Proof of Plancherel formula

Question

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.

Answer 1

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.

Answer 2

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)

Answer 3

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.

Answer 4

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: $$ \lim_{R\uparrow\infty}\frac{1}{\pi}\int_{-\infty}^{\infty}f(t)\frac{\sin(R(x-t))}{x-t}dt = f(x). $$