= Solution
Insert the transform definitions and use Fubini:
$$
\frac1{2\pi}\int\tilde f(k)\tilde g(k)e^{ikx}dk
=\int f(u)\left[\frac1{2\pi}\int\tilde g(k)e^{ik(x-u)}dk\right]du
=\int f(u)g(x-u)du.
$$
For the indicator of $(-1,1)$,
$$
\tilde f(k)=\int_{-1}^1e^{-ikx}dx=\frac{2\sin k}{k}.
$$
Putting $g=f$ and $x=0$ in the <convolution theorem> gives
$$
2=\frac1{2\pi}\int_{-\infty}^{\infty}\frac{4\sin^2k}{k^2}dk,
$$
and hence
$$
\boxed{\int_{-\infty}^{\infty}\frac{\sin^2k}{k^2}dk=\pi.}
$$
Back to article page
