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I want to express $$\sum_{n=-\infty}^\infty \dfrac{1}{(z+n)^2+a^2}$$ in closed form. What comes to mind is the formula $$\pi\cot\pi z = \dfrac{1}{z}+\sum_{n\ne 0}\left(\dfrac{1}{z-n}+\dfrac1n\right)=\dfrac{1}{z}+\sum_{n=1}^\infty\dfrac{2z}{z^2-n^2}$$ and also $$\dfrac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^\infty \dfrac{1}{(z-n)^2}.$$ But neither of these gives the term $(z+n)^2+a^2$ that we want. Perhaps we can adjust somehow?

Mika H.
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2 Answers2

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You can use the residue theorem, based on the following formula (not going to prove here):

$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \operatorname*{Res}_{\zeta=\zeta_k} [\pi \cot{(\pi \zeta)} f(\zeta)]$$

where $\zeta_k$ are the non-integer poles of $f$. Here,

$$f(\zeta) = \frac1{(z+\zeta)^2+a^2}$$

so that the poles $\zeta_{\pm} = -z\pm i a$. The sum is therefore

$$\frac{\pi \cot{\pi (z-i a)}}{i 2 a} - \frac{\pi \cot{\pi (z+i a)}}{i 2 a}= \frac{\pi}{a} \Im{[\cot{\pi(z-i a)}]}$$

Now,

$$\cot{\pi(z-i a)} = \frac{\cos{\pi z} \cosh{\pi a} + i \sin{\pi z} \sinh{\pi a}}{\sin{\pi z} \cosh{\pi a} - i \cos{\pi z} \sinh{\pi a}} $$

Therefore, then sum is

$$\sum_{n=-\infty}^{\infty} \frac1{(z+n)^2+a^2} = \frac{\pi}{2 a} \frac{\sinh{2 \pi a}}{\sin^2{\pi z} \cosh^2{\pi a} + \cos^2{\pi z} \sinh^2{\pi a}}$$

EDIT

This may be simplified even further to

$$\sum_{n=-\infty}^{\infty} \frac1{(z+n)^2+a^2} = \frac{\pi}{2 a} \frac{\sinh{2 \pi a}}{\sin^2{\pi z}+\sinh^2{\pi a}}$$

Ron Gordon
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  • You again with another beautiful answer ! Thanks and cheers. – Claude Leibovici Dec 15 '13 at 18:46
  • Thank you, Ron. It's okay that you're not proving that formula (since you already typed a lot), but could you please give some direction on how the proof goes? – Mika H. Dec 15 '13 at 19:42
  • @MikaH.: consider the integral $$\oint_{C_N} dz , \pi , \cot{(\pi z)} f(z) $$ where $C_N$ is a square contour with vertices $(\pm N \pm 1/2) (1+i)$, $N \in \mathbb{Z}$. Show that the integral vanishes as $N \to \infty$ by considering the magnitude of the integrand along the contour. Then apply the residue theorem. – Ron Gordon Dec 15 '13 at 19:45
  • @RonGordon are there any restrictions on $f(n)$ so that the first result holds? It seems nonsensical if $f(n)$ is entire. – Meow Dec 27 '13 at 19:31
  • @Alyosha: yes - that the magnitude of the contour integral in the comment above vanish as $N\to\infty$. You are right that it would be nonsensical for $f$ entire. – Ron Gordon Dec 27 '13 at 19:54
  • Another solution can be derived by applying the Poisson summation formula for the Fourier transform. – Ryszard Szwarc Mar 04 '23 at 21:26
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This directly answers the OP's question: "Perhaps we can adjust somehow?". We have

\begin{align} & \sum_{n=-\infty}^\infty \frac{1}{(z+n)^2+a^2} = \nonumber \\ & = \frac{1}{z^2+a^2} + \frac{1}{2 ai} \sum_{n=1}^\infty \left( \frac{1}{z-ia+n} - \frac{1}{z+ia+n} \right) \nonumber \\ & + \frac{1}{2 ai} \sum_{n=1}^\infty \left( \frac{1}{-z-ia+n} - \frac{1}{-z+ia+n} \right) \nonumber \\ & = \frac{1}{z^2+a^2} + \frac{1}{2 ai} \sum_{n=1}^\infty \left( \frac{1}{z-ia+n} - \frac{1}{-z+ia+n} \right) \nonumber \\ & + \frac{1}{2 ai} \sum_{n=1}^\infty \left( \frac{1}{-z-ia+n} - \frac{1}{z+ia+n} \right) \nonumber \\ & = \frac{1}{z^2+a^2} + \frac{1}{2ai} \sum_{n=1}^\infty \left( \frac{2(z-ia)}{(z-ia)^2-n^2} \right) - \frac{1}{2ai} \sum_{n=1}^\infty \left( \frac{2(z+ia)}{(z+ia)^2-n^2} \right) \nonumber \\ & = \frac{1}{z^2+a^2} + \frac{1}{2ai} \left( \pi \cot \pi (z-ia) - \frac{1}{z-ia} \right) - \frac{1}{2ai} \left( \pi \cot \pi (z+ia) - \frac{1}{z+ia} \right) \nonumber \\ & = \frac{1}{2ai} \pi \cot \pi (z-ia) - \frac{1}{2ai} \pi \cot \pi (z+ia) \nonumber \\ & = \frac{\pi}{a} \Im [\cot \pi (z-ia)] \end{align}

where we have used $\pi \cot \pi \xi - \frac{1}{\xi} = \sum_{n=1}^\infty \frac{2\xi}{\xi^2-n^2}$. This is the same answer as derived from first principles by @RonGordon.

Dave77
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