fourier transform - Derive Frequency Representation of Impulse Train Function

I want to walk through the derivation of the frequency representation of an impulse train.

The definition of the impulse train function with period $T$ and the frequency representation with sampling frequency $\Omega_s = 2\pi/T$ that I would like to derive is:

\begin{align*} s(t) &= \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) \\ S(j\Omega) &= \frac{2\pi}{T} \sum\limits_{k=-\infty}^{\infty} \delta(\Omega - k\Omega_s) \\ \end{align*}

Using the exponential Fourier series representation of the impulse function and applying the Fourier transform from there results in:

\begin{align*} s(t) &= \frac{1}{T} \sum\limits_{n=-\infty}^{\infty} e^{-jn\Omega_s t} \\ S(j\Omega) &= \int_{-\infty}^\infty s(t) e^{-j\Omega t} dt \\ S(j\Omega) &= \int_{-\infty}^\infty \frac{1}{T} \sum\limits_{n=-\infty}^{\infty} e^{-jn\Omega_s t} e^{-j\Omega t} dt \\ S(j\Omega) &= \frac{1}{T} \int_{-\infty}^\infty \sum\limits_{k=-\infty}^{\infty} e^{-j(k\Omega_s + \Omega) t} dt \\ \end{align*}

To get from there to the end result, it would seem that the integration would need to be over a period of $2\pi$. Where $\Omega = -k\Omega_s$, the exponent would be $e^0$ and integrate to $2\pi$ and for other values of $\Omega$, there would be a full sine wave that would integrate to zero. However, the limits of integration are negative infinity to positive infinity. Can someone explain this? Thanks!

Answer

You correctly figured out that the occurring integrals don't converge in the conventional sense. The easiest (and definitely non-rigorous) way to see the result is by noting the Fourier transform relation

$$1\Longleftrightarrow 2\pi\delta(\Omega)$$

By the shifting/modulation property we have

$$e^{j\Omega_0t}\Longleftrightarrow 2\pi\delta(\Omega-\Omega_0)$$

So each term $e^{jn\Omega_s t}$ in the Fourier series transforms to $2\pi\delta(\Omega-n\Omega_s)$, and the result follows.