Euler Product Expression of Triple Zeta Functions

Hirotaka Akatsuka

Abstract

We construct multiple zeta functions considered as absolute tensor products of usual zeta functions. We establish Euler product expressions for triple zeta functions ζ(s, 𝔽_{p}) ⊗ ζ(s, 𝔽_{q}) ⊗ ζ(s, 𝔽_{r}) with p, q, r distinct primes, via multiple sine functions by using the signatured Poisson summation formula.

1. Introduction and Main Theorem

Let

Z_{j}(s) = ∏ ∐_{ρ∈ℂ}(s − ρ)^{m_{j}(ρ)}

be “zeta functions” expressed as regularized products in the notation of Deninger [D1], [D2], where m_{j} : ℂ → ℤ denotes the multiplicity function for j = 1, …, r. (Later we will specify “zeta functions” to be treated.) Then, as in the paper [K] we define the absolute tensor product Z_{1}(s) ⊗ ⋯ ⊗ Z_{r}(s) as

Z_{1}(s) ⊗ ⋯ ⊗ Z_{r}(s) = ∏ ∐_{ρ_{1},...,ρ_{r}∈ℂ}(s − (ρ_{1} + ⋯ + ρ_{r}))^{m(ρ_{1},...,ρ_{r})},

where

m(ρ_{1}, …, ρ_{r}) := m_{1}(ρ_{1}) ⋯ m_{r}(ρ_{r}) × \begin{cases}1 & if Im(ρ_{1}), …, Im(ρ_{r}) ⩾ 0, \\ (−1)^{r−1} & if Im(ρ_{1}), …, Im(ρ_{r}) < 0, \\ 0 & otherwise. \end{cases}

We define Hasse zeta functions ζ(s, A) for commutative rings A as

ζ(s, A) := ∏_{𝔪}(1 − N(𝔪)^{−s})^{−1},

where 𝔪 runs over maximal ideals of A, and N(𝔪) := #(A/𝔪).

The following result about the absolute tensor product of Hasse zeta functions is known:

Theorem 1.1. ([KoKu] Theorem 1.1) The following expressions hold in Re(s) > 0 with some polynomial Q(s) of degree at most two: (1)When p ≠ q, it holds that

ζ(s, 𝔽_{p}) ⊗ ζ(s, 𝔽_{q}) = e^{Q_{p,q}(s)}(1 − p^{−s})^{\frac{1}{2}} (1 − q^{−s})^{\frac{1}{2}} ×

exp (\frac{1}{2i} ∑_{k=1}^{∞} \frac{cot (πk \frac{log p}{log q})}{k}p^{−ks} + \frac{1}{2i} ∑_{n=1}^{∞} \frac{cot (πn\frac{log q}{log p})}{n}q^{−ns}) .

2000 Mathematics Subject Classification 11M06

Hirotaka Akatsuka

(2)When p = q, it holds that

ζ(s, 𝔽_{p}) ⊗ ζ(s, 𝔽_{p}) = e^{Q_{p,p}(s)}(1 − p^{−s})^{1− \frac{is log p}{2π}} exp (−\frac{Li_{2}(p^{−s})}{2πi}) ,

where

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

In this paper we treat ζ(s, 𝔽_{p}) ⊗ ζ(s, 𝔽_{q}) ⊗ ζ(s, 𝔽_{r}) when p, q, r are distinct primes. We prove the following theorem, which is similar to Theorem 1.1(1): Theorem 1.2. Suppose that p, q, r are distinct primes and s ∈ ℂ satisfies Re(s) > 0. Then,

ζ(s, 𝔽_{p}) ⊗ ζ(s, 𝔽_{q}) ⊗ ζ(s, 𝔽_{r})

= e^{Q_{p,q,r}(s)}(1 − p^{−s})^{− \frac{1}{4}} (1 − q^{−s})^{− \frac{1}{4}} (1 − r^{−s})^{− \frac{1}{4}}
exp (−\frac{1}{4} ∑_{n_{1}=1}^{∞} \frac{1}{n_{1}} cot (πn_{1} \frac{log p}{log q}) cot (πn_{1} \frac{log p}{log r}) p^{−n_{1}s}

− \frac{1}{4} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} cot (πn_{2} \frac{log q}{log p}) cot (πn_{2} \frac{log q}{log r} ) q^{−n_{2}s}

− \frac{1}{4} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} cot (πn_{3} \frac{log r}{log p}) cot (πn_{3} \frac{log r}{log q} ) r^{−n_{3}s}

+ \frac{i}{4} ∑_{n_{1}=1}^{∞} \frac{1}{n_{1}} (cot (πn_{1} \frac{log p}{log q}) + cot (πn_{1} \frac{log p}{log r})) p^{−n_{1}s}

+ \frac{i}{4} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} (cot (πn_{2} \frac{log q}{log p}) + cot (πn_{2} \frac{log q}{log r} )) q^{−n_{2}s}

+\frac{i}{4} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} (cot (πn_{3} \frac{log r}{log p}) + cot (πn_{3} \frac{log r}{log q} )) r^{−n_{3}s}) ,

where Q_{p,q,r}(s) is a polynomial of degree at most three, which depends on p, q, r.

2. Triple Poisson Summation Formula with Signature

Definition 2.1. α ∈ ℝ is said to be generic, if

lim_{m→∞} ∥mα∥^{\frac{1}{m}} = 1,

where we put ∥x∥ := min{|x − n| : n ∈ ℤ} for x ∈ ℝ. Lemma 2.2. If α ∈ ℚ, then α is not generic. Proof. Let α = a/b with a ∈ ℤ and b ∈ ℤ_{>0}. Then ∥(bm)α∥ = 0 for all m ∈ ℤ. Hence α is not generic. Lemma 2.3. (Baker, [B]Theorem 3.1) Let α and β be in \overline{ℚ} ∩ ℝ_{>0}. If \frac{log α}{log β} ∉ ℚ, then \frac{log α}{log β} is generic. Lemma 2.4. Let α and β be generic. Then

∑_{n=1}^{∞} cot(πnα) cot(πnβ)x^{n}  (2.1)

Euler Product Expression of Triple Zeta Functions

converges absolutely in |x| < 1.

Proof. Since α, β are generic, ∥nα∥^{−1}, ∥nβ∥^{−1} = O(e^{εn}) as n → ∞ for any ε > 0. Since cot(πx) ∼ 1/(πx) (x → 0), cot(πnα) cot(πnβ) = O(e^{2εn}) (n → ∞). Hence the radius of convergence of (2.1), say R, satisfies

R = lim sup_{n→∞} |\frac{1}{cot(πnα) cot(πnβ)}|^{\frac{1}{n}} ⩾ e^{−2ε}.

Since ε > 0 is arbitary, R ⩾ 1.

Lemma 2.5. Let n, m be positive integers and a, b be positive real numbers such that a/b is generic. Then, for any ε > 0 we have

|na − mb|^{−1} ≪_{a,b,ε} m^{−1}e^{εn}

Proof. For all n, m ∈ ℤ_{>0} we have

|na − mb| = mb |\frac{na}{mb} − 1| ⩾ mb ([\frac{na}{b}] + 1)^{−1} ∥\frac{na}{b}∥ ,

where [x] := max{k ∈ ℤ : k ⩽ x}. Since a/b is generic, for any ε > 0 we have

∥n\frac{a}{b}∥^{−1} ≪_{a,b,ε} e^{εn}.

These show the lemma.

For f ∈ L^{1}(ℝ), we denote by f̃ the Fourier transform of f:

f̃(x) := ∫_{−∞}^{∞} f(t)e^{itx}dt.

Let H(t) be an even regular function on {z = x + iy : x ∈ ℝ, y ∈ (−R, R)}, which satisfies (i) and (ii): (i)There exists δ > 0 such that H(t) = O(t^{−3−δ}) (|t| → ∞) (ii)There exists μ ∈ (0, 1) such that H̃(x) = O(μ^{x}) (x → ∞)

h(t) and H_{α}(t) denote h(t) := H(t/i), H_{α}(t) := h(α + it), respectively.

Lemma 2.6. Let α ∈ (−R, R) and x ∈ ℝ. Then, (1)H̃_{α}(x) = e^{−αx}H̃(x).

(2){ \atop tH_{α}(t)}(0) = iαH̃(0).

Proof. (1)Applying Cauchy’s theorem to H(t)e^{itx}, we have

∫_{C_{T}} H(t)e^{itx}dt = 0,

where

C_{T} := ∂{z ∈ ℂ| − T < Re(z) < T, min{−α, 0} < Im(z) < max{−α, 0}}.

Considering the limit T → ∞, (1) follows.

(2)Considering the same as (1), we have

∫_{−∞}^{∞} tH(t)dt = ∫_{−∞}^{∞}(t − iα)H_{α}(t)dt.  (2.2)

Since tH(t) is an odd function, left side of (2.2) equals to 0. So

{ \atop tH_{α}(t)}(0) = iαH̃_{α}(0)

Hirotaka Akatsuka

= iαH̃(0).

Theorem 2.7. Let a, b, c ∈ ℝ be positive real numbers such that a/b, b/c, c/a and their inverses are generic. Then we have

∑_{n_{1},n_{2},n_{3}>0} H (2π ( \frac{n_{1}}{a} + \frac{n_{2}}{b} + \frac{n_{3}}{c} )) + \frac{1}{2} ( ∑_{n_{1},n_{2}>0} H (2π ( \frac{n_{1}}{a} + \frac{n_{2}}{b} ))

+ ∑_{n_{2},n_{3}>0} H (2π ( \frac{n_{2}}{b} + \frac{n_{3}}{c} )) + ∑_{n_{1},n_{3}>0} H (2π ( \frac{n_{1}}{a} + \frac{n_{3}}{c} )))

+ \frac{1}{4} ( ∑_{n_{1}>0} H (2π \frac{n_{1}}{a} ) + ∑_{n_{2}>0} H (2π \frac{n_{2}}{b} ) + ∑_{n_{3}>0} H (2π \frac{n_{3}}{c} )) + \frac{1}{8}H(0)

= −\frac{a}{8π} ∑_{n_{1}>0} cot (π \frac{n_{1}a}{b}) cot (π \frac{n_{1}a}{c}) H̃(n_{1}a)

− \frac{b}{8π} ∑_{n_{2}>0} cot (π \frac{n_{2}b}{c}) cot (π \frac{n_{2}b}{a}) H̃(n_{2}b)

− \frac{c}{8π} ∑_{n_{3}>0} cot (π \frac{n_{3}c}{a}) cot (π \frac{n_{3}c}{b}) H̃(n_{3}c)

+ \frac{abc}{48π} (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} ) H̃(0) − \frac{abc}{32π^{3}} H̃′′(0).  (2.3)

Proof. Put Z_{k}(s) = sinh (\frac{ks}{2}) (k = a, b, c). Let D_{T} be the region defined by

D_{T} := {s ∈ ℂ : |s| > α, | Re(s)| < α, 0 < Im(s) < T },

where 0 < α < min \{\frac{2π}{a} , \frac{2π}{b} , \frac{2π}{c} \} .

By Cauchy’s theorem we have

∑_{0<Im(ρ_{1}),Im(ρ_{2}),Im(ρ_{3})<T} h(ρ_{a} + ρ_{b} + ρ_{c})

= \frac{1}{(2πi)^{3}} ∫_{∂D_{T}} ∫_{∂D_{T}} ∫_{∂D_{T}} h(s_{1} + s_{2} + s_{3})\frac{Z′_{a}}{Z_{a}} (s_{1})\frac{Z′_{b}}{Z_{b}} (s_{2})\frac{Z′_{c}}{Z_{c}} (s_{3})ds_{1}ds_{2}ds_{3},  (2.4)

where ρ_{k} denotes the zeros of Z_{k}(s)(k = a, b, c), and the contour ∂D_{T} is taken counterclockwise. Considering T → ∞ in (2.4), we have

∑_{0<Im(ρ_{1}),Im(ρ_{2}),Im(ρ_{3})} h(ρ_{1} + ρ_{2} + ρ_{3})

= \frac{1}{(2πi)^{3}} ∫_{∂D}∫_{∂D}∫_{∂D} h(s_{1} + s_{2} + s_{3})\frac{Z′_{a}}{Z_{a}} (s_{1})\frac{Z′_{b}}{Z_{b}} (s_{2})\frac{Z′_{c}}{Z_{c}} (s_{3})ds_{1}ds_{2}ds_{3},  (2.5)

where

D := {s ∈ ℂ : | Re(s)| < α, |s| > α, Im(s) > 0}.

We decompose ∂D = C_{1} ∪ C_{2} ∪ C_{3} with

C_{1} := {s ∈ ∂D : Re(s) = −α},
C_{2} := {s ∈ ∂D : |s| = α},

Euler Product Expression of Triple Zeta Functions C_{3} := {s ∈ ∂D : Re(s) = α}.

We compute each triple integral I_{i_{1}i_{2}i_{3}} = ∫_{C_{i_{1}}}∫_{C_{i_{2}}}∫_{C_{i_{3}}} in (2.5).

First we calculate I_{i_{1}i_{2}i_{3}} with (i_{1}, i_{2}, i_{3}) ∈ {1, 3}^{3}.

I_{333} = \frac{1}{(2π)^{3}} ∫_{0}^{∞}∫_{0}^{∞}∫_{0}^{∞} h(3α + i(t_{1} + t_{2} + t_{3}))

\frac{Z′_{a}}{Z_{a}} (α + it_{1})\frac{Z′_{b}}{Z_{b}} (α + it_{2})\frac{Z′_{c}}{Z_{c}} (α + it_{3})dt_{1}dt_{2}dt_{3}.  (2.6)

Since

\frac{Z′_{k}}{Z_{k}} (s) = \frac{k}{2} + k ∑_{n=1}^{∞} e^{−kns}  (2.7)

for k = a, b, c and Re(s) > 0, (2.6) turns to

I_{333} = ∑_{n_{1},n_{2},n_{3}⩾0} ε_{n_{1},n_{2},n_{3}} ∫_{0}^{∞}∫_{0}^{∞}∫_{0}^{∞} H_{3α}(t_{1} + t_{2} + t_{3})

e^{−n_{1}a(α+it_{1})}e^{−n_{2}b(α+it_{2})}e^{−n_{3}c(α+it_{3})}dt_{1}dt_{2}dt_{3},

where

ε_{n_{1},n_{2},n_{3}} := (\frac{1}{2})^{♯\{i∈\{1,2,3\}:n_{i}=0\}} .

We replace t_{3} with t = t_{1} + t_{2} + t_{3} to get

I_{333} = \frac{1}{8π^{3}} ∑_{n_{1},n_{2},n_{3}⩾0} ε_{n_{1},n_{2},n_{3}} ∫_{0}^{∞} H_{3α}(t) ×

(∫∫\begin{matrix}t_{1},t_{2}⩾0 \\ t_{1}+t_{2}⩽t \end{matrix} e^{−n_{1}a(α+it_{1})}e^{−n_{2}b(α+it_{2})}e^{−n_{3}c(α+i(t−t_{1}−t_{2}))}dt_{1}dt_{2}) dt.  (2.8)

By

∫∫\begin{matrix}t_{1},t_{2}⩾0 \\ t_{1}+t_{2}⩽t \end{matrix} ⋯ dt_{1}dt_{2} = ∫_{0}^{t}∫_{0}^{t−t_{1}} ⋯ dt_{2}dt_{1} (t > 0)

and

n_{1}a = n_{2}b ⇔ (n_{1}, n_{2}) = (0, 0), n_{2}b = n_{3}c ⇔ (n_{2}, n_{3}) = (0, 0),
n_{1}a = n_{3}c ⇔ (n_{1}, n_{3}) = (0, 0),  (2.9)

which are introduced by Lemma 2.2, we calculate that

I_{333} = −\frac{abc}{8π^{3}} ∑_{(n_{1},n_{2},n_{3})∈Γ} ε_{n_{1},n_{2},n_{3}}e^{−(n_{1}a+n_{2}b+n_{3}c)α} ×

(∫_{0}^{∞} \frac{H_{3α}(t)e^{−n_{1}ait}}{(n_{1}a − n_{2}b)(n_{1}a − n_{3}c)}dt + ∫_{0}^{∞} \frac{H_{3α}(t)e^{−n_{2}bit}}{(n_{2}b − n_{1}a)(n_{2}b − n_{3}c)}dt

+ ∫_{0}^{∞} \frac{H_{3α}(t)e^{−n_{3}cit}}{(n_{3}c − n_{1}a)(n_{3}c − n_{2}b)}dt)

+ \frac{abc}{32π^{3}} ∑_{n_{1}=1}^{∞} e^{−n_{1}aα} (∫_{0}^{∞} \frac{tH_{3α}(t)}{n_{1}ai}dt − ∫_{0}^{∞} \frac{H_{3α}(t)e^{−n_{1}ait}}{n_{1}^{2}a^{2}} dt + ∫_{0}^{∞} \frac{H_{3α}(t)}{n_{1}^{2}a^{2}}dt)

+ \frac{abc}{32π^{3}} ∑_{n_{2}=1}^{∞} e^{−n_{2}bα} (∫_{0}^{∞} \frac{tH_{3α}(t)}{n_{2}bi}dt − ∫_{0}^{∞} \frac{H_{3α}(t)e^{−n_{2}bit}}{n_{2}^{2}b^{2}} dt + ∫_{0}^{∞} \frac{H_{3α}(t)}{n_{2}^{2}b^{2}}dt)

+ \frac{abc}{32π^{3}} ∑_{n_{3}=1}^{∞} e^{−n_{3}cα} (∫_{0}^{∞} \frac{tH_{3α}(t)}{n_{3}ci}dt − ∫_{0}^{∞} \frac{H_{3α}(t)e^{−n_{3}cit}}{n_{3}^{2}c^{2}} dt + ∫_{0}^{∞} \frac{H_{3α}(t)}{n_{3}^{2}c^{2}}dt)

+ \frac{abc}{128π^{3}} ∫_{0}^{∞} t^{2}H_{3α}(t)dt.  (2.10)

where Γ := {(n_{1}, n_{2}, n_{3}) ∈ (ℤ_{⩾0})^{3} : the number of i such that n_{i} = 0 is at most one}. Similarly

I_{111} = −\frac{abc}{8π^{3}} ∑_{(n_{1},n_{2},n_{3})∈Γ} ε_{n_{1},n_{2},n_{3}}e^{−(n_{1}a+n_{2}b+n_{3}c)α} ×

(∫_{−∞}^{0} \frac{H_{3α}(t)e^{−n_{1}ait}}{(n_{1}a − n_{2}b)(n_{1}a − n_{3}c)}dt + ∫_{−∞}^{0} \frac{H_{3α}(t)e^{−n_{2}bit}}{(n_{2}b − n_{1}a)(n_{2}b − n_{3}c)}dt

+ ∫_{−∞}^{0} \frac{H_{3α}(t)e^{−n_{3}cit}}{(n_{3}c − n_{1}a)(n_{3}c − n_{2}b)}dt)

+ \frac{abc}{32π^{3}} ∑_{n_{1}=1}^{∞} e^{−n_{1}aα} (∫_{−∞}^{0} \frac{tH_{3α}(t)}{n_{1}ai}dt − ∫_{−∞}^{0} \frac{H_{3α}(t)e^{−n_{1}ait}}{n_{1}^{2}a^{2}} dt + ∫_{−∞}^{0} \frac{H_{3α}(t)}{n_{1}^{2}a^{2}}dt)

+ \frac{abc}{32π^{3}} ∑_{n_{2}=1}^{∞} e^{−n_{2}bα} (∫_{−∞}^{0} \frac{tH_{3α}(t)}{n_{2}bi}dt − ∫_{−∞}^{0} \frac{H_{3α}(t)e^{−n_{2}bit}}{n_{2}^{2}b^{2}} dt + ∫_{−∞}^{0} \frac{H_{3α}(t)}{n_{2}^{2}b^{2}}dt)

+ \frac{abc}{32π^{3}} ∑_{n_{3}=1}^{∞} e^{−n_{3}cα} (∫_{−∞}^{0} \frac{tH_{3α}(t)}{n_{3}ci}dt − ∫_{−∞}^{0} \frac{H_{3α}(t)e^{−n_{3}cit}}{n_{3}^{2}c^{2}} dt + ∫_{−∞}^{0} \frac{H_{3α}(t)}{n_{3}^{2}c^{2}}dt)

+ \frac{abc}{128π^{3}} ∫_{−∞}^{0} t^{2}H_{3α}(t)dt.  (2.11)

By (2.10) and (2.11) we have I_{111} + I_{333}

= −\frac{abc}{8π^{3}} ∑_{(n_{1},n_{2},n_{3})∈Γ} ε_{n_{1},n_{2},n_{3}}e^{−(n_{1}a+n_{2}b+n_{3}c)α} ×

(\frac{H̃_{3α}(−n_{1}a)}{(n_{1}a − n_{2}b)(n_{1}a − n_{3}c)} + \frac{H̃_{3α}(−n_{2}b)}{(n_{2}b − n_{1}a)(n_{2}b − n_{3}c)} + \frac{H̃_{3α}(−n_{3}c)}{(n_{3}c − n_{1}a)(n_{3}c − n_{2}b)})

+ \frac{abc}{32π^{3}} ∑_{n_{1}=1}^{∞} e^{−n_{1}aα} (\frac{{ \atop tH_{3α}(t)}(0)}{n_{1}ai} − \frac{H̃_{3α}(−n_{1}a)}{n_{1}^{2}a^{2}} + \frac{H̃_{3α}(0)}{n_{1}^{2}a^{2}})

+ \frac{abc}{32π^{3}} ∑_{n_{2}=1}^{∞} e^{−n_{2}bα} (\frac{{ \atop tH_{3α}(t)}(0)}{n_{2}bi} − \frac{H̃_{3α}(−n_{2}b)}{n_{2}^{2}b^{2}} + \frac{H̃_{3α}(0)}{n_{2}^{2}b^{2}})

+ \frac{abc}{32π^{3}} ∑_{n_{3}=1}^{∞} e^{−n_{3}cα} (\frac{{ \atop tH_{3α}(t)}(0)}{n_{3}ci} − \frac{H̃_{3α}(−n_{3}c)}{n_{3}^{2}c^{2}} + \frac{H̃_{3α}(0)}{n_{3}^{2}c^{2}})
+ \frac{abc}{128π^{3}} { \atop t^{2}H_{3α}(t)}(0)

By Lemma 2.6 and H̃(−x) = H̃(x), we have I_{111} + I_{333}

= −\frac{abc}{8π^{3}} ∑_{(n_{1},n_{2},n_{3})∈Γ} ε_{n_{1},n_{2},n_{3}}e^{−(n_{1}a+n_{2}b+n_{3}c)α} ×

(\frac{e^{3n_{1}aα}H̃(n_{1}a)}{(n_{1}a − n_{2}b)(n_{1}a − n_{3}c)} + \frac{e^{3n_{2}bα}H̃(n_{2}b)}{(n_{2}b − n_{1}a)(n_{2}b − n_{3}c)} + \frac{e^{3n_{3}cα}H̃(n_{3}c)}{(n_{3}c − n_{1}a)(n_{3}c − n_{2}b)})

+ \frac{abc}{32π^{3}} (3αH̃(0) ∑_{n_{1}=1}^{∞} \frac{e^{−n_{1}aα}}{n_{1}a} − ∑_{n_{1}=1}^{∞} \frac{e^{2n_{1}aα}H̃(n_{1}a)}{n_{1}^{2}a^{2}} + H̃(0) ∑_{n_{1}=1}^{∞} \frac{e^{−n_{1}aα}}{n_{1}^{2}a^{2}} )

+ \frac{abc}{32π^{3}} (3αH̃(0) ∑_{n_{2}=1}^{∞} \frac{e^{−n_{2}bα}}{n_{2}b} − ∑_{n_{2}=1}^{∞} \frac{e^{2n_{2}bα}H̃(n_{2}b)}{n_{2}^{2}b^{2}} + H̃(0) ∑_{n_{2}=1}^{∞} \frac{e^{−n_{2}bα}}{n_{2}^{2}b^{2}} )

+ \frac{abc}{32π^{3}} (3αH̃(0) ∑_{n_{3}=1}^{∞} \frac{e^{−n_{3}cα}}{n_{3}c} − ∑_{n_{3}=1}^{∞} \frac{e^{2n_{3}cα}H̃(n_{3}c)}{n_{3}^{2}c^{2}} + H̃(0) ∑_{n_{3}=1}^{∞} \frac{e^{−n_{3}cα}}{n_{3}^{2}c^{2}} )

+ \frac{abc}{128π^{3}} { \atop t^{2}H_{3α}(t)}(0).  (2.12)

Similarly we compute I_{113} + I_{331}, I_{131} + I_{313} and I_{311} + I_{133}:

I_{113} + I_{331}

= −\frac{abc}{8π^{3}} ∑_{(n_{1},n_{2},n_{3})∈Γ} ε_{n_{1},n_{2},n_{3}}e^{−(n_{1}a+n_{2}b+n_{3}c)α} ×

(\frac{e^{−n_{1}aα}H̃(n_{1}a)}{(n_{1}a + n_{2}b)(n_{1}a + n_{3}c)} + \frac{e^{n_{2}bα}H̃(n_{2}b)}{(n_{2}b + n_{1}a)(n_{2}b − n_{3}c)} + \frac{e^{n_{3}cα}H̃(n_{3}c)}{(n_{3}c + n_{1}a)(n_{3}c − n_{2}b)})

+ \frac{abc}{32π^{3}} (αH̃(0) ∑_{n_{1}=1}^{∞} \frac{e^{−n_{1}aα}}{n_{1}a} − ∑_{n_{1}=1}^{∞} \frac{e^{−2n_{1}aα}H̃(n_{1}a)}{n_{1}^{2}a^{2}} + H̃(0) ∑_{n_{1}=1}^{∞} \frac{e^{−n_{1}aα}}{n_{1}^{2}a^{2}} )

+ \frac{abc}{32π^{3}} (αH̃(0) ∑_{n_{2}=1}^{∞} \frac{e^{−n_{2}bα}}{n_{2}b} − ∑_{n_{2}=1}^{∞} \frac{H̃(n_{2}b)}{n_{2}^{2}b^{2}} + H̃(0) ∑_{n_{2}=1}^{∞} \frac{e^{−n_{2}bα}}{n_{2}^{2}b^{2}} )

+ \frac{abc}{32π^{3}} (αH̃(0) ∑_{n_{3}=1}^{∞} \frac{e^{−n_{3}cα}}{n_{3}c} − ∑_{n_{3}=1}^{∞} \frac{H̃(n_{3}c)}{n_{3}^{2}c^{2}} + H̃(0) ∑_{n_{3}=1}^{∞} \frac{e^{−n_{3}cα}}{n_{3}^{2}c^{2}} )

+ \frac{abc}{128π^{3}} { \atop t^{2}H_{α}(t)}(0),  (2.13)

I_{131} + I_{313}

= −\frac{abc}{8π^{3}} ∑_{(n_{1},n_{2},n_{3})∈Γ} ε_{n_{1},n_{2},n_{3}}e^{−(n_{1}a+n_{2}b+n_{3}c)α} ×

(\frac{e^{n_{1}aα}H̃(n_{1}a)}{(n_{1}a + n_{2}b)(n_{1}a − n_{3}c)} + \frac{e^{−n_{2}bα}H̃(n_{2}b)}{(n_{2}b + n_{1}a)(n_{2}b + n_{3}c)} + \frac{e^{n_{3}cα}H̃(n_{3}c)}{(n_{3}c − n_{1}a)(n_{3}c + n_{2}b)})

+ \frac{abc}{32π^{3}} (αH̃(0) ∑_{n_{1}=1}^{∞} \frac{e^{−n_{1}aα}}{n_{1}a} − ∑_{n_{1}=1}^{∞} \frac{H̃(n_{1}a)}{n_{1}^{2}a^{2}} + H̃(0) ∑_{n_{1}=1}^{∞} \frac{e^{−n_{1}aα}}{n_{1}^{2}a^{2}} )

+ \frac{abc}{32π^{3}} (αH̃(0) ∑_{n_{2}=1}^{∞} \frac{e^{−n_{2}bα}}{n_{2}b} − ∑_{n_{2}=1}^{∞} \frac{e^{−2n_{2}bα}H̃(n_{2}b)}{n_{2}^{2}b^{2}} + H̃(0) ∑_{n_{2}=1}^{∞} \frac{e^{−n_{2}bα}}{n_{2}^{2}b^{2}} )

+ \frac{abc}{32π^{3}} (αH̃(0) ∑_{n_{3}=1}^{∞} \frac{e^{−n_{3}cα}}{n_{3}c} − ∑_{n_{3}=1}^{∞} \frac{H̃(n_{3}c)}{n_{3}^{2}c^{2}} + H̃(0) ∑_{n_{3}=1}^{∞} \frac{e^{−n_{3}cα}}{n_{3}^{2}c^{2}} )

+ \frac{abc}{128π^{3}} { \atop t^{2}H_{α}(t)}(0),  (2.14)

I_{311} + I_{133}

= −\frac{abc}{8π^{3}} ∑_{(n_{1},n_{2},n_{3})∈Γ} ε_{n_{1},n_{2},n_{3}}e^{−(n_{1}a+n_{2}b+n_{3}c)α} ×

(\frac{e^{n_{1}aα}H̃(n_{1}a)}{(n_{1}a − n_{2}b)(n_{1}a + n_{3}c)} + \frac{e^{n_{2}bα}H̃(n_{2}b)}{(n_{2}b − n_{1}a)(n_{2}b + n_{3}c)} + \frac{e^{−n_{3}cα}H̃(n_{3}c)}{(n_{3}c + n_{1}a)(n_{3}c + n_{2}b)})

+ \frac{abc}{32π^{3}} (αH̃(0) ∑_{n_{1}=1}^{∞} \frac{e^{−n_{1}aα}}{n_{1}a} − ∑_{n_{1}=1}^{∞} \frac{H̃(n_{1}a)}{n_{1}^{2}a^{2}} + H̃(0) ∑_{n_{1}=1}^{∞} \frac{e^{−n_{1}aα}}{n_{1}^{2}a^{2}} )

+ \frac{abc}{32π^{3}} (αH̃(0) ∑_{n_{2}=1}^{∞} \frac{e^{−n_{2}bα}}{n_{2}b} − ∑_{n_{2}=1}^{∞} \frac{H̃(n_{2}b)}{n_{2}^{2}b^{2}} + H̃(0) ∑_{n_{2}=1}^{∞} \frac{e^{−n_{2}bα}}{n_{2}^{2}b^{2}} )

+ \frac{abc}{32π^{3}} (αH̃(0) ∑_{n_{3}=1}^{∞} \frac{e^{−n_{3}cα}}{n_{3}c} − ∑_{n_{3}=1}^{∞} \frac{e^{−2n_{3}cα}H̃(n_{3}c)}{n_{3}^{2}c^{2}} + H̃(0) ∑_{n_{3}=1}^{∞} \frac{e^{−n_{3}cα}}{n_{3}^{2}c^{2}} )

+ \frac{abc}{128π^{3}} { \atop t^{2}H_{α}(t)}(0).  (2.15)

Considering the limit α → 0 in the sum of (2.12), (2.13), (2.14) and (2.15), we have

lim_{α→0}(I_{111} + I_{113} + I_{131} + I_{133} + I_{311} + I_{313} + I_{331} + I_{333} + \frac{abc}{8π^{3}} (K_{1}(α) + K_{2}(α) + K_{3}(α)))

= −\frac{abc}{8π^{3}} ( ∑_{n_{1}=1}^{∞} \frac{H̃(n_{1}a)}{n_{1}^{2}a^{2}} + ∑_{n_{2}=1}^{∞} \frac{H̃(n_{2}b)}{n_{2}^{2}b^{2}} + ∑_{n_{3}=1}^{∞} \frac{H̃(n_{3}c)}{n_{3}^{2}c^{2}})

+ \frac{abc}{8π^{3}} (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} ) ζ(2)H̃(0) − \frac{abc}{32π^{3}} { \atop t^{2}H(t)}(0),  (2.16)

where

K_{1}(α) = ∑_{(n_{1},n_{2},n_{3})∈Γ} ε_{n_{1},n_{2},n_{3}}e^{−(n_{2}b+n_{3}c)α} ×

(\frac{e^{n_{1}aα}(e^{n_{1}aα} + e^{−n_{1}aα})^{2}n_{1}^{2}a^{2}}{(n_{1}^{2}a^{2} − n_{2}^{2}b^{2})(n_{1}^{2}a^{2} − n_{3}^{2}c^{2})} + \frac{(e^{3n_{1}aα} − e^{−n_{1}aα})n_{1}n_{2}ab}{(n_{1}^{2}a^{2} − n_{2}^{2}b^{2})(n_{1}^{2}a^{2} − n_{3}^{2}c^{2})}

+ \frac{(e^{3n_{1}aα} − e^{−n_{1}aα})n_{1}n_{3}ac}{(n_{1}^{2}a^{2} − n_{2}^{2}b^{2})(n_{1}^{2}a^{2} − n_{3}^{2}c^{2})} + \frac{e^{n_{1}aα}(e^{n_{1}aα} − e^{−n_{1}aα})^{2}n_{2}n_{3}bc}{(n_{1}^{2}a^{2} − n_{2}^{2}b^{2})(n_{1}^{2}a^{2} − n_{3}^{2}c^{2})}) H̃(n_{1}a),

K_{2}(α) = ∑_{(n_{1},n_{2},n_{3})∈Γ} ε_{n_{1},n_{2},n_{3}}e^{−(n_{1}a+n_{3}c)α} ×

(\frac{e^{n_{2}bα}(e^{n_{2}bα} + e^{−n_{2}bα})^{2}n_{2}^{2}b^{2}}{(n_{2}^{2}b^{2} − n_{1}^{2}a^{2})(n_{2}^{2}b^{2} − n_{3}^{2}c^{2})} + \frac{(e^{3n_{2}bα} − e^{−n_{2}bα})n_{1}n_{2}ab}{(n_{2}^{2}b^{2} − n_{1}^{2}a^{2})(n_{2}^{2}b^{2} − n_{3}^{2}c^{2})}

+ \frac{(e^{3n_{2}bα} − e^{−n_{2}bα})n_{2}n_{3}bc}{(n_{2}^{2}b^{2} − n_{1}^{2}a^{2})(n_{2}^{2}b^{2} − n_{3}^{2}c^{2})} + \frac{e^{n_{2}bα}(e^{n_{2}bα} − e^{−n_{2}bα})^{2}n_{1}n_{3}ac}{(n_{2}^{2}b^{2} − n_{1}^{2}a^{2})(n_{2}^{2}b^{2} − n_{3}^{2}c^{2})}) H̃(n_{2}b),

K_{3}(α) = ∑_{(n_{1},n_{2},n_{3})∈Γ} ε_{n_{1},n_{2},n_{3}}e^{−(n_{1}a+n_{2}b)α} ×

Euler Product Expression of Triple Zeta Functions

(\frac{e^{n_{3}cα}(e^{n_{3}cα} + e^{−n_{3}cα})^{2}n_{3}^{2}c^{2}}{(n_{3}^{2}c^{2} − n_{1}^{2}a^{2})(n_{3}^{2}c^{2} − n_{2}^{2}b^{2})} + \frac{(e^{3n_{3}cα} − e^{−n_{3}cα})n_{1}n_{3}ac}{(n_{3}^{2}c^{2} − n_{1}^{2}a^{2})(n_{3}^{2}c^{2} − n_{2}^{2}b^{2})}

+ \frac{(e^{3n_{3}cα} − e^{−n_{3}cα})n_{2}n_{3}bc}{(n_{3}^{2}c^{2} − n_{1}^{2}a^{2})(n_{3}^{2}c^{2} − n_{2}^{2}b^{2})} + \frac{e^{n_{3}cα}(e^{n_{3}cα} − e^{−n_{3}cα})^{2}n_{1}n_{2}ab}{(n_{3}^{2}c^{2} − n_{1}^{2}a^{2})(n_{3}^{2}c^{2} − n_{2}^{2}b^{2})}) H̃(n_{3}c),

and ζ(s) is the Riemann zeta function. Estimating |K_{j}(α) − K_{j}(0)| by using Lemma 2.5 and using H̃(x) ≪ μ^{x} for some μ ∈(0,1), we have

lim_{α→0} K_{j}(α) = K_{j}(0)  (2.17)

for j = 1, 2, 3. By (2.16) and (2.17), we have

lim_{α→0}(I_{111} + I_{113} + I_{131} + I_{133} + I_{311} + I_{313} + I_{331} + I_{333})

= −\frac{abc}{8π^{3}} ∑_{(n_{1},n_{2},n_{3})∈Γ} ε_{n_{1},n_{2},n_{3}} ×

(\frac{4n_{1}^{2}a^{2}}{(n_{1}^{2}a^{2} − n_{2}^{2}b^{2})(n_{1}^{2}a^{2} − n_{3}^{2}c^{2})}H̃(n_{1}a) + \frac{4n_{2}^{2}b^{2}}{(n_{2}^{2}b^{2} − n_{1}^{2}a^{2})(n_{2}^{2}b^{2} − n_{3}^{2}c^{2})}H̃(n_{2}b)

+ \frac{4n_{3}^{2}c^{2}}{(n_{3}^{2}c^{2} − n_{1}^{2}a^{2})(n_{3}^{2}c^{2} − n_{2}^{2}b^{2})}H̃(n_{3}c))

− \frac{abc}{8π^{3}} ( ∑_{n_{1}=1}^{∞} \frac{H̃(n_{1}a)}{n_{1}^{2}a^{2}} + ∑_{n_{2}=1}^{∞} \frac{H̃(n_{2}b)}{n_{2}^{2}b^{2}} + ∑_{n_{3}=1}^{∞} \frac{H̃(n_{3}c)}{n_{3}^{2}c^{2}})
+ \frac{abc}{48π} (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} ) H̃(0) − \frac{abc}{32π^{3}} { \atop t^{2}H(t)}(0).  (2.18)

By

∑_{n>0} \frac{2ku}{k^{2}u^{2} − n^{2}v^{2}} + \frac{1}{ku} = \frac{π}{v} cot (π \frac{ku}{v}) ,

(2.18) turns to

lim_{α→0}(I_{111} + I_{113} + I_{131} + I_{133} + I_{311} + I_{313} + I_{331} + I_{333})

= −\frac{a}{8π} ∑_{n_{1}>0} cot (π \frac{n_{1}a}{b}) cot (π \frac{n_{1}a}{c}) H̃(n_{1}a)

− \frac{b}{8π} ∑_{n_{2}>0} cot (π \frac{n_{2}b}{c}) cot (π \frac{n_{2}b}{a}) H̃(n_{2}b)

− \frac{c}{8π} ∑_{n_{3}>0} cot (π \frac{n_{3}c}{a}) cot (π \frac{n_{3}c}{b}) H̃(n_{3}c)

+ \frac{abc}{48π} (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} ) H̃(0) − \frac{abc}{32π^{3}} H̃′′(0).  (2.19)

Next we calculate I_{i_{1}i_{2}i_{3}} with i_{1} = 2, i_{2} = 2 or i_{3} = 2. Put

I^{(1)} := ∑_{j,k=1}^{3} I_{2jk}, I^{(2)} := ∑_{j,k=1}^{3} I_{j2k}, I^{(3)} := ∑_{j,k=1}^{3} I_{jk2},

I^{(4)} := ∑_{j=1}^{3} I_{22j}, I^{(5)} := ∑_{j=1}^{3} I_{2j2}, I^{(6)} := ∑_{j=1}^{3} I_{j22}.

Then we have

I^{(1)} = \frac{1}{(2πi)^{3}} ∫_{C_{2}}∫_{∂D}∫_{∂D} h(s_{1} + s_{2} + s_{3})\frac{Z′_{a}}{Z_{a}} (s_{1})\frac{Z′_{b}}{Z_{b}} (s_{2})\frac{Z′_{c}}{Z_{c}} (s_{3})ds_{1}ds_{2}ds_{3}

= \frac{1}{2πi} ∫_{C_{2}} ∑_{ρ_{a},ρ_{b}} h(ρ_{a} + ρ_{b} + s_{3})\frac{Z′_{c}}{Z_{c}} (s_{3})ds_{3}

= \frac{1}{2πi} ∫_{π}^{0} ∑_{ρ_{a},ρ_{b}>0} h(ρ_{a} + ρ_{b} + αe^{iθ})\frac{Z′_{c}}{Z_{c}} (αe^{iθ})αie^{iθ}dθ,

where ρ_{a}, ρ_{b} run through the zeros of Z_{a}(s), Z_{b}(s) with Im(ρ_{a}) > 0, Im(ρ_{b}) > 0, respectively. As α → 0, we have

lim_{α→0} I^{(1)} = \frac{1}{2πi} ∫_{π}^{0} ∑_{ρ_{a},ρ_{b}} h(ρ_{a} + ρ_{b})idθ = −\frac{1}{2} ∑_{ρ_{a},ρ_{b}} h(ρ_{a} + ρ_{b}).  (2.20)

We similarly deal with I^{(k)}(k = 2, 3, …, 6) and I_{222} to get

lim_{α→0} I^{(2)} = −\frac{1}{2} ∑_{ρ_{a},ρ_{c}} h(ρ_{a} + ρ_{c}),  (2.21)
lim_{α→0} I^{(3)} = −\frac{1}{2} ∑_{ρ_{b},ρ_{c}} h(ρ_{b} + ρ_{c}).  (2.22)
lim_{α→0} I^{(4)} = \frac{1}{4} ∑_{ρ_{a}} h(ρ_{a}),  (2.23)
lim_{α→0} I^{(5)} = \frac{1}{4} ∑_{ρ_{b}} h(ρ_{b}),  (2.24)
lim_{α→0} I^{(6)} = \frac{1}{4} ∑_{ρ_{a}} h(ρ_{a}),  (2.25)
lim_{α→0} I_{222} = −\frac{1}{8}h(0),  (2.26)

where ρ_{k} runs over the zeros of Z_{k}(s) with Im(ρ_{k}) > 0. By (2.20), (2.21), (2.22), (2.23), (2.24), (2.25) and (2.26), we have

lim_{α→0} ∑_{(i_{1},i_{2},i_{3})∈{1,2,3}^{3}\{1,3}^{3}} I_{i_{1}i_{2}i_{3}}

= lim_{α→0}(I^{(1)} + I^{(2)} + I^{(3)} − (I^{(4)} + I^{(5)} + I^{(6)}) + I_{222})

= −\frac{1}{2} ( ∑_{ρ_{a},ρ_{b}} h(ρ_{a} + ρ_{b}) + ∑_{ρ_{a},ρ_{c}} h(ρ_{a} + ρ_{c}) + ∑_{ρ_{b},ρ_{c}>0} h(ρ_{b} + ρ_{c}))

− \frac{1}{4} (∑_{ρ_{a}} h(ρ_{a}) + ∑_{ρ_{b}} h(ρ_{b}) + ∑_{ρ_{c}} h(ρ_{c})) − \frac{1}{8}h(0).  (2.27)

We apply (2.19) and (2.27) to the limit α → 0 in (2.5). This completes the proof.

3. Expression of Triple Sine

We define the multiple Hurwitz zeta function due to Barnes as

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

for \underline{ω} = (ω_{1}, …, ω_{r}) ∈ (ℝ_{>0})^{r}, the multiple gamma and the multiple sine as

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

S_{r}(z, \underline{ω}) := Γ_{r}(z, \underline{ω})^{−1}Γ_{r}(ω_{1} + ⋯ + ω_{r} − z, \underline{ω})^{(−1)^{r}}.

S_{3}(z, \underline{ω}) has a following expression ([KuKo] Proposition 2.4):

S_{3}(z, \underline{ω})
= e^{Q\underline{ω}(z)}z\begin{matrix}∏ & ′ \\ n_{1},n_{2},n_{3}⩾0 \end{matrix} P_{3} (−\frac{z}{n_{1}ω_{1} + n_{2}ω_{2} + n_{3}ω_{3}} ) ∏_{n_{1},n_{2},n_{3}⩾1} P_{3} (\frac{z}{n_{1}ω_{1} + n_{2}ω_{2} + n_{3}ω_{3}} ) , (3.1)

where P_{r}(u) := (1 − u) exp(u + \frac{u^{2}}{2} + ⋯ + \frac{u^{r}}{r}), and Q\underline{ω}(z) ∈ ℂ[z] satisfies deg Q\underline{ω}(z) ⩽ 3.

Lemma 3.1.

\frac{d^{3}}{dz^{3}} log(1 − e^{iaz}) = 2 ∑_{n=−∞}^{∞} \frac{1}{(z − \frac{2πn}{a})^{3}} .

Proof. By

log(1 − e^{iaz}) = −\frac{πi}{2} + \frac{az}{2} i + log (2 sin \frac{az}{2} )

and

2 sin \frac{az}{2} = az ∏_{n=1}^{∞} (1 − (\frac{az}{2πn})^{2})

we have

\frac{d^{3}}{dz^{3}} log(1 − e^{iaz}) = \frac{2}{z^{3}} + 2 ∑_{n=1}^{∞} (\frac{1}{(z − \frac{2πn}{a})^{3}} + \frac{1}{(z + \frac{2πn}{a})^{3}} )

= 2 ∑_{n=−∞}^{∞} \frac{1}{(z − \frac{2πn}{a})^{3}} .

Lemma 3.2. ([KoKu] Theorem 2.3) Let I(t) be odd function in L^{1}(ℝ) with satisfying following (i), (ii): (i)I(t) = O(t^{−2}) (|t| → ∞) (ii)There exists μ ∈ (0, 1) such that Ĩ(x) = O(μ^{x}) (x → ∞). Let a, b be positive real numbers such that a/b is generic. Then we have

∑_{k,n>0} I (2π (\frac{k}{a} + \frac{n}{b})) + \frac{1}{2} (∑_{k>0} I (2π \frac{k}{a}) + ∑_{n>0} I (2π \frac{n}{b}))

= −\frac{ia}{4π} ∑_{k>0} cot (π \frac{ka}{b}) Ĩ(ka) − \frac{ib}{4π} ∑_{n>0} cot (π \frac{nb}{a}) Ĩ(nb) − \frac{iab}{8π^{2}} Ĩ′(0).  (3.2)

Lemma 3.3. ([KoKu] Theorem 3.2) Assume ω_{1}/ω_{2} is generic, then the double sine function has the following expression in Im(z) > 0:

S_{2}(z, (ω_{1}, ω_{2})) = exp(\frac{1}{2i} ∑_{k=1}^{∞} \frac{1}{k} cot (πk \frac{ω_{2}}{ω_{1}} ) e^{2πik \frac{z}{ω_{1}}}

+ \frac{1}{2i} ∑_{n=1}^{∞} \frac{1}{n} cot (πn\frac{ω_{1}}{ω_{2}} ) e^{2πin \frac{z}{ω_{2}}}

+ \frac{1}{2} log(1 − e^{2πi \frac{z}{ω_{1}}} ) + \frac{1}{2} log(1 − e^{2πi \frac{z}{ω_{2}}} )

+ \frac{πiz^{2}}{2ω_{1}ω_{2}} − \frac{πi}{2} (\frac{1}{ω_{1}} + \frac{1}{ω_{2}} ) z + \frac{πi}{12} (\frac{ω_{2}}{ω_{1}} + \frac{ω_{1}}{ω_{2}} + 3)).

Theorem 3.4. Let \underline{ω} = (ω_{1}, ω_{2}, ω_{3}) be in (ℝ_{>0})^{3} such that ω_{i}/ω_{j} are generic (i, j = 1, 2, 3; i ≠ j). Then the triple sine function has the following expression in Im(z) > 0:

S_{3}(z, \underline{ω}) = exp(\frac{1}{4} ∑_{n_{1}=1}^{∞} \frac{1}{n_{1}} cot (π \frac{n_{1}ω_{2}}{ω_{1}} ) cot (π \frac{n_{1}ω_{3}}{ω_{1}} ) e^{2πin_{1} \frac{z}{ω_{1}}}

+ \frac{1}{4} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} cot (π \frac{n_{2}ω_{1}}{ω_{2}} ) cot (π \frac{n_{2}ω_{3}}{ω_{2}} ) e^{2πin_{2} \frac{z}{ω_{2}}}

+ \frac{1}{4} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} cot (π \frac{n_{3}ω_{1}}{ω_{3}} ) cot (π \frac{n_{3}ω_{2}}{ω_{3}} ) e^{2πin_{3} \frac{z}{ω_{3}}}

+ \frac{1}{4i} ∑_{n_{1}=1}^{∞} \frac{1}{n_{1}} (cot (π \frac{n_{1}ω_{2}}{ω_{1}} ) + cot (π \frac{n_{1}ω_{3}}{ω_{1}} )) e^{2πin_{1} \frac{z}{ω_{1}}}

+ \frac{1}{4i} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} (cot (π \frac{n_{2}ω_{1}}{ω_{2}} ) + cot (π \frac{n_{2}ω_{3}}{ω_{2}} )) e^{2πin_{2} \frac{z}{ω_{2}}}

+ \frac{1}{4i} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} (cot (π \frac{n_{3}ω_{1}}{ω_{3}} ) + cot (π \frac{n_{3}ω_{2}}{ω_{3}} )) e^{2πin_{3} \frac{z}{ω_{3}}}

+ \frac{1}{4} log(1 − e^{2πi \frac{z}{ω_{1}}} ) + \frac{1}{4} log(1 − e^{2πi \frac{z}{ω_{2}}} )

+ \frac{1}{4} log(1 − e^{2πi \frac{z}{ω_{3}}} ) − \frac{πiz^{3}}{6ω_{1}ω_{2}ω_{3}} + \frac{πi}{4} (\frac{1}{ω_{1}ω_{2}} + \frac{1}{ω_{2}ω_{3}} + \frac{1}{ω_{3}ω_{1}} ) z^{2}

− \frac{πi}{12} (\frac{3}{ω_{1}} + \frac{3}{ω_{2}} + \frac{3}{ω_{3}} + \frac{ω_{1}}{ω_{2}ω_{3}} + \frac{ω_{2}}{ω_{3}ω_{1}} + \frac{ω_{3}}{ω_{1}ω_{2}} ) z

+ \frac{πi}{24} (\frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{1}} + \frac{ω_{2}}{ω_{3}} + \frac{ω_{3}}{ω_{2}} + \frac{ω_{3}}{ω_{1}} + \frac{ω_{1}}{ω_{3}} + 3))

Proof. By (3.1) we have

\frac{d^{3}}{dz^{3}} (log S_{3}(z, \underline{ω}))

= C\underline{ω} + \frac{2}{z^{3}}

+2 ∑_{n_{1},n_{2},n_{3}⩾1} (\frac{1}{(z + (n_{1}ω_{1} + n_{2}ω_{2} + n_{3}ω_{3}))^{3}} + \frac{1}{(z − (n_{1}ω_{1} + n_{2}ω_{2} + n_{3}ω_{3}))^{3}} )

+2 ( ∑_{n_{1},n_{2}⩾1} \frac{1}{(z + (n_{1}ω_{1} + n_{2}ω_{2}))^{3}} + ∑_{n_{2},n_{3}⩾1} \frac{1}{(z + (n_{2}ω_{2} + n_{3}ω_{3}))^{3}}

+ ∑_{n_{1},n_{3}⩾1} \frac{1}{(z + (n_{1}ω_{1} + n_{3}ω_{3}))^{3}} )

+2 ( ∑_{n_{1}⩾1} \frac{1}{(z + n_{1}ω_{1})^{3}} + ∑_{n_{2}⩾1} \frac{1}{(z + n_{2}ω_{2})^{2}} + ∑_{n_{3}⩾1} \frac{1}{(z + n_{3}ω_{3})^{3}} ) ,  (3.3)

where C\underline{ω} is a constant depending on \underline{ω}.

Put

H(t) = \frac{1}{(z + t)^{3}} + \frac{1}{(z − t)^{3}} (Im(z) > 0).

As we have

H̃(x) = 2πi Res_{t=z}(H(t)e^{itx}) = πix^{2}e^{ixz} (x ⩾ 0)

and H(t) = O(t^{−4})(|t| → ∞), H(t) satisfies (i) and (ii) in §2. As H̃′′(0) = 2πi, by putting

F (z) := ∑_{n_{1},n_{2},n_{3}⩾1} (\frac{1}{(z + (n_{1}ω_{1} + n_{2}ω_{2} + n_{3}ω_{3}))^{3}} + \frac{1}{(z − (n_{1}ω_{1} + n_{2}ω_{2} + n_{3}ω_{3}))^{3}} )

+ \frac{1}{2} ∑_{n_{1},n_{2}⩾1} (\frac{1}{(z + (n_{1}ω_{1} + n_{2}ω_{2}))^{3}} + \frac{1}{(z − (n_{1}ω_{1} + n_{2}ω_{2}))^{3}} )

+ \frac{1}{2} ∑_{n_{2},n_{3}⩾1} (\frac{1}{(z + (n_{2}ω_{2} + n_{3}ω_{3}))^{3}} + \frac{1}{(z − (n_{2}ω_{2} + n_{3}ω_{3}))^{3}} )

+ \frac{1}{2} ∑_{n_{1},n_{3}⩾1} (\frac{1}{(z + (n_{1}ω_{1} + n_{3}ω_{3}))^{3}} + \frac{1}{(z − (n_{1}ω_{1} + n_{3}ω_{3}))^{3}} )

+ \frac{1}{4} ∑_{n_{1}⩾1} (\frac{1}{(z + n_{1}ω_{1})^{3}} + \frac{1}{(z − n_{1}ω_{1})^{3}} )

+ \frac{1}{4} ∑_{n_{2}⩾1} (\frac{1}{(z + n_{2}ω_{2})^{3}} + \frac{1}{(z − n_{2}ω_{2})^{3}} )

+ \frac{1}{4} ∑_{n_{3}⩾1} (\frac{1}{(z + n_{3}ω_{3})^{3}} + \frac{1}{(z − n_{3}ω_{3})^{3}} ) + \frac{1}{4z^{3}} ,

the summation formula (2.3) shows

F (z) = −\frac{a}{8} ∑_{n_{1}=1}^{∞} cot (π \frac{n_{1}a}{b}) cot (π \frac{n_{1}a}{b}) i(n_{1}a)^{2}e^{in_{1}az}

− \frac{b}{8} ∑_{n_{2}=1}^{∞} cot (π \frac{n_{2}b}{a}) cot (π \frac{n_{2}b}{c}) i(n_{2}b)^{2}e^{in_{2}bz}

− \frac{c}{8} ∑_{n_{3}=1}^{∞} cot (π \frac{n_{3}c}{a}) cot (π \frac{n_{3}c}{b}) i(n_{3}c)^{2}e^{in_{3}cz} − \frac{abc}{16π^{2}} i

= \frac{d^{3}}{dz^{3}} (\frac{1}{8} ∑_{n_{1}=1}^{∞} \frac{1}{n_{1}} cot (π \frac{n_{1}a}{b}) cot (π \frac{n_{1}a}{c}) e^{in_{1}az}

+ \frac{1}{8} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} cot (π \frac{n_{2}b}{a}) cot (π \frac{n_{2}b}{c}) e^{in_{2}bz}

+ \frac{1}{8} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} cot (π \frac{n_{3}c}{a}) cot (π \frac{n_{3}c}{b}) e^{in_{3}cz}) − \frac{abc}{16π^{2}} i

By (3.1) with ω_{1} = \frac{2π}{a}, ω_{2} = \frac{2π}{b} and ω_{3} = \frac{2π}{c}, we have

\frac{d^{3}}{dz^{3}} (log S_{3}(z, \underline{ω}))

= C\underline{ω} + 2F (z) + \frac{3}{2z^{3}} + ∑_{n_{1},n_{2}⩾1} (\frac{1}{(z + (n_{1}ω_{1} + n_{2}ω_{2}))^{3}} − \frac{1}{(z − (n_{1}ω_{1} + n_{2}ω_{2}))^{3}} )

+ ∑_{n_{2},n_{3}⩾1} (\frac{1}{(z + (n_{2}ω_{2} + n_{3}ω_{3}))^{3}} − \frac{1}{(z − (n_{2}ω_{2} + n_{3}ω_{3}))^{3}} )

+ ∑_{n_{1},n_{3}⩾1} (\frac{1}{(z + (n_{1}ω_{1} + n_{3}ω_{3}))^{3}} − \frac{1}{(z − (n_{1}ω_{1} + n_{3}ω_{3}))^{3}} )

+\frac{3}{2} ( ∑_{n_{1}⩾1} \frac{1}{(z + n_{1}ω_{1})^{3}} + ∑_{n_{2}⩾1} \frac{1}{(z + n_{2}ω_{2})^{3}} + ∑_{n_{3}⩾1} \frac{1}{(z + n_{3}ω_{3})^{3}} )

−\frac{1}{2} ( ∑_{n_{1}⩾1} \frac{1}{(z − n_{1}ω_{1})^{3}} + ∑_{n_{2}⩾1} \frac{1}{(z − n_{2}ω_{2})^{3}} + ∑_{n_{3}⩾1} \frac{1}{(z − n_{3}ω_{3})^{3}} ) .  (3.4)

Putting

I(t) = \frac{1}{(z + t)^{3}} − \frac{1}{(z − t)^{3}} ,

I(t) is a odd function in L^{1}(ℝ) and satisties I(t) = O(t^{−2}) (|t| → ∞). Besides,

Ĩ(x) = 2πi Res_{t=z}(I(t)e^{itx}) = −πix^{2}e^{ixz} (x ⩾ 0).

Hence I(t) satisfies the assumption of Lemma 3.2. By putting

G_{1}(z) := ∑_{n_{1},n_{2}⩾1} A (n_{1}ω_{1} + n_{2}ω_{2}) + \frac{1}{2} ( ∑_{n_{1}⩾1} A(n_{1}ω_{1}) + ∑_{n_{2}⩾1} A(n_{2}ω_{2})) ,

G_{2}(z) := ∑_{n_{2},n_{3}⩾1} A (n_{2}ω_{2} + n_{3}ω_{3}) + \frac{1}{2} ( ∑_{n_{2}⩾1} A(n_{2}ω_{2}) + ∑_{n_{3}⩾1} A(n_{3}ω_{3})) ,

G_{3}(z) := ∑_{n_{1},n_{3}⩾1} A (n_{1}ω_{1} + n_{3}ω_{3}) + \frac{1}{2} ( ∑_{n_{1}⩾1} A(n_{1}ω_{1}) + ∑_{n_{3}⩾1} A(n_{3}ω_{3})) ,

Lemma 3.2 shows

G_{1}(z) = \frac{1}{4i} ∑_{n_{1}=1}^{∞} cot (π \frac{n_{1}a}{b}) (in_{1}a)^{3}e^{in_{1}az} + \frac{1}{4i} ∑_{n_{2}=1}^{∞} cot (π \frac{n_{2}b}{a}) (in_{2}b)^{3}e^{in_{2}bz}

= \frac{d^{3}}{dz^{3}} (\frac{1}{4i} ∑_{n_{1}=1}^{∞} \frac{1}{n_{1}} cot (π \frac{n_{1}a}{b}) e^{in_{1}az} + \frac{1}{4i} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} cot (π \frac{n_{2}b}{a}) e^{in_{2}bz}) .  (3.5)

Euler Product Expression of Triple Zeta Functions

Similarly we compute

G_{2}(z) = \frac{d^{3}}{dz^{3}} (\frac{1}{4i} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} cot (π \frac{n_{2}b}{c}) e^{in_{2}bz} + \frac{1}{4i} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} cot (π \frac{n_{3}c}{b}) e^{in_{3}cz}) ,  (3.6)
G_{3}(z) = \frac{d^{3}}{dz^{3}} (\frac{1}{4i} ∑_{n_{1}=1}^{∞} \frac{1}{n_{1}} cot (π \frac{n_{1}a}{c}) e^{in_{1}az} + \frac{1}{4i} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} cot (π \frac{n_{3}c}{a}) e^{in_{3}cz}) .  (3.7)

We express (3.4) using G_{1}, G_{2} and G_{3}:

\frac{d^{3}}{dz^{3}} (log S_{3}(z, \underline{ω}))

= C\underline{ω} + 2F (z) + G_{1}(z) + G_{2}(z) + G_{3}(z)

+\frac{1}{2} ( ∑_{n_{1}=−∞}^{∞} \frac{1}{(z − n_{1}ω_{1})^{3}} + ∑_{n_{2}=−∞}^{∞} \frac{1}{(z − n_{2}ω_{2})^{3}} + ∑_{n_{3}=−∞}^{∞} \frac{1}{(z − n_{3}ω_{3})^{3}} ) .  (3.8)

By (3.4), (3.5), (3.6), (3.7) and Lemma 3.1,

Q(z) := log S_{3}(z, \underline{ω}) − \frac{1}{4} ∑_{n_{1}=1}^{∞} \frac{1}{n_{1}} cot (π \frac{n_{1}ω_{2}}{ω_{1}} ) cot (π \frac{n_{1}ω_{3}}{ω_{1}} ) e^{2πin_{1} \frac{z}{ω_{1}}}

− \frac{1}{4} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} cot (π \frac{n_{2}ω_{1}}{ω_{2}} ) cot (π \frac{n_{2}ω_{3}}{ω_{2}} ) e^{2πin_{2} \frac{z}{ω_{2}}}

− \frac{1}{4} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} cot (π \frac{n_{3}ω_{1}}{ω_{3}} ) cot (π \frac{n_{3}ω_{2}}{ω_{3}} ) e^{2πin_{3} \frac{z}{ω_{3}}}

− \frac{1}{4i} ∑_{n_{1}=1}^{∞} \frac{1}{n_{1}} (cot (π \frac{n_{1}ω_{2}}{ω_{1}} ) + cot (π \frac{n_{1}ω_{3}}{ω_{1}} )) e^{2πin_{1} \frac{z}{ω_{1}}}

− \frac{1}{4i} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} (cot (π \frac{n_{2}ω_{1}}{ω_{2}} ) + cot (π \frac{n_{2}ω_{3}}{ω_{2}} )) e^{2πin_{2} \frac{z}{ω_{2}}}

− \frac{1}{4i} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} (cot (π \frac{n_{3}ω_{1}}{ω_{3}} ) + cot (π \frac{n_{3}ω_{2}}{ω_{3}} )) e^{2πin_{3} \frac{z}{ω_{3}}}

− \frac{1}{4} log(1 − e^{2πi \frac{z}{ω_{1}}} ) − \frac{1}{4} log(1 − e^{2πi \frac{z}{ω_{2}}} ) − \frac{1}{4} log(1 − e^{2πi \frac{z}{ω_{3}}} )  (3.9)

is a polynomial of degree at most three. Thus we put Q(z) = a + bz + cz^{2} + dz^{3} and will compute a, b, c and d. First we compute b, c and d by considering

Q(z + ω_{1}) − Q(z) = (bω_{1} + cω_{1}^{2} + dω_{1}^{3}) + (2cω_{1} + 3dω_{1}^{2})z + 3dω_{1}z^{2}.  (3.10)

By (3.9) we have Q(z + ω_{1}) − Q(z)

= log \frac{S_{3}(z + ω_{1}, \underline{ω})}{S_{3}(z, \underline{ω})}

−\frac{1}{4} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} cot (π \frac{n_{2}ω_{1}}{ω_{2}} ) cot (π \frac{n_{2}ω_{3}}{ω_{2}} ) e^{2πin_{2} \frac{z}{ω_{2}}} (e^{2πin_{2} \frac{ω_{1}}{ω_{2}}} − 1)

−\frac{1}{4} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} cot (π \frac{n_{3}ω_{1}}{ω_{3}} ) cot (π \frac{n_{3}ω_{2}}{ω_{3}} ) e^{2πin_{3} \frac{z}{ω_{3}}} (e^{2πin_{3} \frac{ω_{1}}{ω_{3}}} − 1)

−\frac{1}{4i} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} (cot (π \frac{n_{2}ω_{1}}{ω_{2}} ) + cot (π \frac{n_{2}ω_{3}}{ω_{2}} )) e^{2πin_{2} \frac{z}{ω_{2}}} (e^{2πin_{2} \frac{ω_{1}}{ω_{2}}} − 1)

−\frac{1}{4i} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} (cot (π \frac{n_{3}ω_{1}}{ω_{3}} ) + cot (π \frac{n_{3}ω_{2}}{ω_{3}} )) e^{2πin_{3} \frac{z}{ω_{3}}} (e^{2πin_{3} \frac{ω_{1}}{ω_{3}}} − 1)

−\frac{1}{4} log(1 − e^{2πi \frac{z+ω_{1}}{ω_{2}}} ) − \frac{1}{4} log(1 − e^{2πi \frac{z+ω_{1}}{ω_{3}}} )

+\frac{1}{4} log(1 − e^{2πi \frac{z}{ω_{2}}} ) + \frac{1}{4} log(1 − e^{2πi \frac{z}{ω_{3}}} ).  (3.11)

Using the formula

cot(x) = i\frac{e^{2πix} + 1}{e^{2πix} − 1},  (3.12)

(3.11) turns to

Q(z + ω_{1}) − Q(z)

= log \frac{S_{3}(z + ω_{1}, \underline{ω})}{S_{3}(z, \underline{ω})}

−\frac{1}{4i} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} cot (π \frac{n_{2}ω_{1}}{ω_{2}} ) e^{2πin_{2} \frac{z}{ω_{2}}} (e^{2πin_{2} \frac{ω_{1}}{ω_{2}}} − 1)

+\frac{1}{2i} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} cot (π \frac{n_{2}ω_{3}}{ω_{2}} ) e^{2πin_{2} \frac{z}{ω_{2}}}

−\frac{1}{4i} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} cot (π \frac{n_{3}ω_{1}}{ω_{3}} ) e^{2πin_{3} \frac{z}{ω_{3}}} (e^{2πin_{3} \frac{ω_{1}}{ω_{3}}} − 1)

+\frac{1}{2i} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} cot (π \frac{n_{3}ω_{2}}{ω_{3}} ) e^{2πin_{3} \frac{z}{ω_{3}}}

−\frac{1}{4} log(1 − e^{2πi \frac{z+ω_{1}}{ω_{2}}} ) − \frac{1}{4} log(1 − e^{2πi \frac{z+ω_{1}}{ω_{3}}} )

+\frac{1}{4} log(1 − e^{2πi \frac{z}{ω_{2}}} ) + \frac{1}{4} log(1 − e^{2πi \frac{z}{ω_{3}}} ).

Using the formula [KuKo, (2.4)] we get

S_{3}(z + ω_{1}, \underline{ω})S_{3}(z, \underline{ω})^{−1} = S_{2}(z, (ω_{2}, ω_{3}))^{−1}.  (3.13)

By Lemma 3.3 and (3.13), we have

Q(z + ω_{1}) − Q(z)

= −\frac{1}{4i} ∑_{n_{2}=1}^{∞} \frac{1}{n_{2}} cot (π \frac{n_{2}ω_{1}}{ω_{2}} ) e^{2πin_{2} \frac{z}{ω_{2}}} (e^{2πin_{2} \frac{ω_{1}}{ω_{2}}} − 1)

−\frac{1}{4i} ∑_{n_{3}=1}^{∞} \frac{1}{n_{3}} cot (π \frac{n_{3}ω_{1}}{ω_{3}} ) e^{2πin_{3} \frac{z}{ω_{3}}} (e^{2πin_{3} \frac{ω_{1}}{ω_{3}}} − 1)

−\frac{1}{4} log(1 − e^{2πi \frac{z+ω_{1}}{ω_{2}}} ) − \frac{1}{4} log(1 − e^{2πi \frac{z+ω_{1}}{ω_{3}}} )

−\frac{1}{4} log(1 − e^{2πi \frac{z}{ω_{2}}} ) − \frac{1}{4} log(1 − e^{2πi \frac{z}{ω_{3}}} )

−\frac{πiz^{2}}{2ω_{2}ω_{3}} + \frac{πi}{2} (\frac{1}{ω_{2}} + \frac{1}{ω_{3}} ) z − \frac{πi}{12} (\frac{ω_{3}}{ω_{2}} + \frac{ω_{2}}{ω_{3}} + 3) .  (3.14)

Euler Product Expression of Triple Zeta Functions

By (3.12) we have

Q(z + ω_{1}) − Q(z) = −\frac{πiz^{2}}{2ω_{2}ω_{3}} + \frac{πi}{2} (\frac{1}{ω_{2}} + \frac{1}{ω_{3}} ) z − \frac{πi}{12} (\frac{ω_{3}}{ω_{2}} + \frac{ω_{2}}{ω_{3}} + 3) .  (3.15)

Comparing the coefficients of (3.15), we have

b = −\frac{πi}{12} (\frac{3}{ω_{1}} + \frac{3}{ω_{2}} + \frac{3}{ω_{3}} + \frac{ω_{1}}{ω_{2}ω_{3}} + \frac{ω_{2}}{ω_{3}ω_{1}} + \frac{ω_{3}}{ω_{1}ω_{2}} ) ,

c = \frac{πi}{4} (\frac{1}{ω_{1}ω_{2}} + \frac{1}{ω_{2}ω_{3}} + \frac{1}{ω_{3}ω_{1}} ) ,

d = −\frac{πi}{6ω_{1}ω_{2}ω_{3}} .

Next we will treat a by considering

Q(z) + Q(z + \frac{ω_{1}}{2} ) + Q(z + \frac{ω_{2}}{2} ) + Q(z + \frac{ω_{3}}{2} ) + Q(z + \frac{ω_{1} + ω_{2}}{2} )

+Q(z + \frac{ω_{2} + ω_{3}}{2} ) + Q(z + \frac{ω_{3} + ω_{1}}{2} ) + Q(z + \frac{ω_{1} + ω_{2} + ω_{3}}{2} )

−Q(2z).  (3.16)

The constant term of (3.16) is

7a − \frac{7πi}{24ω_{1}ω_{2}ω_{3}} (ω_{1}^{2}ω_{2} + ω_{1}ω_{2}^{2} + ω_{2}^{2}ω_{3} + ω_{2}ω_{3}^{2} + ω_{3}^{2}ω_{1} + ω_{3}ω_{1}^{2} + 3ω_{1}ω_{2}ω_{3}).  (3.17)

On the other hand we will compute (3.16) by using (3.9). Putting

A_{0} := log \frac{∏_{k_{1},k_{2},k_{3}=0}^{1} S_{3} (z + \frac{k_{1}ω_{1} + k_{2}ω_{2} + k_{3}ω_{3}}{2} , \underline{ω})}{S_{3}(2z, \underline{ω})},

A_{1} := −\frac{1}{4} ∑_{n_{1}=1}^{∞} \{\frac{1}{n_{1}} cot (π \frac{n_{1}ω_{2}}{ω_{1}} ) cot (π \frac{n_{1}ω_{3}}{ω_{1}} )

× ( ∑_{k_{1},k_{2},k_{3}=0}^{1} e^{\frac{2πin_{1}(z+ \frac{k_{1}ω_{1}+k_{2}ω_{2}+k_{3}ω_{3}}{2} )}{ω_{1}}} − e^{\frac{4πin_{1}}{ω_{1}}} )\}

− \frac{1}{4i} ∑_{n_{1}=1}^{∞} \{\frac{1}{n_{1}} (cot (π \frac{n_{1}ω_{2}}{ω_{1}} ) + cot (π \frac{n_{1}ω_{3}}{ω_{1}} ))

× ( ∑_{k_{1},k_{2},k_{3}=0}^{1} e^{\frac{2πin_{1}(z+ \frac{k_{1}ω_{1}+k_{2}ω_{2}+k_{3}ω_{3}}{2} )}{ω_{1}}} − e^{\frac{4πin_{1}}{ω_{1}}} )\} ,

A_{2} := −\frac{1}{4} ∑_{n_{2}=1}^{∞} \{\frac{1}{n_{2}} cot (π \frac{n_{2}ω_{1}}{ω_{2}} ) cot (π \frac{n_{2}ω_{3}}{ω_{2}} )

× ( ∑_{k_{1},k_{2},k_{3}=0}^{1} e^{\frac{2πin_{2}(z+ \frac{k_{1}ω_{1}+k_{2}ω_{2}+k_{3}ω_{3}}{2} )}{ω_{2}}} − e^{\frac{4πin_{2}}{ω_{2}}} )\}

− \frac{1}{4i} ∑_{n_{2}=1}^{∞} \{\frac{1}{n_{2}} (cot (π \frac{n_{2}ω_{1}}{ω_{2}} ) + cot (π \frac{n_{2}ω_{3}}{ω_{2}} ))

Hirotaka Akatsuka

× ( ∑_{k_{1},k_{2},k_{3}=0}^{1} e^{\frac{2πin_{2}(z+ \frac{k_{1}ω_{1}+k_{2}ω_{2}+k_{3}ω_{3}}{2} )}{ω_{2}}} − e^{\frac{4πin_{2}}{ω_{2}}} )\} ,

A_{3} := −\frac{1}{4} ∑_{n_{3}=1}^{∞} \{\frac{1}{n_{3}} cot (π \frac{n_{3}ω_{1}}{ω_{3}} ) cot (π \frac{n_{3}ω_{2}}{ω_{3}} )

× ( ∑_{k_{1},k_{2},k_{3}=0}^{1} e^{\frac{2πin_{3}(z+ \frac{k_{1}ω_{1}+k_{2}ω_{2}+k_{3}ω_{3}}{2} )}{ω_{3}}} − e^{\frac{4πin_{3}}{ω_{3}}} )\}

− \frac{1}{4i} ∑_{n_{3}=1}^{∞} \{\frac{1}{n_{3}} (cot (π \frac{n_{3}ω_{1}}{ω_{3}} ) + cot (π \frac{n_{3}ω_{2}}{ω_{3}} ))

× ( ∑_{k_{1},k_{2},k_{3}=0}^{1} e^{\frac{2πin_{3}(z+ \frac{k_{1}ω_{1}+k_{2}ω_{2}+k_{3}ω_{3}}{2} )}{ω_{3}}} − e^{\frac{4πin_{3}}{ω_{3}}} )\} ,

A_{4} := −\frac{1}{4} log \frac{∏_{k_{1},k_{2},k_{3}=0}^{1} (1 − e^{\frac{2πi(z+ \frac{k_{1}ω_{1}+k_{2}ω_{2}+k_{3}ω_{3}}{2} )}{ω_{1}}} )}{1 − e^{\frac{4πiz}{ω_{1}}}},

A_{5} := −\frac{1}{4} log \frac{∏_{k_{1},k_{2},k_{3}=0}^{1} (1 − e^{\frac{2πi(z+ \frac{k_{1}ω_{1}+k_{2}ω_{2}+k_{3}ω_{3}}{2} )}{ω_{2}}} )}{1 − e^{\frac{4πiz}{ω_{2}}}},

A_{6} := −\frac{1}{4} log \frac{∏_{k_{1},k_{2},k_{3}=0}^{1} (1 − e^{\frac{2πi(z+ \frac{k_{1}ω_{1}+k_{2}ω_{2}+k_{3}ω_{3}}{2} )}{ω_{3}}} )}{1 − e^{\frac{4πiz}{ω_{3}}}},

we write (3.16) as ∑_{j=0}^{6} A_{j}. The formula [KuKo, (2.5)] gives A_{0} = 0. Computing A_{1} by (3.12), we have

A_{1} = −\frac{1}{4} ∑_{n_{1}=1}^{∞} \{\frac{1}{n_{1}} (e^{\frac{4πin_{1}(z+ \frac{ω_{2}+ω_{3}}{2} )}{ω_{1}}} + e^{\frac{4πin_{1}(z+ \frac{ω_{2}}{2} )}{ω_{1}}} + e^{\frac{4πin_{1}(z+ \frac{ω_{3}}{2} )}{ω_{1}}} )\}

= \frac{1}{4} log\{(1 − e^{\frac{4πi(z+ \frac{ω_{2}+ω_{3}}{2} )}{ω_{1}}} )(1 − e^{\frac{4πi(z+ \frac{ω_{2}}{2} )}{ω_{1}}} )(1 − e^{\frac{4πi(z+ \frac{ω_{3}}{2} )}{ω_{1}}} )\}.

A_{4} is easily computed as

A_{4} = −\frac{1}{4} log\{(1 − e^{\frac{4πi(z+ \frac{ω_{2}+ω_{3}}{2} )}{ω_{1}}} )(1 − e^{\frac{4πi(z+ \frac{ω_{2}}{2} )}{ω_{1}}} )(1 − e^{\frac{4πi(z+ \frac{ω_{3}}{2} )}{ω_{1}}} )\}.

Therefore A_{1} + A_{4} = 0. Similarly computing we have A_{2} + A_{5} = A_{3} + A_{6} = 0. Hence ∑_{j=0}^{6} A_{j} = 0. Therefore its constant term (3.17) vanishes, which lead to

a = \frac{πi}{24} (\frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{1}} + \frac{ω_{2}}{ω_{3}} + \frac{ω_{3}}{ω_{2}} + \frac{ω_{3}}{ω_{1}} + \frac{ω_{1}}{ω_{3}} + 3) .

Euler Product Expression of Triple Zeta Functions 4. Proof of Theorem 1.2

Let Z_{j}(j = 1, …, r) be meromorphic functions of order μ_{j}. We put the Hadamard product as

Z_{j}(s) = s^{k_{j}}e^{Q_{j}(s)} \begin{matrix}∏ & ′ \\ ρ∈ℂ \end{matrix} P_{μ_{j}} (\frac{s}{ρ})^{m_{j}(ρ)} ,

where P_{r}(u) := (1−u) exp(u+ \frac{u^{2}}{2} +⋯+ \frac{u^{r}}{r}), m_{j} denotes the multiplicity function for Z_{j}, k_{j} := m_{j}(0) and Q_{j} is a polynomial with deg Q_{j}(s) ⩽ μ_{j}. The product \begin{matrix}∏ & ′ \\ ρ∈ℂ \end{matrix} means lim_{R→∞} ∏_{0<|ρ|<R}. Then, we have

Z_{1}(s) ⊗ ⋯ ⊗ Z_{r}(s) = s^{k_{1}···k_{r}}e^{Q(s)}

× \begin{matrix}∏ & ′ \\ ρ_{1},...,ρ_{r}∈ℂ \end{matrix}P_{μ_{1}+···+μ_{r}} (\frac{s}{ρ_{1} + ⋯ + ρ_{r}} )^{m(ρ_{1},··· ,ρ_{r})} ,

where Q(s) is a polynomial with deg Q(s) ⩽ μ_{1} + ⋯ + μ_{r} and

m(ρ_{1}, …, ρ_{r}) := m_{1}(ρ_{1}) ⋯ m_{r}(ρ_{r}) × \begin{cases}1 & if Im(ρ_{1}), …, Im(ρ_{r}) ⩾ 0, \\ (−1)^{r−1} & if Im(ρ_{1}), …, Im(ρ_{r}) < 0, \\ 0 & otherwise. \end{cases}

Lemma 4.1. The absolute tensor product of Hasse zeta function for finite fields is given as follows:

ζ(s, 𝔽_{p}) ⊗ ζ(s, 𝔽_{q}) ⊗ ζ(s, 𝔽_{r}) = e^{Q(s)}S_{3} (is, (\frac{2π}{log p}, \frac{2π}{log q} , \frac{2π}{log r} ))^{−1} ,

where p, q, r are powers of prime and Q(s) ∈ ℂ[s] satisfies deg Q(s) ⩽ 3 depending on p, q, r. Proof. We compute that the Hadamard product for Hasse zeta function is given by

ζ(s, 𝔽_{p}) = s^{−1}e^{Q̃_{p}(s)}\begin{matrix}∞ \\ ∏ & ′ \\ n=−∞ \end{matrix} P_{1} ( \frac{s}{\frac{2πi}{log p}n})^{−1} ,

where Q̃_{j}(s) is a linear polynomial depending on p. Thus by the definition of the absolute tensor product,

ζ(s, 𝔽_{p}) ⊗ ζ(s, 𝔽_{q}) ⊗ ζ(s, 𝔽_{r}) = s^{−1}e^{Q̃_{p,q,r}(s)} ×

\begin{matrix}∏ & ′ \\ n_{1},n_{2},n_{3}∈ℤ \end{matrix}P_{3} ( \frac{s}{\frac{2πi}{log p}n_{1} + \frac{2πi}{log q} n_{2} + \frac{2πi}{log r} n_{3}} )^{m_{n_{1},n_{2},n_{3}}} ,

where Q̃_{p,q,r}(s) ∈ ℂ[s] satisfies deg Q̃_{p,q,r}(s) ⩽ 3 and

m_{n_{1},n_{2},n_{3}} = m (\frac{2πi}{log p}n_{1}, \frac{2πi}{log q} n_{2}, \frac{2πi}{log r} n_{3})

= \begin{cases}−1 & if n_{1}, n_{2}, n_{3} ⩾ 0 or n_{1}, n_{2}, n_{3} < 0, \\ 0 & otherwise. \end{cases}

Hence ζ(s, 𝔽_{p}) ⊗ ζ(s, 𝔽_{q}) ⊗ ζ(s, 𝔽_{r})

= s^{−1}e^{Q̃_{p,q,r}(s)}\begin{matrix}∏ & ′ \\ n_{1},n_{2},n_{3}⩾0 \end{matrix} P_{3} ( \frac{s}{\frac{2πi}{log p}n_{1} + \frac{2πi}{log q} n_{2} + \frac{2πi}{log r} n_{3}} )^{−1} ×

∏_{n_{1},n_{2},n_{3}⩾1} P_{3} (− \frac{s}{\frac{2πi}{log p}n_{1} + \frac{2πi}{log q} n_{2} + \frac{2πi}{log r} n_{3}} )^{−1} .

By (3.1) the result follows.

Proof of Theorem 1.2. Applying Theorem 3.4 with z = is and (ω_{1}, ω_{2}, ω_{3}) = (\frac{2π}{log p}, \frac{2π}{log q} , \frac{2π}{log r}), Theorem 1.2 follows.

References

B A. Baker: Transcendental Number Theory, Cambridge University Press, 1975. D1 C. Deninger: Local L−factors of motives and regularized determinants, Invent. Math. 107 (1992) 135-

150. D2 C. Deninger: Some analogues between number theory systems on foliated spaces, Proceedings of ICM,

vol I (Berlin 1998) Doc. Math. 1998, Extra vol I, 162-186. K N. Kurokawa: Multiple zeta functions: an example, Adv. in Pure Math. 21, (1992) 219-226. KoKu S. Koyama and N. Kurokawa: Multiple Zeta Functions (preprint) KuKo N. Kurokawa and S. Koyama: Multiple Sine Functions, Forum Math. 15, (2003) 839-876.

Hirotaka Akatsuka akatsuka@math.keio.ac.jp Department of Mathematics, Keio University, , 3-14-1 Hiyoshi, Kohoku-ku, Yokohama, 223-8522, Japan.