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On the regularized products of some Dirichlet series

Mounir Hajli
Abstract.

In this paper, we show that the regularized determinants of some Dirichlet series are multiplicative. As an application, we give generalizations of Lerch’s formula for the classical gamma function and we determine the sum of some Dirichlet series generalizing Euler’s formula on the sum of the reciprocal of squares. We recover the results of Kurokawa and Wakayama, and give a new proof for some Euler’s formulas.

MSC  2000: 11M36.

Keywords: Lerch’s formula, Hurwitz zeta function, zeta regularized product.

1. Introduction


Let 0<b1b20<b_{1}\leq b_{2}\leq\ldots be a sequence of real numbers, and assume that

ζ(s)=k=1bks\zeta(s)=\sum_{k=1}^{\infty}b_{k}^{-s}

converges absolutely on Re(s)0\mathrm{Re}(s)\gg 0, and can be continued meromorphically to the whole complex plane and holomorphic at s=0s=0. The regularized determinant of b1,b2,b_{1},b_{2},\ldots is defined as follows

k=1bk:=exp(ζ(0)).\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=1}^{\infty}b_{k}:=\exp(-\zeta^{\prime}(0)).

When (bk)k=1,2,(b_{k})_{k=1,2,\ldots} is the spectrum of a positive self-adjoint operator AA, then using techniques of heat-kernels, one can show that ζA(s)=k=1bks\zeta_{A}(s)=\sum_{k=1}^{\infty}b_{k}^{-s} can be continued analytically to s=0s=0, and k=1bk\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=1}^{\infty}b_{k} is called the regularized determinant of the operator AA (see [16]).

The zeta regularization method can be applied to the sequence (n+x)n(n+x)_{n\in\mathbb{N}} where x>0x>0 (Here, ={0,1,2,}\mathbb{N}=\{0,1,2,\ldots\}). We obtain the regularized product n=0(n+x)\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{n=0}^{\infty}(n+x), which can be viewed as a function in the variable xx, and it can be continued analytically to \mathbb{C}. The classical Gamma function Γ(x)\Gamma(x) can be defined from the Hurwitz zeta function via the Lerch formula [15],

(1) k=0(k+x):=exp(ζHs(x,0))=2πΓ(x)for all x,\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=0}^{\infty}(k+x):=\exp(-\frac{\partial\zeta_{H}}{\partial s}(x,0))=\frac{\sqrt{2\pi}}{\Gamma(x)}\quad\text{for all $x\in\mathbb{R}$},

where ζH(s,x)=k=0(k+x)s\zeta_{H}(s,x)=\sum_{k=0}^{\infty}(k+x)^{-s} for x>0x>0 (the Hurwitz zeta function).

Only a few general methods for the exact evaluation of k=1bk\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=1}^{\infty}b_{k} are available. Many authors aimed to give complete results for a number of cases where it was possible to find formulas for the zeta invariants in terms of some known special functions, see for instance [3, 4, 12, 14, 19]. In [3, 4], the authors determine the regularized product of the spectrum of the Laplacian on the DD-dimensional ball.

In the case of the complex projective space n()\mathbb{P}^{n}(\mathbb{C}) endowed with its Fubini-Study metric, the computation of its holomorphic analytic torsion (see [16]), which is by definition a linear combination of the regularized determinants of the Laplacians Δq\Delta^{q} acting on the space of smooth (0,q)(0,q)-forms on n()\mathbb{P}^{n}(\mathbb{C}) for q=0,,nq=0,\ldots,n , was an important step toward the formulation of an arithmetic Riemann-Roch theorem in the context of Arakelov Geometry (see [6] for the computation of the holomorphic analytic torsion of n()\mathbb{P}^{n}(\mathbb{C}), and [17, Chapter VIII] for the general formulation of the arithmetic Riemann-Roch theorem). In [11, Section 2], we gave a new method for the computation of the regularized determinant of Δ0\Delta^{0} on n()\mathbb{P}^{n}(\mathbb{C}) endowed with the Fubini-Study metric. This approach can be adapted easily to get the regularized determinant of Δq\Delta^{q} for q=1,,nq=1,\ldots,n. One can use the arithmetic Riemann-Roch theorem to compute the analytic torsion of the line bundles 𝒪(m)\mathcal{O}(m), with m0m\geq 0, on n()\mathbb{P}^{n}(\mathbb{C}) endowed with their Fubini-Study metrics, this was done by L. Weng in [21]. In [8], we defined a new class of singular Laplacians Δ𝒪(m)¯\Delta_{\overline{\mathcal{O}(m)}_{\infty}} associated with 𝒪(m)\mathcal{O}(m) on 1()\mathbb{P}^{1}(\mathbb{C}) endowed with their canonical metrics. We showed that the spectrum of Δ𝒪(m)¯\Delta_{\overline{\mathcal{O}(m)}_{\infty}} can be determined explicitly in terms of Bessel functions (see [8, Theorem 1.4]). In [7], we showed that the zeta function ζΔ𝒪(m)¯(s)\zeta_{\Delta_{\overline{\mathcal{O}(m)}_{\infty}}}(s) associated with the spectrum of Δ𝒪(m)¯\Delta_{\overline{\mathcal{O}(m)}_{\infty}} can be continued analytically to s=0s=0, and using [7, Theorem 3.6] we obtained that

ζΔ𝒪(m)¯(0)=23m2,\zeta_{\Delta_{\overline{\mathcal{O}(m)}_{\infty}}}(0)=-\frac{2}{3}-\frac{m}{2},

and, we computed the regularized determinant

λSpec(Δ𝒪(m)¯){0}λ=(m+2)m+1((m+1)!)2exp(4ζ(1)+16),\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{\lambda\in\mathrm{Spec}(\Delta_{\overline{\mathcal{O}(m)}_{\infty}})\setminus\{0\}}\lambda=\frac{(m+2)^{m+1}}{((m+1)!)^{2}}\exp\left(-4\zeta_{\mathbb{Q}}^{\prime}(-1)+\frac{1}{6}\right),

(see [7, Theorem 1.1]). In [9], we developed a generalized spectral theory explaining these results.

Let 0<c1c20<c_{1}\leq c_{2}\leq\ldots and 0<b1b20<b_{1}\leq b_{2}\leq\ldots be two sequences of positive real numbers. We assume that k=1bks\sum_{k=1}^{\infty}b_{k}^{-s} and k=1cks\sum_{k=1}^{\infty}c_{k}^{-s} converge for Re(s)0\mathrm{Re}(s)\gg 0, and admit meromorphic continuations to \mathbb{C} which are holomorphic at s=0s=0. An interesting question is to evaluate the following ratio:

(2) k=1(bkck)k=1bkk=1ck=?\frac{\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=1}^{\infty}(b_{k}c_{k})}{\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=1}^{\infty}b_{k}\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=1}^{\infty}c_{k}}=?

This question was raised in [17, p. 100]. In [10], we give a partial answer to this question. Garate and Friedman in [5] develop a different method which allows them to compute some special cases.

In this paper, we use the main result of [10] to evaluate the following regularized products

(3) k=0((k+x)mεym),\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=0}^{\infty}((k+x)^{m}-\varepsilon y^{m}),

where mm is a positive integer and ε2=1\varepsilon^{2}=1. Our main result is,

Theorem 1.1.

Let mm be a positive integer 2\geq 2. Let L(x):=k=0(k+x)L(x):=\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=0}^{\infty}(k+x), with xx\in\mathbb{C}. We have

k=0((k+x)mεym)=ξm=1L(xξε1my)x,y,\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=0}^{\infty}\left((k+x)^{m}-\varepsilon y^{m}\right)=\prod_{\xi^{m}=1}L(x-\xi\varepsilon^{\frac{1}{m}}y)\quad\forall x,y\in\mathbb{C},

where ε\varepsilon\in\mathbb{R} such that ε2=1\varepsilon^{2}=1, ε1m\varepsilon^{\frac{1}{m}} is an mm-th root of ε\varepsilon.

In this situation, we see that the regularized determinant is multiplicative, i.e. (2) is equal to 11. In general, (2) may be 1\neq 1, see [10, Eq. 11].

Let us review some basic facts about the classical Gamma function. Let Γ(z)\Gamma(z) be the classical Gamma function. This function can be defined as follows

(4) Γ(z)=limnn!nzz(z+1)(z+n)forz.\Gamma(z)=\underset{n\rightarrow\infty}{\lim}\frac{n!n^{z}}{z(z+1)\cdots(z+n)}\quad\text{for}\;z\in\mathbb{C}\setminus\mathbb{Z}.

Bohr and Mollerup used this definition to give a characterization for the Gamma function (see [2, p.14] for more details). We know that

(5) Γ(z)=0tz1et𝑑tforz.\Gamma(z)=\int_{0}^{\infty}t^{z-1}e^{-t}dt\quad\text{for}\;z\in\mathbb{C}\setminus\mathbb{Z}.

The Gamma function is a meromorphic function in the complex plane, with simple poles in 0\mathbb{Z}_{\leq 0}. Weierstrass theory shows that

(6) 1Γ(s)=seγsn=1(1+sn)esn,\frac{1}{\Gamma(s)}=se^{\gamma s}\prod_{n=1}^{\infty}\left(1+\frac{s}{n}\right)e^{-\frac{s}{n}},

where γ\gamma is the Euler-Mascheroni constant. Using the Lerch formula, which is proved in Appendix 5, our theorem recovers the result of Kurokawa and Wakayama [13]. Our approach is new and self-contained. Kurokawa and Wakayama use of the infinite product expression (6) for the Gamma function and the Lerch’s formula (1) which is a crucial step in their proof.

As an application of Theorem 1.1, we obtain the following identities:

12πk2yk2+y2=coth(πy),\frac{1}{2\pi}\sum_{k\in\mathbb{Z}}\frac{2y}{k^{2}+y^{2}}=\coth(\pi y),
1π2k4y3k4+y4=sinh(2πy)+sin(2πy)cosh(2πy)cos(2πy),\frac{1}{\pi\sqrt{2}}\sum_{k\in\mathbb{Z}}^{\infty}\frac{4y^{3}}{k^{4}+y^{4}}=\frac{\sinh(\sqrt{2}\pi y)+\sin(\sqrt{2}\pi y)}{\cosh(\sqrt{2}\pi y)-\cos(\sqrt{2}\pi y)},

and

(7) nπky2n1k2n+y2n=l=0n1eπilncot(πeπilny) for n3\frac{n}{\pi}\sum_{k\in\mathbb{Z}}\frac{y^{2n-1}}{k^{2n}+y^{2n}}=\sum_{l=0}^{n-1}e^{\frac{\pi il}{n}}\cot(\pi e^{\frac{\pi il}{n}}y)\;\text{ for $n\geq 3$, }

for any y{0}y\in\mathbb{R}\setminus\{0\} (see Theorem 4.1). From the first identity, we obtain a new proof for Euler’s theorem:

ζ(2j)=(1)j+122j1(2j)!π2jB2jj=1,2,\zeta_{\mathbb{Q}}(2j)=\frac{(-1)^{j+1}2^{2j-1}}{(2j)!}\pi^{2j}B_{2j}\quad j=1,2,\ldots

By considering the Laurent expansion near y=0y=0 of the third identity, we obtain a second proof for Euler’s theorem.

Motivated by the above results, we established the following theorem which generalizes identity (7).

Corollary 1.2.

(see Theorem 3.2) Let x,yx,y\in\mathbb{C} with Re(x)>1\mathrm{Re}(x)>-1 and Re(y)>0\mathrm{Re}(y)>0. For any positive integer m2m\geq 2, we have

(8) k=11(k+x)m+y=1mξm=1ξ(1)1my1m1Ψ(xξ(1)1my1m+1),\sum_{k=1}^{\infty}\frac{1}{(k+x)^{m}+y}=-\frac{1}{m}\sum_{\xi^{m}=1}\xi(-1)^{\frac{1}{m}}y^{\frac{1}{m}-1}\Psi(x-\xi(-1)^{\frac{1}{m}}y^{\frac{1}{m}}+1),

where (1)1m(-1)^{\frac{1}{m}} denotes an mm-th root of 1-1 and Ψ\Psi is the logarithmic derivative of the Gamma function.

Note that the left hand side of (8) is Mellin’s transform of

θm(t,x;y)=k=1e((k+x)m+y)t,\theta_{m}(t,x;y)=\sum_{k=1}^{\infty}e^{-((k+x)^{m}+y)t},

evaluated at s=1s=1. This function is studied Section 3. When m=2m=2 and x=y=0x=y=0, this is the classical Jacobi theta series

k=1ek2t.\sum_{k=1}^{\infty}e^{-k^{2}t}.

It follows from the Poisson summation formula that θ2(t,0,0)=πtθ2(1t,0,0)12.\theta_{2}(t,0,0)=\tfrac{\sqrt{\pi}}{\sqrt{t}}\theta_{2}(\tfrac{1}{t},0,0)-\frac{1}{2}. Hence, we obtain

θ2(t,0,0)=π2t12+O(t)(t0+).\theta_{2}(t,0,0)=\tfrac{\sqrt{\pi}}{2\sqrt{t}}-\tfrac{1}{2}+O(\sqrt{t})\quad(t\rightarrow 0^{+}).

2. On the regularized determinant of some Dirichlet series


Let 0<b1b20<b_{1}\leq b_{2}\leq\ldots be an unbounded and nondecreasing sequence of positive real numbers. To this data, we associate the following series

θ(t):=k=1ebkt for t>0 (whenever it converges).\theta(t):=\sum_{k=1}^{\infty}e^{-b_{k}t}\quad\text{ for $t>0$ (whenever it converges)}.

We suppose that θ\theta satisfies the following three conditions:

  1. (Θ\Theta1):

    The series kebkt\sum_{k}e^{-b_{k}t} converges for any t>0t>0,

  2. (Θ\Theta2):

    The series θ(t)\theta(t) admits an asymptotic expansion for t0+t\rightarrow 0^{+}

    θ(t)mcimtim,\theta(t)\sim\sum_{m\in\mathbb{N}}c_{i_{m}}t^{i_{m}},

    where (in)n(i_{n})_{n\in\mathbb{N}} is an unbounded and non-decreasing sequence of real numbers. Moreover, we assume that ci00c_{i_{0}}\neq 0,

  3. (Θ\Theta3):

    kebkt=O(eκt)\sum_{k}e^{-b_{k}t}=O(e^{-\kappa t}) for t+t\rightarrow+\infty, for some positive real number κ\kappa.

We set

(9) i0:=i(θ),i_{0}:=i(\theta),

and call it the index of θ\theta. The zeta function ξ\xi associated with θ\theta, is by definition the function

ξ(s):=1Γ(s)0ts1θ(t)𝑑t=k=11bksfor Re(s)>i(θ).\xi(s):=\frac{1}{\Gamma(s)}\int_{0}^{\infty}t^{s-1}\theta(t)dt=\sum_{k=1}^{\infty}\frac{1}{b_{k}^{s}}\quad\text{for }\mathrm{Re}(s)>-i(\theta).

The properties (Θ1),(Θ2)(\Theta 1),(\Theta 2) and (Θ3)(\Theta 3) imply that ξ\xi can be continued into a meromorphic function on the whole complex plane, which is holomorphic at s=0s=0.

In the sequel, we consider the following sequence (Q(k))k(Q(k))_{k\in\mathbb{N}}, where QQ is a monic polynomial of degree \ell having only distinct roots with positive real part, and Q(k)>0Q(k)>0 for any k=1,2,k=1,2,\ldots It is known that the Dirichlet series

k=11Q(k)s,\sum_{k=1}^{\infty}\frac{1}{Q(k)^{s}},

converges on any compact subset of Re(s)>1/\mathrm{Re}(s)>1/\ell, and has a meromorphic continuation to the whole complex plane (see [10, Theorem 3.2]). The regularized product of the sequence (Q(k))k=1,2,(Q(k))_{k=1,2,\ldots} is by definition

k=1Q(k)=exp(s(k=11Q(k)s)|s=0).\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=1}^{\infty}Q(k)=\exp\left(-\frac{\partial}{\partial s}\left(\sum_{k=1}^{\infty}\frac{1}{Q(k)^{s}}\right)_{|_{s=0}}\right).

We set

L(x):=n=0(n+x)x.L(x):=\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{n=0}^{\infty}(n+x)\quad\forall x\in\mathbb{R}.

(see Appendix 5 for the properties of this function).

We have

Theorem 2.1.

Under the conditions above,

k=0Q(k)=i=1L(di)=(2π)/2i=1Γ(di),\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=0}^{\infty}Q(k)=\prod_{i=1}^{\ell}L(d_{i})=\frac{(2\pi)^{\ell/2}}{\prod_{i=1}^{\ell}\Gamma(d_{i})},

where d1,,dd_{1},\ldots,d_{\ell} are the roots of QQ.

Proof.

Let Z(s)=k=11Q(k)sZ(s)=\sum_{k=1}^{\infty}\frac{1}{Q(k)^{s}}. By definition,

k=1Q(k):=exp(sZ(s)|s=0).\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=1}^{\infty}Q(k):=\exp(-\frac{\partial}{\partial s}Z(s)_{|_{s=0}}).

Let

gi(s)=k=11(k+di)swith Re(s)>1, for i=1,,l1,.g_{i}(s)=\sum_{k=1}^{\infty}\frac{1}{(k+d_{i})^{s}}\quad\text{with $\mathrm{Re}(s)>1$, for $i=1,\ldots,l-1,\ell$}.

We put f=gf=g_{\ell}. We have

f(s)=1Γ(s)0ts1θ(t)𝑑t,f(s)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}t^{s-1}\theta(t)dt,

where θ(t)=k=1e(k+d)t\theta(t)=\sum_{k=1}^{\infty}e^{-(k+d_{\ell})t} for t>0t>0. We have

θ(t)=e(1+d)t1et=1t+(an analytic series) (for 0<t1).\theta(t)=\frac{e^{-(1+d_{\ell})t}}{1-e^{-t}}=\frac{1}{t}+\text{(an analytic series) (for $0<t\ll 1$)}.

Then, it is easy to see that i(θ)=1i(\theta)=-1.

By [10, Theorem 3.4], there exists a polynomial FF in \ell variables such that

(10) exp(Z(0))=i=1exp(gi(0))exp(F(d1,,d)),\exp(-Z^{\prime}(0))=\prod_{i=1}^{\ell}\exp(-g_{i}^{\prime}(0))\exp(F(d_{1},\ldots,d_{\ell})),

(see also [10, eq. (11)]).

But i(θ)=1i(\theta)=-1, then

F(d1,,d)=0.F(d_{1},\ldots,d_{\ell})=0.

On the other hand,

gi(s)=k=11(k+di)s=ζH(s;di)1disfor i=1,,l,g_{i}(s)=\sum_{k=1}^{\infty}\frac{1}{(k+d_{i})^{s}}=\zeta_{H}(s;d_{i})-\frac{1}{d_{i}^{s}}\quad\text{for $i=1,\ldots,l$},

So,

exp(gi(0))=L(di+1)for i=1,,l\exp(-g_{i}^{\prime}(0))=L(d_{i}+1)\quad\text{for $i=1,\ldots,l$}

By Lerch’s formula [15] (see Appendix 5 for a different proof), we get

exp(gi(0))=2πdiΓ(di)for i=1,,l\exp(-g_{i}^{\prime}(0))=\frac{\sqrt{2\pi}}{d_{i}\Gamma(d_{i})}\quad\text{for $i=1,\ldots,l$}

Gathering all these computations, (10) becomes

k=1Q(k)=(2π)degQ/2i=1diΓ(di).\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=1}^{\infty}Q(k)=\frac{(2\pi)^{\deg Q/2}}{\prod_{i=1}^{\ell}d_{i}\Gamma(d_{i})}.

So,

k=0Q(k)=(2π)degQ/2i=1Γ(di).\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=0}^{\infty}Q(k)=\frac{(2\pi)^{\deg Q/2}}{\prod_{i=1}^{\ell}\Gamma(d_{i})}.

Proof of Theorem 1.1.

Let mm be a positive integer. We consider the polynomial in tt:

Q(t)=(tn+x)mεym,Q(t)=(t^{n}+x)^{m}-\varepsilon y^{m},

with ε2=1\varepsilon^{2}=1, and x,yx,y\in\mathbb{C}. We choose an mm-th root of ε\varepsilon, which we denote by ε1m\varepsilon^{\frac{1}{m}}. Note that

Q(t)=ξm=1(t+xξε1my).Q(t)=\prod_{\xi^{m}=1}\left(t+x-\xi\varepsilon^{\frac{1}{m}}y\right).

We conclude the proof by using Theorem 2.1. ∎

Using Lerch’s formula, we obtain

Corollary 2.2.

With the same assumptions as in Theorem 1.1, we have

k=0((k+x)mεym)=(2π)m/2ξm=1Γ(xξε1my).\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=0}^{\infty}\left((k+x)^{m}-\varepsilon y^{m}\right)=\frac{(2\pi)^{m/2}}{\prod_{\xi^{m}=1}\Gamma(x-\xi\varepsilon^{\frac{1}{m}}y)}.

3. On the variation of the regularized product k=0((k+x)m+y)\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=0}^{\infty}\left((k+x)^{m}+y\right)

For m=1,2,m=1,2,\ldots, x0x\geq 0 and yy\in\mathbb{C}, we set

(10) θm(t,x;y)=k=1e((k+x)m+y)tfort>0.\theta_{m}(t,x;y)=\sum_{k=1}^{\infty}e^{-((k+x)^{m}+y)t}\quad\text{for}\;t>0.

The following proposition is important in our study of the variation of

k=0((k+x)m+y)\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=0}^{\infty}((k+x)^{m}+y)

as a function in the variable yy.

Proposition 3.1.

The function tθm(t,x;y)t\mapsto\theta_{m}(t,x;y) is smooth on (0,)(0,\infty) and has an asymptotic expansion for small positive tt of the form

θm(t,x;y)1Γ(1m+1)t1m+k=0ck(x;y)tk,\theta_{m}(t,x;y)\sim\frac{1}{\Gamma(\frac{1}{m}+1)}t^{-\frac{1}{m}}+\sum_{k=0}^{\infty}c_{k}(x;y)t^{k},

(where ck(x;y)c_{k}(x;y) are constants which depend on xx and yy), and which decays exponentially at infinity, more precisely, we have for tt sufficiently large, |θm(t,x;y)|C(y)et|\theta_{m}(t,x;y)|\leq C(y)e^{-t} for t1t\gg 1, where C(y)C(y) is a constant which depends continuously on yy).

Moreover, for y0y\geq 0 and x0x\geq 0,

ζm(s,x;y):=1Γ(s)0ts1θm(t,x;y)𝑑t=k=11((k+x)m+y)sforRe(s)>1m,\zeta_{m}(s,x;y):=\frac{1}{\Gamma(s)}\int_{0}^{\infty}t^{s-1}\theta_{m}(t,x;y)dt=\sum_{k=1}^{\infty}\frac{1}{((k+x)^{m}+y)^{s}}\quad\text{for}\;\mathrm{Re}(s)>\frac{1}{m},

which admits a meromorphic continuation to \mathbb{C} and has only one simple pole at s=1ms=\frac{1}{m}. In particular,

i(θm(t,x;y))=1m.i(\theta_{m}(t,x;y))=-\frac{1}{m}.
Proof.

It is clear that tθm(t,x;y)t\mapsto\theta_{m}(t,x;y) is a smooth function.

Let σ1<σ2\sigma_{1}<\sigma_{2} be two real numbers. On the closed strip σ1Re(s)σ2\sigma_{1}\leq\mathrm{Re}(s)\leq\sigma_{2}, we have for any r>0r>0,

Γ(s)=O(|s|r),\Gamma(s)=O(|s|^{-r}),

as |s||s|\rightarrow\infty in the strip. This decay follows from the complex version of Stirling’s formula (see [1, p. 257]). On the other hand, it is known that for σ<1\sigma<1, one has

ζH(σ+it,x)=O(|t|1σ)for t.\zeta_{H}(\sigma+it,x)=O(|t|^{1-\sigma})\quad\text{for\;}t\in\mathbb{R}.

Since θm(t,x;0)\theta_{m}(t,x;0) is the inverse Mellin transform of Γ(s)ζH(ms,,x+1)\Gamma(s)\zeta_{H}(ms,,x+1), and knowing that ζH(ms,x+1)\zeta_{H}(ms,x+1) can be continued analytically to Re(s)<1/m\mathrm{Re}(s)<1/m, we can use the Mellin inversion theorem to show that θm(t,x;0)\theta_{m}(t,x;0) has an asymptotic expansion for tt small enough,

θm(t,x;0)1Γ(1m+1)t1m+k=0ck(x)tkfor 0<t1,\theta_{m}(t,x;0)\sim\frac{1}{\Gamma(\frac{1}{m}+1)}t^{-\frac{1}{m}}+\sum_{k=0}^{\infty}c_{k}(x)t^{k}\quad\text{for}\;0<t\ll 1,

where the dominant term corresponds to the unique pole of ζH(ms,x+1)\zeta_{H}(ms,x+1). In fact, the coefficients of the asymptotic expansion can be determined in terms of the special values of ζH(ms,x+1)\zeta_{H}(ms,x+1) at the non-negative integers (see [10, Proposition 2.1]). Since θm(t,x;y)=θm(t,x;0)eyt\theta_{m}(t,x;y)=\theta_{m}(t,x;0)e^{-yt}, we can deduce the asymptotic expansion (10). That is the existence of a sequence of constants (ck(x;y))k(c_{k}(x;y))_{k\in\mathbb{N}} such that

(11) k=1e((k+x)m+y)t1Γ(1m+1)t1m+k=0ck(x;y)tk.\sum_{k=1}^{\infty}e^{-((k+x)^{m}+y)t}\sim\frac{1}{\Gamma(\frac{1}{m}+1)}t^{-\frac{1}{m}}+\sum_{k=0}^{\infty}c_{k}(x;y)t^{k}.

We have, for x0x\geq 0 and yy\in\mathbb{C}

|k=1e(k+x)mtyt|k=1ekteRe(y)tt>0.\left|\sum_{k=1}^{\infty}e^{-(k+x)^{m}t-yt}\right|\leq\sum_{k=1}^{\infty}e^{-kt}e^{-\mathrm{Re}(y)t}\quad\forall t>0.

So, we deduce the existence of a constant C(y)C(y) which depends continuously on yy such that

(12) |k=1e(k+x)mtyt|C(y)ett1.\left|\sum_{k=1}^{\infty}e^{-(k+x)^{m}t-yt}\right|\leq C(y)e^{-t}\quad\forall t\geq 1.

From (11) and (12), and using [10, Proposition 2.1], we conclude that ζm(s,x;y)\zeta_{m}(s,x;y) admits a meromorphic continuation to \mathbb{C} with only one pole at s=1ms=\frac{1}{m}. The asymptotic expansion shows that this pole is simple.

Theorem 3.2.

Let x,y>0x,y>0. For any m=2,3,,m=2,3,\ldots, we have

ξm=1Γ(xξ(1)1my1m+1)y(1ξm=1Γ(xξ(1)1my1m+1))=k=11(k+x)m+y,\sum_{\xi^{m}=1}\Gamma(x-\xi(-1)^{\frac{1}{m}}y^{\frac{1}{m}}+1)\frac{\partial}{\partial y}\left(\frac{1}{\prod_{\xi^{m}=1}\Gamma(x-\xi(-1)^{\frac{1}{m}}y^{\frac{1}{m}}+1)}\right)=\sum_{k=1}^{\infty}\frac{1}{(k+x)^{m}+y},

where (1)1m(-1)^{\frac{1}{m}} denotes an mm-th root of 1-1.

Proof.

By Theorem 1.1, it is enough to prove the following identity

(13) ylogk=1((k+x)m+y)=k=11((k+x)m+y),\frac{\partial}{\partial y}\log\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=1}^{\infty}\left((k^{\ell}+x)^{m}+y\right)=\sum_{k=1}^{\infty}\frac{1}{((k+x)^{m}+y)},

for x>0x>0 and yy\in\mathbb{C} sufficiently small, and to notice that the left hand side of this equality is ζms(s,x;εy)|s=0-\frac{\partial\zeta_{m}}{\partial s}(s,x;-\varepsilon y)_{|_{s=0}} (see the notations in Proposition 3.1).

We have

yζm(s,x;y)=1Γ(s)0tsθm(t,x;y)𝑑t=sΓ(s)0ts1Θm(t,x;y)𝑑ty,\begin{split}\frac{\partial}{\partial y}\zeta_{m}(s,x;y)=&-\frac{1}{\Gamma(s)}\int_{0}^{\infty}t^{s}\theta_{m}(t,x;y)dt\\ =&\frac{s}{\Gamma(s)}\int_{0}^{\infty}t^{s-1}\Theta_{m}(t,x;y)dt\quad\forall y\in\mathbb{C},\end{split}

where Θm(t,x;y):=tk=1e((k+x)m+y)udu\Theta_{m}(t,x;y):=\int_{t}^{\infty}\sum_{k=1}^{\infty}e^{-((k+x)^{m}+y)u}du, which decays exponentially as tt\rightarrow\infty.

We set

g(t)=θm(t,x;y)Γ(1+1m)1t1mt>0.g(t)=\theta_{m}(t,x;y)-\Gamma(1+\frac{1}{m})\frac{1}{t^{\frac{1}{m}}}\quad\forall t>0.

By Proposition 3.1, gg is a bounded function for tt small enough. Using the asymptotic expansion of θm(t,x;y),\theta_{m}(t,x;y), we have

ζm(s,x;y)=1Γ(s)Γ(1+1m)01ts+1m1𝑑t+1Γ(s)01ts1g(t)𝑑t+1Γ(s)1ts1θ(t)𝑑t.\begin{split}\zeta_{m}(s,x;y)=&\frac{1}{\Gamma(s)}\Gamma(1+\frac{1}{m})\int_{0}^{1}t^{s+\frac{1}{m}-1}dt+\frac{1}{\Gamma(s)}\int_{0}^{1}t^{s-1}g(t)dt\\ &+\frac{1}{\Gamma(s)}\int_{1}^{\infty}t^{s-1}\theta(t)dt.\end{split}

It follows that ζm(s,x;y)\zeta_{m}(s,x;y) can be continued into a holomorphic function in the variables ss and yy, for any ss in an open neighborhood of 0 and yy in any bounded subset of \mathbb{C}.

In the sequel, we shall use the same notation ζm(s,x;y)\zeta_{m}(s,x;y) to denote its holomorphic continuation. We have

2ysζm(s,x;y)=2syζm(s,x;y),\frac{\partial^{2}}{\partial y\partial s}\zeta_{m}(s,x;y)=\frac{\partial^{2}}{\partial s\partial y}\zeta_{m}(s,x;y),

which holds on an open neighborhood of s=0s=0.

We obtain

2ζmysζm(0,x;y)=s(sΓ(s)0ts1Θm(t,x;y)𝑑t)|s=0.\frac{\partial^{2}\zeta_{m}}{\partial y\partial s}\zeta_{m}(0,x;y)=\frac{\partial}{\partial s}\left(\frac{s}{\Gamma(s)}\int_{0}^{\infty}t^{s-1}\Theta_{m}(t,x;y)dt\right)_{|_{s=0}}.

But,

Θm(t,x,;y)=k=11((k+x)m+y)e((k+x)m+y)tfort>0,\Theta_{m}(t,x,;y)=\sum_{k=1}^{\infty}\frac{1}{((k+x)^{m}+y)}e^{-((k+x)^{m}+y)t}\quad\text{for}\;t>0,

So,

s(sΓ(s)0ts1Θm(t,x;y)𝑑t)|s=0=s(sk=11((k+x)m+y)s+1)|s=0=s(sζm(s+1,x;y))|s=0.\begin{split}\frac{\partial}{\partial s}\left(\frac{s}{\Gamma(s)}\int_{0}^{\infty}t^{s-1}\Theta_{m}(t,x;y)dt\right)_{|_{s=0}}=&\frac{\partial}{\partial s}\left(s\sum_{k=1}^{\infty}\frac{1}{((k+x)^{m}+y)^{s+1}}\right)_{|_{s=0}}\\ =&\frac{\partial}{\partial s}(s\zeta_{m}(s+1,x;y))_{|_{s=0}}.\end{split}

Since ζm(s,x;y)\zeta_{m}(s,x;y) is holomorphic at s=1s=1 (use Proposition 3.1). Then

s(sζm(s+1,x;y))|s=0=ζm(1,x;y).\frac{\partial}{\partial s}(s\zeta_{m}(s+1,x;y))_{|_{s=0}}=\zeta_{m}(1,x;y).

Hence

2ζmys(0;x,y)=ζm(1,x;y).\frac{\partial^{2}\zeta_{m}}{\partial y\partial s}(0;x,y)=\zeta_{m}(1,x;y).

Therefore,

yexp(sζm(s,x;y)|s=0)=sζm(s,x;y)|s=0ζm(1,x;y).\frac{\partial}{\partial y}\exp\left(-\frac{\partial}{\partial s}\zeta_{m}(s,x;y)_{|_{s=0}}\right)=-\frac{\partial}{\partial s}\zeta_{m}(s,x;y)_{|_{s=0}}\zeta_{m}(1,x;y).

This concludes the proof of the theorem.

4. Some applications

We have

k=1(k4+y)=2y12(cosh(2πy14)cos(2πy14))y>0,\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=1}^{\infty}(k^{4}+y)=2y^{-\frac{1}{2}}(\cosh(\sqrt{2}\pi y^{\frac{1}{4}})-\cos(\sqrt{2}\pi y^{\frac{1}{4}}))\quad\forall y>0,

(see [13, p. 943]) which can be proved using the reflection formula for the Gamma function.

Theorem 4.1.

We have

  1. (1)
    12πk2yk2+y2=coth(πy),\frac{1}{2\pi}\sum_{k\in\mathbb{Z}}\frac{2y}{k^{2}+y^{2}}=\coth(\pi y),
  2. (2)
    1π2k4y3k4+y4=sinh(2πy)+sin(2πy)cosh(2πy)cos(2πy),\frac{1}{\pi\sqrt{2}}\sum_{k\in\mathbb{Z}}^{\infty}\frac{4y^{3}}{k^{4}+y^{4}}=\frac{\sinh(\sqrt{2}\pi y)+\sin(\sqrt{2}\pi y)}{\cosh(\sqrt{2}\pi y)-\cos(\sqrt{2}\pi y)},
  3. (3)

    For any n3n\geq 3,

    nπky2n1k2n+y2n=l=0n1eπilntan(πeπilky),\frac{n}{\pi}\sum_{k\in\mathbb{Z}}\frac{y^{2n-1}}{k^{2n}+y^{2n}}=\sum_{l=0}^{n-1}e^{\frac{\pi il}{n}}\tan(\pi e^{\frac{\pi il}{k}}y),

for any y{0}y\in\mathbb{R}\setminus\{0\}.

Proof.
  1. (1)

    For any y>0y>0, we have

    k=0(k2+y)=2y12sinh(πy12).\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=0}^{\infty}(k^{2}+y)=2y^{\frac{1}{2}}\sinh(\pi y^{\frac{1}{2}}).

    (this is identity is a direct consequence of (3) of Theorem 5.1). So, from (13), we get the following identity

    (14) k=11k2+y=12y+π2y12coth(πy12).\sum_{k=1}^{\infty}\frac{1}{k^{2}+y}=-\frac{1}{2y}+\frac{\pi}{2y^{\frac{1}{2}}}\coth(\pi y^{\frac{1}{2}}).

    That is

    12πk2yk2+y2=coth(πy).\frac{1}{2\pi}\sum_{k\in\mathbb{Z}}^{\infty}\frac{2y}{k^{2}+y^{2}}=\coth(\pi y).
  2. (2)

    From [13], we know that

    (2π)2ζ4=1Γ(ζ1+i2y)=2y2(cosh(2πy)cos(2πy)),\frac{(2\pi)^{2}}{\underset{\zeta^{4}=1}{\prod}\Gamma(-\zeta\frac{1+i}{\sqrt{2}}y)}=2y^{2}(\cosh(\sqrt{2}\pi y)-\cos(\sqrt{2}\pi y)),

    (The authors uses the reflection formula to prove this equation. See Reflection formula for a proof of this formula). Using again (13), we get for any y>0y>0,

    (y12(cosh(2πy14)cos(2πy14)))1ddy(y12(cosh(2πy14)cos(2πy14)))=k=11k4+y.\left(y^{-\frac{1}{2}}\bigl{(}\cosh(\sqrt{2}\pi y^{\frac{1}{4}})-\cos(\sqrt{2}\pi y^{\frac{1}{4}})\bigr{)}\right)^{-1}\frac{d}{dy}\left(y^{-\frac{1}{2}}\bigl{(}\cosh(\sqrt{2}\pi y^{\frac{1}{4}})-\cos(\sqrt{2}\pi y^{\frac{1}{4}})\bigr{)}\right)=\sum_{k=1}^{\infty}\frac{1}{k^{4}+y}.

    That is,

    k=11k4+y=12y+2π4y34(sinh(2πy14)+sin(2πy14)cosh(2πy14)cos(2πy14)).\sum_{k=1}^{\infty}\frac{1}{k^{4}+y}=-\frac{1}{2y}+\frac{\sqrt{2}\pi}{4y^{\frac{3}{4}}}\left(\frac{\sinh(\sqrt{2}\pi y^{\frac{1}{4}})+\sin(\sqrt{2}\pi y^{\frac{1}{4}})}{\cosh(\sqrt{2}\pi y^{\frac{1}{4}})-\cos(\sqrt{2}\pi y^{\frac{1}{4}})}\right).

    Then

    4π2ky3k4+y4=(sinh(2πy)+sin(2πy)cosh(2πy)cos(2πy)).\frac{4}{\pi\sqrt{2}}\sum_{k\in\mathbb{Z}}\frac{y^{3}}{k^{4}+y^{4}}=\left(\frac{\sinh(\sqrt{2}\pi y)+\sin(\sqrt{2}\pi y)}{\cosh(\sqrt{2}\pi y)-\cos(\sqrt{2}\pi y)}\right).
  3. (3)

    Using the reflection formula, we obtain

    (2π)2nξ2n=1Γ(ξz)=(1)k2kzkeπi2n(n1)l=0n1sin(πeπilneπi2kz).\frac{(2\pi)^{2n}}{\prod_{\xi^{2n}=-1}\Gamma(\xi z)}=(-1)^{k}2^{k}z^{k}e^{\frac{\pi i}{2}n(n-1)}\prod_{l=0}^{n-1}\sin(\pi e^{\frac{\pi il}{n}}e^{\frac{\pi i}{2k}}z).

    Then,

    k=1(k2n+z2k)=(1)k2kzkeπi2n(n1)l=0n1sin(πeπilneπi2kz).\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{k=1}^{\infty}(k^{2n}+z^{2k})=(-1)^{k}2^{k}z^{-k}e^{\frac{\pi i}{2}n(n-1)}\prod_{l=0}^{n-1}\sin(\pi e^{\frac{\pi il}{n}}e^{\frac{\pi i}{2k}}z).

    Taking the logarithmic derivative, we get

    neπi2nπkz2n1k2n+z2n=l=0n1eπilncot(πeπilkeπi2nz).\frac{ne^{-\frac{\pi i}{2n}}}{\pi}\sum_{k\in\mathbb{Z}}\frac{z^{2n-1}}{k^{2n}+z^{2n}}=\sum_{l=0}^{n-1}e^{\frac{\pi il}{n}}\cot(\pi e^{\frac{\pi il}{k}}e^{\frac{\pi i}{2n}}z).

    By letting y=eπi2nzy=e^{\frac{\pi i}{2n}}z, we obtain the desired identity.

Corollary 4.2.

For any j=1,2,j=1,2,\ldots,

ζ(2j)=(1)j+122j1(2j)!π2jB2j,\zeta_{\mathbb{Q}}(2j)=\frac{(-1)^{j+1}2^{2j-1}}{(2j)!}\pi^{2j}B_{2j},

where B2jB_{2j} is the 2j2j-th Bernoulli number.

Proof.

Recall that

12y+π2y12coth(πy12)=j=1B2j(2j)!(2π)2jyj,\begin{split}\frac{1}{2y}+\frac{\pi}{2y^{\frac{1}{2}}}\coth(\pi y^{\frac{1}{2}})=\sum_{j=1}^{\infty}\frac{B_{2j}}{(2j)!}(2\pi)^{2j}y^{j},\end{split}

where B2jB_{2j} is the 2j2j-th Bernoulli number. From (14), it is easy to see that

k=11k2j=(1)j+122j1(2j)!π2jB2j.\sum_{k=1}^{\infty}\frac{1}{k^{2j}}=\frac{(-1)^{j+1}2^{2j-1}}{(2j)!}\pi^{2j}B_{2j}.

Corollary 4.3.

We have

  1. (i)
    (15) k1k2+y2=πy(e2πy+1e2πy1),\sum_{k\in\mathbb{Z}}\frac{1}{k^{2}+y^{2}}=\frac{\pi}{y}\left(\frac{e^{2\pi y}+1}{e^{2\pi y}-1}\right),
  2. (ii)
    (Euler) ζ(2k)==112k=(1)k1(2π)2kB2k2(2k)!,\zeta_{\mathbb{Q}}(2k)=\sum_{\ell=1}^{\infty}\frac{1}{\ell^{2k}}=(-1)^{k-1}\frac{(2\pi)^{2k}B_{2k}}{2(2k)!},
  3. (iii)
    k1k4+y4=π24y3(sinh(2πy)+sin(2πy)cosh(2πy)cos(2πy)).\sum_{k\in\mathbb{Z}}^{\infty}\frac{1}{k^{4}+y^{4}}=\frac{\pi\sqrt{2}}{4y^{3}}\left(\frac{\sinh(\sqrt{2}\pi y)+\sin(\sqrt{2}\pi y)}{\cosh(\sqrt{2}\pi y)-\cos(\sqrt{2}\pi y)}\right).
  4. (iv)
    k=01(k+x)m=ζH(m,x)=(1)m1Ψ(m1)(x)(m1)!,\sum_{k=0}^{\infty}\frac{1}{(k+x)^{m}}=\zeta_{H}(m,x)=(-1)^{m-1}\frac{\Psi^{(m-1)}(x)}{(m-1)!},

    where ζH\zeta_{H} is Hurwitz’s zeta function.

5. Appendix : On Lerch’s formula

The reflection formula was first proved by Euler. In order to obtain his formula, Euler has to prove the following product formula for the sine function:

sin(πx)=πxk=1(1x2k2),\sin(\pi x)=\pi x\prod_{k=1}^{\infty}\left(1-\frac{x^{2}}{k^{2}}\right),

which can be deduced from the following identity

πcot(πx)=ne1n+x=1x+2xn=11x2n2x,\pi\cot(\pi x)=\sum_{n\in\mathbb{Z}}^{e}\frac{1}{n+x}=\frac{1}{x}+2x\sum_{n=1}^{\infty}\frac{1}{x^{2}-n^{2}}\quad\forall x\in\mathbb{R}\setminus\mathbb{Z},

where 𝑒\overset{e}{\sum} stands for the Eisenstein summation, see [20] for the proof of this identity. Now,

Γ(x+1)=k=1k1x(k+1)xx+k.\Gamma(x+1)=\prod_{k=1}^{\infty}\frac{k^{1-x}(k+1)^{x}}{x+k}.

Therefore, replacing xx by x-x:

Γ(1x)=k=1k1+x(k+1)xkx.\Gamma(1-x)=\prod_{k=1}^{\infty}\frac{k^{1+x}(k+1)^{-x}}{k-x}.

Therefore,

Γ(1+x)Γ(1x)=k=1k2k2x2=k=111x2k2.\Gamma(1+x)\Gamma(1-x)=\prod_{k=1}^{\infty}\frac{k^{2}}{k^{2}-x^{2}}=\prod_{k=1}^{\infty}\frac{1}{1-\frac{x^{2}}{k^{2}}}.

So,

Γ(x+1)Γ(1x)=πxsin(πx).\Gamma(x+1)\Gamma(1-x)=\frac{\pi x}{\sin(\pi x)}.

In the sequel, we give a different proof of Lerch’s formula and the reflection formula for the Gamma function. We do not claim that our proof is new. Probably this method is known, but we could not find a reference containing this proof.

We set

L(x)=n=0(n+x)forx0.L(x)=\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{n=0}^{\infty}(n+x)\;\text{for}\;x\geq 0.
Theorem 5.1.

The function \ell extends to an analytic function on \mathbb{C} which possesses the following properties:

  1. (1)

    L(x)>0L(x)>0 for any x>0x>0,

  2. (2)

    L(x+1)=1xL(x)L(x+1)=\frac{1}{x}L(x) for any xx\in\mathbb{R},

  3. (3)

    For xx\in\mathbb{C},

    L(x)L(x)=2xsin(πx),L(x)L(-x)=-2x\sin(\pi x),
  4. (4)

    L(1)=n=1n=2πL(1)=\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{n=1}^{\infty}n=\sqrt{2\pi}.

Moreover,

L(x)=(2π)12Γ(x)x,L(x)=\frac{(2\pi)^{\frac{1}{2}}}{\Gamma(x)}\quad\forall x\in\mathbb{C},

and

(Reflection formula) 1Γ(x)Γ(x)=1πxsin(πx)x.\frac{1}{\Gamma(x)\Gamma(-x)}=-\frac{1}{\pi}x\sin(\pi x)\quad\forall x\in\mathbb{C}.
Proof.

Let x>0x>0. ζH(s,x)\zeta_{H}(s,x) has the following integral representation for any Re(s)>1\mathrm{Re}(s)>1,

ζH(s,x)=1Γ(s)01ts2ext𝑑t+1Γ(s)01ts1(11et1t)ext𝑑t+1Γ(s)1ts1ext1et𝑑t\begin{split}\zeta_{H}(s,x)=&\frac{1}{\Gamma(s)}\int_{0}^{1}t^{s-2}e^{-xt}dt+\frac{1}{\Gamma(s)}\int_{0}^{1}t^{s-1}\left(\frac{1}{1-e^{-t}}-\frac{1}{t}\right)e^{-xt}dt\\ &+\frac{1}{\Gamma(s)}\int_{1}^{\infty}t^{s-1}\frac{e^{-xt}}{1-e^{-t}}dt\end{split}

It is clear that the first and the third integrals have a analytic continuation to s=0s=0, and their derivatives with respect to ss at 0 is an analytic function in xx.

11et1t\frac{1}{1-e^{-t}}-\frac{1}{t}

is an analytic function for any tt. We conclude that L(x)L(x) is an analytic function on (0,+)(0,+\infty), and hence on Re(x)>0\mathrm{Re}(x)>0.

Let xx\in\mathbb{C}. Let nx=[Re(x)]+1n_{x}=[-\mathrm{Re}{(x)}]+1. The series

nnx1(n+x)s=n=01(n+nx+x)s,\sum_{n\geq n_{x}}\frac{1}{(n+x)^{s}}=\sum_{n=0}^{\infty}\frac{1}{(n+n_{x}+x)^{s}},

is well defined. By definition,

L(nx+x)=n=0(n+nx+x)L(n_{x}+x)=\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{n=0}^{\infty}(n+n_{x}+x)

We set

n=0(n+x)=L(nx+x)n=0nx1(n+x).\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{n=0}^{\infty}(n+x)=L(n_{x}+x)\prod_{n=0}^{n_{x}-1}(n+x).

So, \ell can be continued analytically to \mathbb{C}.

Note that this result is a special case of theory developed in [19].

  1. (1)

    By definition, we have for any x>0x>0, L(x)=exp(sζH(s,x)|s=0)L(x)=\exp\left(-\frac{\partial}{\partial s}\zeta_{H}(s,x)_{|_{s=0}}\right). It is clear that ζH(s,x)\zeta_{H}(s,x)\in\mathbb{R} when Re(s)>1\mathrm{Re}(s)>1, and its meromorphic continuation is a real-valued function on \mathbb{R}. So,

    sζH(s,x)|s=0 for any x>0,{\frac{\partial}{\partial s}\zeta_{H}(s,x)_{|_{s=0}}}\in\mathbb{R}\;\text{ for any $x>0$},

    and hence

    L(x)>0,x>0.L(x)>0,\quad\forall x>0.
  2. (2)

    Since,

    ζH(s,x+1)=ζH(s,x)1xs,x>0,\zeta_{H}(s,x+1)=\zeta_{H}(s,x)-\frac{1}{x^{s}},\quad\forall x>0,

    Then,

    L(x+1)=1xL(x)x>0.L(x+1)=\frac{1}{x}L(x)\quad\forall x>0.

    By the uniqueness of the analytic continuation, the above identity holds on \mathbb{C}.

    We know that

    L(x)L(x)=n=0(n2x2).L(x)L(-x)=\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{n=0}^{\infty}(n^{2}-x^{2}).
  3. (3)

    By Bohr-Mollerup theorem (see [2, Theorem 2.1 p.14]),

    L(x)=n=1nΓ(x)x,L(x)=\frac{\prod_{n=1}^{\infty}n}{\Gamma(x)}\quad\forall x\in\mathbb{C},

    (we recall that L(1)=n=1nL(1)=\prod_{n=1}^{\infty}n).

    According to [19], L(z)L(z) is a holomorphic function of the complex variable zz whose zeros (counted with multiplicity) are the numbers 1,2,-1,-2,\ldots, which is by bounded by exp(a+b|z|N)\exp(a+b|z|^{N}) for some constants a,ba,b and NN (i.e. L(z)L(z) is a function of finite order, see [18, p. 138]). The sine function is an entire function of finite order.

    We can conclude that the following function

    f(z)=zsin(πz)L(z)L(z),f(z)=\frac{z\sin(\pi z)}{L(z)L(-z)},

    is entire of finite order without zeros on \mathbb{C}. By Hadamard’s factorization theorem, there exists a polynomial PP such that

    f(z)=eP(z).f(z)=e^{P(z)}.

    We can verify easily that f(z+1)=f(z)f(z+1)=f(z) for any zz\in\mathbb{C}, and since PP is continuous, then P(z+1)P(z)P(z+1)-P(z) is the constant polynomial. It follows that P(z)=αz+βP(z)=\alpha z+\beta for some α,β\alpha,\beta\in\mathbb{C}. Taking the logarithmic derivative of ff, we get

    a=f(z)f(z)=f(0)f(0)z.a=\frac{f^{\prime}(z)}{f(z)}=\frac{f^{\prime}(0)}{f(0)}\quad\forall z\in\mathbb{C}.

    But f(z)=f(z)f(z)=f(-z) for any zz\in\mathbb{C}. Then, a=0a=0, and hence f(z)=f(0)f(z)=f(0) for any z.z\in\mathbb{C}. That is,

    zsin(πz)L(z)L(z)=πn=1n2z.\frac{z\sin(\pi z)}{L(z)L(-z)}=-\frac{\pi}{\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{n=1}^{\infty}n^{2}}\quad\forall z\in\mathbb{C}.

    Recall that L(z)=(n=1n)/Γ(z)L(z)=(\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{n=1}^{\infty}n)/\Gamma(z). Then, we obtain the reflection formula for the Gamma function:

    (16) zsin(πz)=πΓ(z)Γ(z)z.-z\sin(\pi z)=\frac{\pi}{\Gamma(z)\Gamma(-z)}\quad\forall z.

    (we have used that n=1n2=(n=1n)2\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{n=1}^{\infty}n^{2}=(\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{n=1}^{\infty}n)^{2}).


Let us evaluate n=1n\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{n=1}^{\infty}n. We have

11xs𝑑x=01ζH(s,x+1)𝑑x,Re(s)>1.\int_{1}^{\infty}\frac{1}{x^{s}}dx=\int_{0}^{1}\zeta_{H}(s,x+1)dx,\quad\forall\ \mathrm{Re}(s)>1.

By a meromorphic continuation, and elementary computations, we obtain,

1=01sζH(0,x+1)𝑑x=log(n=1n)1+01logΓ(x)𝑑x.-1=\int_{0}^{1}\frac{\partial}{\partial s}\zeta_{H}(0,x+1)dx=-\log(\prod_{n=1}^{\infty}n)-1+\int_{0}^{1}\log\Gamma(x)dx.

That is,

01logΓ(x)𝑑x=log(n=1n).\int_{0}^{1}\log\Gamma(x)dx=\log(\prod_{n=1}^{\infty}n).

The computation of 01logΓ(x)𝑑x\int_{0}^{1}\log\Gamma(x)dx is well known, and uses 16. For reader’s convenience, we recall the proof here. We have

01logΓ(x)𝑑x=01logΓ(1x)𝑑x=logπ01logΓ(x)𝑑x01logsin(πx)dx(by 16).\begin{split}\int_{0}^{1}\log\Gamma(x)dx=&\int_{0}^{1}\log\Gamma(1-x)dx\\ =&\log\pi-\int_{0}^{1}\log\Gamma(x)dx-\int_{0}^{1}\log\sin(\pi x)dx\quad\text{(by \ref{reflection2})}.\end{split}

Using an elementary change of variables, 01logsin(πx)dx=log2-\int_{0}^{1}\log\sin(\pi x)dx=-\log 2. We conclude that

01logΓ(x)𝑑x=12log(2π).\int_{0}^{1}\log\Gamma(x)dx=\frac{1}{2}\log(2\pi).

Then,

n=1n=(2π)12.\operatorname*{\mathchoice{\ooalign{$\displaystyle\prod$\cr$\displaystyle\coprod$\cr}}{\ooalign{$\textstyle\prod$\cr$\textstyle\coprod$\cr}}{\ooalign{$\scriptstyle\prod$\cr$\scriptstyle\coprod$\cr}}{\ooalign{$\scriptscriptstyle\prod$\cr$\scriptscriptstyle\coprod$\cr}}}_{n=1}^{\infty}n=(2\pi)^{\frac{1}{2}}.

We conclude that

L(z)L(z)=2zsin(πz)z.L(z)L(-z)=2z\sin(\pi z)\quad\forall z\in\mathbb{C}.

This ends the proof of the theorem.

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