Closed form for $\sum_{n=-\infty}^\infty \frac{1}{(z+n)^2+a^2}$

Question

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?

Answer 1

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

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