Use $\widehat{f*f}(x)=2\sqrt{2/\pi}\left( \sin x/x \right)^2$ to show that $\int_{-\infty}^{+\infty}\left( \frac{\sin x}{x} \right)^2 dx=\pi. $

Question

I have already showen that the Fourier transform of $ f=\chi_{[-1,1]} $ is $$ \widehat{f}(\omega)=\sqrt{\frac{2}{\pi}}\frac{\sin \omega}{\omega} $$ Then I showed that $$ \widehat{f*f}(x)=2\sqrt{\frac{2}{\pi}}\left( \frac{\sin x}{x} \right)^2 .$$ I would now like to use the above to show that $$\int_{-\infty}^{+\infty}\left( \frac{\sin x}{x} \right)^2 dx=\pi. $$ I was thinking maybe I can use the inverse Fourier transform $$ f(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\widehat{f}(x)e^{i\omega x}dx\\ $$ (do I have the definition right?) or maybe the relation $ \widehat{\widehat{f}}(x)=f(-x) $, but I don't see how neither of these does the trick. Can anyone help me out?

Answer 1

Your work looks fine so far.

To finish off, note that the Fourier inversion formula tells you that

$$ f \ast f (t) = \frac 1 {\sqrt{2\pi}}\int_{-\infty}^\infty \widehat{f \ast f} (\omega) e^{i\omega t}d\omega.$$

So if you substitute $t = 0$ on both sides, you get

$$ f \ast f (0) = \frac 1 {\sqrt{2\pi}}\int_{-\infty}^\infty \widehat{f \ast f} (\omega) d\omega,$$

i.e.

$$ \int_{-\infty}^\infty \chi_{[-1,1]}(u) \chi_{[-1,1]}(-u) du = \frac 2 \pi \int_{-\infty}^{\infty} \left( \frac{\sin \omega}{\omega}\right)^2 d\omega $$

and it should be simple to finish off from here.

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By the way, it's worth pointing out that the final result can be thought of as a special case of Parseval's theorem:

$$ \int_{-\infty}^\infty |f(x)|^2 dx = \int_{\infty}^{\infty} | \hat f(\omega) |^2 d\omega$$