We have the function $$f\left(x\right)=\begin{cases} 1+x, & -1\leq x\leq0\\ 1-x, & 0\leq x\leq1\\ 0, & \textrm{otherwise} \end{cases} $$ we computed its Fourier transform:
$$\hat f(t) = {\sin^2(\pi t) \over \pi^2 t^2}$$
we are asked to deduce the value of the integral
$$\int_0^{\infty} {\sin^4 x \over x^4} dx$$
how to do this?
Help is much appreciated
By Parseval's theorem we have $$\begin{align} \int_{-\infty}^{\infty}\frac{\sin^{4}\left(x\right)}{x^{4}}dx= & \pi\int_{-\infty}^{\infty}\frac{\sin^{4}\left(\pi t\right)}{\left(\pi t\right)^{4}}dt \\ = & \pi\int_{-1}^{0}\left(1+x\right)^{2}dx+\pi\int_{0}^{1}\left(1-x\right)^{2}dx \\ = & \frac{2}{3}\pi \end{align} $$ and now observe that $$\int_{-\infty}^{\infty}\frac{\sin^{4}\left(x\right)}{x^{4}}dx=2\int_{0}^{\infty}\frac{\sin^{4}\left(x\right)}{x^{4}}dx$$ so