TITLE: Truth of the Poisson summation formula
QUESTION [39 upvotes]: The Poisson summation says, roughly, that summing a smooth $L^1$-function of a real variable at integral points is the same as summing its Fourier transform at integral points(after suitable normalization). Here is the wikipedia link.
For many years I have wondered why this formula is true. I have seen more than one proof, I saw the overall outline, and I am sure I could understand each step if I go through them carefully. But still it wouldn't tell me anything about why on earth such a thing should be true.
But this formula is exceedingly important in analytic number theory. For instance, in the book of Iwaniec and Kowalski, it is praised to high heavens. So I wonder what is the rationale of why such a result should be true.

REPLY [3 votes]: The Poisson formula is just the decomposition in Fourier series of the Dirac measure on the circle. 
It happens that the theory of distributions on the circle is far more simpler than the one on the real line since everything is compactly supported. In his book, Laurent Schwartz starts the chapter about the Fourier transform with the circle and shows that any distribution has a convergent Fourier series (VII.1.3 "theorie des distributions").
Moreover the space of distributions is the union of all the negative Sobolev spaces and these spaces can be caracterized in terms of the decay of the Fourier coefficients. I think that the Dirac is in $H^{-1}$. Anyway the usual formula holds and we have
$$
\delta_0(x) = \sum_n \langle \delta_0, e^{inx} \rangle e^{inx} = {1\over 2\pi}\sum_n e^{inx}, \quad x\in {\bf R}/{\bf Z}.
$$
Note that with this choice of normalization, the scalar product is ${1\over 2\pi}\int_0^{2\pi} f(x)\overline{g(x)} \,dx$ so that the $e^{inx}$ are unit vectors.This gives the ${1\over 2\pi}$ on the right hand side.
Now we compose this with the projection $\pi : {\bf R} \rightarrow {\bf R} /2\pi{\bf Z}$ to get the Poisson formula on ${\bf R}$. One must be a bit cautious because although we can always compose a function with another as soon as the domains match, it is not true that composition is always possible for distributions. 
In our particular case, the Dirac measure is the limit of an approximate identity, and so it is easy to guess what is happening just by composing the approximate identity and passing to the limit. Without surprise, the pullback of the Dirac measure at $\{0\}$ is given by the Dirac measure on the pullback of $\{0\}$, which is just the sum of all the Dirac measures on the points $2\pi{\bf Z}$.
$$
\sum_{n\in {\bf Z}} \delta_{2\pi n}(x) = {1\over 2\pi} \sum_{n\in {\bf Z}} e^{inx}, \quad x\in {\bf R}.
$$
We can evaluate this against a function in Schwartz space or use a convolution to obtain the original formula of Poisson. And we got the $2\pi$ at the correct locations without even thinking about it. So that's the truth.