Integration using Fourier Transform How to integrate the function $(\sin x)^2/x^2$ using Fourier transform of function $g(x)=1$ if $|x|<1$ else $g(x)=0$ which is $(\sin w/w)*2\pi$?!thank you in advance.
 A: Parseval/Plancherel's identity/theorem (note that Wikipedia uses a different convention) says: if $f,g$ are square-integrable functions, $ \tilde{f}(k)= \int_{-\infty}^{\infty} e^{-ikx} f(x) \, dx $, similarly $\tilde{g}(k)$ are their Fourier transforms, then
$$ 2\pi\int_{-\infty}^{\infty} \overline{f(x)}g(x) \, dx = \int_{-\infty}^{\infty} \overline{\tilde{f}(k)}\tilde{g}(k) \, dk \tag{1} $$
(where the presence of the $2\pi$ depends upon the convention used in the Fourier transform). Taking $f(x)=g(x)= \begin{cases} 1 & \lvert x \rvert <1 \\ 0 & \text{else} \end{cases}$ gives
$$ \tilde{f}(k) = \int_{-1}^1 e^{ikx} \, dx = \frac{e^{ik}-e^{-ik}}{ik} = \frac{2\sin{k}}{k}. $$
Then using (1),
$$ 2\pi\int_{-\infty}^{\infty} \lvert f(x)\rvert^2 \, dx = \int_{-\infty}^{\infty} \lvert \tilde{f}(k) \rvert^2 \, dk \\
2\pi \int_{-1}^1 1 \, dx = \int_{-\infty}^{\infty} \left(\frac{2\sin{k}}{k}\right)^2 \, dk,  $$
and cancelling factors gives
$$ \int_{-\infty}^{\infty} \left(\frac{\sin{k}}{k}\right)^2 \, dk = \pi. $$
