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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?

Community
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Cyclone
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4 Answers4

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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.

Jack D'Aurizio
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  • Thank you very much for your reply, Jack. So Herglotz's trick is to prove that some properties of the functions on RHS and LHS are the same to conclude that they're equal. Here, the only singularity on both sides occurs for $y=0$, in the limit $x=0$, right? I'm not sure I understand the proof you suggest. I've managed to prove the identity using Poisson's formula and another sum, $\sum_{n=0}^\infty \cos(\lambda n) \exp(-an) = \frac{e^a-\cos(\lambda)}{2(\cosh(a)-\cos(\lambda))}$ (from Albert D. Wheelon's "Tables of Summable Series"). But there again I'm unsure how that is proved itself. – Cyclone Jan 21 '18 at 16:43
  • Could you sketch how the proof with Weierstrass products would work? And maybe, if you happen to know how Wheelon's formula comes about, that would also be great. – Cyclone Jan 21 '18 at 16:46
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    I realised the identity I quoted is simply a consequence of the complex extension of the geometric series. – Cyclone Jan 21 '18 at 17:23
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Using the formula $$ \sum_{n\in\mathbb{Z}}\frac1{n+z}=\pi\cot(\pi z) $$ proven in this answer, we get $$\newcommand{\Re}{\operatorname{Re}} \begin{align} \sum_{n=-\infty}^\infty\frac{-x}{x^2+(y-2\pi n)^2} &=\frac i{4\pi}\sum_{n=-\infty}^\infty\left(\frac1{n-\frac{y}{2\pi}-i\frac{x}{2\pi}}-\frac1{n-\frac{y}{2\pi}+i\frac{x}{2\pi}}\right)\\ &=-\frac i4\cot\left(\frac y2+i\frac x2\right)+\frac i4\cot\left(\frac y2-i\frac x2\right)\\ &=\frac12\Re\left(\frac{e^{i\frac y2-\frac x2}+e^{-i\frac y2+\frac x2}}{e^{i\frac y2-\frac x2}-e^{-i\frac y2+\frac x2}}\right)\\ &=\frac12\Re\left(\frac{e^{i\frac y2-\frac x2}+e^{-i\frac y2+\frac x2}}{e^{i\frac y2-\frac x2}-e^{-i\frac y2+\frac x2}}\frac{e^{-i\frac y2-\frac x2}-e^{i\frac y2+\frac x2}}{e^{-i\frac y2-\frac x2}-e^{i\frac y2+\frac x2}}\right)\\ &=\frac12\Re\left(\frac{e^{-x}-e^x-e^{iy}+e^{-iy}}{e^x+e^{-x}-e^{iy}-e^{-iy}}\right)\\[3pt] &=-\frac12\Re\left(\frac{\sinh(x)+i\sin(y)}{\cosh(x)-\cos(y)}\right)\\[6pt] &=-\frac12\frac{\sinh(x)}{\cosh(x)-\cos(y)} \end{align} $$

robjohn
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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.
Cyclone
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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.

Claude Leibovici
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