Sum of infinite series of Sech: $\sum_n \text{sech}(x(n-1/2))\text{sech}(x(n+p-1/2))$

Question

I am wondering if anyone recognises the sum

$\sum_{n=-\infty}^\infty \frac{1}{\cosh\left(x \; (n+p-\frac{1}{2}) \right) \cosh \left(x\; (n-\frac{1}{2}) \right) }$ ?

I am trying to evaluate sums of this (and related) forms, but I appear to be stuck...

(The full form I eventually want to find is $ \sum_{n , \vec{p} \; \in \;\mathbb{Z} \;} \; \frac{ \prod_{i} p_i ^{i-1} \; \prod_{i< j } \; \; [p_j - p_i] \; \; }{ \prod_{i} 2 \cosh \left[ x \; ( p_i - \frac{1}{2} )\right] \; 2\cosh \left[ x \; ( p_i+ n - \frac{1}{2} )\right] }\; $, but any help would be highly appreciated! )

I have been looking through the papers by Zucker -78 (http://epubs.siam.org/doi/abs/10.1137/0510019) and related works, (such as for example this paper by Bruckman, http://www.fq.math.ca/Scanned/15-4/bruckman.pdf), and know how to evaluate sums of the form $\sum_{n=-\infty}^\infty \frac{1}{ \cosh^2 \left(x\; (n-\frac{1}{2}) \right) }$, but I run into problems as soon as one of the arguments is shifted.

The answer to the sum above is given in terms of complete elliptic integrals, and I suspect that that will be the case for the shifted one as well, assuming there is a closed form.

And, on another note, this is my first post here so hope I haven't made any massive mistakes :)

Thanks a lot!

Answer 1

It is essentially a discrete convolution between $\frac{1}{\cosh}$ and $\frac{1}{\cosh}$. It is interesting to point out that $\frac{1}{\cosh}$ is more or less a fixed point of the Fourier transform: $$ \mathscr{F}\left(\frac{1}{\cosh x}\right)=\frac{\pi}{\cosh\frac{\pi s}{2}},$$ and its inverse Laplace transform is given by $$ \mathscr{L}^{-1}\left(\frac{1}{\cosh x}\right) = 2\sum_{n\geq 0}(-1)^n \delta(s-2n-1)=f(s),$$ so: $$ \frac{1}{\cosh\left(x\left(\frac{1}{2}-n\right)\right)}=\int_{0}^{+\infty}f(s) e^{-sx(n-1/2)}\,ds$$ and: $$ \frac{1}{\cosh\left(x\left(n-\frac{1}{2}\right)\right)\cosh\left(x\left(n+p-\frac{1}{2}\right)\right)}=\int_{0}^{+\infty}\int_{0}^{+\infty} f(s)\,f(t)\,e^{-ptx}\,e^{-(s+t)x(n+1/2)}\,ds\,dt$$ so: $$ \sum_{n\geq 0}\frac{1}{\cosh\left(x\left(n-\frac{1}{2}\right)\right)\cosh\left(x\left(n+p-\frac{1}{2}\right)\right)}=\int_{0}^{+\infty}\int_{0}^{+\infty}\frac{ 2\, f(s)\,f(t)\,e^{-ptx}}{\sinh\left((s+t)\frac{x}{2}\right)}\,ds\,dt$$ equals:

$$ \sum_{a\geq 0}\sum_{b\geq 0}\frac{8(-1)^{a+b} e^{-px(2b+1)}}{\sinh((a+b+1)x)}=e^{-px}\sum_{h=0}^{+\infty}\frac{8(-1)^h}{\sinh((h+1)x)}\cdot\frac{1-e^{-2(h+1)px}}{1-e^{-2px}} \tag{1}$$

that is a pretty fast convergent series.

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Addendum. We may notice that the degenerate case $p=0$ is way easier to deal with.

Given the Weierstrass product for the $\cosh$ function, we have: $$ \frac{1}{\cosh^2(x)}=\frac{d^2}{dx^2}\log\cosh(x) = -\sum_{m\geq 0}\left(\frac{1}{\left(x+\pi i\frac{2m+1}{2}\right)^2}+\frac{1}{\left(x-\pi i\frac{2m+1}{2}\right)^2}\right)$$ so we may recognize in the RHS a meromorphic function with equally-spaced and equally-behaving double poles along the imaginary axis. It follows that the meromorphic function defined through $$ f(x)=\sum_{n\in\mathbb{Z}}\frac{1}{\cosh^2(x+n)} $$ is a constant multiple of a Weierstrass $\wp$ function, whose connection with elliptic integrals is well-known. It is interesting to point out that the RHS of $(1)$, as $p\to 0$, gives a fast-converging expansion for a Weierstrass $\wp$ function. On the other hand, the meromorphic function $$ g(x)=\sum_{n\in\mathbb{Z}}\frac{1}{\cosh(x+n)\cosh(x+n+t)} $$ has only simple poles, due to: $$\frac{1}{\cosh x}=\sum_{n\geq 0}(-1)^n\left(\frac{i}{x+(2m+1)\frac{\pi i}{2}}-\frac{i}{x-(2m+1)\frac{\pi i}{2}}\right)$$ so $g(x)$ is related with a Jacobi elliptic function; a useful identity (already exploited in a slightly disguised form through the inverse Laplace transform) is:

$$ \frac{1}{\cosh x} = \frac{2}{\pi}\int_{0}^{+\infty}\frac{\cos\left(\frac{2x}{\pi}u\right)}{\cosh u}\,du\tag{2} $$

that makes us aware of the following fact: a discrete convolution of $\frac{1}{\cosh}$ is the integral over $\mathbb{R}^+$ of $\frac{1}{\cosh u}$ times a discrete convolution of $\cos$, way easier to deal with. It turns out that the RHS of $(1)$ is related with $\frac{1}{\cosh(px)}$, too.

Answer 2

I can evaluate your first sum in terms of the log-derivative of the theta function$$\vartheta(p,x)=e^{-p^2x}\prod_{k\ge1}(1+e^{2px-(2k-1)x})(1+e^{-2px-(2k-1)x}).\tag1$$ First, split the sum between positive and negative $n$:\begin{align}\sum_{n\in\mathbb{Z}}\frac1{\cosh[(n-\tfrac12\!+p)x]\cosh[(n-\tfrac12)x]}&=\Bigg[\sum_{n\le0}+\sum_{n\ge1}\Bigg]\frac1{\cosh[(n-\tfrac12\!+p)x]\cosh[(n-\tfrac12)x]}\\[2ex] &=\sum_{n\ge1}\frac1{\cosh[(n-\tfrac12\!+p)x]\cosh[(n-\tfrac12)x]}\\[1ex] &\quad+\sum_{n\ge1}\frac1{\cosh[(-n+\tfrac12\!+p)x]\cosh[(-n+\tfrac12)x]}\\[2ex] &=\sum_{n\ge1}\frac1{\cosh[(n-\tfrac12\!+p)x]\cosh[(n-\tfrac12)x]}\\ &\quad+\sum_{n\ge1}\frac1{\cosh[(n-\tfrac12\!-p)x]\cosh[(n-\tfrac12)x]} \end{align} Then, rewrite each $\cosh$ as its exponential form.... $$\sum_{n\ge1}\frac1{\cosh[(n-\tfrac12\!+p)x]\cosh[(n-\tfrac12)x]}+\sum_{n\ge1}\frac1{\cosh[(n-\tfrac12\!-p)x]\cosh[(n-\tfrac12)x]} $$ \begin{align}&=4\sum_{n\ge1}\frac1{(e^{(n-1/2+p)x}+e^{-(n-1/2+p)x})(e^{(n-1/2)x}+e^{-(n-1/2)x})}\\[1ex] &\quad\;+4\sum_{n\ge1}\frac1{(e^{(n-1/2-p)x}+e^{-(n-1/2-p)x})(e^{(n-1/2)x}+e^{-(n-1/2)x})}\\[2ex] &=4\sum_{n\ge1}\frac{e^{(2n-1)x+px}}{(e^{(2n-1+2p)x}+1)(e^{(2n-1)x}+1)}+4\sum_{n\ge1}\frac{e^{(2n-1)x-px}}{(e^{(2n-1-2p)x}+1)(e^{(2n-1)x}+1)}\\[3ex] &=4\sum_{n\ge1}\frac{e^{(2n-1)x}}{e^{(2n-1)x}+1}\bigg(\frac{e^{px}}{e^{(2n-1)x+2px}+1}+\frac{e^{-px}}{e^{(2n-1)x-2px}+1}\bigg)\\[4ex] &=4\sum_{n\ge1}\frac{e^{px}}{e^{2px}-1}\bigg(\frac1{e^{(2n-1)x}+1}-\frac1{e^{(2n-1)x+2px}+1}\bigg)\\[1ex] &\quad+4\sum_{n\ge1}\frac{e^{-px}}{e^{-2px}-1}\bigg(\frac1{e^{(2n-1)x}+1}-\frac1{e^{(2n-1)x-2px}+1}\bigg)\\[4ex] &=2\operatorname{csch}px\cdot-\sum_{n\ge1}\frac1{e^{(2n-1)x+2px}+1}+2\operatorname{csch}px\sum_{n\ge1}\frac1{e^{(2n-1)x-2px}+1}\tag{*} \end{align} Now take the $p$-log-derivative of $(1)$: \begin{align}\frac{\vartheta'}\vartheta(p,x)=D_p\ln\vartheta(p,x)&=D_p\Big(-p^2x+\sum_{k\ge1}\ln\left(1+e^{2px-(2k-1)x}\right)\Big)+\sum_{k\ge1}\ln\left(1+e^{-2px-(2k-1)x}\right)\Big)\\[2ex] &=-2px+2x\sum_{k\ge1}\left(\frac{e^{2px-(2k-1)x}}{1+e^{2px-(2k-1)x}}-\frac{e^{-2px-(2k-1)x}}{1+e^{-2px-(2k-1)x}}\right)\\[2ex] &=-2px+2x\sum_{k\ge1}\left(\frac1{e^{(2k-1)x-2px}+1}-\frac1{e^{2px+(2k-1)x}+1}\right), \end{align} and finally plug the above result into $(\,^*)$ to get that your sum equals$$\frac2x\operatorname{csch}px\bigg[\frac{\vartheta'}\vartheta(p,x)+2px\bigg].$$ As for your general sum, you could rearrange the summations, then try to evaluate $\sum_{n\in\mathbb{Z}}\operatorname{sech}[x(n-1/2+p_i)]$ (which should have a simple relation to a Jacobi elliptic function); the $p_i, p_j$ terms may force some inevitable complexities later...