Showing $\int_{-\infty}^{\infty}\frac{\sin^2(\omega)}{\omega^2}d\omega=\pi$ Given the function $f$ with $f(t)=1$ for $|t|<1$ and $f(t)=0$ otherwise, I have to calculate its Fourier-transform, the convolution of $f$ with itself
and from that I have to show that 
$$\int_{-\infty}^{\infty}\frac{\sin^2(\omega)}{\omega^2}d\omega=\pi$$ and
$$\int_{-\infty}^{\infty}\frac{\sin^4(\omega)}{\omega^4}d\omega
=\frac{2\pi}{3}$$
( $\tilde{f}(\omega)=\frac{1}{\sqrt{2\pi}}$$\int_{-t}^{t}1\cdot 
e^{-i\omega t}dt$ be the Fourier-transform of $f$ )
For the first two parts I have:
$\tilde{f}(\omega)=\frac{2}{\sqrt{2\pi}}\frac{\sin(\omega t)}{\omega}$ and
$(f*f)(\omega)=\frac{2}{\pi}\frac{\sin^2(\omega t)}{\omega^2}$.
But from here I dont know how to compute the integrals.
My idea for the first one was using Fourier-Inversion of $(f*f)$ and then
putting $t=1$. But that gives me $0$ for the integral.
Does someone has another idea?
I would be grateful for any hint or advice!
Thank you.
 A: Hint: Use Plancherel theorem
$$\int _{-\infty }^{\infty }|f(x)|^{2}\,dx=\int_{-\infty }^{\infty}|{\hat {f}}(\omega)|^{2}\,d\omega$$
then
$$\int _{-1}^{1}dx=\int_{-\infty }^{\infty}\frac{4}{2\pi}\dfrac{\sin^2\omega}{\omega^2}\,d\omega$$
A: Your expression for $\tilde{f}\left(\omega\right)$ is incorrect; you should have $$\tilde{f}\left(\omega\right)=\frac{1}{\sqrt{2\pi}}\int_{-1}^{1}e^{-i\omega t}dt=\sqrt{\frac{2}{\pi}}\frac{\sin\omega}{\omega}.$$The integrals you seek are then $$\int_{\Bbb R}\frac{\pi}{2}\left|\tilde{f}\left(\omega\right)\right|^{2}d\omega,\,\int_{\Bbb R}\frac{\pi^{2}}{4}\left|\tilde{f}\left(\omega\right)\right|^{4}d\omega.$$By Plancherel, the first integral is $\frac{\pi}{2}\int_{-1}^1dt=\pi$, while the second is $\frac{1}{2\pi}\frac{\pi^2}{4}\int_{\Bbb R}|(f\ast f)(t)|^2dt$. In terms of Iverson brackets, $$(f\ast f)(t)=\int_{\Bbb R}[u\in[-1,\,1]][t-u\in[-1,\,1]]du.$$I'll leave it as an exercise to verify $(f\ast f)(t)=(2-|t|)[t\in[-2,\,2]]$, so$$\frac{\pi}{8}\int_{\Bbb R}|(f\ast f)(t)|^2dt=\frac{\pi}{4}\int_0^2(2-t)^2dt=\frac{2\pi}{3}.$$
