Poisson summation formula (in general)

Question

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 $$ P_r (2\pi x) = \sum_{n=-\infty}^\infty P_y (x+n)\;\;\;\;\text{(here $r = e^{-2 \pi y} $}) $$

Answer 1

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.