I have an assignment which is:
Show that $\int_{-\infty}^\infty (\frac{\sin x}{x})^2dx=\pi$ by calculate the Fourier Transformation of $ f(t) = \left\{ \begin{array} /1, |t| \leq 1 \\ 0, |t| > 1 \end{array} \right.\ $
So I calculated the Fourier Transformation which will be:
$$ f\hat(t) = 2 \int_0^1e^{-iwt}dt = 2\frac{i(e^{-iw}-1)}{w}= \frac{2((icosv+sinv)-1)}{w}$$
But I don't understand how I could use that to prove the first statement... What am I supposed to do next?
You can use Parseval's Theorem which we reproduce below in terms of $\omega$ $$\int_{-\infty}^{\infty}\left|f(t)\right|^2dt=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left|F(\omega)\right|^2d\omega$$ where $F(\omega)$ is the continuous Fourier Transform of $f(t)$.
Now, for the given $f(t)$ the Fourier Transform is $$F(\omega)=\int_{-1}^{1}e^{-i\omega t}dt=\frac{1}{iw}(e^{i\omega}-e^{-i\omega})=\frac{2\sin(\omega)}{\omega}$$ Substituting this into Parseval's theorem results in $$\int_{-1}^{+1}1.dt=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{4\sin^2(\omega)}{\omega^2}d\omega$$ Replacing $\omega$ by $x$ and rearranging, we have $$\int_{-\infty}^{\infty}\left(\frac{\sin(x)}{x}\right)^2dx=\pi$$
