This is a problem from a past qualifying exam:
The Fourier transform of the characteristic function $h=\chi_{[-1,1]}$ of the interval $[-1,1]$ is
$$\hat h(\xi) =\sqrt{\frac{2}{\pi}} \frac{\sin \xi}{\xi}.$$
Using various properties of the Fourier transform, calculate
$$\int_0^\infty \frac{\sin x}{x}dx$$
$$\int_0^\infty \big(\frac{\sin x}{x}\big)^2dx$$
$$\int_0^\infty \big(\frac{\sin x}{x}\big)^4dx.$$
Note: Here we are using the definition of the Fourier transform
$$\hat f(\xi) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-ix\xi} f(x) dx.$$
To evaluate the third integral, you may make use of the formula
$$ (h\ast h)(x) = \begin{cases}
2-|x| & \ |x|<2 \\
0 & \ |x| \geq 2
\end{cases}$$.
I figured out the $\int_0^\infty \big(\frac{\sin x}{x}\big)^2dx$ portion by simply using Plancherel, i.e. $||h||_2^2 = ||\hat h||_2^2$. I am still struggling with the other cases, however. I have tried the Fourier Inversion as well as multiplication formula. Because of the hint, I would gather that in the last case you are also supposed to use the fact that $\widehat{h\ast h} = \hat h \cdot \hat h$.
Best Answer
Applying the inversion theorem, we may write
$$h(x) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{i\xi x}\sqrt{\frac{2}{\pi}}\frac{\sin(\xi)}{\xi} \mathrm{d}\xi$$ which leads to $$\int_{-\infty}^{\infty} \frac{1}{\pi}\frac{\sin(\xi)}{\xi} \mathrm{d}\xi = h(0) = 1. $$ This in turn implies
To calculate the second integral it is sufficient to use Plancherel. You should get $\pi/2$ as well if I remember correctly. The third one can be calculated in a similar fashion: $$\int_{-\infty}^{\infty} \left(\sqrt{\frac{2}{\pi}}\frac{\sin(\xi)} {\xi}\right)^4 \mathrm{d}\xi =\|\hat{h}^2\|_{L^2}^2 = \frac{1}{2\pi} \|\widehat{h\ast h}\|_{L^2}^2 = \frac{1}{2\pi}\|h \ast h\|_{L^2}^2 = \frac{1}{2\pi}\frac{16}{3} $$ so we obtain