I have a problem. I'm studing Fourier analysis and i saw a metode to calculate $\int_0^{\infty}\frac{\sin(x)}{x}dx$ using Fourier transform but I'm asked to calculate $\int_0^{\infty}(\frac{\sin(x)}{x})^2dx$ . The general idea is to find a function, that has the Fourier transform exactly the function that I have to integrate and then to apply the inverse theorem or something else? Any tip on understanding the general method on integrating this kind of function using Fourier transform?
Also I would be happy to know if this is true for the Fourier series method. I saw that is possible to calculate the sum of the series $\sum_{n\ge1}\frac{1}{n^2}$. Using Fourier series, is it possible to calculate other interesting series? And what is the general method?
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 $$\int_{0}^{\infty} \frac{\sin(x)}{x} \mathrm{d}x = \frac{\pi}{2}$$
To calculate other integrals of other powers, it is sufficient to use Plancherel. The fourth power 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
$$\int_{0}^{\infty}\frac{\sin^4(x)}{x^4} \mathrm{d}x = \frac{\pi}{3}$$
This will hold also for other powers. I leave you the second power as an exercise.