The Fourier transform of a "comb function" is a comb function? Let $f(x) = \sum_{n=-\infty}^{\infty} \delta(x - n)$, where $\delta$ is the Dirac delta function. This function $f$ (a "comb function") is important in signal processing because evenly sampling a function $g$ can be viewed as multiplying $g$ pointwise with $f$. This idea is the first step towards understanding how to approximate the Fourier transform of $g$, given evenly spaced samples of $g$. The next step is to note that $\hat{gf} = \hat{g}*\hat{f}$, where $\hat{f}$ is the Fourier transform of $f$ and $*$ denotes convolution.
Is it true that $\hat f$ is also a comb function? If that statement isn't quite right, what is the correct statement?
I would like to see: 1) an informal, intuitive, nonrigorous but easy derivation of the Fourier transform of $f$ . 2) A rigorous version of the same calculation.  (How would you even define $f$ rigorously, is it a distribution?)
(I'd also be interested in recommendations of math textbooks that cover this topic, including the Nyquist sampling theorem, even if it's only an exercise or series of exercises in an analysis textbook.)
 A: Intuitive Explanation 
The Comb is a sum of Time Shifted Dirac Delta. 
The Fourier Transform of a Dirac Delta is known to be a constant. 
The Fourier Transform of a Time Shifted Function is known to be Fourier Transform of the function multiplied by a complex exponential factor which is $ \exp(-i 2 \pi f T) $ 
Just apply this points to the Comb Function considered as a sum of Time Shifted Dirac Delta with distance $ kT $ and you get a sum of Frequency Shifted exponential functions, each of which multiplied by a constant. 
Finally use Euler’s Formula to consider complex exponentials as a periodic sinusoidal function and observe that you have constructive interference only in frequencies which are integer multiple of $ \frac{1}{T} $ 
A: Up to constants and normalizations, this Dirac comb is its own Fourier transform: the assertion of this evaluated on (for example) a Schwartz function, $\sum_{n\in \mathbb Z} f(n) \;=\; \sum_{n\in\mathbb Z} \widehat{f}(n)$, is the Poisson summation formula.
A: Assume an arbitrary periodic function $f(t)$ with period $T$. Consider the Fourier series representation of $f(t)$, in which $\omega_0=\frac{2\pi}{T}$:
$$f(t)=\sum_{n=-\infty}^{+\infty}c_n e^{i n \omega_0 t}$$
Take the Fourier transform of the sides:
\begin{align}
\mathcal{F}\{f(t)\}=&\mathcal{F}\{\sum_{n=-\infty}^{+\infty}c_n e^{i n \omega_0 t}\}\\
=&\sum_{n=-\infty}^{+\infty}c_n\mathcal{F}\{ e^{i n \omega_0 t}\}\\
F(\omega)=&2\pi\sum_{n=-\infty}^{+\infty}c_n\delta(\omega-n\omega_0)
\end{align}
This means that the Fourier transform of a periodic signal is an impulse train where the impulse amplitudes are $2\pi$ times the Fourier coefficients of that signal.
With $f(t)=\delta(t)$, the Fourier series coefficients are $c_n=\frac{1}{T}$ for all $n$.
Hence, 
$$\mathcal{F}\{\sum_{n=-\infty}^{+\infty}\delta(t-nT)\}=\frac{2\pi}{T} \sum_{n=-\infty}^{+\infty}\delta(\omega-n\omega_0)$$
or in comb notation:
$$\boxed{\mathcal{F}\{\text{comb}_T(t)\}=\omega_0\ \text{comb}_{\omega_0}(\omega)}$$
where 
$$\text{comb}_A(x)\triangleq\sum_{n=-\infty}^{+\infty}\delta(x-nA)$$
