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I've been playing with lots of Poisson sums lately, and I thought this one to be interesting: $$\sum_{k\in\mathbb{Z}}\left(\frac{1}{(k+x)\sinh{[(k+z)\pi q}]}-\frac{1}{\pi q (k+z)^2}\right)$$I want to find a closed form for this sum and its derivatives over $x$ when $x=0$ and $q=1$. Since its poles are of the form $k+\frac{in}{q}\,(k,n\text { integers})$ with double-order poles at the integers, I figure its expression may include trigonometric and theta functions...but I can't figure anything beyond its singularities. Any help would be appreciated.

I've managed to turn the sum into a Fourier series $\left(-4\sum_{k\ge 1}\ln(1+e^{-2k\pi/q})\cos{2k\pi z}-2\ln2\right)$, but even with its simplicity, I haven't been able to crack it.

(Edit) I think I have a way to evaluate the Fourier series: if I expand the cosines into Taylor series, then I just have to sum series of the form $$\sum_{k \ge 1}k^{2n}\ln(1+e^{-2k\pi/q})$$ which I can rewrite as $$\sum_{m\ge 1}\frac{(-1)^{m-1}}m\sum_{k\ge 1}k^{2n}e^{-2km\pi/q}$$and since $\displaystyle{\sum_{k\ge 1}e^{-2km\pi/q}=\frac1{e^{2m\pi/q}-1}}$, $\displaystyle{\sum_{k\ge 1}k^2e^{-2km\pi/q}=\frac14\frac{\cosh{\frac{m\pi}q}}{\sinh^3{\frac{m\pi}q}}}$ and subsequent sums consist of a hyperbolic cosine times an odd reciprocal polynomial in the hyperbolic sine, I've reduced my problem to evaluating sums of the form $$\sum_{m \ge 1}\frac{(-1)^{m-1}}m \frac{\cosh{\frac{m\pi}q}}{\sinh^{2n+1}{\frac{m\pi}q}}$$which is also $\displaystyle{\int{\sum_{k\ge 1}\frac{(-1)^{k-1}\sinh{kz}}{\sinh^{2n+1}{\frac{k\pi}q}}\, dz}}$ with $z=\frac{i\pi}q$. But here I am stuck.

I've found something promising: four theta functions I've been studying lately, \begin{align}\vartheta(z,q)&=q^{-1/2}\prod_{k\ge1}(1-e^{-2k\pi/q})(1+e^{-(2k-1)\pi/q+2i\pi z})(1+e^{-(2k-1)\pi/q-2i\pi z})\\ \theta(z,q)&=q^{-1/2}\prod_{k\ge1}(1-e^{-2k\pi/q})(1-e^{-(2k-1)\pi/q+2i\pi z})(1-e^{-(2k-1)\pi/q-2i\pi z})\\ \varphi(z,q)&=2q^{-1/2}e^{-\pi/(4q)}\sin\pi z\prod_{k\ge1}(1-e^{-2k\pi/q})(1-e^{-2k\pi/q+2i\pi z})(1-e^{-2k\pi/q-2i\pi z})\\ \phi(z,q)&=2q^{-1/2}e^{-\pi/(4q)}\cos\pi z\prod_{k\ge1}(1-e^{-2k\pi/q})(1+e^{-2k\pi/q+2i\pi z})(1+e^{-2k\pi/q-2i\pi z}) \end{align} satisfy useful quasi-periodic identities: \begin{align}\vartheta(z+k+in/q,q)&=e^{n^2\pi/q-2in\pi z}\vartheta(z,q)\\[1ex] \theta(z+k+in/q,q)&=(-1)^ne^{n^2\pi/q-2in\pi z}\theta(z,q)\\[1ex] \varphi(z+k+in/q,q)&=(-1)^{k+n}e^{n^2\pi/q-2in\pi z}\varphi(z,q)\\[1ex] \phi(z+k+in/q,q)&=(-1)^ke^{n^2\pi/q-2in\pi z}\phi(z,q) \end{align}thus with $[\vartheta',\theta',\varphi',\phi'](z,q)=\partial_z[\vartheta,\theta,\varphi,\phi](z,q)$, \begin{align}\vartheta'(k+in/q,q)&=-2in\pi e^{n^2\pi/q}\vartheta(0,q)\\[1ex] \theta'(k+in/q,q)&=-2in\pi(-1)^ne^{n^2\pi/q}\theta(0,q)\\[1ex] \varphi'(k+in/q,q)&=(-1)^{k+n}e^{n^2\pi/q}\varphi'(0,q)\\[1ex] \varphi''(k+in/q,q)&=(-1)^{k+n}\cdot-4in\pi e^{n^2\pi/q}\varphi'(0,q)\\[1ex] \phi'(k+in/q,q)&=-2in\pi(-1)^ke^{n^2\pi/q}\phi(0,q). \end{align} And my sum has poles at $x=-k+in/q$ with residues equal to $$\frac1{-in/q\cdot\pi q\cosh(-in\pi)}=\frac{i(-1)^n}{n\pi} (n\neq0),$$so it could feature $1/\varphi(z,q)$...and while I'd be tempted to guess the remaining poles would be covered by $\sin\pi z$, that won't produce the right denominator—but $\phi'(z,q)$ can! Specifically, I'll get $-2in\pi (-1)^ne^{2n^2\pi/q}(\varphi'\cdot\phi)(0,q)$ as the denominator, but I can cancel out the exponential by adding $\vartheta^2(z,q)$ to the numerator—leaving$$\frac{(-1)^{-n}\vartheta^2(0,q)}{-2in\pi\varphi'\phi(0,q)}=\frac i{2\pi}\frac{(-1)^n}n\frac{\vartheta^2}{\varphi'\phi}(0,q)$$ as my residue...

So my sum now may have the form$$2\frac{\varphi'\phi}{\vartheta^2}(0,q)\frac{\vartheta^2}{\varphi\phi'}(z,q)$$but I don't know if $\phi'(z,q)$ has any more zeros; and there's an additional series of double poles at the integers, but that's easy to resolve:$$\lim_{z\to n}2\frac{\varphi'\phi}{\vartheta^2}(0,q)(z-n)^2\frac{\vartheta^2}{\varphi\phi'}(z,q)=2\frac{\varphi'\phi}{\vartheta^2}(0,q)\frac{\vartheta^2}{\varphi'\phi''}(0,q)=2\frac\phi{\phi''}(0,q)$$I calculated the residues at the integers and got $0$ for all of them, so now I have \begin{align}2\frac{\varphi'\phi}{\vartheta^2}(0,q)\frac{\vartheta^2}{\varphi\phi'}(z,q)&=2\pi^2\frac\phi{\phi''}(0,q)\csc^2\pi z\\ &\quad+2\frac{\varphi'\phi}{\vartheta^2}(0,q)\sum_{k\in\mathbb Z\\n\ge1}\left(\frac1{z-k-in/q}\frac{\vartheta^2}{\varphi'\phi'}(k+in/q,q)+\frac1{z-k+in/q}\frac{\vartheta^2}{\varphi'\phi'}(k-in/q,q)\right)\\ &=2\pi^2\frac\phi{\phi''}(0,q)\csc^2\pi z\\ &+2\frac{\varphi'\phi}{\vartheta^2}(0,q)\sum_{k\in\mathbb Z\\n\ge1}\left(\frac1{z-k-in/q}\frac1{-2in\pi(-1)^n}\frac{\vartheta^2}{\varphi'\phi}(0,q)\\ \qquad+\frac1{z-k+in/q}\frac1{2in\pi(-1)^{-n}}\frac{\vartheta^2}{\varphi'\phi}(0,q)\right)\\ &=2\pi^2\frac\phi{\phi''}(0,q)\csc^2\pi z\\ &\quad+\frac i\pi\sum_{k\in\mathbb Z\\n\ge1}\frac{(-1)^n}n\frac{2in/q}{(z-k)^2+n^2/q^2}\\ &=2\pi^2\frac\phi{\phi''}(0,q)\csc^2\pi z\\ &\quad-\frac2{\pi q}\sum_{k\in\mathbb Z}\left(\frac{\pi q}{2(z-k)}\operatorname{csch}[\pi q(z-k)]-\frac1{2(z-k)^2}\right)\\ &=2\pi^2\frac\phi{\phi''}(0,q)\csc^2\pi z+\frac\pi q\csc^2\pi z-\sum_{k\in\mathbb Z}\frac1{(z+k)\sinh[\pi q(z+k)]}\end{align} There'll have to be an extra entire function (in $z$) in the final formulation to correct my result, since it fails at $z=1/2$: $$2\frac{\varphi'\phi}{\vartheta^2}(0,q)\frac{\vartheta^2}{\varphi'\phi}(1/2,q)\neq\frac\pi q+2\pi^2\frac\phi{\phi''}(0,q)-2\ln\frac\vartheta\phi(0,q/2)$$but I know that that extra function must be $1$-periodic and even...

Antonio DJC
  • 674

1 Answers1

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I've found an integral representation for my series: With $$S(z,a;q):=\frac{\pi q\varphi(a,q)}{\varphi'(0,q)}\sum_{k\in\mathbb Z}\frac{e^{-2(a+k)\pi qz}}{\sinh(a+k)\pi q},$$ \begin{align}\int_0^{i/q}S(z,a;q)\,dz&=\frac{\pi q\varphi(a,q)}{\varphi'(0,q)}\int_0^{i/q}\sum_{k\in\mathbb Z}\frac{e^{-2(a+k)\pi qz}}{\sinh(a+k)\pi q}\,dz\\[2ex] &=\frac{\pi q\varphi(a,q)}{\varphi'(0,q)}\sum_{k\in\mathbb Z}\frac1{\sinh(a+k)\pi q}\frac{e^{-2(a+k)i\pi}-1}{-2(a+k)\pi q}\\[2ex] &=\frac{\varphi(a,q)}{2\varphi'(0,q)}\sum_{k\in\mathbb Z}\frac{e^{-2ai\pi-2ki\pi}-1}{-(a+k)\sinh(a+k)\pi q}\\[2ex] &=\frac{\varphi(a,q)}{2\varphi'(0,q)}(1-e^{-2ai\pi})\sum_{k\in\mathbb Z}\frac1{(a+k)\sinh(a+k)\pi q}\tag1. \end{align}And $S(z,a;q)$ just happens to be the quotient of two theta functions (see proof in the appendix below): $S=\phi(z+a,q)/\phi(z,q)$, where \begin{align}\phi(z,q):&=\sum_{k\in\mathbb Z}(-1)^ke^{-\pi q(k+z)^2}\\[2ex] &=e^{-\pi qz^2}\prod_{k\ge1}(1-e^{-2k\pi q})(1-e^{-(2k-1)\pi q+2\pi qz})(1-e^{-(2k-1)\pi q-2\pi qz})\tag2\\[3ex] \phi(z,q)&=\varphi(z\!+\!1/2,q) \end{align} The identity $\phi(z+i/q,q)=e^{\pi/q-2i\pi z}\phi(z,q)$ (which can be derived directly from $(2)$) leads to $S(z+i/q,a;q)=e^{-2ai\pi}S(z,a;q)$, implying double periodicity for $a\in\mathbb Q$. Thus the expression $(1)$, in principle, will have a closed form in terms of log-theta functions when $a$ is rational.

Proof that $S(z,a;q)$ is a quotient of theta functions

To prove $S$ has such a representation, create a quotient product with $(2)$:\begin{align}\frac{\phi(z+a,q)}{\phi(z,q)}&=e^{-\pi q(z+a)^2+\pi qz^2}\prod_{k\ge1}\bigg(\frac{1-e^{-(2k-1)\pi q+2\pi q(z+a)}}{1-e^{-(2k-1)\pi q+2\pi qz}}\bigg)\bigg(\frac{1-e^{-(2k-1)\pi q-2\pi q(z+a)}}{1-e^{-(2k-1)\pi q-2\pi qz}}\bigg)\\[3ex] &=e^{-2a\pi qz-a^2\pi q}\prod_{k\ge1}\bigg(\frac{1-e^{-(2k-1)\pi q+2\pi qz}\cdot e^{2a\pi q}}{1-e^{-(2k-1)\pi q+2\pi qz}}\bigg)\bigg(\frac{1-e^{-(2k-1)\pi q-2\pi qz}\cdot e^{-2a\pi q}}{1-e^{-(2k-1)\pi q-2\pi qz}}\bigg)\\[3ex] &=e^{-2a\pi qz-a^2\pi q}\prod_{k\ge1}\bigg(e^{2a\pi q}+\frac{1-e^{2a\pi q}}{1-e^{-(2k-1)\pi q+2\pi qz}}\bigg)\bigg(e^{-2a\pi q}+\frac{1-e^{-2a\pi q}}{1-e^{-(2k-1)\pi q-2\pi qz}}\bigg) \end{align}Expanding the product will produce a constant (w.r.t. $z$) and terms with denominators $(1-e^{-(2k-1)\pi q+2\pi qz})(1-e^{-(2n-1)\pi q-2\pi qz})$, and using partial fraction decomposition,\begin{align}\frac1{(1-e^{-(2k-1)\pi q+2\pi qz})(1-e^{-(2n-1)\pi q-2\pi qz})}&=\frac{e^{(2n-1)\pi q+2\pi qz}}{(1-e^{-(2k-1)\pi q+2\pi qz})(e^{(2n-1)\pi q+2\pi qz}-1)}\\[4ex] &=\frac1{1-e^{-(2k-1)\pi q+2\pi qz}}+\frac1{(1-e^{-(2k-1)\pi q+2\pi qz})(e^{(2n-1)\pi q+2\pi qz}-1)}\\[3ex] &=\frac1{1-e^{-(2k-1)\pi q+2\pi qz}}\\ &\quad+\frac1{e^{-(2n-1)\pi q}-e^{(2k-1)\pi q}}\bigg(\frac{e^{-(2n-1)\pi q}}{1-e^{-(2k-1)\pi q+2\pi qz}}-\frac{e^{(2k-1)\pi q}}{1-e^{(2n-1)\pi q+2\pi qz}}\bigg)\\[2ex] &\cong A+\frac B{1-e^{-(2k-1)\pi q+2\pi qz}}+\frac C{1-e^{-(2n-1)\pi q-2\pi qz}}; \end{align}thus the quotient function can be written as \begin{align}&e^{-2a\pi qz-a^2\pi q}\Bigg[D(a,q)+\sum_{k\ge1}\bigg(\frac{E_k}{1-e^{-(2k-1)\pi q+2\pi qz}}+\frac{F_k}{1-e^{-(2k-1)\pi q-2\pi qz}}\bigg)\Bigg]\\ &=e^{-2a\pi qz-a^2\pi q}\Bigg[D(a,q)+\sum_{k\ge1}\bigg(E_k+\frac{E_k}{e^{(2k-1)\pi q-2\pi qz}-1}+F_k+\frac{F_k}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg)\Bigg]\\ &=e^{-2a\pi qz-a^2\pi q}\Bigg[D_1(a,q)+\sum_{k\ge1}\bigg(\frac{E_k}{e^{(2k-1)\pi q-2\pi qz}-1}+\frac{F_k}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg)\Bigg]. \end{align}Now I need to consult a functional identity for $\phi(z,q)$:$$\phi(z+1,q)=-\phi(z,q)$$(it can be derived from $\phi$'s series definition).

So \begin{align}\frac{\phi(z+1+a,q)}{\phi(z+1,q)}&=e^{-2a\pi q(z+1)-a^2\pi q}\Bigg[D_1(a,q)+\sum_{k\ge1}\bigg(\frac{E_k}{e^{(2k-1)\pi q-2\pi q(z+1)}-1}+\frac{F_k}{e^{(2k-1)\pi q+2\pi q(z+1)}-1}\bigg)\Bigg]\\[2ex] &=e^{-2a\pi qz-a^2\pi q-2a\pi q}\Bigg[D_1(a,q)+\sum_{k\ge1}\bigg(\frac{E_k}{e^{(2k-3)\pi q-2\pi qz}-1}+\frac{F_k}{e^{(2k+1)\pi q+2\pi qz}-1}\bigg)\Bigg]\\[2ex] &=e^{-2a\pi qz-a^2\pi q-2a\pi q}\Bigg[D_1(a,q)+\frac{E_1}{e^{-\pi q-2\pi qz}-1}+\frac{E_2}{e^{\pi q-2\pi qz}-1}\\ &\quad+\sum_{k\ge3}\bigg(\frac{E_k}{e^{(2k-3)\pi q-2\pi qz}-1}+\frac{F_{k-1}}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg)\Bigg]\\[2ex] &=e^{-2a\pi qz-a^2\pi q-2a\pi q}\Bigg[D_1(a,q)-\frac{E_1}{e^{\pi q+2\pi qz}-1}-E_1+\frac{E_2}{e^{\pi q-2\pi qz}-1}\\ &\quad+\sum_{k\ge2}\bigg(\frac{E_{k+1}}{e^{(2k-1)\pi q-2\pi qz}-1}+\frac{F_{k-1}}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg)\Bigg] \end{align} and since $\frac{\phi(z+1+a,q)}{\phi(z+1,q)}=\frac{\phi(z+a,q)}{\phi(z,q)}$, you can equate terms with like denominators:$$D_1(a,q)=e^{-2a\pi q}(D_1(a,q)-E_1)$$ $$E_1=e^{-2a\pi q}E_2$$ $$F_1=-e^{-2a\pi q}E_1$$ $$E_{k\ge2}=e^{-2a\pi q}E_{k+1}$$ $$F_{k\ge2}=e^{-2a\pi q}F_{k-1}$$ So now\begin{align}e^{a^2\pi q+2a\pi qz}\frac{\phi(z+a,q)}{\phi(z,q)}&=\frac{E_1}{1-e^{2a\pi q}}+\sum_{k\ge1}\bigg(\frac{e^{(2k-2)a\pi q}E_1}{e^{(2k-1)\pi q-2\pi qz}-1}+\frac{e^{-(2k-2)a\pi q}\cdot-e^{-2a\pi q}E_1}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg)\\ &=\frac{E_1}{1-e^{2a\pi q}}+E_1\sum_{k\ge1}\bigg(\frac{e^{(2k-2)a\pi q}}{e^{(2k-1)\pi q-2\pi qz}-1}-\frac{e^{-2ak\pi q}}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg) \end{align}and $E_1$ requires only the calculation of one residue: \begin{align}\lim_{z\to1/2}e^{a^2\pi q+2a\pi qz}(e^{\pi q-2\pi qz}-1)\frac{\phi(z+a,q)}{\phi(z,q)}&=e^{a^2\pi q+a\pi q}\cdot-\varphi(a,q)\frac{-2\pi qe^{\pi q-\pi q}}{\phi'(1/2,q)}\\ &=-2\pi qe^{a^2\pi q+a\pi q}\frac{\varphi(a,q)}{\varphi'(0,q)} \end{align}After some minor simplifications, we have$$e^{2a\pi qz}\frac{\phi(z+a,q)}{\phi(z,q)}=\frac{\pi q\varphi(a,q)}{\varphi'(0,q)}\Bigg[\operatorname{csch}a\pi q-2\sum_{k\ge1}\bigg(\frac{e^{(2k-1)a\pi q}}{e^{(2k-1)\pi q-2\pi qz}-1}-\frac{e^{(1-2k)a\pi q}}{e^{(2k-1)\pi q+2\pi qz}-1}\bigg)\Bigg]$$ Now I'm going to write the summand as geometric series: I get$$e^{(2k-1)a\pi q}\sum_{n\ge1}e^{-(2k-1)n\pi q+2n\pi qz}-e^{-(2k-1)a\pi q}\sum_{n\ge1}e^{-(2k-1)n\pi q-2n\pi qz}$$Then I swap the indices and sum over $k$:\begin{align}e^{2a\pi qz}\frac{\phi(z+a,q)}{\phi(z,q)}&=\frac{\pi q\varphi(a,q)}{\varphi'(0,q)}\Bigg[\operatorname{csch}a\pi q\\ &\quad-2\sum_{n\ge1}\bigg(e^{(-a+n)\pi q+2n\pi qz}\frac{e^{2\pi q(a-n)}}{1-e^{2\pi q(a-n)}}-e^{(a+n)\pi q-2n\pi qz}\frac{e^{2\pi q(-a-n)}}{1-e^{2\pi q(-a-n)}}\bigg)\Bigg]\\[3ex] &=\frac{\pi q\varphi(a,q)}{\varphi'(0,q)}\Bigg[\operatorname{csch}a\pi q+\sum_{n\ge1}\bigg(\frac{e^{2n\pi qz}}{\sinh \pi q(a-n)}+\frac{e^{-2n\pi qz}}{\sinh\pi q(a+n)}\bigg)\Bigg]\tag{$\square$} \end{align}

Antonio DJC
  • 674