About Fourier transform of periodic signal In Fourier transform for periodic signal, I checked different books and I found a different explanation in each book. Let's take the explanation in Signals and Systems by Rajeshwari &amp; Rao:


  The resulting Fourier transform for a periodic signal consist of a train of impulses in frequency, with areas of impulses proportional to the Fourier series coefficients. 
  
  To suggest the general result, let us consider $x(t)$ with Fourier transform $X(\omega)$ which is a single impulse of area $2\pi$ at $\omega=\omega_0$, that is,  $$X(\omega)=2\pi\delta(\omega-\omega_0)$$
  To determine $x(t)$ for which this is Fourier transform we can apply  the inverse Fourier transform to obtain 
  $$ x(t) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} 2\pi \delta(\omega-\omega_0)e^{j\omega t}d\omega$$


The things I want to ask is:  


If we have Fourier series of a periodic signal which will be one impulse, then the Fourier transform of that impulse will be the same single impulse? That's what it is explained above? 
Why we used shifted impulse? Why we can't take $\delta(\omega)$
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.