Multiple Euler Products

Shin-ya Koyama and Nobushige Kurokawa

Dedicated to Professor Sergei Vostokov

Running title. Multiple Euler Products Abstract. We construct a new zeta function having zeros or poles at sums of zeros or poles of Euler products. Multiple Euler factors and multiple Euler products are explicitly calculated in basic cases.

2000 Mathematics Subject Classification: 11M06

1 Introduction

In 1737, Leonhard Euler [E] made a quite important discovery at St Petersburg on the existence of the Euler product expression for the zeta function ζ(s) = ∑_{n=1}^{∞} n^{−s}:

ζ(s) = ∏_{p}(1 − p^{−s})^{−1},

where p runs over the prime numbers. A typical consequence of this Euler product is the explicit formula for the number π(x) of primes under x given by Riemann [R] using the zeros and the pole of ζ(s). Up to now it is known that there exist various zeta functions having Euler product expressions.

The purpose of this paper is to study multiple Euler factors (MEF) and multiple Euler products (MEP) via multiple explicit formulas (MEF) naturally associated to “absolute tensor products” of several zeta functions.

We recall the construction of the absolute tensor product (ATP). We refer to [KK1] and [KW1] for details.

Let

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

= exp (−\frac{∂}{∂w}|_{w=0} ∑_{ρ∈C} \frac{m_{j}(ρ)}{(s − ρ)^{w}} )

be “zeta functions” expressed as regularized product, where

m_{j} : C → Z

denotes the multiplicity function for j = 1, ..., r. The absolute tensor product (Z_{1} ⊗ ⋯ ⊗ Z_{r})(s) is defined as

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

with

m(ρ_{1}, ⋯, ρ_{r}) = m_{1}(ρ_{1}) ⋯ m_{r}(ρ_{r}) × \{ \begin{matrix}1 & Im(ρ_{j}) ≥ 0, & (j = 1, ..., r) \\ (−1)^{r−1} & Im(ρ_{j}) < 0, & (j = 1, ..., r) \\ 0 & otherwise. \end{matrix}

This definition originates from [K]. We refer to the excellent survey of Manin [M]. The notation of the regularized product is due to Deninger [Den1]. See [HKW] concerning the needed regularized products. The absolute tensor product was studied by Schröter [S] in the name of the “Kurokawa tensor product.”

We are especially interested in the case of Hasse zeta functions Z_{j}(s) = ζ(s, A_{j}) for commutative rings A_{1}, ..., A_{r} of finite type over Z. We recall that the Hasse zeta function ζ(s, A) of a commutative ring A is defined to be

ζ(s, A) = ∏_{m}(1 − N(m)^{−s})^{−1}

= exp (∑_{m} ∑_{k=1}^{∞} \frac{1}{k}N(m)^{−ks}) ,

where m runs over maximal ideals of A and N(m) = #(A/m). This is also written as

ζ(s, A) = exp ( ∑_{p:primes} ∑_{m=1}^{∞} \frac{#Hom_{ring}(A, F_{p^{m}})}{m}p^{−ms}) .

For simplicity we write

ζ(s, A_{1} ⊗ ⋯ ⊗ A_{r}) = ζ(s, A_{1}) ⊗ ⋯ ⊗ ζ(s, A_{r}).

Actually, as was explained by Manin [M], we expect that our multiple zeta function would be the zeta function of the “absolute tensor product”

A_{1} ⊗_{F_{1}} ⋯ ⊗_{F_{1}} A_{r}

that is the tensor product over the (virtual) “one element field” F_{1}. See [KOW] and [Dei] for the absolute mathematics over “F_{1}”. In any way, we notice that ζ(s, A_{1} ⊗ ⋯ ⊗ A_{r}) has the following additive structure on zeros and poles: if ζ(s_{j}, A_{j}) = 0 or ∞ and Im(s_{j}) (j = 1, ..., r) have the same signature, then ζ(s_{1} + ⋯ + s_{r}, A_{1} ⊗ ⋯ ⊗ A_{r}) = 0 or ∞.

Such an additive structure was crucial in the study of Hasse zeta functions of positive characteristic (congruence zeta functions) pursued by Grothendieck [G] and Deligne [D], where Euler products were important to restrict the region of zeros and poles for our reaching to the analogue of the Riemann Hypothesis.

We expect that our multiple zeta functions also have Euler products of the following form:

ζ(s, A_{1}) ⊗ ⋯ ⊗ ζ(s, A_{r}) = ∏_{(m_{1},...,m_{r})} H_{(m_{1},...,m_{r})}(N(m_{1})^{−s}, ..., N(m_{1})^{−s})

where m_{i} runs over the maximal ideals of A_{i} and H_{(m_{1},...,m_{r})}(T_{1}, ..., T_{r}) is a power series in T_{1},...,T_{r} of the constant term 1 with a possible degeneration at (m_{1}, ..., m_{r}), where N(m_{i}) = N(m_{j}) for some i ≠ j. More generally we expect that the multiple zeta function Z_{1}(s) ⊗ ⋯ ⊗ Z_{r}(s) has an Euler product

Z_{1}(s) ⊗ ⋯ ⊗ Z_{r}(s) = ∏_{(p_{1},...,p_{r})∈P_{1}×···×P_{r}} H_{(p_{1},...,p_{r})}(N(p_{1})^{−s}, ..., N(p_{r})^{−s})

when each zeta function Z_{j}(s) has an Euler product

Z_{j}(s) = ∏_{p∈P_{j}} H_{p}^{j}(N(p)^{−s})

and a functional equation; here H_{p}^{j}(T) is a power series in T and H_{(p_{1},...,p_{r})}(T_{1}, ..., T_{r}) is a power series in (T_{1}, ..., T_{r}) with a possible degeneration at (p_{1}, ..., p_{r}), where N(p_{i}) = N(p_{j}) for some i ≠ j.

In a previous paper [KK1] we investigated the absolute tensor product ζ(s, F_{p})⊗ζ(s, F_{q}) for primes p and q by using a signed double Poisson summation formula, where ζ(s, F_{p}) = (1 − p^{−s})^{−1}. In other words we constructed a new zeta function having zeros (or poles) at sums of poles of ζ(s, F_{p}) and ζ(s, F_{q}). We state the result as follows in refining and complementing the main theorem of [KK1]. For simplicity we use the notation F (s) ≅ G(s) for functions F (s) and G(s) to indicate that F (s) = e^{Q(s)}G(s) for some polynomial Q(s).

Theorem 1 Let p and q be distinct prime numbers. Define the function ζ_{p,q}(s) in Re(s) > 0 as follows:

ζ_{p,q}(s) := exp (−\frac{i}{2} ∑_{n=1}^{∞} \frac{cot (πn\frac{log p}{log q})}{n}p^{−ns} − \frac{i}{2} ∑_{n=1}^{∞} \frac{cot (πn\frac{log q}{log p})}{n}q^{−ns}

−\frac{1}{2} ∑_{n=1}^{∞} \frac{1}{n}p^{−ns} − \frac{1}{2} ∑_{n=1}^{∞} \frac{1}{n}q^{−ns}) .

Then the function ζ_{p,q}(s) has the following properties:

(0) It converges absolutely in Re(s) > 0.

(1) The function ζ_{p,q}(s) has an analytic continuation to all s ∈ C as a meromorphic function of order two.

(2) All zeros and poles of ζ_{p,q}(s) are simple and located at

s = 2πi (\frac{m}{log p} + \frac{n}{log q} ) ,

where (m, n) is a pair of nonnegative integers or a pair of negative integers. Indeed it gives a zero or pole according as they are nonnegative or negative.

(3) We have the identification

ζ_{p,q}(s) ≅ ζ(s, F_{p}) ⊗ ζ(s, F_{q}).

(4) The function ζ_{p,q}(s) satisfies a functional equation:

ζ_{p,q}(−s) = ζ_{p,q}(s)^{−1}(pq)^{\frac{s}{2}} (1 − p^{−s})(1 − q^{−s})

× exp (\frac{i log p log q}{4π} s^{2} − \frac{πi}{6} (\frac{log q}{log p} + \frac{log p}{log q} + 3)) .

When p = q the result is as follows: Theorem 2 Let

ζ_{p,p}(s) := exp (\frac{i}{2π} ∑_{n=1}^{∞} \frac{1}{n^{2}} p^{−ns} − (1 − \frac{i log p}{2π}s) ∑_{n=1}^{∞} \frac{1}{n}p^{−ns})

in Re(s) > 0. Then the function ζ_{p,p}(s) has the following properties:

(0) It converges absolutely in Re(s) > 0.

(1) The function ζ_{p,p}(s) has an analytic continuation to all s ∈ C as a meromorphic function of order two.

(2) All zeros and poles of ζ_{p,p}(s) are located at

s = \frac{2πin}{log p},

which gives a zero or pole of order |n + 1|, according as n is a nonnegative or negative integer.

(3) We have the identification

ζ_{p,p}(s) ≅ ζ(s, F_{p}) ⊗ ζ(s, F_{p}).

(4) The function ζ_{p,p}(s) satisfies a functional equation:

ζ_{p,p}(−s) = ζ_{p,p}(s)^{−1}p^{s}(1 − p^{−s})^{2} exp (\frac{i(log p)^{2}}{4π} s^{2} − \frac{5πi}{6}) .

Remark 1.1 These theorems are natural generalizations of the following well-known facts on the simplest zeta function

ζ_{p}(s) = exp (∑_{m=1}^{∞} \frac{1}{m}p^{−ms})

defined in Re(s) > 0.

(0) It converges absolutely in Re(s) > 0.

(1) The function ζ_{p}(s) has an analytic continuation to all s ∈ C as a meromorphic function of order one. In fact, ζ_{p}(s) = (1 − p^{−s})^{−1} for all s ∈ C.

(2) The function ζ_{p}(s) has no zeros. All poles of ζ_{p}(s) are simple and located at s = 2πin/ log p for n ∈ Z.

(3) We have the identification

ζ_{p}(s) = ζ(s, F_{p}).

(4) The function ζ_{p}(s) satisfies a functional equation ζ_{p}(−s) = ζ_{p}(s)(−p^{−s}). Remark 1.2 The functional equation in Theorem 2(4) “coincides” with Theorem 1(4) when p = q. This is remarkable considering the quite different appearance of ζ_{p,q}(s) and ζ_{p,p}(s).

Remark 1.3 Let

Li_{r}(u) = ∑_{n=1}^{∞} \frac{u^{n}}{n^{r}}

be the polylogarithm function of order r. Then the function ζ_{p}(s) in Remark 1.1 is written as

ζ_{p}(s) = exp (Li_{1}(p^{−s}))

in Re(s) > 0 since Li_{1}(u) = log(\frac{1}{1−u}) for |u| < 1. Similarly the function ζ_{p,p}(s) in Theorem 2 is written as

ζ_{p,p}(s) = exp (\frac{i}{2π} Li_{2}(p^{−s}) − (1 − \frac{i log p}{2π}s) Li_{1}(p^{−s}))

in Re(s) > 0. Appearance of such a polylogarithm is characteristic in our multiple Euler factors. The function

∑_{m=1}^{∞} \frac{cot(πmα)}{m}u^{m}

appearing in Theorem 1 is considered as a variation of a polylogarithm (of order 2); see [KW1] for such a “multiple polylogarithm” (multi-q-log) and its relation to Appell’s O-function.

We prove Theorems 1 and 2 in §2 supplementing [KK1], where the signed double Poisson summation formula and the theory of multiple sine functions developed in [KK2] are essential. The main difference from [KK1] is the functional equation not stated there.

From our viewpoint, it is very interesting to see the nature of ∏_{p,q} ζ_{p,q}(s). Unfortunately, however, it does not converge even for sufficiently large Re(s). Our “α-version” ζ_{p,q}^{α}(s) treated below remedies the situation. In passing we notice on the analyticity of the diagonal Euler product shown in §2.

Theorem 3 Let

Z(s) = ∏_{p} ζ_{p,p}(s).

Then, Z(s) is absolutely convergent in Re(s) > 1, and it has an analytic continuation with singularities to Re(s) > 0 with the natural boundary Re(s) = 0.

In the later half of this paper we study “the double Riemann zeta function” ζ(s, Z) ⊗ ζ(s, Z) by establishing the signed double explicit formula in the following theorem, which generalizes the signed double Poisson summation formula used in the proof of Theorems 1 and 2.

For simplicity put ξ(s) = ζ̂(s + \frac{1}{2}) = ζ̂(s + \frac{1}{2}, Z) for ζ̂(s) = Γ_{R}(s)ζ(s) with Γ_{R}(s) = π^{−s/2}Γ(s/2). The functional equation of ζ(s) is written as ξ(s) = ξ(−s). We recall that nontrivial zeros of ζ(s) are zeros in the strip |Re(s − \frac{1}{2})| < 1/2. We denote by \frac{1}{2} + iγ such a zero, where γ is a complex number in −1/2 < Im(γ) < 1/2.

Hereafter let h(t) be an odd regular function in |Re(t)| < 1 satisfying h(t) = O(|t|^{−3}) as |t| → ∞. We put H_{α}(t) := h(2α + it) and

H̃(u) := ∫_{−∞}^{∞} H(t)e^{itu}dt.

Theorem 4 Let 1/2 < α < 1. We have

∑_{Re(γ_{1}),Re(γ_{2})>0} H_{0}(γ_{1} + γ_{2}) = ∑_{{p,q \atop p≠q}} H_{p,q}^{α} + ∑_{p} H_{p,p}^{α} + ∑_{p} H_{p,∞}^{α} + H_{∞,∞}^{α} + H_{0}^{α},

where the sum in the left hand side is taken over pairs (\frac{1}{2} + iγ_{1}, \frac{1}{2} + iγ_{2}) of nontrivial zeros of the Riemann zeta function, the sum in the right hand side is taken over pairs of distinct primes p, q or primes p, and we define for pairs of distinct primes p, q as

H_{p,q}^{α} = \frac{i}{4π^{2}} ∑_{m,n} \frac{log p log q}{log(p^{m}q^{n})} \frac{1}{p^{m(α+ \frac{1}{2} )}q^{n(α+ \frac{1}{2} )}} (H̃_{0}(−m log p) + H̃_{0}(−n log q))  (1.1)

+\frac{i}{4π^{2}} ∑_{m,n} \frac{log p log q}{log(\frac{p^{m}}{q^{n}})} \frac{1}{p^{m(α+ \frac{1}{2} )}q^{n(α+ \frac{1}{2} )}} (H̃_{α}(−m log p) − H̃_{α}(−n log q)) , (1.2)

and for a prime p,

H_{p,p}^{α} = \frac{i}{4π^{2}} ∑_{m,n} \frac{log p}{(m + n)p^{(m+n)(α+ \frac{1}{2} )}} (H̃_{0}(−m log p) + H̃_{0}(−n log p))  (1.3)

+\frac{i}{4π^{2}} ∑_{m≠n} \frac{log p}{(m − n)p^{(m+n)(α+ \frac{1}{2} )}} (H̃_{α}(−m log p) − H̃_{α}(−n log p))  (1.4)

+\frac{1}{4π^{2}} (log p)^{2} ∑_{m=1}^{∞} p^{−2m(α+ \frac{1}{2} )}{ \atop tH_{α}(t)}(−m log p),  (1.5)

H_{p,∞}^{α} = −\frac{1}{2π^{2}} ∑_{m=1}^{∞} \frac{log p}{p^{m(α+ \frac{1}{2} )}} ∫_{−∞}^{∞} p^{−imt}H_{α}(t) ∫_{0}^{t} p^{imt′} \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} + it′) dt′dt  (1.6)

−\frac{1}{2π^{2}} ∑_{m=1}^{∞} \frac{log p}{p^{m(α+ \frac{1}{2} )}} ∫_{−∞}^{∞} H_{0}(t) ∫_{0}^{t} Re (p^{im(t−t′)} \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} + it′)) dt′dt, (1.7)

H_{∞,∞}^{α} = \frac{1}{4π^{2}} ∫_{−∞}^{∞} H_{α}(t) ∫_{0}^{t} \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} + it_{1}) \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} + i(t − t_{1})) dt_{1}dt (1.8)
+\frac{1}{4π^{2}} ∫_{−∞}^{∞} H_{0}(t) ∫_{0}^{t} \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} + it_{1}) \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} − i(t − t_{1})) dt_{1}dt, (1.9)

and

H_{0}^{α} = −\frac{α}{π} ∫_{0}^{π} ∑_{Re(γ_{1})>0} h(iγ_{1} + αe^{iθ})\frac{ξ′}{ξ} (αe^{iθ})e^{iθ}dθ  (1.10)

−\frac{α^{2}}{4π^{2}} ∫_{0}^{π} ∫_{0}^{π} h(αe^{iθ_{1}} + αe^{iθ_{2}})\frac{ξ′}{ξ} (αe^{iθ_{1}})\frac{ξ′}{ξ} (αe^{iθ_{2}})e^{i(θ_{1}+θ_{2})}dθ_{1}dθ_{2},  (1.11)

where m, n ∈ Z, m, n ≥ 1.

Notice that only pairs of zeros in the upper (or lower) half plane are counted in the left hand side of Theorem 4. The method of Cramér [C] is important in the proof: see Deninger [Den2] and Voros [V] around Cramér’s method.

For defining the (p, q)-Euler factors, we put

h(t) = \frac{1}{(t + s)^{2}} − \frac{1}{(t − s)^{2}}  (1.12)

in Theorem 4. We denote by p, q any (finite or infinite) places. The (p, q)-Euler factor of the double Riemann zeta function ζ(s, Z) ⊗ ζ(s, Z) is defined as follows:

ζ_{p,q}^{α}(s + 1) = exp (∫∫ H_{p,q}^{α}(s)dsds) .

We also denote the remainder factor by

ζ_{0}^{α}(s + 1) = exp (∫∫ H_{0}^{α}(s)dsds) .

Here we notice that these definitions of ζ_{p,q}^{α}(s) and ζ_{0}^{α}(s) have ambiguities emerged from the integral constants, that is the factor exp(Q(s)) with Q(s) a polynomial with deg(Q) ≤ 2. We do not normalize them at first. Rather we normalize them a posteriori after the calculation; see the remark to Theorem 5 below. The double Riemann zeta function ζ(s, Z) ⊗ ζ(s, Z) is expressed by an Euler product over the pairs of places (p, q).

Theorem 5 The (p, q)-Euler factors of the double Riemann zeta function ζ(s, Z) ⊗ ζ(s, Z) are described as follows:

(1) For distinct prime numbers p and q, we have

ζ_{p,q}^{α}(s) ≅ exp (\frac{1}{πi} ∑_{m,n} \frac{(log p)(log q)}{(m log p)^{2} − (n log q)^{2}}

(\frac{cosh(mα log p)}{q^{n(α+ \frac{1}{2} )}}p^{−m(s− \frac{1}{2} )} + \frac{n log q}{m log p} \frac{sinh(mα log p)}{q^{n(α+ \frac{1}{2} )}}p^{−m(s− \frac{1}{2} )}

−\frac{m log p}{n log q} \frac{sinh(nα log q)}{p^{m(α+ \frac{1}{2} )}}q^{−n(s− \frac{1}{2} )} − \frac{cosh(nα log q)}{p^{m(α+ \frac{1}{2} )}}q^{−n(s− \frac{1}{2} )}) )

in Re(s) > α + \frac{1}{2}, where the sum is taken over all pairs of all positive integers m and n. It has an analytic continuation to the entire plane.

(2) For a prime number p, we have in Re(s) > α + \frac{1}{2}

ζ_{p,p}^{α}(s) ≅ exp (\frac{2}{πi} ∑_{m≠n} \frac{p^{−m(s− \frac{1}{2} )−n(α+ \frac{1}{2} )}}{m^{2} − n^{2}} (cosh(mα log p) + \frac{n}{m} sinh(mα log p)) (1.13)

+\frac{1}{2πi} ((log p)(s − 1 − 2α) log(1 − p^{−s}) − Li_{2} (p^{−s}) + Li_{2} (p^{−s−2α}))),(1.14)

which has an analytic continuation to the entire plane.

(3) The (p, ∞)-factor ζ_{p,∞}^{α}(s) of the double Riemann zeta function has an analytic continuation to the entire plane, and moreover ∏_{p} ζ_{p,∞}^{α}(s) has an analytic continuation to the entire plane with possible singularities at s = \frac{1}{2} − 2k ± α (k ≥ 0), 1 − 2k (k ≥ 0), −2k (k ≥ 1), ρ − 2k (k ≥ 0), \frac{1}{2} + ρ ± α, \frac{3}{2} ± α with ρ any nontrivial zero of ζ(s).

(4) The (∞, ∞)-factor ζ_{∞,∞}^{α}(s) of the double Riemann zeta function is analytic with possible singularities at s = −2n, −2n + α + \frac{1}{2} with n = 0, 1, 2, ...

After completing the proof of (1)-(3) of Theorem 5 in §4-§6 of the text, we fix ζ_{p,q}^{α}(s) as calculated there “with the integral constant 0”; in other words we fix ζ_{p,q}^{α}(s) (p ≠ q) and ζ_{p,p}^{α}(s) as in the right hand sides of (1) and (2) of Theorem 5; see §6 concerning ζ_{p,∞}^{α}(s) and ∏_{p} ζ_{p,∞}^{α}(s). In particular we will use these normalized Euler factors for the statement of Theorem 7 below.

Remark 1.4 To compare ζ_{p,q}^{α}(s) (p ≠ q) and ζ_{p,p}^{α}(s) with ζ_{p,q}(s) (p ≠ q) and ζ_{p,p}(s) treated in Theorem 1 and Theorem 2 respectively, it would be suggestive to take α = −1/2 formally. For this parameter α, the sums over sinh-terms diverge. We look at the non-sinh parts:

ζ_{p,q}^{− \frac{1}{2}} (s)_{non-sinh} = exp (\frac{1}{πi} ∑_{m,n} \frac{log p log q}{(m log p)^{2} − (n log q)^{2}}

(cosh (−\frac{m}{2} log p) p^{−m(s− \frac{1}{2} )} − cosh (−\frac{n}{2} log q) q^{−n(s− \frac{1}{2} )}))

and

ζ_{p,p}^{− \frac{1}{2}} (s)_{non-sinh} = exp (\frac{2}{πi} ∑_{m≠n} \frac{cosh(−\frac{m}{2} log p)}{m^{2} − n^{2}}p^{−m(s− \frac{1}{2} )}

+ \frac{1}{2πi} ((log p)s log(1 − p^{−s}) − Li_{2}(p^{−s}) + Li_{2}(p^{−(s−1)}))).

Then, the elementary summation

∑_{{n=1 \atop n≠±a}}^{∞} \frac{1}{a^{2} − n^{2}} = \{ \begin{matrix}\frac{π}{2a} cot(πa) − \frac{1}{2a^{2}} & (a ∈ C \ Z) \\ −\frac{3}{4a^{2}} & (a ∈ Z \ \{0\}) \end{matrix}

implies the followings:

ζ_{p,q}^{− \frac{1}{2}} (s)_{non-sinh} = exp(−\frac{i}{4} ∑_{m=1}^{∞} \frac{cot(πm\frac{log p}{log q})}{m}p^{−ms} − \frac{i}{4} ∑_{m=1}^{∞} \frac{cot(πm\frac{log p}{log q})}{m}p^{−m(s−1)}

−\frac{i}{4} ∑_{n=1}^{∞} \frac{cot(πn\frac{log q}{log p})}{n}q^{−ns} − \frac{i}{4} ∑_{n=1}^{∞} \frac{cot(πn\frac{log q}{log p})}{n}q^{−n(s−1)}

+\frac{i}{4π} \frac{log q}{log p} (Li_{2}(p^{−s}) + Li_{2}(p^{−(s−1)})) + \frac{i}{4π} \frac{log p}{log q} (Li_{2}(q^{−s}) + Li_{2}(q^{−(s−1)})))

and

ζ_{p,p}^{− \frac{1}{2}} (s)_{non-sinh} = exp (\frac{5i}{4π} Li_{2}(p^{−s}) + \frac{i}{4π} Li_{2}(p^{−(s−1)}) − \frac{is log p}{2π}Li_{1}(p^{−s})) .

Theorem 6 The remaining factor ζ_{0}^{α}(s) of the double Riemann zeta function is an analytic function on C with possible singularities at s = ρ + \frac{1}{2} + sgn(Im(ρ))αe^{iθ} with 0 ≤ θ ≤ π for any nontrivial zero ρ of ζ(s) and at s belonging to |s − 1| ≤ 2α.

In the next theorem we use the half Riemann zeta function ζ_{+}(s) studied in [HKW] (see §4 of the text) and the multiple gamma function Γ_{r}(s) of Barnes [Bar] (see §2 of the text).

Theorem 7 The Euler product for the double Riemann zeta function

ζ(s, Z) ⊗ ζ(s, Z) ≅ (∏_{p,q} ζ_{p,q}^{α}(s)) ζ_{0}^{α}(s) (\frac{∏_{m=1}^{∞}ζ_{+}(s + 2m)}{ζ_{+}(s − 1)})^{2} Γ_{2} (\frac{s}{2})^{−1} Γ_{1}(s)^{2}s(s − 2)

is absolutely convergent in Re(s) > α + \frac{3}{2}, where (p, q) runs through pairs of all (finite or infinite) places. It has an analytic continuation (with singularities) to the entire plane and satisfies a functional equation between s and 2 − s.

The construction of the later half of this paper is as follows. Theorem 4 is proved in §3. Then as an application of Theorem 4, Theorem 5 will be proved in §4-§7: ζ_{p,q}^{α}(s) (p ≠ q) is treated in §4, ζ_{p,p}^{α}(s) in §5, ζ_{p,∞}^{α}(s) in §6, and ζ_{∞,∞}^{α}(s) in §7. Next, Theorem 6 is shown in §8. Consequently we prove Theorem 7 in §9. Lastly we briefly notice on remaining problems in §10. Acknowledgement This paper is dedicated to Professor Sergei Vostokov for his 60th birthday. We express our hearty thanks to him for indicating the way of mathematics. We thank Professor Christopher Deninger for his interests in our multiple Euler products from the beginning and supplying the book [S]. We also thank Professor Ivan Fesenko for his enthusiastic treatment of our paper. We thank Professor Miki Hirano as well, for his thorough reading of the manuscript with indicating important errors. Some parts of this work were done while the first author was staying at The Isaac Newton Institute for Mathematical Sciences, participating in the programme “Random Matrix Approaches in Number Theory” in June/July of 2004. He deeply appreciates the efforts of the organizers and all the staff members for providing the perfect environment for research. The second author would like to express his hearty thanks to the Euler International Mathematical Institute at St. Petersburg for inviting to the excellent conference “Arithmetic Geometry” in June of 2004, where this paper was read.

2 Proof of Theorems 1, 2 and 3

We recall the multiple sine functions studied in [KK2]. For this purpose we use the multiple Hurwitz zeta function investigated by Barnes [Bar] just 100 years ago:

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

for ω = (ω_{1}, ..., ω_{r}). The definitions of the multiple gamma function and the multiple sine function are as follows:

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

= (∏ ∐_{n≥0}(n · ω + z))^{−1} ,

S_{r}(z, ω) = Γ_{r}(z, ω)^{−1}Γ_{r}(ω_{1} + ⋯ + ω_{r} − z, ω)^{(−1)^{r}}
= (∏ ∐_{n≥0}(n · ω + z)) (∏ ∐_{n≥1}(n · ω − z))^{(−1)^{r−1}} .

We write Γ_{r}(z) = Γ_{r}(z, (1, ..., 1)) and S_{r}(z) = S_{r}(z, (1, ..., 1)) for simplicity. When r = 2, we have ω = (ω_{1}, ω_{2}) and

S_{2}(z, (ω_{1}, ω_{2})) = Γ_{2}(z, (ω_{1}, ω_{2}))^{−1}Γ_{2}(ω_{1} + ω_{2} − z, (ω_{1}, ω_{2})).

The study of this double sine function was originated by Shintani [Sh] towards the Kronecker’s Jugendtraum for a real quadratic field. We obtained the following proposition in [KK2, Proposition 2.4].

Proposition 2.1 We have an expression:

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

with Q_{ω}(z) a polynomial with deg Q_{ω} ≤ r, P_{r}(u) := (1 − u) exp(u + \frac{u^{2}}{2} + ⋯ + \frac{u^{r}}{r}) and 1 := (1, ⋯, 1).

Lemma 2.2

S_{2}(z, ω)S_{2}(−z, ω) = −4 sin (\frac{πz}{ω_{1}} ) sin (\frac{πz}{ω_{2}} ) .

Proof. By the preceding proposition we have

S_{2}(z, ω)S_{2}(−z, ω) ≅ \frac{−z^{2}}{−z^{2} + (ω_{1} + ω_{2})^{2}}

\begin{matrix}∏ & ′ \\ n≥0 \end{matrix} \frac{P_{2} (−\frac{z}{n_{1}ω_{1}+n_{2}ω_{2}} ) P_{2} (\frac{z}{n_{1}ω_{1}+n_{2}ω_{2}} )}{P_{2} (\frac{z}{(n_{1}+1)ω_{1}+(n_{2}+1)ω_{2}} ) P_{2} (−\frac{z}{(n_{1}+1)ω_{1}+(n_{2}+1)ω_{2}} )}

= \frac{−z^{2}}{−z^{2} + (ω_{1} + ω_{2})^{2}} P_{2} (\frac{z}{ω_{1} + ω_{2}} ) P_{2} (−\frac{z}{ω_{1} + ω_{2}} )

∏_{n_{1}=1}^{∞} P_{2} (\frac{z}{n_{1}ω_{1}} ) P_{2} (−\frac{z}{n_{1}ω_{1}} ) ∏_{n_{2}=1}^{∞} P_{2} (\frac{z}{n_{2}ω_{2}} ) P_{2} (−\frac{z}{n_{2}ω_{2}} )

≅ z^{2} ∏_{n_{1}=1}^{∞} (1 − \frac{z}{n_{1}ω_{1}} ) (1 + \frac{z}{n_{1}ω_{1}} ) ∏_{n_{2}=1}^{∞} (1 − \frac{z}{n_{2}ω_{2}} ) (1 + \frac{z}{n_{2}ω_{2}} )

≅ sin (\frac{πz}{ω_{1}} ) sin (\frac{πz}{ω_{2}} ) .

Thus we put

S_{2}(z, ω)S_{2}(−z, ω) = e^{Q(z)}4 sin (\frac{πz}{ω_{1}} ) sin (\frac{πz}{ω_{2}} )  (2.1)

and will first prove that the polynomial Q(z) is a constant. The periodic property of S_{2}(z, ω) ([KK2, Theorem 2.1(a)])

S_{2}(z + ω_{1}, ω) = S_{2}(z, ω)S_{1}(z, ω_{2})^{−1} = S_{2}(z, ω) (2 sin \frac{πz}{ω_{2}} )^{−1}

shows that

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

Substituting z + ω_{1} for z in (2.1), we have

S_{2}(z, ω) (2 sin \frac{πz}{ω_{2}} )^{−1} S_{2}(−z, ω) (−2 sin \frac{π(z + ω_{1})}{ω_{2}}) = −e^{Q(z+ω_{1})}4 sin (\frac{πz}{ω_{1}} ) sin (\frac{π(z + ω_{1})}{ω_{2}}) ,

namely

S_{2}(z, ω)S_{2}(−z, ω) = e^{Q(z+ω_{1})}4 sin (\frac{πz}{ω_{1}} ) sin (\frac{πz}{ω_{2}} ) .

Thus we have e^{Q(z)} = e^{Q(z+ω_{1})} for any z, which states that Q(z) is a constant.

Next we put

S_{2}(z, ω)S_{2}(−z, ω) = 4C sin (\frac{πz}{ω_{1}} ) sin (\frac{πz}{ω_{2}} )  (2.2)

and will prove that the constant C is equal to −1. We consider the special value at z = ω_{2}/2. First by [KK2, Remark 2.2],

S_{2} (\frac{ω_{2}}{2} , ω) = \sqrt{2},  (2.3)

and again by the periodicity we have

S_{2} (−\frac{ω_{2}}{2} , ω) = S_{2} (\frac{ω_{2}}{2} , ω) S_{1} (−\frac{ω_{2}}{2} , ω_{1}) = 2\sqrt{2} sin (−\frac{πω_{2}}{2ω_{1}} ) .

Hence (2.2) shows that C = −1.

Remark 2.3 The formula (2.3) is an example of algebraic division values of multiple sine functions. Since the proof of (2.3) is omitted in [KK2, Remark 2.2] we give a proof here. By definition

S_{2} (\frac{ω_{2}}{2} , ω) = \frac{Γ_{2}(\frac{ω_{2}}{2} + ω_{1}, ω)}{Γ_{2}(\frac{ω_{2}}{2} , ω)}.

Here the periodicity of Γ_{2}(x, ω) in the form

Γ_{2}(x + ω_{1}, ω) = Γ_{2}(x, ω)Γ_{1}(x, ω_{2})^{−1}

gives

S_{2} (\frac{ω_{2}}{2} , ω) = Γ_{1} (\frac{ω_{2}}{2} , ω_{2})^{−1} .

Then the formula

Γ_{1}(x, ω) = \frac{Γ(\frac{x}{ω})}{\sqrt{2π}}ω^{\frac{x}{ω} − \frac{1}{2}}

originally due to Lerch implies (2.3).

Proof of Theorem 1. The statements (1)-(3) are proved in [KK1], since the main theorem of [KK1] shows that ζ(s, F_{p}) ⊗ ζ(s, F_{q}) ≅ ζ_{p,q}(s) indeed. We recall that the main ingredient for the proof of (1)-(3) given in [KK1] is the following signed double Poisson summation formula.

Proposition 2.4 Let H(t) be an odd function in L^{1}(R) with H(t) = O(t^{−2}) as |t| → ∞, and put

H̃(u) = ∫_{−∞}^{∞} H(t)e^{itu}dt.

Assume both a/b and b/a are generic and that the test function H(t) satisfies

H̃(x) = O(μ^{x})  (2.4)

as x → ∞ for some 0 < μ < 1, then we have

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

= −\frac{ia}{4π} ∑_{m>0} cot (π \frac{ma}{b}) H̃(ma) − \frac{ib}{4π} ∑_{n>0} cot (π \frac{nb}{a}) H̃(nb) − \frac{iab}{8π^{2}} H̃′(0). (2.5)

Here we say that a real number α is generic if and only if

lim_{m→∞} ∥mα∥^{\frac{1}{m}} = 1,

where we put ∥x∥ := min{|x − n| : n ∈ Z} for x ∈ R. For example:

(1) If α ∈ (\overline{Q} ∩ R) \ Q, then α is generic.

(2) Let x, y ∈ \overline{Q} ∩ R. If α = \frac{log x}{log y} ∉ Q, then α is transcendental and generic (Baker [B, Theorem 3.1], Baker-Wüstholz [BW]).

For our purpose we use the genericity of \frac{log p}{log q} and \frac{log q}{log p} for distinct primes p and q. We

notice that this property is used even for the proof of convergence in Theorem 1(0): see [KK1].

Now we prove (4). By [KK1, Theorem 2], which is obtained via the above signed double Poisson summation formula, we have for Re(s) > 0

S_{2} (is, (\frac{2π}{log p}, \frac{2π}{log q} ))

= exp (\frac{1}{2i} ∑_{m=1}^{∞} \frac{1}{m} cot (πm\frac{log p}{log q}) p^{−ms} + \frac{1}{2i} ∑_{n=1}^{∞} \frac{1}{n} cot (πn\frac{log q}{log p}) q^{−ns}

+ \frac{1}{2} log (1 − p^{−s}) + \frac{1}{2} log (1 − q^{−s})

+\frac{is^{2}(log p)(log q)}{8π} + \frac{1}{4}(log p + log q)s + \frac{πi}{12} (\frac{log p}{log q} + \frac{log q}{log p} + 3))

= ζ_{p,q}(s) exp (−\frac{is^{2}(log p)(log q)}{8π} + \frac{1}{4}(log p + log q)s + \frac{πi}{12} (\frac{log p}{log q} + \frac{log q}{log p} + 3)) .

As ζ_{p,q}(s) has a meromorphic continuation to all s ∈ C, we also have

S_{2} (−is, (\frac{2π}{log p}, \frac{2π}{log q} ))

= ζ_{p,q}(−s) exp (−\frac{is^{2}(log p)(log q)}{8π} − \frac{1}{4}(log p + log q)s + \frac{πi}{12} (\frac{log p}{log q} + \frac{log q}{log p} + 3)) .

Therefore

ζ_{p,q}(s)ζ_{p,q}(−s) = S_{2} (is, (\frac{2π}{log p}, \frac{2π}{log q} )) S_{2} (−is, (\frac{2π}{log p}, \frac{2π}{log q} ))
exp (\frac{is^{2}(log p)(log q)}{4π} − \frac{πi}{6} (\frac{log p}{log q} + \frac{log q}{log p} + 3))
= S_{2} (is, (\frac{2π}{log p}, \frac{2π}{log q} )) S_{2} (−is, (\frac{2π}{log p}, \frac{2π}{log q} ))

exp (\frac{i(log p)(log q)}{4π} (s^{2} − \frac{2π^{2}}{3} (\frac{1}{(log q)^{2}} + \frac{1}{(log p)^{2}} + \frac{3}{(log p)(log q)}))) .

Hence it suffices to show that

S_{2} (is, (\frac{2π}{log p}, \frac{2π}{log q} )) S_{2} (−is, (\frac{2π}{log p}, \frac{2π}{log q} )) = (p^{\frac{s}{2}} − p^{− \frac{s}{2}} )(q^{\frac{s}{2}} − q^{− \frac{s}{2}} ).

This is proved in Lemma 2.2.

Proof of Theorem 2. The idea of the proof is the same as the previous one. Properties (1)-(3) are shown in [KK1]; they easily follow from [KK2] also. Hence it is sufficient to prove (4).

We first have an expression

ζ_{p,p}(s) = S_{2} (\frac{i log p}{2π}s) exp (\frac{i(log p)^{2}}{8π} s^{2} − \frac{log p}{2}s − \frac{5πi}{12}) .

Then we have

ζ_{p,p}(s)ζ_{p,p}(−s) = S_{2} (\frac{i log p}{2π}s) S_{2} (−\frac{i log p}{2π}s) exp (\frac{i(log p)^{2}}{4π} s^{2} − \frac{5πi}{6}) .

So using

S_{2}(x)S_{2}(−x) = (S_{2}(1 + x)S_{1}(x))(S_{2}(1 − x)S_{1}(−x)) = −S_{1}(x)^{2},

we have

S_{2} (\frac{i log p}{2π}s) S_{2} (−\frac{i log p}{2π}s) = −S_{1} (\frac{i log p}{2π}s)^{2} = p^{s}(1 − p^{−s})^{2}.

Proof of Theorem 3. We use the following result of [KW2]. Proposition 2.5 Let ζ_{r}(s) = ∏_{p} exp (Li_{r}(p^{−s})) for an integer r ≥ 2, where Li_{r}(s) = ∑_{n=1}^{∞} \frac{u^{n}}{n^{r}} is the polylogarithm of order r. Then ζ_{r}(s) is analytic in Re(s) > 0 with the natural boundary Re(s) = 0. Remark 2.6 ζ_{1}(s) is the Riemann zeta function. Lemma 2.7 The relation between Z(s) and ζ_{2}(s) is given by

\frac{Z′}{Z} (s) = −\frac{is}{2π} (\frac{ζ′_{2}}{ζ_{2}} )′ (s) − \frac{ζ′}{ζ} (s).

Proof. From

log ζ_{2}(s) = ∑_{p} ∑_{m} \frac{1}{m^{2}} p^{−ms}

we obtain

\frac{ζ′_{2}}{ζ_{2}} (s) = − ∑_{p} ∑_{m} \frac{log p}{m}p^{−ms}

and

(\frac{ζ′_{2}}{ζ_{2}} )′ (s) = ∑_{p} ∑_{m}(log p)^{2}p^{−ms}.

On the other hand, from

log Z(s) = \frac{i}{2π} ∑_{p} ∑_{m} \frac{1}{m^{2}} p^{−ms} − ∑_{p} ∑_{m} (1 − \frac{i log p}{2π}s) \frac{1}{m}p^{−ms}

we get

\frac{Z′}{Z} (s) = −\frac{i}{2π} ∑_{p} ∑_{m} \frac{log p}{m}p^{−ms}

+\frac{i}{2π} ∑_{p} ∑_{m} \frac{log p}{m}p^{−ms}

+ ∑_{p} ∑_{m} (1 − \frac{i log p}{2π}s) (log p)p^{−ms}

= ∑_{p} ∑_{m} (1 − \frac{i log p}{2π}s) (log p)p^{−ms}.

Hence we have

\frac{Z′}{Z} (s) = −\frac{is}{2π} ∑_{p} ∑_{m}(log p)^{2}p^{−ms} + ∑_{p} ∑_{m}(log p)p^{−ms}

= −\frac{is}{2π} (\frac{ζ′_{2}}{ζ_{2}} )′ (s) − \frac{ζ′}{ζ} (s).

From this lemma and the result of [KW2], we see that Z(s) is analytic in Re(s) > 0 with the natural boundary Re(s) = 0 since

(\frac{ζ′_{2}}{ζ_{2}} )′ (s) = ∑_{m=1}^{∞} (∑_{n|m} μ(n)n) (\frac{ζ′}{ζ} )′ (ms)

has double poles at essential zeros of ζ(ms). The above equality is proved as follows. Let φ(s) = ∑_{p} p^{−s}. Then it is well-known that

φ(s) = ∑_{n=1}^{∞} \frac{μ(n)}{n} log ζ(ns).

Hence

log ζ_{2}(s) = ∑_{m} ∑_{p} \frac{1}{m^{2}} p^{−ms}
= ∑_{m} \frac{1}{m^{2}} φ(ms)
= ∑_{m} \frac{1}{m^{2}} ∑_{n} \frac{μ(n)}{n} log ζ(mns)
= ∑_{m} \frac{1}{m^{2}} (∑_{n|m} μ(n)n) log ζ(ms).

Thus

\frac{ζ′_{2}}{ζ_{2}} (s) = ∑_{m=1}^{∞} \frac{1}{m} (∑_{n|m} μ(n)n) \frac{ζ′}{ζ} (ms)

and

(\frac{ζ′_{2}}{ζ_{2}} )′ (s) = ∑_{m=1}^{∞} (∑_{n|m} μ(n)n) (\frac{ζ′}{ζ} )′ (ms).

3 Signed Double Explicit Formula: Proof of Theorem

4

Lemma 3.1 (1) Let M and N be distinct integers larger than 1. Then | log M − log N|^{−1} < 2 min(M, N) ≤ 2\sqrt{MN}.

(2) For prime numbers p, q and positive integers m, n satisfying p^{m} ≠ q^{n} we have |m log p − n log q|^{−1} < 2 min(p^{m}, q^{n}) ≤ 2p^{m/2}q^{n/2}.

(3) Let M and N be distinct positive integers. Then

| log M − log N|^{−1} < \frac{max(M, N)}{|M − N|}.

Proof. Since

log \frac{M}{N} ≥ log \frac{N + 1}{N} = log (1 + \frac{1}{N} )

for M ≥ N + 1 and

log \frac{M}{N} ≤ log \frac{N − 1}{N} = − log (1 + \frac{1}{N − 1})

for M ≤ N − 1, we have

|log \frac{M}{N} | ≥ min \{log (1 + \frac{1}{N} ) , log (1 + \frac{1}{N − 1})\}
= log (1 + \frac{1}{N} )
= ∫_{0}^{1/N} \frac{du}{1 + u}
> ∫_{0}^{1/N} \frac{du}{2}
= \frac{1}{2N} .

Hence

| log M − log N|^{−1} < 2N.

By symmetry we have also

| log M − log N|^{−1} < 2M.

This proves (1). Then (2) is the case M = p^{m} and N = q^{n} in (1). Finally we show (3) in case of M < N:

| log M − log N|^{−1} = − (log (1 − \frac{N − M}{N} ))^{−1}

= (∑_{k=1}^{∞} \frac{1}{k} (\frac{N − M}{N} )^{k})^{−1}
< \frac{N}{N − M} ,

by taking the term with k = 1, since all other terms are positive.

Proof of Theorem 4. Let D_{T} be the region defined by

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

By Cauchy’s theorem we have

∑_{0<Re(γ_{1}),Re(γ_{2})<T} h(iγ_{1} + iγ_{2}) = \frac{1}{(2πi)^{2}} ∫_{∂D_{T}} ∫_{∂D_{T}} h(s_{1} + s_{2})\frac{ξ′}{ξ} (s_{1})\frac{ξ′}{ξ} (s_{2})ds_{1}ds_{2},  (3.1)

where the integrals along ∂D_{T} are taken counter clockwise. Considering the limits as T → ∞ in the both sides of (3.1), we have

∑_{0<Re(γ_{1}),Re(γ_{2})} h(iγ_{1} + iγ_{2}) = \frac{1}{(2πi)^{2}} ∫_{∂D} ∫_{∂D} h(s_{1} + s_{2})\frac{ξ′}{ξ} (s_{1})\frac{ξ′}{ξ} (s_{2})ds_{1}ds_{2},  (3.2)

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},
C_{3} = {s ∈ ∂D | Re(s) = α}.

We compute each double integral I_{ij} = \frac{1}{(2πi)^{2}} ∫_{C_{i}} ∫_{C_{j}} in (3.2).

First we treat the integral along the vertical lines. We compute

I_{33} = \frac{1}{4π^{2}} ∫_{0}^{∞} ∫_{0}^{∞} h(2α + i(t_{1} + t_{2}))\frac{ξ′}{ξ} (α + it_{1})\frac{ξ′}{ξ} (α + it_{2})dt_{1}dt_{2}
= \frac{1}{4π^{2}} ∫_{0}^{∞} h(2α + it) ∫_{0}^{t} \frac{ξ′}{ξ} (α + it_{1})\frac{ξ′}{ξ} (α + i(t − t_{1}))dt_{1}dt

and

I_{11} = \frac{1}{4π^{2}} ∫_{∞}^{0} ∫_{∞}^{0} h(−2α + i(t_{1} + t_{2}))\frac{ξ′}{ξ} (−α + it_{1})\frac{ξ′}{ξ} (−α + it_{2})dt_{1}dt_{2}
= \frac{1}{4π^{2}} ∫_{−∞}^{0} h(2α + it) ∫_{0}^{t} \frac{ξ′}{ξ} (α + it_{1})\frac{ξ′}{ξ} (α + i(t − t_{1}))dt_{1}dt.

Hence

I_{11} + I_{33} = \frac{1}{4π^{2}} ∫_{−∞}^{∞} h(2α + it) ∫_{0}^{t} \frac{ξ′}{ξ} (α + it_{1})\frac{ξ′}{ξ} (α + i(t − t_{1}))dt_{1}dt  (3.3)
= \frac{1}{4π^{2}} ∫_{−∞}^{∞} h(2α + it)
(∫_{0}^{t} \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} + it_{1}) \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} + i(t − t_{1})) dt_{1}  (3.4)
+ ∫_{0}^{t} \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} + it_{1}) \frac{ζ′}{ζ} (α + \frac{1}{2} + i(t − t_{1})) dt_{1}  (3.5)
+ ∫_{0}^{t} \frac{ζ′}{ζ} (α + \frac{1}{2} + it_{1}) \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} + i(t − t_{1})) dt_{1}  (3.6)
+ ∫_{0}^{t} \frac{ζ′}{ζ} (α + \frac{1}{2} + it_{1}) \frac{ζ′}{ζ} (α + \frac{1}{2} + i(t − t_{1})) dt_{1}) dt.  (3.7)

The contribution from (3.4) is equal to (1.8). Next we compute that both (3.5) and (3.6) are equal to

− ∑_{p} ∑_{m=1}^{∞} \frac{log p}{p^{m(α+ \frac{1}{2} )}} p^{−imt} ∫_{0}^{t} p^{imt′} \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} + it′) dt′.

Thus we have (1.6). The contribution from (3.7) is calculated as

\frac{1}{4π^{2}} ∑_{p,q,m,n} \frac{log p log q}{p^{m(α+ \frac{1}{2} )}q^{n(α+ \frac{1}{2} )}} ∫_{−∞}^{∞} H_{α}(t)q^{−nit} ∫_{0}^{t} p^{−mit_{1}}q^{nit_{1}}dt_{1}dt.

When p, q, m, n satisfy that p^{m} ≠ q^{n}, the integral on t_{1} is equal to \frac{i(p^{−mit}q^{nit} − 1)}{m log p − n log q}, and

the summand is

\frac{i log p log q}{p^{m(α+ \frac{1}{2} )}q^{n(α+ \frac{1}{2} )}(m log p − n log q)} ∫_{−∞}^{∞} H_{α}(t) (p^{−mit} − q^{−nit}) dt.  (3.8)

In case p^{m} = q^{n}, which means p = q and m = n, the integral on t_{1} is equal to t, and the summand is

\frac{(log p)^{2}}{p^{2m(α+ \frac{1}{2} )}} ∫_{−∞}^{∞} tH_{α}(t)p^{−mit}dt.  (3.9)

Thus the total contribution from (3.7) to (3.3) is

\frac{i}{4π^{2}} ∑_{{p,q,m,n \atop p^{m}≠q^{n}}} \frac{log p log q}{log(\frac{p^{m}}{q^{n}})p^{m(α+ \frac{1}{2} )}q^{n(α+ \frac{1}{2} )}} (H̃_{α}(−m log p) − H̃_{α}(−n log q))

+ \frac{1}{4π^{2}} ∑_{p,m} \frac{(log p)^{2}}{p^{2m(α+ \frac{1}{2} )}} { \atop tH_{α}(t)}(−m log p). (3.10)

Thus we obtain (1.2), (1.4) and (1.5). Here we notice that the absolute convergence of (1.2) follows from Lemma 3.1.

We calculate I_{13} and I_{31} to have

I_{13} + I_{31} = \frac{1}{4π^{2}} ∫_{−∞}^{∞} h(it) ∫_{0}^{t} \frac{ξ′}{ξ} (α + it_{1})\frac{ξ′}{ξ} (α − i(t − t_{1}))dt_{1}dt
= \frac{1}{4π^{2}} ∫_{−∞}^{∞} h(it)
(∫_{0}^{t} \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} + it_{1}) \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} − i(t − t_{1})) dt_{1}  (3.11)
+ ∫_{0}^{t} \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} + it_{1}) \frac{ζ′}{ζ} (α + \frac{1}{2} − i(t − t_{1})) dt_{1}  (3.12)
+ ∫_{0}^{t} \frac{ζ′}{ζ} (α + \frac{1}{2} + it_{1}) \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} − i(t − t_{1})) dt_{1}  (3.13)
+ ∫_{0}^{t} \frac{ζ′}{ζ} (α + \frac{1}{2} + it_{1}) \frac{ζ′}{ζ} (α + \frac{1}{2} − i(t − t_{1})) dt_{1}) dt.  (3.14)

We find that (3.11) is equal to (1.9). Next (3.12) and (3.13) are computed as

(3.12) = −\frac{1}{4π^{2}} ∑_{m=1}^{∞} ∑_{p} \frac{log p}{p^{m(α+ \frac{1}{2} )}} ∫_{−∞}^{∞} H_{0}(t) ∫_{0}^{t} p^{im(t−t′)} \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} + it′) dt′dt

and

(3.13) = −\frac{1}{4π^{2}} ∑_{m=1}^{∞} ∑_{p} \frac{log p}{p^{m(α+ \frac{1}{2} )}} ∫_{−∞}^{∞} H_{0}(t) ∫_{0}^{t} p^{−im(t−t′)} \frac{Γ′_{R}}{Γ_{R}} (α + \frac{1}{2} − it′) dt′dt.

Thus (3.12) + (3.13) is equal to (1.7). Finally by taking into account that H_{0}(t) is an odd function, we have (1.1) and (1.3) from (3.14).

We write the remaining integrals as I_{2} + I′_{2} − I_{22} with I_{2} := I_{21} + I_{22} + I_{23} and I′_{2} := I_{12} + I_{22} + I_{32}, and first calculate I_{2}:

I_{2} = \frac{1}{2πi} ∫_{C_{2}} (\frac{1}{2πi} ∫_{∂D} h(s_{1} + s_{2})\frac{ξ′}{ξ} (s_{1})ds_{1}) \frac{ξ′}{ξ} (s_{2})ds_{2}
= \frac{1}{2πi} ∫_{C_{2}} ∑_{γ_{1}} h(iγ_{1} + s_{2})\frac{ξ′}{ξ} (s_{2})ds_{2},

where γ_{1} runs through the nontrivial zeros of ζ(s) with Re(γ_{1}) > 0 and |Im(γ_{1})| < α. Putting s_{2} = αe^{iθ}, this together with I′_{2} makes the term (1.10). Finally the integral I_{22} becomes (1.11).

In the next corollary we split (1.3) into partial sums over m ≠ n and m = n, and combine the former with (1.1). We also combine (1.2) and (1.4). Put α = \frac{1}{2} + δ with δ > 0.

Corollary 1 For δ > 0 it holds that

∑_{Re(γ_{1}),Re(γ_{2})>0} H_{0}(γ_{1} + γ_{2})

= \frac{i}{4π^{2}} ∑_{{p,q,m,n \atop p^{m}≠q^{n}}} \frac{log p log q}{log(p^{m}q^{n})} \frac{1}{p^{m(1+δ)}q^{n(1+δ)}} (H̃_{0}(−m log p) + H̃_{0}(−n log q))  (3.15)

+ \frac{i}{4π^{2}} ∑_{p} ∑_{m=1}^{∞} \frac{log p}{mp^{2m(1+δ)}} H̃_{0}(−m log p)  (3.16)

+ \frac{i}{4π^{2}} ∑_{{p,q,m,n \atop p^{m}≠q^{n}}} \frac{log p log q}{log(\frac{p^{m}}{q^{n}})} \frac{1}{p^{m(1+δ)}q^{n(1+δ)}} ({ \atop H_{\frac{1+2δ}{2}}}(−m log p) − { \atop H_{\frac{1+2δ}{2}}}(−n log q))  (3.17)

+ \frac{1}{4π^{2}} ∑_{p}(log p)^{2} ∑_{m=1}^{∞} p^{−2m(1+δ)}{ \atop (tH_{\frac{1+2δ}{2}} (t))}(−m log p)  (3.18)

− \frac{1}{2π^{2}} ∑_{p} ∑_{m=1}^{∞} \frac{log p}{p^{m(1+δ)}} ∫_{−∞}^{∞} H_{\frac{1+2δ}{2}} (t) ∫_{0}^{t} p^{im(t′−t)} \frac{Γ′_{R}}{Γ_{R}} (1 + δ + it′)dt′dt  (3.19)

− \frac{1}{2π^{2}} ∑_{p} ∑_{m=1}^{∞} \frac{log p}{p^{m(1+δ)}} ∫_{−∞}^{∞} H_{0}(t) ∫_{0}^{t} Re (p^{im(t−t′)} \frac{Γ′_{R}}{Γ_{R}} (1 + δ + it′)) dt′dt  (3.20)

+ \frac{1}{4π^{2}} ∫_{−∞}^{∞} H_{\frac{1+2δ}{2}} (t) ∫_{0}^{t} \frac{Γ′_{R}}{Γ_{R}} (1 + δ + it_{1})\frac{Γ′_{R}}{Γ_{R}} (1 + δ + i(t − t_{1}))dt_{1}dt  (3.21)
+ \frac{1}{4π^{2}} ∫_{−∞}^{∞} H_{0}(t) ∫_{0}^{t} \frac{Γ′_{R}}{Γ_{R}} (1 + δ + it_{1})\frac{Γ′_{R}}{Γ_{R}} (1 + δ − i(t − t_{1}))dt_{1}dt  (3.22)
− \frac{1 + 2δ}{2π} ∫_{0}^{π} ∑_{Re(γ_{1})>0} h(iγ_{1} + (\frac{1}{2} + δ)e^{iθ})\frac{ξ′}{ξ} ((\frac{1}{2} + δ)e^{iθ})e^{iθ}dθ  (3.23)

− \frac{(1 + 2δ)^{2}}{16π^{2}} ∫_{0}^{π} ∫_{0}^{π} h((\frac{1}{2} + δ)e^{iθ_{1}} + (\frac{1}{2} + δ)e^{iθ_{2}})\frac{ξ′}{ξ} ((\frac{1}{2} + δ)e^{iθ_{1}})\frac{ξ′}{ξ} ((\frac{1}{2} + δ)e^{iθ_{2}})e^{i(θ_{1}+θ_{2})}dθ_{1}dθ_{2},

(3.24)

where the sum in the left hand side is taken over pairs (\frac{1}{2} + iγ_{1}, \frac{1}{2} + iγ_{2}) of nontrivial zeros of the Riemann zeta function, p, q denote prime numbers, and m, n ∈ Z, m, n ≥ 1.

4 ζ_{p,q}^{α}(s): Proof of Theorem 5(1)

The goal of this section is to calculate the factor ζ_{p,q}^{α}(s) of the double Riemann zeta function for distinct prime numbers p and q. The case p = q and those involving infinite places will be treated in §5-§7.

We assume Re(s) > 1 + 2δ for δ = α − \frac{1}{2} and take the test function h(t) as in (1.12). Then

H_{\frac{1+2δ}{2}} (t) = \frac{1}{(it + 1 + 2δ + s)^{2}} − \frac{1}{(it + 1 + 2δ − s)^{2}}
= \frac{−1}{(t − i(1 + 2δ + s))^{2}} + \frac{1}{(t − i(1 + 2δ − s))^{2}} ,  (4.1)

and we compute for x < 0 that

{ \atop H_{\frac{1+2δ}{2}}}(x) = ∫_{−∞}^{∞} H_{\frac{1+2δ}{2}} (t)e^{ixt}dt
= −2πiRes_{t=(1+2δ−s)i}(H_{\frac{1+2δ}{2}} (t)e^{ixt})
= 2πxe^{−x(1+2δ−s)},

where we considered the integral along the lower half circle since |e^{ixt}| = e^{−xImt} and x < 0.

Lemma 4.1 Let 1/2 < α < 1. For distinct prime numbers p and q, the (p, q)-Euler factor of ζ̂(s + \frac{1}{2}, Z) ⊗ ζ̂(s + \frac{1}{2}, Z) is given by

exp (\frac{1}{πi} ∑_{m, n} \frac{(log p)(log q)p^{−m/2}q^{−n/2}}{(m log p)^{2} − (n log q)^{2}}

(\frac{cosh(mα log p)}{p^{ms}q^{nα}} + \frac{n log q}{m log p} \frac{sinh(mα log p)}{p^{ms}q^{nα}} − \frac{m log p}{n log q} \frac{sinh(nα log q)}{p^{mα}q^{ns}} − \frac{cosh(nα log q)}{p^{mα}q^{ns}}))

where the sum is taken over all pairs of positive integers m and n, and is absolutely convergent for Re(s) > α − \frac{1}{2}.

Proof. Put

L(s) = ∏_{{ρ_{1},ρ_{2} \atop Im(ρ_{1}),Im(ρ_{2})>0}} (1 − \frac{s}{ρ_{1} + ρ_{2}} ) exp (\frac{s}{ρ_{1} + ρ_{2}} + \frac{s^{2}}{2(ρ_{1} + ρ_{2})^{2}} ) ,

where ρ_{1} and ρ_{2} run through the upper half zeros of ξ(s) = ζ̂(s + \frac{1}{2}). Then we have by definition

(ξ ⊗ ξ)(s) ≅ \frac{L(s)}{L(−s)}ζ_{+}(s)^{−2}ζ_{+}(s + 1)^{−2}(s − 1)s^{2}(s + 1)  (4.2)

with

ζ_{+}(s) = ∏_{{ρ:ζ̂(ρ)=0 \atop Im(ρ)>0}} (1 − \frac{s}{ρ}) e^{s/ρ}

being the half Riemann zeta function introduced in [HKW]. In fact, from

ξ(s) ≅ \frac{Z^{1}(s)}{Z^{0}(s)Z^{2}(s)}

with

Z^{0}(s) = s + \frac{1}{2},
Z^{1}(s) = ∏ ∐_{ζ̂(ρ)=0} (s − (ρ − \frac{1}{2})) ,
Z^{2}(s) = s − \frac{1}{2}

we have

ξ(s) ⊗ ξ(s) ≅ (Z^{1}(s) ⊗ Z^{1}(s)) × (Z^{2}(s) ⊗ Z^{1}(s))^{−2} × (Z^{0}(s) ⊗ Z^{1}(s))^{−2}

× (Z^{2}(s) ⊗ Z^{2}(s)) × (Z^{2}(s) ⊗ Z^{0}(s))^{2} × (Z^{0}(s) ⊗ Z^{0}(s))

with

Z^{1}(s) ⊗ Z^{1}(s) ≅ \frac{L(s)}{L(−s)},
Z^{2}(s) ⊗ Z^{1}(s) ≅ ζ_{+}(s),
Z^{0}(s) ⊗ Z^{1}(s) ≅ ζ_{+}(s + 1),
Z^{2}(s) ⊗ Z^{2}(s) ≅ s − 1,
Z^{2}(s) ⊗ Z^{0}(s) ≅ s,
Z^{0}(s) ⊗ Z^{0}(s) ≅ s + 1.

If we put (ξ ⊗ ξ)(s) = e^{F (s)} and

M(s) := ζ_{+}(s)^{−2}ζ_{+}(s + 1)^{−2}(s − 1)s^{2}(s + 1),

it holds that

F′′(s) = \frac{d}{ds} (\frac{L′}{L} (−s) + \frac{L′}{L} (s) + \frac{M′}{M} (s)) .

In other words we have

\frac{d}{ds} (\frac{L′}{L} (−s) + \frac{L′}{L} (s)) = F′′(s) − \frac{d}{ds} \frac{M′}{M} (s).  (4.3)

Then the left hand side of Corollary 1 is equal to that of (4.3). Thus it is equal to the right hand side of Corollary 1, among which the terms (3.15) and (3.17) produce the desired factors in the theorem. Now we compute that

(3.15) + (3.17)

= \frac{i}{4π^{2}} ∑_{{p,m,q,n \atop p^{m}≠q^{n}}} \frac{(log p)(log q)}{p^{m( \frac{1}{2} +α)}q^{n( \frac{1}{2} +α)}} (\frac{H̃_{α}(−m log p) − H̃_{α}(−n log q)}{m log p − n log q} + \frac{H̃_{0}(−m log p) + H̃_{0}(−n log q)}{m log p + n log q}) .

The terms in the big parentheses are equal to

\frac{1}{(m log p)^{2} − (n log q)^{2}} × ( (H̃_{α}(−m log p) + H̃_{0}(−m log p)) m log p
+ (H̃_{α}(−m log p) − H̃_{0}(−m log p)) n log q
− (H̃_{α}(−n log q) − H̃_{0}(−n log q)) m log p
− (H̃_{α}(−n log q) + H̃_{0}(−n log q)) n log q).

Hence

(3.15) + (3.17) = \frac{1}{2πi} ∑_{{p,m,q,n \atop p^{m}≠q^{n}}} \frac{(log p)(log q)p^{−m( \frac{1}{2} +α)}q^{−n( \frac{1}{2} +α)}}{(m log p)^{2} − (n log q)^{2}}

(m^{2}(log p)^{2}p^{−ms}(p^{2mα} + 1) + m(log p)n(log q)p^{−ms}(p^{2mα} − 1)

−m(log p)n(log q)q^{−ns}(q^{2nα} − 1) − n^{2}(log q)^{2}q^{−ns}(q^{2nα} + 1))

= \frac{1}{πi} ∑_{{p,q,m,n \atop p^{m}≠q^{n}}} \frac{(log p)(log q)p^{−m/2}q^{−n/2}}{(m log p)^{2} − (n log q)^{2}}

(m^{2}(log p)^{2} \frac{cosh(mα log p)}{p^{ms}q^{nα}} + m(log p)n(log q)\frac{sinh(mα log p)}{p^{ms}q^{nα}}

−m(log p)n(log q)\frac{sinh(nα log q)}{p^{mα}q^{ns}} − n^{2}(log q)^{2} \frac{cosh(nα log q)}{p^{mα}q^{ns}}).

Integrating twice leads to

\frac{1}{πi} ∑_{{p,q,m,n \atop p^{m}≠q^{n}}} \frac{(log p)(log q)p^{−m/2}q^{−n/2}}{(m log p)^{2} − (n log q)^{2}}

(\frac{cosh(mα log p)}{p^{ms}q^{nα}} + \frac{n log q}{m log p} \frac{sinh(mα log p)}{p^{ms}q^{nα}} − \frac{m log p}{n log q} \frac{sinh(nα log q)}{p^{mα}q^{ns}} − \frac{cosh(nα log q)}{p^{mα}q^{ns}}) (4.4)

which is a factor of the double zeta function.

If we shift the variable as s ↦ s − 1 in Lemma 4.1 we get the factor in Theorem 5(1). Now we show the analytic continuation of ζ_{p,q}^{α}(s) for all s ∈ C. For Re(s) > 0 and β > 0,

let

φ_{p,q}(s, β) = ∑_{m,n} \frac{log p log q}{(m log p)^{2} − (n log q)^{2}} p^{−ms}q^{−nβ},

and

ψ_{p,q}(s, β) = ∑_{m,n} \frac{n}{m} \frac{(log q)^{2}}{(m log p)^{2} − (n log q)^{2}} p^{−ms}q^{−nβ}.

Then we have

ζ_{p,q}^{α}(s) = exp(\frac{1}{2πi} (φ_{p,q} (s + α − \frac{1}{2}, α + \frac{1}{2}) + φ_{p,q} (s − α − \frac{1}{2}, α + \frac{1}{2})

−ψ_{p,q} (s + α − \frac{1}{2}, α + \frac{1}{2}) + ψ_{p,q} (s − α − \frac{1}{2}, α + \frac{1}{2}) + φ_{q,p} (s + α − \frac{1}{2}, α + \frac{1}{2})

+φ_{q,p} (s − α − \frac{1}{2}, α + \frac{1}{2}) − ψ_{q,p} (s + α − \frac{1}{2}, α + \frac{1}{2}) + ψ_{q,p} (s − α − \frac{1}{2}, α + \frac{1}{2}))).

Hence it is sufficient to show that φ_{p,q}(s, β) and ψ_{p,q}(s, β) are analytic in s ∈ C. For a generic α ∈ R and any β > 0, we put

L_{p,q}(s; α, β) = ∑_{m,n} \frac{p^{−ms}q^{−nβ}}{m − nα}

in Re(s) > 0. The absolute convergence is easily seen from the genericity of α. Moreover, the integral expression

L_{p,q}(s; α, β) = ∫_{0}^{1} \frac{dv}{(p^{s} − v)(q^{β}v^{α} − 1)}

gives the analytic continuation to all s ∈ C. This equality is shown as follows:

∫_{0}^{1} \frac{dv}{(p^{s} − v)(q^{β}v^{α} − 1)} = ∫_{0}^{1} (∑_{m=1}^{∞} p^{−ms}v^{m−1}) (∑_{n=1}^{∞} q^{−nβ}v^{−nα}) dv

= ∑_{m,n} p^{−ms}q^{−nβ} ∫_{0}^{1} v^{m−nα−1}dv

= ∑_{m,n} \frac{p^{−ms}q^{−nβ}}{m − nα} .

Hence if we see that

φ′_{p,q}(s, β) = −\frac{log q}{2} (L_{p,q} (s; \frac{log q}{log p}, β) + L_{p,q} (s; −\frac{log q}{log p}, β))

and

ψ′_{p,q}(s, β) = −\frac{log q}{2} (L_{p,q} (s; \frac{log q}{log p}, β) − L_{p,q} (s; −\frac{log q}{log p}, β)) ,

then we would have the analytic continuation of φ_{p,q}(s, β) and ψ_{p,q}(s, β) for all s ∈ C. To prove these identities notice first that:

L_{p,q} (s; α, β) + L_{p,q} (s; −α, β) = ∑_{m,n} (\frac{1}{m − nα} + \frac{1}{m + nα}) p^{−ms}q^{−nβ}

= ∑_{m,n} \frac{2m}{m^{2} − n^{2}α^{2}} p^{−ms}q^{−nβ}

and

L_{p,q} (s; α, β) − L_{p,q} (s; −α, β) = ∑_{m,n} (\frac{1}{m − nα} − \frac{1}{m + nα}) p^{−ms}q^{−nβ}

= ∑_{m,n} \frac{2nα}{m^{2} − n^{2}α^{2}} p^{−ms}q^{−nβ}.

Thus

φ′_{p,q}(s, β) = − ∑_{m,n} \frac{m(log p)^{2} log q}{(m log p)^{2} − (n log q)^{2}} p^{−ms}q^{−nβ}

= −\frac{log q}{2} ∑_{m,n} \frac{2m}{m^{2} − n^{2}(\frac{log q}{log p})^{2}} p^{−ms}q^{−nβ}

= −\frac{log q}{2} (L_{p,q} (s; \frac{log q}{log p}, β) + L_{p,q} (s; −\frac{log q}{log p}, β))

and

ψ′_{p,q}(s, β) = − ∑_{m,n} \frac{n(log q)^{2} log p}{(m log p)^{2} − (n log q)^{2}} p^{−ms}q^{−nβ}

= −\frac{log q}{2} ∑_{m,n} \frac{2n(\frac{log q}{log p})}{m^{2} − n^{2}(\frac{log q}{log p})^{2}} p^{−ms}q^{−nβ}

= −\frac{log q}{2} (L_{p,q} (s; \frac{log q}{log p}, β) − L_{p,q} (s; −\frac{log q}{log p}, β)) .

This completes the proof of the analytic continuation of ζ_{p,q}^{α}(s).

5 ζ_{p,p}^{α}(s): Proof of Theorem 5(2)

The (p, p)-factors are the sum of the contributions of four terms (3.15)-(3.18) in case of p = q. For (3.15) and (3.17), we compute in the same way as in the previous theorem to get by putting p = q

exp (\frac{1}{πi} ∑_{m≠n} \frac{p^{−(m+n)/2}}{m^{2} − n^{2}}

(\frac{cosh(mα log p)}{p^{ms+nα}} + \frac{n}{m} \frac{sinh(mα log p)}{p^{ms+nα}} − \frac{m}{n} \frac{sinh(nα log p)}{p^{mα+ns}} − \frac{cosh(nα log p)}{p^{mα+ns}}))

= exp (\frac{2}{πi} ∑_{m≠n} \frac{p^{−m( \frac{1}{2} +s)−n( \frac{1}{2} +α)}}{m^{2} − n^{2}} (cosh(mα log p) + \frac{n}{m} sinh(mα log p))) .

The shift s ↦ s − 1 gives the first line (1.13).

Next we calculate the contributions from (3.16) and (3.18). By taking the test function (4.1), we compute

(3.16) = \frac{1}{2πi} ∑_{p} ∑_{m=1}^{∞}(log p)^{2}p^{−m(s+2(1+δ))}.

Thus

∫∫ (3.16)dsds = \frac{1}{2πi} ∑_{p} Li_{2} (p^{−(s+1+2α)}) .

The shift s ↦ s − 1 leads to the last term in (1.14).

Next the calculation for (3.18) shows

(3.18) = \frac{1}{2πi} ∑_{p}(log p)^{2} ∑_{m=1}^{∞}(1 + (s − 1 − 2δ)(−m log p))p^{−m(s+1)},

where we compute

H̃′_{α}(x) = it̃H_{α}(x) = 2π(1 + (s − 2α)x)e^{x(s−2α)}.

By partial integration

∫ (3.18)ds = \frac{1}{2πi} ∑_{p}(log p)^{2} ∑_{m=1}^{∞} ((\frac{1}{−m log p} + s − 1 − 2δ) p^{−m(s+1)} − ∫ p^{−m(s+1)}ds)

= \frac{1}{2πi} ∑_{p}(log p)^{2} ∑_{m=1}^{∞} (s − 1 − 2δ) p^{−m(s+1)}.

∫∫ (3.18)dsds = \frac{1}{2πi} ∑_{p}(log p)^{2} ∑_{m=1}^{∞} (\frac{s − 1 − 2δ}{−m log p} p^{−m(s+1)} − ∫ \frac{1}{−m log p}p^{−m(s+1)}ds)

= \frac{1}{2πi} ∑_{p} ∑_{m=1}^{∞} (−\frac{log p}{m}p^{−m(s+1)}(s − 1 − 2δ) − \frac{1}{m^{2}} p^{−m(s+1)})

= \frac{1}{2πi} ∑_{p} ((log p) (log (1 − p^{−(s+1)})) (s − 1 − 2δ) − Li_{2} (p^{−(s+1)})) .

Thus we determined the form of ζ_{p,p}^{α}(s). The analytic continuation of ζ_{p,p}^{α}(s) is given as follows. We decompose ζ_{p,p}^{α}(s) into two parts

ζ_{p,p}^{α}(s) = Z_{1}(s)Z_{2}(s)

with

Z_{1}(s) = exp (\frac{2}{πi} ∑_{m≠n} \frac{p^{−m(s− \frac{1}{2} )−n(α+ \frac{1}{2} )}}{m^{2} − n^{2}} (cosh(mα log p) + \frac{n}{m} sinh(mα log p)))

and

Z_{2}(s) = exp (\frac{1}{2πi} ((s − 1 − 2α) log p log(1 − p^{−s}) − Li_{2}(p^{−s}) + Li_{2}(p^{−s−2α}))) .

We start with Z_{2}(s) since it is treated directly by Theorem 2. In fact,

exp (−\frac{1}{2πi}Li_{2}(p^{−s})) = ζ_{p,p}(s) exp (− (1 + \frac{log p}{2πi}s) log(1 − p^{−s}))

and

exp (\frac{1}{2πi}Li_{2}(p^{−s−2α})) = ζ_{p,p}(s + 2α)^{−1} exp ((1 + \frac{log p}{2πi}(s + 2α)) log(1 − p^{−s−2α}))

imply that

Z_{2}(s) = ζ_{p,p}(s)ζ_{p,p}(s + 2α)^{−1} exp (− (1 + \frac{log p}{2πi}(1 + 2α)) log(1 − p^{−s})

+ (1 + \frac{log p}{2πi}(s + 2α)) log(1 − p^{−s−2α})) ,

so Z_{2}(s) is analytic in s ∈ C from Theorem 2(1).

Next we look at Z_{1}(s). For Re(s) > 0 and β > 0, let

φ_{p}(s, β) = ∑_{m≠n} \frac{p^{−ms−nβ}}{m^{2} − n^{2}}

and

ψ_{p}(s, β) = ∑_{m≠n} \frac{n}{m} \frac{p^{−ms−nβ}}{m^{2} − n^{2}} .

Then we have

Z_{1}(s) = exp (\frac{2}{πi} ∑_{m≠n} \frac{p^{−m(s− \frac{1}{2} )−n(α+ \frac{1}{2} )}}{m^{2} − n^{2}} (\frac{p^{mα} + p^{−mα}}{2} + \frac{n}{m} \frac{p^{mα} − p^{−mα}}{2} ))

= exp(\frac{1}{πi} (φ_{p} (s + α − \frac{1}{2}, α + \frac{1}{2}) + φ_{p} (s − α − \frac{1}{2}, α + \frac{1}{2})

+ ψ_{p} (s − α − \frac{1}{2}, α + \frac{1}{2}) − ψ_{p} (s + α − \frac{1}{2}, α + \frac{1}{2}))).

Hence it is sufficient to show the analyticity of φ_{p}(s, β) and ψ_{p}(s, β) in s ∈ C. Now, putting

L_{p}^{−}(s, β) = ∑_{m≠n} \frac{p^{−ms−nβ}}{m − n}

and

L_{p}^{+}(s, β) = ∑_{m≠n} \frac{p^{−ms−nβ}}{m + n}

we see that

φ′_{p}(s, β) = ∑_{m≠n} \frac{−m(log p)p^{−ms−nβ}}{m^{2} − n^{2}}

= −\frac{log p}{2} (L_{p}^{−}(s, β) + L_{p}^{+}(s, β))

and

ψ′_{p}(s, β) = ∑_{m≠n} \frac{−n(log p)p^{−ms−nβ}}{m^{2} − n^{2}}

= −\frac{log p}{2} (L_{p}^{−}(s, β) − L_{p}^{+}(s, β)) .

Thus our task is to show that L_{p}^{−}(s, β) and L_{p}^{+}(s, β) have analytic continuations to all s ∈ C. We show the needed analyticity by giving the following expressions:

L_{p}^{−}(s, β) = \frac{1}{1 − p^{s+β}} log (\frac{1 − p^{−s}}{1 − p^{−β}} )

and

L_{p}^{+}(s, β) = \frac{1}{2} log(1 − p^{−s−β}) + \frac{1}{1 − p^{β−s}} log(1 − p^{−s}) + \frac{1}{1 − p^{s−β}} log(1 − p^{−β}).

These expressions are obtained by direct calculations:

L_{p}^{−}(s, β) = ∑_{m>n} \frac{p^{−ms−nβ}}{m − n} + ∑_{m<n} \frac{p^{−ms−nβ}}{m − n}

= ∑_{n=1}^{∞} ∑_{k=1}^{∞} \frac{p^{−ks−n(s+β)}}{k} − ∑_{m=1}^{∞} ∑_{l=1}^{∞} \frac{p^{−m(s+β)−lβ}}{l}

= − log(1 − p^{−s})\frac{p^{−(s+β)}}{1 − p^{−(s+β)}} + log(1 − p^{−β})\frac{p^{−(s+β)}}{1 − p^{−(s+β)}}
= \frac{1}{1 − p^{s+β}} log (\frac{1 − p^{−s}}{1 + p^{−β}} ) .

and

L_{p}^{+}(s, β) = ∑_{m,n≥1} \frac{p^{−ms−nβ}}{m + n} − ∑_{m=1}^{∞} \frac{p^{−m(s+β)}}{2m}

= ∑_{l=2}^{∞} \frac{1}{l} (p^{−(l−1)s−β} + ⋯ + p^{−s−(l−1)β}) − \frac{1}{2} ∑_{m=1}^{∞} \frac{1}{m}p^{−m(s+β)}

= ∑_{l=2}^{∞} \frac{1}{l}p^{−s−β} \frac{p^{−(l−1)s} − p^{−(l−1)β}}{p^{−s} − p^{−β}} + \frac{1}{2} log(1 − p^{−s−β})

= ∑_{l=1}^{∞} \frac{1}{l}p^{−s−β} \frac{p^{−(l−1)s} − p^{−(l−1)β}}{p^{−s} − p^{−β}} + \frac{1}{2} log(1 − p^{−s−β})

= −\frac{p^{−β}}{p^{−s} − p^{−β}} log(1 − p^{−s}) + \frac{p^{−s}}{p^{−s} − p^{−β}} log(1 − p^{−β}) + \frac{1}{2} log(1 − p^{−s−β})
= \frac{1}{2} log(1 − p^{−s−β}) + \frac{1}{1 − p^{β−s}} log(1 − p^{−s}) + \frac{1}{1 − p^{s−β}} log(1 − p^{−β}).

Thus we proved the analyticity of ζ_{p,p}^{α}(s) for all s ∈ C.

Remark 5.1 We must distinguish L_{p}^{−}(s, β) from “L_{p,p}(s; 1, β)” carefully. In fact “L_{p,p}(s; 1, β)” diverges.

6 ζ_{p,∞}^{α}(s): Proof of Theorem 5(3)

We calculate ζ_{p,∞}^{α}(s), and moreover we describe the analyticity and possible singularities of ∏_{p} ζ_{p,∞}^{α}(s).

We first deal with (3.19) by using the formula

\frac{Γ′_{R}}{Γ_{R}} (s) = −\frac{1}{2} log π − \frac{γ}{2} − \frac{1}{s} − ∑_{k=1}^{∞} (\frac{1}{s + 2k} − \frac{1}{2k} )

with γ the Euler constant. We begin with the calculation of the integral on t′ in (3.19):

∫_{0}^{t} p^{imt′} \frac{Γ′_{R}}{Γ_{R}} (1 + δ + it′)dt′

= ∫_{0}^{t} p^{imt′} (−\frac{1}{2} log π − \frac{γ}{2} − \frac{1}{1 + δ + it′} − ∑_{k=1}^{∞} (\frac{1}{1 + δ + it′ + 2k} − \frac{1}{2k} )) dt′. (6.1)

We divide the integral (6.1) into three parts. First we compute

∫_{0}^{t} p^{imt′} (−\frac{1}{2} log π − \frac{γ}{2} ) dt′ = (−\frac{1}{2} log π − \frac{γ}{2} ) \frac{p^{imt} − 1}{im log p}
= i (\frac{1}{2} log π + \frac{γ}{2} ) \frac{p^{imt} − 1}{m log p}.  (6.2)

Secondly we have by putting t′′ = 1 + δ + it′

− ∫_{0}^{t} \frac{p^{imt′}}{1 + δ + it′} dt′ = − ∫_{1+δ}^{it+1+δ} \frac{p^{m(t′′−1−δ)}}{t′′} \frac{dt′′}{i}
= i ∫_{1+δ}^{it+1+δ} ∑_{n=0}^{∞} a_{n}(t′′)^{n−1}dt′′

= i (a_{0} log \frac{it + 1 + δ}{1 + δ} + ∑_{n=0}^{∞} a_{n+1} \frac{(it + 1 + δ)^{n+1} − (1 + δ)^{n+1}}{n + 1} )(6.3)

with the expansion p^{m(t′′−1−δ)} = ∑_{n=0}^{∞} a_{n}(t′′)^{n} with a_{n} = p^{m(−1−δ)}(m log p)^{n}/n!.

Next we calculate by putting t′′ = 1 + δ + it′ + 2k

− ∫_{0}^{t} p^{imt′} ∑_{k=1}^{∞} (\frac{1}{1 + δ + it′ + 2k} − \frac{1}{2k} ) dt′

= − ∑_{k=1}^{∞} p^{−2mk} ∫_{1+δ+2k}^{it+1+δ+2k} p^{m(t′′−1−δ)} (\frac{1}{t′′} − \frac{1}{2k} ) \frac{dt′′}{i}
= i ∑_{k=1}^{∞} p^{−2mk} ∑_{n=0}^{∞} a_{n} ∫_{1+δ+2k}^{it+1+δ+2k} ((t′′)^{n−1} − \frac{(t′′)^{n}}{2k} ) dt′′
= i ∑_{k=1}^{∞} p^{−2mk} (a_{0} log \frac{it + 1 + δ + 2k}{1 + δ + 2k} + ∑_{n=0}^{∞} (a_{n+1} − \frac{a_{n}}{2k} ) ∫_{1+δ+2k}^{it+1+δ+2k}(t′′)^{n}dt′′)
= i ∑_{k=1}^{∞} p^{−2mk} (a_{0} log \frac{it + 1 + δ + 2k}{1 + δ + 2k} + ∑_{n=0}^{∞} (a_{n+1} − \frac{a_{n}}{2k} ) \frac{(it + 1 + δ + 2k)^{n+1} − (1 + δ + 2k)^{n+1}}{n + 1} ) .

(6.4)

By (6.2), (6.3), (6.4), we deduce that (3.19) is equal to

− \frac{i}{2π^{2}} ∑_{p} ∑_{m=1}^{∞} \frac{log p}{p^{m(1+δ)}} ∫_{−∞}^{∞} H_{\frac{1+2δ}{2}} (t)p^{−imt} ((\frac{1}{2} log π + \frac{γ}{2} ) \frac{p^{imt} − 1}{m log p}

+ ∑_{k=0}^{∞} p^{−2mk} (a_{0} log \frac{it + 1 + δ + 2k}{1 + δ + 2k} + ∑_{n=0}^{∞} a_{n,k} \frac{(it + 1 + δ + 2k)^{n+1} − (1 + δ + 2k)^{n+1}}{n + 1} )) dt

with a_{n,0} = a_{n+1} and a_{n,k} = a_{n+1} − \frac{a_{n}}{2k} for k ≥ 1. For avoiding the singular point t = i(1 + δ + 2k) in the log term, we calculate the integral by integrating along the lower half circle. It encircles the double pole t = −i(s − 1 − 2δ) of (4.1). Thus (3.19) is equal to

\frac{1}{π} ∑_{p} ∑_{m=1}^{∞} \frac{log p}{p^{m(1+δ)}} Res_{t=−i(s−1−2δ)} \frac{p^{−imt}}{(t + i(s − 1 − 2δ))^{2}} ((\frac{1}{2} log π + \frac{γ}{2} ) \frac{p^{imt} − 1}{m log p}

+ ∑_{k=0}^{∞} p^{−2mk} (a_{0} log \frac{it + 1 + δ + 2k}{1 + δ + 2k} + ∑_{n=0}^{∞} a_{n,k} \frac{(it + 1 + δ + 2k)^{n+1} − (1 + δ + 2k)^{n+1}}{n + 1} ))

= \frac{1}{π} ∑_{p} ∑_{m=1}^{∞} \frac{log p}{p^{m(1+δ)}} \frac{d}{dt}|_{t=−i(s−1−2δ)} p^{−imt} ((\frac{1}{2} log π + \frac{γ}{2} ) \frac{p^{imt} − 1}{m log p}

+ ∑_{k=0}^{∞} p^{−2mk} (a_{0} log \frac{it + 1 + δ + 2k}{1 + δ + 2k} + ∑_{n=0}^{∞} a_{n,k} \frac{(it + 1 + δ + 2k)^{n+1} − (1 + δ + 2k)^{n+1}}{n + 1} ))

= \frac{1}{π} ∑_{p} ∑_{m=1}^{∞} \frac{log p}{p^{m(s−δ)}} (−im(log p) ((\frac{1}{2} log π + \frac{γ}{2} ) \frac{p^{m(s−1−2δ)} − 1}{m log p}

+ ∑_{k=0}^{∞} p^{−2mk} (a_{0} log \frac{s − δ + 2k}{1 + δ + 2k} + ∑_{n=0}^{∞} a_{n,k} \frac{(s − δ + 2k)^{n+1} − (1 + δ + 2k)^{n+1}}{n + 1} ))

+i ((\frac{1}{2} log π + \frac{γ}{2} ) p^{m(s−1−2δ)} + ∑_{k=0}^{∞} p^{−2mk} (\frac{a_{0}}{s − δ + 2k} + ∑_{n=0}^{∞} a_{n,k}(s − δ + 2k)^{n})))

= \frac{1}{π} ∑_{p} ∑_{m=1}^{∞} \frac{log p}{p^{m(s−δ)}} (i (\frac{1}{2} log π + \frac{γ}{2} ) − im(log p)a_{0} ∑_{k=0}^{∞} p^{−2mk} log \frac{s − δ + 2k}{1 + δ + 2k}  (6.5)

−im(log p) ∑_{k=0}^{∞} p^{−2mk} ∑_{n=0}^{∞} a_{n,k} \frac{(s − δ + 2k)^{n+1} − (1 + δ + 2k)^{n+1}}{n + 1}  (6.6)

+i ∑_{k=0}^{∞} p^{−2mk} (\frac{a_{0}}{s − δ + 2k} + ∑_{n=0}^{∞} a_{n,k}(s − δ + 2k)^{n})) .  (6.7)

We compute the sum over p and m in (6.5)-(6.7). Since

\frac{ζ′}{ζ} (s) = − ∑_{p} ∑_{m=1}^{∞} \frac{log p}{p^{ms}}

and

(\frac{ζ′}{ζ} )′ (s) = ∑_{p} ∑_{m=1}^{∞} \frac{m(log p)^{2}}{p^{ms}} ,

(6.5) is equal to

−\frac{i}{π} ((\frac{1}{2} log π + \frac{γ}{2} ) \frac{ζ′}{ζ} (s − δ) + ∑_{k=0}^{∞} (\frac{ζ′}{ζ} )′ (s + 2k + 1) log \frac{s − δ + 2k}{1 + δ + 2k} ) .  (6.8)

As we also have

\frac{d^{r}}{ds^{r}} \frac{ζ′}{ζ} (s) = (−1)^{r−1} ∑_{p} ∑_{m=1}^{∞} \frac{m^{r}(log p)^{r+1}}{p^{ms}} ,

(6.6) is equal to

\frac{−i}{π} ∑_{p} ∑_{m=1}^{∞} \frac{m(log p)^{2}}{p^{m(s−δ)}} ∑_{k=0}^{∞} p^{−2mk} ∑_{n=0}^{∞} (a_{n+1} − \frac{a_{n}}{2k} ) \frac{(s − δ + 2k)^{n+1} − (1 + δ + 2k)^{n+1}}{n + 1}

= \frac{−i}{π} ∑_{k=0}^{∞} ∑_{n=0}^{∞} \frac{(s − δ + 2k)^{n+1} − (1 + δ + 2k)^{n+1}}{n + 1}

(\frac{1}{(n + 1)!} ∑_{p} ∑_{m=1}^{∞} \frac{m^{n+2}(log p)^{n+3}}{p^{m(s+2k+1)}} − \frac{1}{2k(n!)} ∑_{p} ∑_{m=1}^{∞} \frac{m^{n+1}(log p)^{n+2}}{p^{m(s+2k+1)}} )

= \frac{−i}{π} ∑_{k=0}^{∞} ∑_{n=0}^{∞} \frac{(s − δ + 2k)^{n+1} − (1 + δ + 2k)^{n+1}}{n + 1}
(\frac{(−1)^{n+1}}{(n + 1)!} (\frac{ζ′}{ζ} )^{(n+2)} (s + 2k + 1) − \frac{(−1)^{n}}{2k(n!)} (\frac{ζ′}{ζ} )^{(n+1)} (s + 2k + 1)) ,  (6.9)

where we used the convention that a_{n}/2k = 0 if k = 0. Finally (6.7) is

\frac{i}{π} ∑_{p} ∑_{m=1}^{∞} \frac{log p}{p^{m(s−δ)}} ∑_{k=0}^{∞} p^{−2mk} (\frac{a_{0}}{s − δ + 2k} + ∑_{n=0}^{∞} a_{n,k}(s − δ + 2k)^{n})

= \frac{i}{π} ∑_{p} ∑_{m=1}^{∞} \frac{log p}{p^{m(s+1)}} ∑_{k=0}^{∞} p^{−2mk} (\frac{1}{s − δ + 2k} + ∑_{n=0}^{∞} (\frac{(m log p)^{n+1}}{(n + 1)!} − \frac{(m log p)^{n}}{2k(n!)} ) (s − δ + 2k)^{n})
= \frac{i}{π} ∑_{k=0}^{∞} (\frac{1}{s − δ + 2k} ∑_{p} ∑_{m=1}^{∞} \frac{log p}{p^{m(s+2k+1)}}

+ ∑_{n=0}^{∞}(s − δ + 2k)^{n} (\frac{1}{(n + 1)!} ∑_{p} ∑_{m=1}^{∞} \frac{m^{n+1}(log p)^{n+2}}{p^{m(s+2k+1)}} − \frac{1}{2k(n!)} ∑_{p} ∑_{m=1}^{∞} \frac{m^{n}(log p)^{n+1}}{p^{m(s+2k+1)}} ))

= \frac{i}{π} ∑_{k=0}^{∞} (\frac{−1}{s − δ + 2k} \frac{ζ′}{ζ} (s + 2k + 1)

+ ∑_{n=0}^{∞}(s − δ + 2k)^{n} (\frac{(−1)^{n}}{(n + 1)!} (\frac{ζ′}{ζ} )^{(n+1)} (s + 2k + 1) − \frac{(−1)^{n−1}}{2k(n!)} (\frac{ζ′}{ζ} )^{(n)} (s + 2k + 1)))

(6.10)

Integrating (6.8)-(6.10) twice gives a function in s which is analytic except at s = δ − 2k (k ≥ 0), −2k (k ≥ 0), −1 − 2k (k ≥ 1), −1 + ρ − 2k (k ≥ 0), δ + ρ and δ + 1. Writing in terms of α = \frac{1}{2} + δ and the shift s ↦ s − 1 lead to possible singularities at s = \frac{1}{2} − 2k + α (k ≥ 0), 1 − 2k (k ≥ 0), −2k (k ≥ 1), ρ − 2k (k ≥ 0), \frac{1}{2} + ρ + α and \frac{3}{2} + α.

Treating (3.20) similarly we have singularities of ∏_{p} ζ_{p,∞}^{α}(s) at s = \frac{1}{2} − 2k − α, \frac{1}{2} + ρ − α and \frac{3}{2} − α.

7 ζ_{∞,∞}^{α}(s): Proof of Theorem 5(4)

Lemma 7.1 We have the following formula on an indefinite integral:

∫ \frac{log x}{ax + b}dx = \begin{cases}\frac{1}{a} (log (1 + \frac{a}{b}x) log x + Li_{2} (−\frac{a}{b}x)) + C & (|x| < |\frac{b}{a}|) \\ \frac{1}{a} (log (1 + \frac{b}{ax}) log x − bLi_{2} (−\frac{b}{ax}) + \frac{(log x)^{2}}{2} ) + C, & (|x| > |\frac{b}{a}|) \end{cases}

where a and b are nonzero complex numbers. In particular it is an analytic function in x except at x = 0, −b/a.

Proof. We first note that for |x| < 1

\frac{d}{dx}Li_{2}(x) = \frac{d}{dx} ∑_{k=1}^{∞} \frac{x^{k}}{k^{2}} = ∑_{k=1}^{∞} \frac{x^{k−1}}{k} = \frac{1}{x} ∑_{k=1}^{∞} \frac{x^{k}}{k} = −\frac{log(1 − x)}{x}.

Hence when |x| < |\frac{b}{a}|, by differentiating the right hand side of Lemma,

\frac{d}{dx}(R.H.S) = \frac{1}{a} (\frac{a/b}{1 + \frac{a}{b}x} log x + log (1 + \frac{a}{b}x) \frac{1}{x} − \frac{log(1 + \frac{a}{b}x)}{−\frac{a}{b}x} (−\frac{a}{b}))
= \frac{log x}{ax + b}.

We similarly compute for the case |x| > |\frac{b}{a}|.

Put

ε_{n} = \{ \begin{matrix}\frac{1}{2n} & (n = 1, 2, 3, ...) \\ −\frac{log π}{2} − \frac{γ}{2} & (n = 0) \end{matrix}

and we use the expansion

\frac{Γ′_{R}}{Γ_{R}} (s) = − ∑_{k=0}^{∞} (\frac{1}{s + 2k} − ε_{k}) .

The integral on t_{1} in (3.21) is

∫_{0}^{t} ∑_{n=0}^{∞} (\frac{1}{2n + 1 + δ + it_{1}} − ε_{n}) ∑_{m=0}^{∞} (\frac{1}{2m + 1 + δ + i(t − t_{1})} − ε_{m}) dt_{1}.

The quadratic term in t_{1} is calculated as follows:

∫_{0}^{t} \frac{1}{2n + 1 + δ + it_{1}} \frac{1}{2m + 1 + δ + i(t − t_{1})}dt_{1}

= \frac{1}{2(m + n + 1 + δ) + it} ∫_{0}^{t} (\frac{1}{2n + 1 + δ + it_{1}} + \frac{1}{2m + 1 + δ + i(t − t_{1})}) dt_{1}
= \frac{−i}{2(m + n + 1 + δ) + it} (log \frac{2n + 1 + δ + it}{2n + 1 + δ} − log \frac{2m + 1 + δ}{2m + 1 + δ + it})
= \frac{−i}{2(m + n + 1 + δ) + it} log ((1 + \frac{it}{2n + 1 + δ} ) (1 + \frac{it}{2m + 1 + δ} )) .

Its contribution to (3.21) is

\frac{1}{4π^{2}} ∫_{−∞}^{∞} H_{\frac{1+2δ}{2}} (t) ∑_{m,n=0}^{∞} \frac{−i}{2(m + n + 1 + δ) + it} log ((1 + \frac{it}{2n + 1 + δ} ) (1 + \frac{it}{2m + 1 + δ} )) dt

= \frac{−i}{2π}Res_{t=−i(s−1−2δ)} \frac{1}{(t − i(1 + 2δ − s))^{2}}
∑_{m,n=0}^{∞} \frac{−i}{2(m + n + 1 + δ) + it} log ((1 + \frac{it}{2n + 1 + δ} ) (1 + \frac{it}{2m + 1 + δ} ))
= \frac{−1}{2π} \frac{d}{dt}|_{t=−i(s−1−2δ)} ∑_{m,n=0}^{∞} \frac{1}{2(m + n + 1 + δ) + it} log ((1 + \frac{it}{2n + 1 + δ} ) (1 + \frac{it}{2m + 1 + δ} ))
= \frac{−1}{2π} ∑_{m,n=0}^{∞} (\frac{−i}{(2(m + n + 1 + δ) + it)^{2}} log ((1 + \frac{it}{2n + 1 + δ} ) (1 + \frac{it}{2m + 1 + δ} ))

+ \frac{1}{2(m + n + 1 + δ) + it} (\frac{i}{2n + 1 + δ + it} + \frac{i}{2m + 1 + δ + it}))|_{t=−i(s−1−2δ)}

= \frac{−1}{2π} ∑_{m,n=0}^{∞} (\frac{−i}{(2(m + n) + 1 + s)^{2}} log ((1 + \frac{s − 1 − 2δ}{2n + 1 + δ} ) (1 + \frac{s − 1 − 2δ}{2m + 1 + δ} ))

+ \frac{1}{2(m + n) + 1 + s} (\frac{i}{s + 2n − δ} + \frac{i}{s + 2m − δ} ))

= \frac{−i}{2π} ∑_{m,n=0}^{∞} \frac{d}{ds} (\frac{1}{2(m + n) + 1 + s} log ((1 + \frac{s − 1 − 2δ}{2n + 1 + δ} ) (1 + \frac{s − 1 − 2δ}{2m + 1 + δ} ))) .

Thus

∫ (the quadratic part in (3.21))ds

= \frac{−i}{2π} ∑_{m,n=0}^{∞} \frac{1}{2(m + n) + 1 + s} log ((1 + \frac{s − 1 − 2δ}{2n + 1 + δ} ) (1 + \frac{s − 1 − 2δ}{2m + 1 + δ} )) .

By putting x = 1 + \frac{s−1−2δ}{2n+1+δ}, it follows that

\frac{1}{2(m + n) + 1 + s} log (1 + \frac{s − 1 − 2δ}{2n + 1 + δ} ) = \frac{log x}{ax + b}  (7.1)

with a = 2n + 1 + δ and b = 2m + 1 + δ. By lemma 7.1 its indefinite integral is analytic in s except at s = −2n + δ and s = −2n − 1. The shift s ↦ s − 1 leads us to the singularities at s = −2n + δ + 1 and s = −2n.

Next we deal with non-quadratic parts in (3.21). Their contributions to (3.21) are

\frac{1}{4π^{2}} ∫_{−∞}^{∞} H_{\frac{1+2δ}{2}} (t) ∫_{0}^{t} ∑_{m,n=0}^{∞} (\frac{−ε_{n}}{2m + 1 + δ + i(t − t_{1})} + \frac{−ε_{m}}{2n + 1 + δ + it_{1}} + ε_{m}ε_{n}) dt_{1}dt

= \frac{1}{4π^{2}} ∫_{−∞}^{∞} H_{\frac{1+2δ}{2}} (t) ∑_{m,n=0}^{∞} (−iε_{n} log \frac{2m + 1 + δ}{2m + 1 + δ + it}

+iε_{m} log \frac{2n + 1 + δ + it}{2n + 1 + δ} + ε_{m}ε_{n}t) dt

= \frac{−i}{2π}Res_{t=−i(s−1−2δ)} \frac{1}{(t − i(1 + 2δ − s))^{2}} ∑_{m,n=0}^{∞} (iε_{n} log \frac{2m + 1 + δ + it}{2m + 1 + δ}

+iε_{m} log \frac{2n + 1 + δ + it}{2n + 1 + δ} + ε_{m}ε_{n}t)

= \frac{−i}{2π} \frac{d}{dt}|_{t=−i(s−1−2δ)} ∑_{m,n=0}^{∞} (iε_{n} log \frac{2m + 1 + δ + it}{2m + 1 + δ}

+iε_{m} log \frac{2n + 1 + δ + it}{2n + 1 + δ} + ε_{m}ε_{n}t)

= \frac{−i}{2π} ∑_{m,n=0}^{∞} (\frac{−ε_{n}}{2m + 1 + δ + it} + \frac{−ε_{m}}{2n + 1 + δ + it} + ε_{m}ε_{n})|_{t=−i(s−1−2δ)}

= \frac{−i}{2π} ∑_{m,n=0}^{∞} (\frac{−ε_{n}}{2m − δ + s} + \frac{−ε_{m}}{2n − δ + s} + ε_{m}ε_{n}) .

By integrating twice, we see that the resulting function is analytic except at s = −2n + δ with n = 0, 1, 2, ...

This completes the proof of Theorem 5(4) and thus Theorem 5.

8 ζ_{0}^{α}(s): Proof of Theorem 6

We first deal with (1.10):

−\frac{α}{π} ∑_{Re(γ)>0} ∫_{0}^{π} (\frac{1}{(iγ + αe^{iθ} + s)^{2}} − \frac{1}{(iγ + αe^{iθ} − s)^{2}} ) \frac{ξ′}{ξ} (αe^{iθ})e^{iθ}dθ.

This function in s is holomorphic in Re(s) > \frac{1}{2} + α, and it has an analytic continuation to all s ∈ C except for possible singularities on ±C(γ) with

C(γ) = {iγ + αe^{iθ} | 0 ≤ θ ≤ π}.

Next we deal with (1.11), which is equal to

−\frac{α^{2}}{4π^{2}} ∫_{0}^{π} ∫_{0}^{π} (\frac{1}{(αe^{iθ_{1}} + αe^{iθ_{2}} + s)^{2}} − \frac{1}{(αe^{iθ_{1}} + αe^{iθ_{2}} − s)^{2}} ) \frac{ξ′}{ξ} (αe^{iθ_{1}})\frac{ξ′}{ξ} (αe^{iθ_{2}})e^{i(θ_{1}+θ_{2})}dθ_{1}dθ_{2}.

The integrand is a bounded function in θ_{1} and θ_{2} for any fixed s, and thus the integral defines an analytic function for all s ∈ C except for possible singularities in |s| ≤ 2α.

The shift s ↦ s − 1 leads to the desired result.

9 ζ(s, Z) ⊗ ζ(s, Z): Proof of Theorem 7

We first prove the absolute convergence of the multiple Euler product. Then other properties follow from our previous discussions noting that

ζ̂(s, Z) ⊗ ζ̂(s, Z) = (ζ(s, Z) ⊗ ζ(s, Z))(ζ(s, Z) ⊗ Γ_{R}(s))^{2}(Γ_{R}(s) ⊗ Γ_{R}(s)).

We see the convergence from the following lemma.

Lemma 9.1 Let σ > α + \frac{3}{2}. Then:

(1)

∑_{p≠q} ∑_{m,n≥1} \frac{(log p)(log q)}{|(m log p)^{2} − (n log q)^{2}|} \frac{p^{mα}}{q^{n(α+ \frac{1}{2} )}} p^{−m(σ− \frac{1}{2} )} < ∞.

(2)

∑_{p≠q} ∑_{m,n≥1} \frac{n}{m}(log q)^{2} \frac{1}{|(m log p)^{2} − (n log q)^{2}|} \frac{p^{mα}}{q^{n(α+ \frac{1}{2} )}} p^{−m(σ− \frac{1}{2} )} < ∞.

In the proof we use the following lemma.

Lemma 9.2 Let a > 0 and w > 0. Then

∑_{l=1}^{∞} \frac{1}{l}(a + l)^{−w} < 2^{−w}a^{−w/2} \frac{w + 2}{w}.

Proof.

∑_{l=1}^{∞} \frac{1}{l}(a + l)^{−w} < ∑_{l=1}^{∞} \frac{1}{l} (2\sqrt{al})^{−w}
= 2^{−w}a^{−w/2} ∑_{l=1}^{∞} l^{−1− \frac{w}{2}}
= 2^{−w}a^{−w/2}ζ (1 + \frac{w}{2})
< 2^{−w}a^{−w/2} \frac{1 + \frac{w}{2}}{\frac{w}{2}}
= 2^{−w}a^{−w/2} \frac{w + 2}{w}.

Here we used the inequality ζ(s) < s/(s − 1) for s > 1. Proof of Lemma 9.1. We first prove (1). From Lemma 3.1(3) and the fact that

|m log p + n log q| > 1,

It suffices to prove for σ > α + \frac{3}{2} that

A := ∑_{{p,q,m,n \atop p^{m}>q^{n}}} \frac{p^{m}}{p^{m} − q^{n}} q^{−n(α+ \frac{1}{2} )}p^{−m(σ−α− \frac{1}{2} )} < ∞  (9.1)

and that

B := ∑_{{p,q,m,n \atop p^{m}<q^{n}}} \frac{q^{n}}{q^{n} − p^{m}} q^{−n(α+ \frac{1}{2} )}p^{−m(σ−α− \frac{1}{2} )} < ∞,  (9.2)

since the factors log p, log q, m and n are easily treated by differentiation (or directly). For proving (9.1) we have by putting l = p^{m} − q^{n} ≥ 1 and by Lemma 9.2

∑_{{p,q,m,n \atop p^{m}>q^{n}}} \frac{p^{m}}{p^{m} − q^{n}} p^{−ma}q^{−nb} < ∑_{q,n} ∑_{l=1}^{∞} \frac{l + q^{n}}{l} (l + q^{n})^{−a}q^{−nb}

= ∑_{q,n} (∑_{l=1}^{∞} \frac{1}{l}(q^{n} + l)^{−(a−1)}) q^{−nb}

< ∑_{q,n} 2^{−(a−1)} \frac{a + 1}{a − 1}q^{− \frac{n}{2} (a−1)}q^{−nb}

= 2^{−(a−1)} \frac{a + 1}{a − 1} ∑_{q,n} q^{−( \frac{a−1}{2} +b)n}
< ∞  (9.3)

if \frac{a−1}{2} + b > 1 and a > 1, which is equivalently a + 2b > 3 and a > 1. We obtain (9.1) by putting a = σ − α − \frac{1}{2} and b = α + \frac{1}{2}.

Next for proving (9.2) we similarly have by putting l = q^{n} − p^{m} ≥ 1

∑_{{p,q,m,n \atop p^{m}<q^{n}}} \frac{q^{n}}{q^{n} − p^{m}} p^{−ma}q^{−nb} < ∑_{p,m} ∑_{l=1}^{∞} \frac{l + p^{m}}{l} (l + p^{m})^{−b}p^{−ma}

= ∑_{p,m} (∑_{l=1}^{∞} \frac{1}{l}(p^{m} + l)^{−(b−1)}) p^{−ma}

≤ ∑_{p,m} 2^{−(b−1)} \frac{b + 1}{b − 1}p^{−( \frac{b−1}{2} +a)m}
< ∞

if \frac{b−1}{2} + a > 1 and b > 1, which is equivalently 2a + b > 3 and b > 1. We obtain (9.2) by putting a = σ − α − \frac{1}{2} and b = α + \frac{1}{2}. This completes the proof of (1).

By considering \frac{∂}{∂b}(9.3), we similarly have (2). Hence Lemma 9.1.

Now we calculate the multiple Euler product expression. From

ζ̂(s, Z) = ζ(s, Z)Γ_{R}(s)

we have

ζ̂(s, Z) ⊗ ζ̂(s, Z) = (ζ(s, Z) ⊗ ζ(s, Z)) (ζ(s, Z) ⊗ Γ_{R}(s))^{2} (Γ_{R}(s) ⊗ Γ_{R}(s))

with

Γ_{R}(s) ⊗ Γ_{R}(s) ≅ ∏ ∐_{m,n≥0}(s + 2m + 2n)
≅ ∏ ∐_{m,n≥0}(\frac{s}{2} + m + n)

≅ Γ_{2} (\frac{s}{2})^{−1}

and

ζ(s, Z) ⊗ Γ_{R}(s) ≅ ( ∏ ∐_{ζ̂(ρ)=0}(s − ρ) ∏ ∐_{n=1}^{∞}(s + 2n) × (s − 1)^{−1}) ⊗ ( ∏ ∐_{m≥0}(s + 2m))^{−1}

≅ ( ∏ ∐_{ζ̂(ρ)=0}(s − ρ) ⊗ ∏ ∐_{m≥0}(s + 2m)^{−1}) (∏ ∐_{n=1}^{∞}(s + 2n) ⊗ ∏ ∐_{m≥0}(s + 2m)^{−1})

((s − 1)^{−1} ⊗ ∏ ∐_{m≥0}(s + 2m)^{−1})

≅ ( ∏ ∐_{{ρ,m \atop Im(ρ)>0}}(s − (ρ − 2m)))^{−1} ( ∏ ∐_{{n≥1 \atop m≥0}}(s + 2(n + m)))^{−1} ∏ ∐_{m≥0}(s + 2m − 1)

≅ ∏_{m=0}^{∞} ζ_{+}(s + 2m)^{−1}Γ_{2} (\frac{s + 2}{2}) Γ_{1} (\frac{s − 1}{2})^{−1} .

Thus

ζ(s, Z) ⊗ ζ(s, Z) = (ζ̂(s, Z) ⊗ ζ̂(s, Z)) (ζ(s, Z) ⊗ Γ_{R}(s))^{−2} (Γ_{R}(s) ⊗ Γ_{R}(s))^{−1}
≅ (ζ̂(s, Z) ⊗ ζ̂(s, Z)) (∏_{m=0}^{∞} ζ_{+}(s + 2m)^{2}) Γ_{2} (\frac{s + 2}{2})^{−2} Γ_{1} (\frac{s − 1}{2})^{2} Γ_{2} (\frac{s}{2})

≅ (ζ̂(s, Z) ⊗ ζ̂(s, Z)) (∏_{m=0}^{∞} ζ_{+}(s + 2m)^{2}) Γ_{2} (\frac{s}{2})^{−1} Γ_{1}(s)^{2}(s − 1)^{−2}

since

Γ_{2} (\frac{s + 2}{2})^{−2} Γ_{1} (\frac{s − 1}{2})^{2} Γ_{2} (\frac{s}{2}) = Γ_{2} (\frac{s}{2} + 1)^{−2} Γ_{2} (\frac{s}{2}) Γ_{1} (\frac{s − 1}{2})^{2}
= (Γ_{2} (\frac{s}{2}) Γ_{1} (\frac{s}{2})^{−1})^{−2} Γ_{2} (\frac{s}{2}) Γ_{1} (\frac{s − 1}{2})^{2}
= Γ_{2} (\frac{s}{2})^{−1} (Γ_{1} (\frac{s}{2}) Γ_{1} (\frac{s − 1}{2}))^{2}
≅ Γ_{2} (\frac{s}{2})^{−1} Γ_{1}(s − 1)^{2}
= Γ_{2} (\frac{s}{2})^{−1} Γ_{1}(s)^{2}(s − 1)^{−2},

where we used the relation

Γ_{r}(x + 1) = Γ_{r}(x)Γ_{r−1}(x)^{−1}

(see [KK2]). Note that

Γ_{1}(x) = \frac{Γ(x)}{\sqrt{2π}}

and

Γ_{0}(x) = x^{−1}.

We used also the duplication formula

Γ_{1} (\frac{x}{2}) Γ_{1} (\frac{x + 1}{2}) ≅ Γ_{1}(x).

On the other hand, from §4 we recall the formula (4.2) (after the translation s → s − 1)

ζ̂(s, Z) ⊗ ζ̂(s, Z) ≅ \frac{L(s − 1)}{L(1 − s)}ζ_{+}(s)^{−2}ζ_{+}(s − 1)^{−2}(s − 2)(s − 1)^{2}s

and

\frac{L(s − 1)}{L(1 − s)} ≅ ∏_{p,q} ζ_{p,q}^{α}(s) × ζ_{0}^{α}(s).

Hence

ζ(s, Z) ⊗ ζ(s, Z) ≅ (∏_{p,q} ζ_{p,q}^{α}(s)) ζ_{0}^{α}(s)ζ_{+}(s − 1)^{−2}ζ_{+}(s)^{−2}(s − 2)(s − 1)^{2}s

× (∏_{m=0}^{∞} ζ_{+}(s + 2m)^{2}) Γ_{2} (\frac{s}{2})^{−1} Γ_{1}(s)^{2}(s − 1)^{−2}

= (∏_{p,q} ζ_{p,q}^{α}(s)) ζ_{0}^{α}(s) (\frac{∏_{m=1}^{∞} ζ_{+}(s + 2m)}{ζ_{+}(s − 1)})^{2} Γ_{2} (\frac{s}{2})^{−1} Γ_{1}(s)^{2}(s − 2)s.

Lastly, the functional equation s → 2 − s is obtained from the invariance

\frac{L(s − 1)}{L(1 − s)} = (\frac{L((2 − s) − 1)}{L(1 − (2 − s))})^{−1} .

Remark 9.3 From the applicational viewpoint (see Problem (2) in §10 below) it is quite interesting to see the convergence region of

ζ_{0}(s, Z) ⊗ ζ_{0}(s, Z) = (∏_{p,q} ζ_{p,q}^{α}(s)) ζ_{0}^{α}(s) (∏_{m=1}^{∞}ζ_{+}(s + 2m))^{2} Γ_{2} (\frac{s}{2})^{−1} Γ_{1}(s)^{2}Γ_{1} (\frac{s + 1}{2})^{−2} s.

Does this converge in Re(s) > α + 1? Here ζ_{0}(s, Z) = (s − 1)ζ(s, Z).

10 Problems

We list some of the remaining problems.

(1) Higher tensor products:

We studied above Z_{1}(s) ⊗ Z_{2}(s) only. It is quite interesting to investigate Z_{1}(s) ⊗ ⋯ ⊗ Z_{r}(s) for r ≥ 3: for example, ζ(s, Z) ⊗ ζ(s, Z) ⊗ ζ(s, Z). Our method is extendable to such cases but the needed calculations become cumbersome. Akatsuka [A] studied ζ(s, F_{p_{1}}) ⊗ ζ(s, F_{p_{2}}) ⊗ ζ(s, F_{p_{3}}) for distinct primes p_{1}, p_{2} and p_{3} by our method, and he obtained the following Euler product (or the “Euler factor”):

Theorem (Akatsuka [A]) Let p, q, r be distinct primes. In Re(s) > 0 we have

ζ(s, F_{p}) ⊗ ζ(s, F_{q}) ⊗ ζ(s, F_{r})
= e^{Q(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{cot (πn_{1} \frac{log p}{log q}) cot (πn_{1} \frac{log p}{log r})}{n_{1}}p^{−n_{1}s}

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

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

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

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

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

where Q(s) is a polynomial of degree at most 3.

Moreover, Akatsuka calculated the degenerate cases (p, p, p) and (p, p, q). After Akatsuka’s paper [A], the general case ζ(s, F_{p_{1}}) ⊗ ⋯ ⊗ ζ(s, F_{p_{r}}) for distinct primes p_{1}, ..., p_{r} is treated in [KW1] by a different method.

(2) Obtaining results on essential zeros:

If we can prove that ζ(s, Z)⊗ζ(s, Z) (or ζ_{0}(s, Z)⊗ζ_{0}(s, Z) in Remark 9.3) has no essential zeros and poles in Re(s) > \frac{3}{2}, then we have \frac{1}{4} ≤ Re(ρ) ≤ \frac{3}{4} for all essential zeros ρ of ζ(s, Z). More generally if we know that there are no essential zeros and poles of ζ(s, Z)^{⊗r} for r ≥ 2 in Re(s) > \frac{r+1}{2}, then we obtain \frac{r−1}{2r} ≤ Re(ρ) ≤ \frac{r+1}{2r} for all essential zeros ρ of ζ(s, Z). Thus letting r → ∞ we would have Re(ρ) = \frac{1}{2}. (See Grothendieck [G] and Deligne [D] for the case of a scheme over F_{p}.)

We may expect that our Euler product for ζ(s, Z)⊗ζ(s, Z) (or ζ_{0}(s, Z)⊗ζ_{0}(s, Z)) should imply such a needed result for ζ(s, Z) ⊗ ζ(s, Z) (or ζ_{0}(s, Z) ⊗ ζ_{0}(s, Z)). But at present we do not have results in this direction. Does this consideration have some relations to the “absolute scheme” Spec(Z ⊗_{F_{1}} Z) as in [KOW] and [Dei]?

(3) Theory of multiple Euler products:

Is there a theory of multiple Euler products implying

ζ(s, Z) ⊗ ζ(s, Z) = ∏_{p,q} ζ(s, F_{p}) ⊗ ζ(s, F_{q})

perfectly with functorial relations? We see some discrepancy from our present results.

Note (May 2006): Recetnly H. Akatsuka (Tokyo Institute of Technology) calculated the case “α = \frac{1}{2}”. His method gives the double Euler product in a direct way; see H. Akatsuka “The double Riemann zeta function” (in preparation).

References

[A] H. Akatsuka: Euler product expression of triple zeta functions Int. J. Math. 16 (2005) 111-136.

[B] A. Baker: Transcendental Number Theory, Cambridge University Press, 1975.

[BW] A. Baker and G. Wüstholz: Logarithmic forms and group varieties, J. reine Angew. Math. 442 (1993) 19-62.

[Bar] E.W. Barnes: On the theory of the multiple gamma function, Trans. Cambridge Philos. Soc., 19 (1904) 374-425.

[C] H. Cramér: Studien über die Nullstellen der Riemannschen Zetafunktion, Math. Z. 4 (1919) 104-130.

[D] P. Deligne: La conjecture de Weil (I), Inst. Hautes Êtudes Sci. Publ. Math. 43 (1974) 273-307.

[Dei] A. Deitmar: Schemes over F_{1} (preprint, 2004).

[Den1] C. Deninger: Local L-factors of motives and regularized determinants, Invent. Math. 107 (1992) 135-150.

[Den2] C. Deninger: Some analogues between number theory and dynamical systems on foliated spaces, Proceedings of ICM, vol. I (Berlin 1998) Doc. Math. 1998, Extra vol. I, 163-186.

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[G] A. Grothendieck: Séminaire de Géométrie Algébrique du Bois-Marie 1965-66 (SGA 5), Lecture Notes in Math. 589 (1977), Springer-Verlag.

[HKW] M. Hirano, N. Kurokawa and M. Wakayama: Half zeta functions, J. Ramanujan Math. Soc. 18 (2003) 195-209.

[K] N. Kurokawa: Multiple zeta functions: an example, Adv. Stud. in Pure Math. 21 (1992) 219-226.

[KK1] S. Koyama and N. Kurokawa: Multiple zeta functions: the double sine function and the signed double Poisson summation formula, Compositio Math. 140 (2004) 1176-1190.

[KK2] N. Kurokawa and S. Koyama: Multiple sine functions, Forum Math. 15 (2003) 839-876.

[KOW] N. Kurokawa, H. Ochiai and M. Wakayama: Absolute derivations and zeta functions, Documenta Math. 2003 (Kato Extra Volume), 565-584.

[KW1] N. Kurokawa and M. Wakayama: Absolute tensor products, International Math. Research Notices 2004-No.5 (2004), 249-260.

[KW2] N. Kurokawa and M. Wakayama: Analyticity of polylogarithmic Euler products, Rendiconti del Circolo Matematico di Palermo, 52 (2003) 382-388.

[M] Yu. I. Manin: Lectures on zeta functions and motives (according to Deninger and Kurokawa), Asterisque 228 (1995) 121-163.

[R] B. Riemann: Ueber die Anzahl der Primzahlen unter einer gegebenen Grösse, Monatsberichte der Berliner Akademie, November 1859, 671-680.

[S] M. Schröter: Über Kurokawa-Tensorprodukte von L-Reihen, Schriftenreihe des Mathematischen Instituts der Universität Münster, 3. Serie, Heft 18, April, 1996, 240 pages.

[Sh] T. Shintani: On a Kronecker limit formula for real quadratic fields, J. Fac. Sci. Univ. Tokyo, 24 (1977) 167-199.

[V] A. Voros: Zeta functions for the Riemann zeros, Ann. Inst. Fourier, Grenoble 53 (2003) 665-699.

The Isaac Newton Institute for Mathematical Sciences, University of Cambridge 20 Clarkson Road, Cambridge, CB3 0EH, United Kingdom S.Koyama@newton.cam.ac.uk Current Address: Zeta Institute 2-5-27 Hayabuchi, Tsuzuki-ku, Yokohama 224-0025, Japan koyama@tmtv.ne.jp

Department of Mathematics, Tokyo Institute of Technology Oo-okayama, Meguro-ku, Tokyo 152-8551, Japan kurokawa@math.titech.ac.jp