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!