IDTFT of delta sum I am trying to compute 
$$
\int_{-1/2}^{+1/2} \sum_{n = -\infty}^{+\infty} \delta(\nu+n) \cdot e^{i\cdot2\pi\nu\cdot k} \,d\nu
$$
My stuck attempts: 
\begin{multline}
\int_{-1/2}^{+1/2} \sum_{n = -\infty}^{+\infty} \delta(\nu+n) \cdot e^{i\cdot2\pi\nu\cdot k} \,d\nu= 
 \sum_{n = -\infty}^{+\infty} \int_{-1/2}^{+1/2}\delta(\nu+n) \cdot e^{i\cdot2\pi\nu\cdot k}  \,d\nu= 1
\end{multline}
Assuming that $n,k\in\mathbb{Z}$ and $\int_{I} \delta(x-x_0) \cdot f(x) = f(x_0)$ if $x$ is interior point of $I$ and $0$ otherwise it seems that integral under the sum is zero always except $n = 0$ where it equals 1.
I feel that something is wrong with this, cause I feel that I should get some discrete sum function, not continuous one 
 A: You are evidently being asked to determine the Fourier transform of the Dirac comb $\sum_{n\in\mathbb Z} \delta_n$. (This is a tempered distribution so such computations can be legitimized...) You can already take the Fourier transform of $\delta_n$, which is a translate of $\delta_0$, whose Fourier transform is $1$. Fourier transform does not preserve translation (so you won't get $1_{x=n}$, but, rather, converts translation to multiplication by an exponential. So the Fourier transform of $\delta_n$ is $e^{2\pi inx}\cdot 1$ (depending on normalizations).
Thus, the Fourier transform of the Dirac comb is $\sum_n e^{2\pi inx}$.
This computation (which can be justified entirely...) gives another proof of the Poisson summation formula.
Yes, the fact that the Fourier transform of the infinite sum of translates of $\delta$ really is the infinite sum of the Fourier transforms depends on continuity properties in the topology of tempered distributions... but the natural heuristic does in fact give the true answer.
