About Fourier transform of periodic signal
We can figure out what's going on if we first understand a simple identity and then just compute the Fourier transform of the periodic function.

A useful identity

First let's prove that
$$D(\omega - \omega') \equiv \int_{-\infty}^\infty dt \, e^{i (\omega-\omega') t} = 2\pi \, \delta(\omega - \omega')\,. $$
We just use a test function $\tilde{g}(\omega)$:
\begin{align}
\int_{-\infty}^\infty \frac{d\omega}{2\pi}
\underbrace{ \left(\int_{-\infty}^\infty dt \, e^{i (\omega-\omega') t} \right)}_{D(\omega - \omega')} \tilde{g}(\omega)
&amp;= \int_{-\infty}^\infty dt \, e^{-i \omega' t}\left( \int_{-\infty}^\infty \frac{d\omega}{2\pi} \tilde{g}(\omega) e^{i \omega t}\right) \\
&amp;= \int_{-\infty}^\infty dt \, g(t) e^{-i \omega' t} \\
&amp;\equiv \tilde{g}(\omega') \, . \\
\end{align}
This shows that by definition $D(\omega-\omega') = 2\pi \, \delta(\omega - \omega')$.
We didn't have to say anything weird about distributions or use any magic.

Fourier transform of periodic function

Now consider a periodic function $x(t)$ with period $T$.
Since $x$ is periodic we can write it as a Fourier series
$$x(t) = \sum_{n=-\infty}^\infty \hat{x}_n e^{i 2\pi n t/T} \, .$$

Now let's compute the Fourier transform,
\begin{align}
\tilde{x}(\omega)
&amp;\equiv \int_{-\infty}^\infty x(t)e^{-i\omega t} \, dt\\
\text{plug in Fourier series} \qquad &amp;= \sum_{n=-\infty}^\infty \hat{x}_n \underbrace{\int_{-\infty}^\infty dt \, e^{i(2\pi n/T -\omega) t}}_{D(2\pi n / T - \omega)} \\
&amp;= \sum_{n=-\infty}^\infty 2\pi \, \hat{x}_n \delta(\omega - 2\pi n / T) \, .
\end{align}
So there we go: the Fourier transform of a periodic function $x(t)$ is a sum of delta functions appearing at $x(t)$'s Fourier series frequencies (i.e. the frequencies $2 \pi n / T$).
The weight of each delta function is $2\pi$ times the Fourier series coefficient $\hat{x}_n$.
The factor of $2\pi$ is there simply because we used the Fourier transform convention with angular frequency $\omega$ but the Fourier series convention with regular frequency.