# 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.

• Are you unclear on why the integral converges, why it converges to that value, how how one would prove the statement? – user121330 May 30 '17 at 19:11
• I don't understand why it converges at that value, given the definition of complex exponential involves sin and cos – tommy1996q May 30 '17 at 19:27
• Do you know what the Delta function is? – user121330 May 30 '17 at 19:35
• Wait, I thought it was Kronecker delta! No, I don't know what the Dirac delta is. I've just heard that another important measure, except for Lebesgue, is the Dirac measure. – tommy1996q May 30 '17 at 19:38
• They are similar (thus the similar notation). Kronecker is 1 when two indices are the same. Dirac's Delta is a special $\infty$ (such that it integrates to 1) when the argument is zero. The wiki article should clear things up for you. – user121330 May 30 '17 at 19:47

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)

• I never used things like Schwartz functions or tempered distributions, I have just started studying Fourier transform. Could you please explain it in a bit more step-by-step way? – tommy1996q May 30 '17 at 19:35
• @tommy1996q : Do you know what is the Dirac delta $\delta$ defined by $$\int_{-\infty}^\infty \delta(t) \varphi(t)dt = \varphi(0)$$ whenever $\varphi$ is continuous ? What is its Fourier transform ? It is not complicated : $$\mathcal{F}[\delta](\omega) = \int_{-\infty}^\infty \delta(t) e^{-i \omega t}dt = e^{-i \omega t}|_{t= 0} = 1$$ thus, by the Fourier inversion theorem of distributions $$\delta = \mathcal{F}^{-1}[1]$$ and you get what I wrote in my answer when $\omega' = 0$ – reuns May 30 '17 at 20:01
• Ok I think I got it now, thanks! – tommy1996q May 30 '17 at 20:40

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

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.

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:

$$\tag{1} \int_{-\infty}^\infty e^{i ( \omega - \omega )t} dt=\int_{-\infty}^\infty 1\, dt = \infty$$

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.

• the integral doesn't converge – reuns May 30 '17 at 20:18
• Are you asking about whether the integral converges when $\omega = \omega'$ or when $\omega \neq \omega'$? – user121330 May 30 '17 at 20:49
• The improper Riemann integral $\int_{-\infty}^\infty e^{i a t}dt$ doesn't converge for any $a \in \mathbb{R}$ – reuns May 30 '17 at 20:56
• Also the Dirac delta distribution $\delta$ is not a function. Its definition is $\int_{-\infty}^\infty \delta(t) \varphi(t)dt = \varphi(0)$ whenever $\varphi$ is continuous – reuns May 30 '17 at 20:58
• @user1952009 But it doesn't converge in different ways. Also, you should edit that title on Wikipedia and Wolfram Mathworld if you're adamant about the function thing. It accepts a number and returns a number; that's the definition of a function. – user121330 May 30 '17 at 21:13