Summing a simple series

Question

Numerically, I find that the following identity is true:

$$ \sum_{n=-\infty}^\infty \frac{-x}{x^2+(y-2\pi n)^2} = \frac{\sinh(x)}{2(\cos(y)-\cosh(x))}. $$

However, I am unsure how to go about proving it. I have tried the Laplace transform method (cf. here eqn. (23)), but this leads to integrals that neither I nor Wolfram Alpha could solve. I thought about using the Taylor series of $\sinh$, $\cos$, and $\cosh$ but then I am facing the problem of dividing by a Taylor series which is nasty.

Does someone have an idea how to prove this identity?

Answer 1

Do you know Herglotz' trick? For a fixed value of $x$, both the LHS and the RHS, regarded as meromorphic functions of the complex variable $y$, have the same singularities (simple poles) and the same residues. The value of the LHS at $y=0$ is simple to compute through Fourier series and this essentially finishes the proof. As an alternative approach, you may just consider the Weierstrass products for the $\sin$ / $\sinh$ functions together with their logarithmic derivatives. Yet another way is to exploit the Poisson summation formula.

Answer 2

Thanks to Jack's hint, I managed to prove the formula using the Poisson summation formula. Here is how it works, for reference.

1. The Poisson summation formula reads: $$\sum_{n=-\infty}^\infty f(n) = \sum_{k=-\infty}^\infty \hat{f}(k),$$ where $\hat{f}(k)=\int_{-\infty}^\infty e^{-2\pi i n k} f(n) \mathrm{d}n$ is the Fourier transform of $f$. In our case, the Fourier transform is readily calculated: $$\hat{f}(k)=\frac{1}{2}\begin{cases} e^{-(x+iy)k)}:k\geq 0 \\ e^{(x-iy)k}: k<0 \end{cases}$$ This leads to $$\sum_{k=-\infty}^\infty \hat{f}(k) = -\frac{1}{2}+\sum_{k=0}^{\infty} e^{-kx} \cos(ky). $$
2. The RHS sum is equal to the real part of $\sum_{k=0}^\infty e^{-(x+iy)k} = \frac{1}{1-\exp(-x-iy))}$ by the geometric series. Taking the real part gives $$ \sum_{k=0}^{\infty} e^{-kx} \cos(ky) = \frac{e^x - \cos(y)}{2(\cosh(x)-\cos(y))}$$ and hence, by reducing to a common denominator, we find the desired result.

Answer 3

Much less cleaver.

Let us consider $$S_m=\sum_{n=-m}^m \frac{-x}{x^2+(y-2\pi n)^2}$$ $${x^2+(y-2\pi n)^2}=4 \pi ^2 n^2-4 \pi yn+x^2+y^2=4 \pi ^2(n-a)(n-b)$$ where $$a=\frac{y+i x}{2 \pi } \qquad \text{and}\qquad b=\frac{y-i x}{2 \pi }$$ making $$S_m=-\frac{x}{4 \pi ^2}\sum_{n=-m}^m \frac 1{(n-a)(n-b)}=-\frac{x}{4 \pi ^2(a-b)}\sum_{n=-m}^m \left(\frac 1{n-a}-\frac 1{n-b} \right)$$ Now, using generalized harmonic numbers $$\sum_{n=-m}^m \left(\frac 1{n-a}-\frac 1{n-b} \right)=H_{m-a}-H_{m+a}-H_{m-b}+H_{m+b}+\pi \left( \cot (\pi b)- \cot (\pi a)\right)$$ Now, expanding as series for infinitely lage values of $m$ $$S_m=\frac{x (\cot (\pi a)-\cot (\pi b))}{4 \pi (a-b)}+\frac{x}{2 \pi ^2 m}+O\left(\frac 1{m^2}\right)\implies \lim_{m\to \infty } \, S_m=\frac{x (\cot (\pi a)-\cot (\pi b))}{4 \pi (a-b)}$$

Replace $a$ and $b$ by their values and simplify the trigonometric expressions to get your result.

Answer 4

Using the formula $$ \sum_{n\in\mathbb{Z}}\frac1{n+z}=\pi\cot(\pi z) $$ proven in this answer, we get $$\newcommand{\Re}{\operatorname{Re}} \begin{align} \sum_{n=-\infty}^\infty\frac{-x}{x^2+(y-2\pi n)^2} &=\frac i{4\pi}\sum_{n=-\infty}^\infty\left(\frac1{n-\frac{y}{2\pi}-i\frac{x}{2\pi}}-\frac1{n-\frac{y}{2\pi}+i\frac{x}{2\pi}}\right)\\ &=-\frac i4\cot\left(\frac y2+i\frac x2\right)+\frac i4\cot\left(\frac y2-i\frac x2\right)\\ &=\frac12\Re\left(\frac{e^{i\frac y2-\frac x2}+e^{-i\frac y2+\frac x2}}{e^{i\frac y2-\frac x2}-e^{-i\frac y2+\frac x2}}\right)\\ &=\frac12\Re\left(\frac{e^{i\frac y2-\frac x2}+e^{-i\frac y2+\frac x2}}{e^{i\frac y2-\frac x2}-e^{-i\frac y2+\frac x2}}\frac{e^{-i\frac y2-\frac x2}-e^{i\frac y2+\frac x2}}{e^{-i\frac y2-\frac x2}-e^{i\frac y2+\frac x2}}\right)\\ &=\frac12\Re\left(\frac{e^{-x}-e^x-e^{iy}+e^{-iy}}{e^x+e^{-x}-e^{iy}-e^{-iy}}\right)\\[3pt] &=-\frac12\Re\left(\frac{\sinh(x)+i\sin(y)}{\cosh(x)-\cos(y)}\right)\\[6pt] &=-\frac12\frac{\sinh(x)}{\cosh(x)-\cos(y)} \end{align} $$