Showing $\int_{-\infty}^{\infty}\frac{\sin^2(\omega)}{\omega^2}d\omega=\pi$

Question

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.

Answer 1

Hint: Use Plancherel theorem $$\int _{-\infty }^{\infty }|f(x)|^{2}\,dx=\int_{-\infty }^{\infty}|{\hat {f}}(\omega)|^{2}\,d\omega$$ then $$\int _{-1}^{1}dx=\int_{-\infty }^{\infty}\frac{4}{2\pi}\dfrac{\sin^2\omega}{\omega^2}\,d\omega$$