Infinite sum related to Hurwitz Zeta

Question

I want to evaluate the following sum: \begin{equation} \sum_{-\infty}^{\infty}\frac{(-1)^n}{(n+a)^2} \end{equation} Where $a\in\mathbb{R}$ is not an integer. Such is similar to $\zeta(2,a)$, but it ranges over the entire line and it's alternating. I know you can just note that this is minus the residue of $f(z)=\pi\csc\pi z$ at $z = −a$, such that:

\begin{equation} \pi \frac{d}{dz} \csc(\pi z) \bigg|_{z = -a} = -\frac{\pi^2 \cos(-\pi a)}{\sin^2(-\pi a)}=-S \end{equation}

Yielding the result. That's fine and all, but I've been wanting a different approach. I've tried using the weierstrass factorization for trig functions, but got nowhere. Also tried Poisson's summation formula, but got nowhere also.

Any help is appreciated.

Answer 1

Here is a derivation using the Fourier series of $\cos ax$: $$\cos ax=\frac{\sin \pi a}{\pi a}+\frac{2a\sin\pi a}{\pi}\sum_{n=1}^\infty\frac{(-1)^n\cos nx}{a^2-n^2}.$$ Hence for $x=0$ we have the identity $$\frac{\pi}{\sin \pi a}=\frac{1}{a}+2a\sum_{n=1}^\infty \frac{(-1)^n}{a^2-n^2}=\sum_{n=-\infty}^\infty \frac{(-1)^n}{a+n}.$$ Take the derivative with respect to $a$ and the desired result follows, $$\sum_{n=-\infty}^\infty \frac{(-1)^n}{(a+n)^2}=\frac{\pi^2 \cos\pi a}{\sin^2 \pi a}.$$