<s>[INST] Poisson summation formula (in general) Define Poisson kernel as $$ P_r ( \theta) := \frac{1}{2\pi} \frac{1-r^2}{1- 2r \cos \theta + r^2} $$ Then I want to prove the Poisson summation formula  which is $$  P_r (2\pi x) = \sum_{n=-\infty}^\infty P_y (x+n)\;\;\;\;\text{(here $r = e^{-2 \pi y}  $}) $$
 [/INST] We will use the result that 
$$\sum_{n=-\infty}^{\infty} \frac{y}{(x+n)^2+y^2}= \frac{1}{2} \frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi  x ) + e^{-4 \pi y}} = P_y(2\pi\,x)\,, $$
Recalling Poisson formula in the upper half plane for $y>0$, 
$$ P_{y}(x) = \frac{y}{x^2+y^2}\,. $$
We construct the sum
$$ \sum_{n=-\infty}^{\infty} P_y(x+n) = \sum_{n=-\infty}^{\infty} \frac{y}{(x+n)^2+y^2} =\frac{1}{2} \frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos( 2 \pi  x ) + e^{-4 \pi y}} = P_y(2\pi\,x) $$
Substituting $r= {\rm e}^{-2\pi y}$ in the above result gives the desired result.
$$ P_r(2\pi x) = \frac{1}{2} \frac{1 - r^2}{1 - 2 r \cos( 2 \pi  x ) + r^2} $$  
The whole idea was to exploit the Poisson integral formula in the upper half plane.
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