a question about an infinite series calculation. I want to prove that for $y >0$, $ x \in \mathbb R$, 
$$ \sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2} = \frac{1}{2} \frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi  x ) + e^{-4 \pi y}}$$
 A: Let's use J.M.'s excellent hint :
$$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}=-\ \Im{\sum_{n=-\infty}^\infty \frac 1{x+iy+n}}$$
Setting $z:=x+iy\ $ we will evaluate :
$$\sum_{n=-\infty}^\infty \frac 1{z+n}=\frac 1z+\sum_{n=1}^\infty \frac 1{z+n}+\frac 1{z-n}=\frac 1z+\sum_{n=1}^\infty \frac {2z}{z^2-n^2}$$
The series at the right may be simplified (this was proved many times at S.E. for example here or here or here) getting :
$$\pi\cot(\pi z)=\frac1{z}+\sum_{n=1}^{\infty}\frac{2z}{z^2-n^2}$$
and the simple result :
$$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}= -\pi\;\Im{(\cot\pi (x+iy))}$$
Your remaining problem is that this simple answer doesn't appear related to your more complicated result. Worse your L.H.S and R.H.S. terms don't numerically correspond so that it seems that you have a problem in your question !
Let's add that your R.H.S. term may be rewritten as half :
$$\frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi  x ) + e^{-4 \pi y}}=\frac{\sinh(2\pi y)}{\cosh(2 \pi y) - \cos ( 2 \pi  x )}$$

UPDATE: In fact your equation should read (simply multiplying your R.H.S. by $2\pi$) :
$$\boxed{\displaystyle \sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2} = \pi\frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi  x ) + e^{-4 \pi y}}}$$
We want :
$$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}= -\pi\Im{(\cot \pi (x+iy))}$$
Let's observe that $\ \sin a-\sin b=2\sin\frac{a-b}2\cos\frac{a+b}2$ and $\ \cos a-\cos b=-2\sin\frac{a+b}2\sin\frac{a-b}2$
implies (dividing these expressions) :
$$\frac {\sin a-\sin b}{\cos a-\cos b}=-\cot \frac{a+b}2$$
Set $a:=2\pi x,\ b:=2\pi i y\ $ to get :
$$-\cot\pi (x+iy)=\frac {\sin 2\pi x-\sin 2\pi i y}{\cos 2\pi x-\cos 2\pi i y}=\frac {\sin 2\pi x-i\sinh 2\pi y}{\cos 2\pi x-\cosh 2\pi y}$$
so that :
$$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}=-\pi\frac {\sinh 2\pi y}{\cos 2\pi x-\cosh 2\pi y}$$
$$= \pi\frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi  x ) + e^{-4 \pi y}}$$
A: For appropriate functions $ƒ$, the Poisson summation formula may be stated as:
$$ \sum_{k=-\infty}^{\infty} f(k+x) {\rm e}^{-ikx} = \sum_{n=-\infty}^\infty  F(n+x)\,, $$
where $F$ is the fourier transform of $f$. Let $ f(\omega) = \frac{1}{2}{\rm e}^{-y|\omega|}\, $ then 
$$ F(t) = \frac{y}{y^2+t^2}\,. $$
Using the above Poisson summation formula, we have,
$$ \sum_{n=-\infty}^{\infty} \frac{y}{y^2+(x+n)^2} =  \sum_{k-\infty}^{\infty} {\rm e}^{-y|k+x|} {\rm e}^{-ikx} \,.$$.
Now, all you need to do is to manipulate the sum on the right hand side in the above equation. 
