4}(0, x; ω) − ∑_{n=1}^{∞} \frac{(−1)^{n}}{n sin(π \frac{n}{τ_{1}} ) sin(π \frac{nτ_{2}}{τ_{1}} ) sin(π \frac{nτ_{3}}{τ_{1}} )}

− ∑_{n=1}^{∞} \frac{(−1)^{n}}{n sin(π \frac{nτ_{1}}{τ_{2}} ) sin(π \frac{n}{τ_{2}} ) sin(π \frac{nτ_{3}}{τ_{2}} )} − ∑_{n=1}^{∞} \frac{(−1)^{n}}{n sin(π \frac{nτ_{1}}{τ_{3}} ) sin(π \frac{nτ_{2}}{τ_{3}} ) sin(π \frac{n}{τ_{3}} )}.

Proof of Theorem 1.4. We compute the product (7.1) for

ω = (ω_{1}, ω_{2}, ω_{3}, ω_{4}) = (τ_{1}, τ_{2}, τ_{3}, 1)

and

x = \frac{τ_{1} + τ_{2} + τ_{3} + 1}{2}.

The factor for j = 4 in the double sum is

∑_{n=1}^{∞} \frac{e(nx)}{n ∏_{l=1}^{3}(1 − e(nτ_{l}))} = ∑_{n=1}^{∞} \frac{e(nx)}{n} ∑_{n_{1}=0}^{∞} ∑_{n_{2}=0}^{∞} ∑_{n_{3}=0}^{∞} e(n(n_{1}τ_{1} + n_{2}τ_{2} + n_{3}τ_{3}))

= − log ∏_{n_{1}=0}^{∞} ∏_{n_{2}=0}^{∞} ∏_{n_{3}=0}^{∞} (1 − e(x + n_{1}τ_{1} + n_{2}τ_{2} + n_{3}τ_{3}))

= − log ∏_{n_{1}=0}^{∞} ∏_{n_{2}=0}^{∞} ∏_{n_{3}=0}^{∞} (1 + e ((n_{1} + \frac{1}{2}) τ_{1} + (n_{2} + \frac{1}{2}) τ_{2} + (n_{3} + \frac{1}{2}) τ_{3}))
= − log F (τ_{1}, τ_{2}, τ_{3}).

For j = 1, 2, 3, we can similarly deal with the terms involving ω_{l}/ω_{j} (l < j) as Im(ω_{l}/ω_{j}) > 0. But in the other cases when l > j, we have Im(ω_{l}/ω_{j}) < 0 and compute as follows:

∑_{n=1}^{∞} \frac{1}{n} \frac{e(\frac{nx}{ω_{j}})}{∏_{l≠j}(1 − e(\frac{nω_{l}}{ω_{j}}))} = ∑_{n=1}^{∞} \frac{e(\frac{nx}{ω_{j}})}{n} \frac{1}{∏_{l>j}(1 − e(\frac{nω_{l}}{ω_{j}})) ∏_{l<j}(1 − e(\frac{nω_{l}}{ω_{j}}))}

= ∑_{n=1}^{∞} \frac{e(\frac{nx}{ω_{j}})}{n} ∏_{l>j} \frac{e(−\frac{nω_{l}}{ω_{j}})}{e(−\frac{nω_{l}}{ω_{j}}) − 1} ∏_{l<j} \frac{1}{1 − e(\frac{nω_{l}}{ω_{j}})}

= ∑_{n=1}^{∞} \frac{e(\frac{nx}{ω_{j}})}{n} ∏_{l>j} (− ∑_{n_{l}=0}^{∞} e (−\frac{n(n_{l} + 1)ω_{l}}{ω_{j}} )) ∏_{l<j} ∑_{n_{l}=0}^{∞} e (\frac{nn_{l}ω_{l}}{ω_{j}} )

= ∑_{n=1}^{∞} \frac{(−1)^{n}}{n} ∏_{l>j} (− ∑_{n_{l}=0}^{∞} e (−\frac{n(n_{l} + \frac{1}{2})ω_{l}}{ω_{j}} )) ∏_{l<j} ∑_{n_{l}=0}^{∞} e (\frac{n(n_{l} + \frac{1}{2})ω_{l}}{ω_{j}} )

= (−1)^{#\{l | l>j\}} ∑_{n=1}^{∞} \frac{(−1)^{n}}{n} ∑_{{n_{l}=0 \atop l≠j}}^{∞} e (− ∑_{l>j} \frac{n(n_{l} + \frac{1}{2})ω_{l}}{ω_{j}} + ∑_{l<j} \frac{n(n_{l} + \frac{1}{2})ω_{l}}{ω_{j}} )

= (−1)^{#\{l | l>j\}+1} ∑_{{n_{l}=0 \atop l≠j}}^{∞} log (1 + e (− ∑_{l>j} \frac{(n_{l} + \frac{1}{2})ω_{l}}{ω_{j}} + ∑_{l<j} \frac{(n_{l} + \frac{1}{2})ω_{l}}{ω_{j}} ))

= (−1)^{#\{l | l>j\}+1} log ∏_{{n_{l}=0 \atop l≠j}}^{∞} (1 + e (− ∑_{l>j} \frac{(n_{l} + \frac{1}{2})ω_{l}}{ω_{j}} + ∑_{l<j} \frac{(n_{l} + \frac{1}{2})ω_{l}}{ω_{j}} ))

= \begin{cases}log F (\frac{τ_{1}}{τ_{3}} , \frac{τ_{2}}{τ_{3}} , −\frac{1}{τ_{3}} ) & (j = 3) \\ − log F (\frac{τ_{1}}{τ_{2}} , −\frac{1}{τ_{2}} , −\frac{τ_{3}}{τ_{2}} ) & (j = 2) \\ log F (−\frac{1}{τ_{1}} , −\frac{τ_{2}}{τ_{1}} , −\frac{τ_{3}}{τ_{1}} ) & (j = 1). \end{cases}

Hence from Theorem 4.4 and (7.1) we have

1 = S_{4}(x; ω) = e^{πiζ_{4}(0,x;ω)} \frac{F (\frac{τ_{1}}{τ_{3}} , \frac{τ_{2}}{τ_{3}} , −\frac{1}{τ_{3}} ) F (−\frac{1}{τ_{1}} , −\frac{τ_{2}}{τ_{1}} , −\frac{τ_{3}}{τ_{1}} )}{F (τ_{1}, τ_{2}, τ_{3})F (\frac{τ_{1}}{τ_{2}} , −\frac{1}{τ_{2}} , −\frac{τ_{3}}{τ_{2}} )}.

8 The graph of S_{4}(x; ω)

Throughout this section we assume ω = (ω_{1}, ω_{2}, ω_{3}, ω_{4}) ∈ R^{4} with 0 < ω_{1} ≤ ω_{2} ≤ ω_{3} ≤ ω_{4}, and put S(x) = S_{4}(x; ω) for simplicity. We recall the notation |ω| := ω_{1} + ω_{2} + ω_{3} + ω_{4}.

Lemma 8.1 At least one of the following ten special values are greater than one:

S (\frac{ω_{j}}{2} )  (j = 1, 2, 3, 4),  S (\frac{ω_{i} + ω_{j}}{2} )  (i, j = 1, 2, 3, 4, i < j).

Proof. By Proposition 2.4 in [KK1] we have for any l = 1, 2, 3, 4

S_{3}(x; ω(l)) = S_{4}(x; ω)S_{4}(x + ω_{l}; ω)^{−1} = S_{4}(x; ω)S_{4}(|ω| − x − ω_{l}; ω).

Taking x = |ω(l)|/2 leads to

S_{3} (\frac{|ω(l)|}{2}; ω(l)) = S_{4} (\frac{|ω(l)|}{2}; ω)^{2} = S (\frac{|ω(l)|}{2})^{2} .

By Kurokawa’s recent result S_{3} (\frac{|ω(l)|}{2}; ω(l)) < 1 proved in [K3], we have

S (\frac{|ω(l)|}{2}) < 1

for any l = 1, 2, 3, 4. Thus in the formula (1.2) with N = 2

(∏_{j=1}^{4} S (\frac{ω_{j}}{2} )) ( ∏_{{1≤i,j≤4 \atop i<j}} S (\frac{ω_{i} + ω_{j}}{2} )) ( ∏_{{1≤i,j,k≤4 \atop i<j<k}} S (\frac{ω_{i} + ω_{j} + ω_{k}}{2} )) = 2,

the third product in the left hand side is less than 1. Thus the remaining part must be greater than 1. Hence we have the desired conclusion.

Lemma 8.2

(\frac{S′}{S} )′′ (\frac{|ω|}{2}) < 0.

Proof. By definition we have

S(x) = Γ_{4}(x; ω)^{−1}Γ_{4}(|ω| − x; ω)

with

Γ_{4}(x; ω) = exp (\frac{∂}{∂s}ζ_{4}(0, x; ω))

for

ζ_{4}(s, x; ω) = ∑_{n_{1},n_{2},n_{3},n_{4}≥0}(n_{1}ω_{1} + n_{2}ω_{2} + n_{3}ω_{3} + n_{4}ω_{4} + x)^{−s}.

We find that ζ_{4}(s, x; ω) is absolutely convergent in Re(s) > 4 and that it has a meromorphic continuation to all s ∈ C. Moreover ζ_{4}(s, x; ω) is holomorphic at s = 0.

Since

log S(x) = − log Γ_{4}(x; ω) + log Γ_{4}(|ω| − x; ω)
= −\frac{∂}{∂s}ζ_{4}(0, x; ω) + \frac{∂}{∂s}ζ_{4}(0, |ω| − x; ω)

we have

\frac{S′}{S} (x) = −\frac{∂^{2}}{∂s∂x}ζ_{4}(0, x; ω) − \frac{∂^{2}}{∂s∂x}ζ_{4}(0, |ω| − x; ω).

Hence

(\frac{S′}{S} )′ (x) = −\frac{∂^{3}}{∂s∂x^{2}} ζ_{4}(0, x; ω) + \frac{∂^{3}}{∂s∂x^{2}} ζ_{4}(0, |ω| − x; ω)

and differentiating repeatedly gives

(\frac{S′}{S} )^{(4)} (x) = −\frac{∂^{6}}{∂s∂x^{5}} ζ_{4}(0, x; ω) − \frac{∂^{4}}{∂s∂x^{3}} ζ_{4}(0, |ω| − x; ω)

= 4! ( ∑_{n_{1},n_{2},n_{3},n_{4}≥0}(n_{1}ω_{1} + n_{2}ω_{2} + n_{3}ω_{3} + n_{4}ω_{4} + x)^{−5}

+ ∑_{m_{1},m_{2},m_{3},m_{4}≥1}(m_{1}ω_{1} + m_{2}ω_{2} + m_{3}ω_{3} + m_{4}ω_{4} − x)^{−5}) ,

where we use the relation

\frac{∂^{5}}{∂x^{5}} ζ_{4}(s, x; ω) = (−s)(−s − 1)(−s − 2)(−s − 3)(−s − 4)ζ_{4}(s + 5, x; ω)

coming from

\frac{∂}{∂x}ζ_{4}(s, x; ω) = \frac{∂}{∂x} ( ∑_{n_{1},n_{2},n_{3},n_{4}≥0}(n_{1}ω_{1} + n_{2}ω_{2} + n_{3}ω_{3} + n_{4}ω_{4} + x)^{−s})
= −s ∑_{n_{1},n_{2},n_{3},n_{4}≥0}(n_{1}ω_{1} + n_{2}ω_{2} + n_{3}ω_{3} + n_{4}ω_{4} + x)^{−s−1}
= −sζ_{4}(s + 1, x; ω).

Thus

(\frac{S′}{S} )^{(4)} (x) > 0  (8.1)

in 0 < x < |ω| and

(\frac{S′}{S} )^{(3)} (\frac{|ω|}{2}) = 0.  (8.2)

x  0 ⋯ \frac{|ω|}{2} ⋯ |ω|
(\frac{S′}{S} )^{(4)} + +
(\frac{S′}{S} )^{(3)}  ↗ 0 ↗
(\frac{S′}{S} )^{(3)}  − 0 +
(\frac{S′}{S} )′′  ↘  ↗

By (8.1) and (8.2) we see as in the above table that (\frac{S′}{S} )^{(3)} (x) < 0 for 0 < x < \frac{|ω|}{2} and

(\frac{S′}{S} )^{(3)} (x) > 0 for \frac{|ω|}{2} < x < |ω|, and consequently that (\frac{S′}{S} )′′ (x) takes the minimum at x = \frac{|ω|}{2}.

For proving (\frac{S′}{S} )′′ (\frac{|ω|}{2}) < 0, we assume (\frac{S′}{S} )′′ (\frac{|ω|}{2}) ≥ 0 to have a contradiction as

below. By the above table, the assumption (\frac{S′}{S} )′′ (\frac{|ω|}{2}) ≥ 0 implies that (\frac{S′}{S} )′′ (x) ≥ 0 for

0 < x < |ω|. Hence (\frac{S′}{S} )′ (x) is increasing throughout the interval. Proposition 2.4 in [KK1] shows that

S(x) = \frac{x}{x − |ω|} × O(1)  (8.3)

with O(1) part being a nonzero holomorphic bounded function as x → 0 and x → |ω|. Thus

lim_{x→+0} (\frac{S′}{S} )′ (x) = −∞

and

lim_{x→|ω|} (\frac{S′}{S} )′ (x) = ∞.

Therefore there exists a unique γ ∈ (0, |ω|) such that (\frac{S′}{S} )′ (γ) = 0. This means that (\frac{S′}{S} ) (x) takes the minimum at x = γ. If this minimum is non-negative, then (\frac{S′}{S} ) (x) ≥ 0 and so S′(x) ≥ 0. This shows S(x) is increasing, which contradicts the previous lemma, because Theorem 1.1 says

S (\frac{|ω|}{2}) = 1.

Hence the minimum is negative. Here we again use (8.3) to get

lim_{x→0,|ω|} \frac{S′}{S} (x) = ∞.

Then there exist γ_{1} ∈ (0, γ) and γ_{2} ∈ (γ, |ω|) such that \frac{S′}{S} (γ_{1}) = \frac{S′}{S} (γ_{2}) = 0, and we see that \frac{S′}{S} (x) is positive in [0, γ_{1}) ∪ (γ_{2}, |ω|], and negative in (γ_{1}, γ_{2}). As S(x) > 0 in the whole interval, we also determine the sigunature of S′(x) to obtain the behavior of S(x).

Since S(x)S(|ω| − x) = 1, it holds that log S(x) is symmtric with respect to the point (|ω|/2, 0). Thus γ_{1} < |ω|/2 < γ_{2}, and consequently there exists some c ∈ (0, |ω|/2) such that 0 < S(x) < 1 in x ∈ (0, c) and 1 < S(x) in x ∈ (c, |ω|/2). This implies that if two points x, x′ ∈ (0, |ω|/2) satisfy that S(x) < 1 and that S(x′) > 1, then it must hold that x < x′. This contradicts the previous lemma, taking into account that

S (\frac{ω_{1} + ω_{2} + ω_{3}}{2} ) = S_{4} (\frac{ω_{1} + ω_{2} + ω_{3}}{2} ; ω) = S_{3} (\frac{ω_{1} + ω_{2} + ω_{3}}{2} ; (ω_{1}, ω_{2}, ω_{3}))^{\frac{1}{2}} < 1,

which is shown by Kurokawa [K3]. Hence the conclusion.

Theorem 8.1 The function S(x) = S_{4}(x; ω) has four extremal values in the interval (0, |ω|). More precisely each of the intervals (0, |ω|/2) and (|ω|/2, |ω|) has both a maximal and a minimal points.

Proof. By the previous lemma and by (8.3) we have the following table.

x  0  ⋯ \frac{|ω|}{2}  ⋯  |ω|
(\frac{S′}{S} )′′ +∞ ↘  −  ↗ +∞
(\frac{S′}{S} )′′  + 0 − − − 0 +
(\frac{S′}{S} )′ −∞ ↗ ↘ 0 ↘ ↗ +∞
(\frac{S′}{S} )′ −∞ − 0 + 0 − 0 + +∞
\frac{S′}{S} +∞ ↘ ↗  ↘ ↗ +∞

As shown in the proof of the previous lemma, S(x) cannot be monotone increasing in (0, |ω|/2). So the minimal value of \frac{S′}{S} in (0, |ω|/2) should be negative. Therefore we have the following table:

x 0  ⋯  \frac{|ω|}{2}  ⋯  |ω|
\frac{S′}{S} +∞ + 0 − 0 + + + 0 − 0 + +∞
S′  + 0 − 0 + + + 0 − 0 +
S 0  ↗ ↘ ↗  1  ↗ ↘ ↗ +∞

The proof is complete.

Theorem 8.2 Let α, β ∈ (0, |ω|/2) satisfy S(α) = S(β) = 1 with α < β as shown in Figure 1 in Section 1. Then for 0 < ω_{1} ≤ ω_{2} ≤ ω_{3} ≤ ω_{4}, the following statements concerning the location of 2-division points are true.

(1) At least one of \frac{ω_{j}}{2} (j = 1, 2, 3, 4) and \frac{ω_{i}+ω_{j}}{2} (i, j = 1, 2, 3, 4, i < j) lie in the interval (α, β).

(2) \frac{ω_{3}+ω_{4}}{2} > α.

(3) Both \frac{ω_{1}+ω_{3}+ω_{4}}{2} and \frac{ω_{2}+ω_{3}+ω_{4}}{2} lie in the interval (β, |ω|/2).

Proof. The first assertion follows from Lemma 8.1 and the previous theorem.

The second assertion follows from the first one, because \frac{ω_{3}+ω_{4}}{2} is the largest among the ten points given in (1), and in particular it is greater than the element in (α, β).

Then by (2), since both \frac{ω_{1}+ω_{3}+ω_{4}}{2} and \frac{ω_{2}+ω_{3}+ω_{4}}{2} are greater than \frac{ω_{3}+ω_{4}}{2} , they are both greater than α. On the other hand by Kurokawa’s result [K3], it holds that S (\frac{ω_{1}+ω_{3}+ω_{4}}{2} ) < 1 and S (\frac{ω_{2}+ω_{3}+ω_{4}}{2} ) < 1, as shown at the end of the proof of Lemma 8.1. Hence we obtain the last assertion.

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[K3] Kurokawa, N.: The central value of the triple sine function. (preprint, 2006)

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Zeta Institute, 2-5-27, Hayabuchi, Tsuzuki-ku, Yokohama 224-0025, Japan koyama@tmtv.ne.jp