5
$\begingroup$

I've been playing with lots of Poisson sums lately, and I thought this one to be interesting: $$\sum_{k\in\mathbb{Z}}\left(\frac{1}{(k+x)\sinh{(k+x)\pi q}}-\frac{1}{\pi q (k+x)^2}\right)$$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 $\left(-4\sum_{k\ge 1}\ln(1+e^{-2k\pi / q})\cos{2k\pi x}\right)\quad\;$, 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.

$\endgroup$
0
$\begingroup$

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}

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.