calculus - Proof of $int_0^infty left(frac{sin x}{x}right)^2 mathrm dx=frac{pi}{2}.$

I am looking for a short proof that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$$
What do you think?

It is kind of amazing that $$\int_0^\infty \frac{\sin x}{x} \mathrm dx$$ is also $\frac{\pi}{2}.$ Many proofs of this latter one are already in this post.

Answer

Let $f(x)=\max\{0,1-|x|\}$. It is easy to calculate the Fourier transform
$$\hat{f}(\xi)=\int_{-\infty}^{\infty}f(x)e^{-ix\xi}dx=\left(\frac{\sin(\xi/2)}{\xi/2}\right)^2.$$
Taking the inverse Fourier transform, we get
$$\int_{-\infty}^{\infty}\left(\frac{\sin(\xi/2)}{\xi/2}\right)^2e^{ix\xi}d\xi=2\pi f(x),$$
and the result follows.

The second integral can be computed in a similar way. Just take $f(x)=\chi_{[-1,1]}(x)$ (the indicator function of the interval $[-1,1]$).

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Edit. It might be interesting to note that there are analogous formulas for the sinc
sums
$$\sum_{n=1}^{\infty}\frac{\sin n}{n}=\sum_{n=1}^{\infty}\left(\frac{\sin n}{n}\right)^2=
\frac{\pi}{2}-\frac{1}{2}.$$

I learned about this from the note "Surprising Sinc Sums and Integrals" by Baillie, Borwein, and Borwein (can be found through a quick web search).