Dirac / Fourier relation

Question

How do I prove the following?

$\frac {1}{2 \pi} \sum_{n=-\infty}^{\infty} e^{in(x-x_{0})} = \delta ( x- x_{0})$

Answer 1

The easiest way might be to use distributions. Take a test function in the appropriate space in order for everything to be defined. The RHS is trivial, so you just need to work out the LHS to find the same thing. Of course the Fourier transform will play a key role, so the right space is probably $\mathcal{S}$.

This is a particular case of the Poisson summation formula, you can in particular look at the "Distributional Formulation" on this page and adapt it to the periodicity of your problem:: https://en.wikipedia.org/wiki/Poisson_summation_formula

Answer 2

Suppose $f$ is an $L^2$ or $L^1$ function on $[0,2\pi]$ such that the Fourier series for $f$ at $x\in [0,2\pi]$ converges to $f(x)$. Then, for that $x$, $$ \lim_{N\rightarrow\infty}\sum_{n=-N}^{N} \frac{1}{2\pi}\int_{0}^{2\pi}f(t)e^{-int}dt e^{inx} = f(x) \\ \lim_{N\rightarrow\infty}\int_{0}^{2\pi}f(t)\frac{1}{2\pi}\sum_{n=-N}^{N}e^{in(x-t)}dt = f(x). $$ That should be enough for you to find a way to interpret your sum as a $\delta$, without belaboring the point of spaces of distributions, etc.. But you do need an interval of $2\pi$ length or, as mentioned in the comments, you end up with other evaluations as well.