I was experimenting and I found the sum $$ \sum_{n=-\infty}^{\infty}{{1}\over{x-\pi n}} $$ As I input higher values for "infinity" (the graphing calculator I'm using doesn't let you put infinity in sums) it seems to get closer and closer to the function $ \cot{x} $. Is this true? How do you prove it?
(The tags might not be correct, please change them if they aren't.)
Notice that forall $ k\in\mathbb{N} :$\begin{aligned}\frac{2x}{x^{2}-\pi^{2}k^{2}}=\frac{2}{\pi\sin{x}\cos{x}}\int_{0}^{\pi}{\cos{\left(\frac{2xy}{\pi}\right)}\cos{\left(2ky\right)}\,\mathrm{d}y}\end{aligned}
Thus, if $n\in\mathbb{N}^{*} :$\begin{aligned}\small\sum_{k=1}^{n}{\frac{2x}{x^{2}-\pi^{2}k^{2}}}&\small=\frac{2}{\pi\sin{x}\cos{x}}\int_{0}^{\pi}{\cos{\left(\frac{2xy}{\pi}\right)}\sum_{k=1}^{n}{\cos{\left(2ky\right)}}\,\mathrm{d}y}\\ &\small=\frac{1}{\pi\sin{x}\cos{x}}\int_{0}^{\pi}{\cos{\left(\frac{2xy}{\pi}\right)}\left(\frac{\sin{\left(\left(2n+1\right)y\right)}}{\sin{y}}-1\right)\mathrm{d}x}\\ &\small=-\frac{1}{x}+\frac{1}{\pi\sin{x}\cos{x}}\int_{0}^{\pi}{\cos{\left(\frac{2xy}{\pi}\right)}\frac{\sin{\left(\left(2n+1\right)y\right)}}{\sin{y}}\,\mathrm{d}y}\\ &\small=-\frac{1}{x}+\frac{1}{\pi\sin{x}\cos{x}}\left(\int_{0}^{\frac{\pi}{2}}{\cos{\left(\frac{2xy}{\pi}\right)}\frac{\sin{\left(\left(2n+1\right)y\right)}}{\sin{y}}\,\mathrm{d}y}+\int_{\frac{\pi}{2}}^{\pi}{\cos{\left(\frac{2xy}{\pi}\right)}\frac{\sin{\left(\left(2n+1\right)y\right)}}{\sin{y}}\,\mathrm{d}y}\right)\\ &\small=-\frac{1}{x}+\frac{1}{\pi\sin{x}\cos{x}}\left(\int_{0}^{\frac{\pi}{2}}{\cos{\left(\frac{2xy}{\pi}\right)}\frac{\sin{\left(\left(2n+1\right)y\right)}}{\sin{y}}\,\mathrm{d}y}+\int_{0}^{\frac{\pi}{2}}{\cos{\left(\frac{2x\left(\pi-y\right)}{\pi}\right)}\frac{\sin{\left(\left(2n+1\right)\right)}}{\sin{y}}\,\mathrm{d}y}\right)\\ &\small=-\frac{1}{x}+\frac{2\cot{x}}{\pi}\int_{0}^{\frac{\pi}{2}}{\frac{\sin{\left(\left(2n+1\right)y\right)}}{\sin{y}}\,\mathrm{d}y}+\frac{1}{\pi\sin{x}\cos{x}}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{\left(\frac{2xy}{\pi}\right)}+\cos{\left(\frac{2x\left(\pi-y\right)}{\pi}\right)}-2\cos^{2}{x}}{\sin{y}}\sin{\left(\left(2n+1\right)y\right)}\,\mathrm{d}x}\\ \small\sum_{k=1}^{n}{\frac{2x}{x^{2}-\pi^{2}k^{2}}}&\small=-\frac{1}{x}+\cot{x}+\int_{0}^{\frac{\pi}{2}}{f_{x}\left(y\right)\sin{\left(\left(2n+1\right)y\right)}\,\mathrm{d}y}\end{aligned}
Where $ f_{x} $ is the function defined as $ f_{x}:y\mapsto\frac{\cos{\left(\frac{2xy}{\pi}\right)}+\cos{\left(\frac{2x\left(\pi-y\right)}{\pi}\right)}-2\cos^{2}{x}}{\pi\sin{x}\cos{x}\sin{y}}$.
$\textbf{Note :}$ In the last line, we used the fact that $ I_{n}=\int_{0}^{\frac{\pi}{2}}{\frac{\sin{\left(\left(2n+1\right)y\right)}}{\sin{y}}\,\mathrm{d}y}=\frac{\pi}{2}$, forall $ n\in\mathbb{N} $, which can be proven by calculating $ I_{n}-I_{n-1} $, for $ n\geq 1$.
Now we have : \begin{aligned}\small\sum_{k=-n}^{n}{\frac{1}{x-\pi k}}&\small=\sum_{k=-n}^{-1}{\frac{1}{x-\pi k}}+\frac{1}{x}+\sum_{k=1}^{n}{\frac{1}{x-\pi k}}\\&\small=\sum_{k=1}^{n}{\frac{1}{x+\pi k}}+\frac{1}{x}+\sum_{k=1}^{n}{\frac{1}{x-\pi k}}\\ &\small=\frac{1}{x}+\sum_{k=1}^{n}{\frac{2x}{x^{2}-\pi^{2}k^{2}}}\\ \small\sum_{k=-n}^{n}{\frac{1}{x-\pi k}}&\small=\cot{x}+\int_{0}^{\frac{\pi}{2}}{f_{x}\left(y\right)\sin{\left(\left(2n+1\right)y\right)}\,\mathrm{d}y} \end{aligned}
Since $ f_{x} $ is piecewise continuous on $\left[0,\frac{\pi}{2}\right] $, then $\int_{0}^{\frac{\pi}{2}}{f_{x}\left(y\right)\sin{\left(\left(2n+1\right)y\right)}\,\mathrm{d}y}\underset{n\to +\infty}{\longrightarrow}0 $
Thus : $$ \sum_{n=-\infty}^{+\infty}{\frac{1}{x-\pi n}}=\lim_{n\to +\infty}{\sum_{k=-n}^{n}{\frac{1}{x-\pi k}}}=\cot{x} $$