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.
 A: A classical way to interpret what you have is through the Fourier transform and its inverse. If $f$ is continuous at $x$ where it has left- and right-hand derivatives, and if $f$ is suitably integrable on $\mathbb{R}$, then
$$
     \lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\hat{f}(s)e^{isx}ds = f(x).
$$
This can be written as
\begin{align}
      f(x)&=\lim_{R\rightarrow\infty}\frac{1}{2\pi}\int_{-R}^{R}\int_{-\infty}^{\infty}f(t)e^{-ist}dt e^{isx}ds \\
    &=\lim_{R\rightarrow\infty}\int_{-\infty}^{\infty}\left(\frac{1}{2\pi}\int_{-R}^{R}e^{is(x-t)}ds\right)f(t)dt
\end{align}
This is being represented in a short-hand form as
$$
       \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{is(x-t)}ds=\delta(x-t).
$$
There are several ways to interpret the above, but none of them including treating the integral by itself.
The symmetric truncated integral is
$$
      \frac{1}{2\pi}\int_{-R}^{R} e^{is(x-t)}ds = \frac{1}{\pi}\frac{\sin(R(x-t))}{x-t}.
$$
So you're really looking at a very classical limit of an integral:
$$
           \lim_{R\uparrow\infty}\frac{1}{\pi}\int_{-\infty}^{\infty}f(t)\frac{\sin(R(x-t))}{x-t}dt = f(x).
$$
A: As you noted $\int _{- \infty}^{+\infty} e^{i(\omega - \omega')t} dt= 2 \pi \delta(\omega - \omega')$ is of course not true. This is an abuse of notation, what it really means is that the Fourier transform of the (tempered) distribution $f(\omega) = e^{i \omega' t}$ is  the (tempered) distribution $\hat{f}(\omega) = 2 \pi \delta(\omega-\omega')$.

In general, if $f$ is a tempered distribution, then its Fourier transform (in the sense of distributions) is the tempered distribution $\hat{f}$ iff for every Schwartz function $\varphi$ : 
$$\int_{-\infty}^\infty f(t) \varphi(t)dt = \int_{-\infty}^\infty \hat{f}(\omega)\hat{\varphi}(\omega)d\omega$$
(where in general $\int_{-\infty}^\infty f(t) \varphi(t)dt$ is not a Riemann integral, but the pairing of a distribution with a test function)
A: I think he used that $$ 1 = \hat{\delta(w)} $$ so, $$\int _{-\infty}^{+\infty} e^{i(\omega-\omega ')t} dt $$ is the antitransform of $\delta$  values in $(\omega - \omega') $ plus $2\pi$ for definition of antitransform.
A: 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.
