Proof of Plancherel formula I was looking at this question posted here some time ago. 
How to Prove Plancherel's Formula?
I get it until in the third line he practically says that $\int _{- \infty}^{+\infty} e^{i(\omega - \omega')t} dt= 2 \pi \delta(\omega - \omega')$.
I mean, I would understand if we were integrating over a period of length $2 \pi$, but here the integration is over $\mathbb{R}$. 
P.S. I would have asked this directly to the author of the post, but it's been over a year since he last logged in.

Supposing that $\omega = \omega'$, it's clear that we don't expect the integral to converge - the Dirac's Delta function is infinite at zero:
\begin{equation}\tag{1}
\int_{-\infty}^\infty e^{i ( \omega - \omega )t} dt=\int_{-\infty}^\infty 1\, dt = \infty
\end{equation}
In the case that $\omega \neq \omega'$, neither the positive nor the negative sides converge separately, but they don't have to. 
$$\begin{eqnarray}
\int_{-\infty}^\infty e^{i u} du &=& i \int_{-\infty}^\infty \sin(u) du + \int_{-\infty}^\infty \cos(u) du \\
\int_{-\infty}^\infty \sin(u) &=& \lim_{a \rightarrow \infty} \int_{-a}^a \sin(u) du \\
&=& \lim_{a \rightarrow \infty} - \cos(a) + \cos(-a)\\
&=& \lim_{a \rightarrow \infty} 0 \\
&=& 0 \\
\int_{-\infty}^\infty \cos(u) &=& \int_{-\infty + \frac{\pi}{2}}^{\infty + \frac{\pi}{2}} \cos \left(x - \frac{\pi}{2} \right) dx \\
&=& \int_{-\infty}^{\infty} \sin(x) dx \\
&=& 0 \\
\int_{-\infty}^\infty e^{i u} du &=& 0 + i 0 = 0
\end{eqnarray}$$
This may feel uncomfortable but it's as rigorous as $a - a + b - b = 0$.
The final ambiguity is the value of $\infty$ from Equation 1. Without getting into the weeds of how one defines frequency (which moves that $2\pi$ all over the place), we know from the definition of the Fourier Transform that both:
$$\begin{eqnarray}
\mathcal{F}(g(t)) &=& G(\omega) \\
\mathcal{F}(G(\omega)) &=& g(t)
\end{eqnarray}$$
Since $\mathcal{F(F}(g(t))) = \mathcal{F}(G(\omega)) = g(t)$, we may equivalently show that
$$\mathcal{F}(2 \pi \delta(\omega - \omega')) = \int_{-\infty}^{\infty} e^{i \omega t} \delta(\omega - \omega') d\omega = e^{-i \omega' t} = g(t)$$
by the sifting property which is what we sought.
