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I have to evaluate $$\int_{-\infty}^{\infty} \bigg(\frac{\sin x}{x}\bigg)^{2}\ dx$$ and hint says that use Plancherel theorem. Now, in my notes Plancherel theorem is just a statement that we can extend fourier transform map which was defined originally from $S(\mathbb R)$ to $S(\mathbb R)$ to a map from $L^2(R)$ to $L^2(R)$. But I am not getting at all how Plancherel theorem is going to help me evaluate this integral.Any help.Thanks.

Andrew Chin
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ogirkar
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  • by $S(\mathbb R)$, i mean schwarz space – ogirkar Nov 16 '19 at 17:07
  • Plancherel’s Theorem usually refers to the fact the the square integral of a nice function is equal to the square integral of its Fourier transform, which seems pretty applicable here. – spaceisdarkgreen Nov 16 '19 at 17:28
  • @spaceisdarkgreen..can you show me first few steps, i am still not clear – ogirkar Nov 16 '19 at 17:36
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    It means that you can compute this integral by computing the square integral of the Fourier transform of $\sin(x)/x$ instead of the square integral of $\sin(x)/x$ itself. So, find the Fourier transform of $\sin(x)/x$ and observe that its square integral is very easy to compute. – spaceisdarkgreen Nov 16 '19 at 17:46
  • If that's really all the Plancherel says in your notes you need to take much better notes - you left out a huge part of the statement, namely that if $\int| f|^2<\infty$ then $\int|\hat f|^2=\int|f|^2$. – David C. Ullrich Nov 16 '19 at 18:01
  • @spaceisdarkgreen ok..now i understood what exactly i have to do...one last question ..do i have to do all the stuff done here(i am sharing link below) to evaluate fourier transform of $\frac{sinx}{x}$..it is pretty long way – ogirkar Nov 16 '19 at 22:24
  • https://math.stackexchange.com/q/281870/409319 – ogirkar Nov 16 '19 at 22:25
  • @DavidC.Ullrich..actually this equality is mentioned as parseval's identity..that's why i got bit confused – ogirkar Nov 16 '19 at 22:26
  • @Believer I don't know what you have to do... that depends on the context of the course. It's certainly a "well-known fact" that the fourier transform of a sinc is a rect, so I'm not sure if you'd be required to justify it here. The main answer you linked is straightforward even if it does go into extreme detail (and there are other answers). As another idea, look at things in reverse: it's easier to show that the fourier transform of a rect is a sinc. (But I don't recall how easy it is to make this shortcut rigorous.. sinc is not $L_1$ so there may be complications with Fourier inversion.) – spaceisdarkgreen Nov 16 '19 at 23:18
  • @spaceisdarkgreen...Thanks. – ogirkar Nov 16 '19 at 23:53

2 Answers2

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Apply integration-by-parts,

$$\int_{-\infty}^{\infty} \bigg(\frac{\sin x}{x}\bigg)^{2}\ dx = -2\int_{0}^{\infty} {\sin^2x}\ d(\frac1x) =2\int_{0}^{\infty}\frac{\sin 2x}{x}dx$$ $$=2\int_{0}^{\infty}\frac{\sin t}{t}dt=2\cdot \frac\pi2=\pi$$

Quanto
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Hint What is the Fourier transform of the rectangular function?

Gribouillis
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  • so we know that the function will be $\mathbb{1}{[-1,1]}$ and the fourier transform of that gives us $\frac{\sin x}{x}$. My question is how do we compute the integral $2\pi \int{-\infty}^{\infty}|f|^2 dx$? – Anon Feb 19 '20 at 13:21
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    @Anon It depends on how you normalize the Fourier transform. If you define it by ${\cal F} f(\xi)= \frac{1}{\sqrt{2\pi}}\int_{\mathbb R} e^{-i x \xi} f(x) d x$, then one has ${\cal F}1_{[-1, 1]}={\sqrt{\frac{2}{\pi}} \frac{\sin(\xi)}{\xi}}$. The Plancherel theorem tells you that $f$ and ${\cal F}f$ have the same $L^2$ norm ($\sqrt{2}$ in our case). This gives the desired integral ($\pi$). – Gribouillis Feb 19 '20 at 22:57