Identify $ \sum\limits_{k=-\infty}^{\infty} \frac{1}{(x+k)^2}$

Note that (see here) for $x\not \in \mathbb{Z}$ $$\pi\cot(\pi x)=\sum_{k\in \mathbb{Z}}\frac{1}{x-k}=\sum_{k\in \mathbb{Z}}\frac{1}{x+k}.$$ Since it is analytic you can differentiate it term by term and we obtain, $$-\frac{\pi^2}{\sin^2{(\pi x)}}=\left(\pi\cot(\pi x)\right)'=\sum_{k\in \mathbb{Z}}\frac{-1}{(x+k)^2}.$$ Finally $$\sum_{k\in \mathbb{Z}}\frac{1}{(x+k)^2}=\frac{\pi^2}{\sin^2{(\pi x)}}.$$

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Note that $\ds{\left.\sum_{k = 0}^{\infty}{1 \over \pars{k + z}^{2}}\right\vert_{\ x\ \not\in\ \mathbb{Z}} = \Psi\,'\pars{z}}$ is a well known identity. $\ds{\Psi}$ is the Digamma Function.

\begin{align} \color{#f00}{\mrm{f}\pars{x}} & \equiv \sum_{k = -\infty}^{\infty}\,\,{1 \over \pars{x + k}^{2}} = {1 \over x^{2}} + \sum_{k = 0}^{\infty} \bracks{{1 \over \pars{k + x}^{2}} + {1 \over \pars{k - x}^{2}}} \\[5mm] & = \overbrace{{1 \over x^{2}} + \Psi\, '\pars{x}}^{\ds{\Psi\, '\pars{1 + x}}}\ +\ \Psi\, '\pars{-x}\qquad\pars{~Recurrence~} \\[5mm] & = -\pi\,\totald{\cot\pars{\pi x}}{x}\qquad \pars{~Euler\ Reflection\ Formula~} \\[5mm] & = \color{#f00}{\pi^{2}\csc^{2}\pars{\pi x}} \end{align}