Prove $\cos(\pi x)=\sinh(\pi)/\pi\sum_{n=0}^{\infty}\left(\frac{\left(-1\right)^{n}}{1+\left(x-n\right)^{2}}+...\right)$ It seems that $\cos(\pi x)$ is given by 
$$\cos(\pi x)=\frac{\sinh(\pi)}{\pi}\sum_{n=0}^{\infty}\left(\frac{\left(-1\right)^{n}}{1+\left(x-n\right)^{2}}-\frac{\left(-1\right)^{n\ }}{1+\left(x+n+1\right)^{2}}\right).$$
I found this by playing around in desmos. Do you know how we could prove this? Has a similar result to this been published elsewhere on the internet? I can't find anything.
 A: The correct value of the sum reads:
$$
\sum_{n=-\infty}^\infty\frac{(-1)^n}{1+(x-n)^2}=
\frac{\pi\sinh(\pi)\cos(\pi x)}{\sinh^2(\pi)+\sin^2(\pi x)},\tag1
$$
which (slightly) deviates from your claim. It would be identical if $\sin^2(\pi x)$ in the denominator were removed. Here the difference of the claimed and actual expressions is shown:

The simplest way to prove the expression (1) is using the Mittag-Leffler's theorem. Indeed both sides of the equality have the same sets of simple poles at $z^\pm_n=n\pm i$ with the same corresponding residues $\operatorname{res}_{z^\pm_n}(f)=\pm\frac{(-1)^n}{2i}$.

A more constructive approach:
$$\begin{align}
\sum_{n=-\infty}^\infty\frac{(-1)^n}{1+(x-n)^2}
&=\frac1{2i}\sum_{n=-\infty}^\infty(-1)^n\left[\frac{1}{x-n-i}-\frac{1}{x-n+i}\right]\tag2\\
&=\frac\pi{2i}\left[\frac{1}{\sin(\pi(x-i))}-\frac{1}{\sin(\pi(x+i))}\right]\tag3\\
&=\frac\pi{2i}\frac{2\cos(\pi x)\sin(\pi i)}
{[\sin(\pi x)\cos(\pi i)]^2-[\cos(\pi x)\sin(\pi i)]^2}\tag4\\
&=\frac{\pi\sinh(\pi)\cos(\pi x)}
{\cosh^2(\pi)\sin^2(\pi x)+\sinh^2(\pi)\cos^2(\pi x)}\tag5\\
&=\frac{\pi\sinh(\pi)\cos(\pi x)}{\sinh^2(\pi)+\sin^2(\pi x)}.\tag6
\end{align}
$$


Explanation:
$(2)\to(3)$: $\displaystyle\sum_{n=-\infty}^\infty\frac{(-1)^n}{z-n}=\frac\pi{\sin(\pi z)}$;
$(3)\to(4)$: $\displaystyle\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$;
$(4)\to(5)$: $\displaystyle\sin(ix)=i\sinh(x),\; \cos(ix)=\cosh(x)$;
$(5)\to(6)$: $\displaystyle \cosh^2(x)=\sinh^2(x)+1$.

