Calculate integral using convolution

Question

I want to calculate $$\int_{-\infty}^{+\infty}\frac{\sin^2x}{x^2}dx$$ and I'm trying to do via convultions and Fourier Transforms. I know that the fourier transform of $1_{[-1,1]}$, the indicator function on the interval $[-1,1]$, is $\frac{2\sin(s)}{s}$ and I have worked out that the convolution $f*f = 2-|x|$ and so was attempting to use the fact that $\mathfrak{F}(f*f) = \mathfrak{F}(f)\mathfrak{F}(f)$. But I am struggling with this last step, any help?

Answer 1

Let $$f: \mathbb R \to \mathbb R, \quad f(x) = \begin{cases} \frac{\sin x}{x} & x \neq 0 \\ 1 & x= 0 \end{cases}.$$ Note that, for $x\neq 0$, $$\mathfrak F^{-1} \mathbb 1_{[-1,,1]}(x) = \frac{1}{\sqrt{2\pi}} \int_\mathbb R e^{ikx} \mathbb 1_{[-1,1]}(k)\, dk = \frac{1}{\sqrt{2\pi}} \int_{-1}^1 e^{ikx} \, dk = \left [\frac{e^{ikx}}{ix} \right ]_{k=-1}^{k=1} = \sqrt{\frac{2}{\pi}}\frac{\sin x}{x},$$ so $$f = \sqrt{\frac{\pi}{2}} \mathfrak F^{-1}\mathbb 1_{[-1,1]}.$$ By Plancherel's theorem we get $$\int_{\mathbb R \setminus \{0\}}\left(\frac{\sin x}{x}\right )^2 \, dx = \lVert f \rVert_2^2 = \frac{\pi}{2}\lVert \mathbb 1_{[-1,1]}\rVert_2^2 = \pi.$$