Let $f(x) = \operatorname{sinc}(x)^2$. Find $(f*f)(x)$?
This is what I tried
$f(x)=\mathrm{sinc}(x)^2$ $$ \begin{align} f\ast f(x) &=\int f(u)\cdot f(x-u)\,\mathrm{d}u\\ &=\int\mathrm{sinc}(u)^2\cdot\mathrm{sinc}(x-u)^2\,\mathrm{d}u\\ \end{align} $$ Since $\mathrm{sinc}(x)=\frac{\sin(x)}{x}$. I have no clue what to do after this?
We can use the Fourier Transform to ease some of the work $$ \begin{align} \int_{-a}^ae^{-2\pi ix\xi}\,\mathrm{d}x &=\frac1{-2\pi i\xi}\left(e^{-2\pi ia\xi}-e^{2\pi ia\xi}\right)\\[6pt] &=\frac{\sin(2\pi a\xi)}{\pi\xi} \end{align} $$ Thus, the Fourier Transform of $\overbrace{\pi\left[|\xi|\le\frac1{2\pi}\right]}^\chi$ is $\mathrm{sinc}(x)$, where $[\cdots]$ are Iverson Brackets.
and the Fourier Transform of $\overbrace{\pi\left(1-\pi|\xi|\right)\left[|\xi|\le\frac1\pi\right]}^{\chi\ast\chi}$ is $\mathrm{sinc}^2(x)$
and the Fourier Transform of $\overbrace{\pi^2\left(1-\pi|\xi|\right)^2\left[|\xi|\le\frac1\pi\right]}^{(\chi\ast\chi)^2}$ is $\mathrm{sinc}^2\!\ast\mathrm{sinc}^2(x)$
Apply the inverse Fourier Transform $$ \begin{align} \pi^2\int_{-1/\pi}^{1/\pi}(1-\pi|\xi|)^2e^{2\pi ix\xi}\,\mathrm{d}\xi &=\pi^2\int_{-1/\pi}^{1/\pi}(1-\pi|\xi|)^2\cos(2\pi x\xi)\,\mathrm{d}\xi\\ &=2\pi^2\int_0^{1/\pi}(1-\pi\xi)^2\cos(2\pi x\xi)\,\mathrm{d}\xi\\ &=2\pi\int_0^1(1-\xi)^2\cos(2x\xi)\,\mathrm{d}\xi\\ &=\frac{2\pi}{x}\int_0^1(1-\xi)\sin(2x\xi)\,\mathrm{d}\xi\\ &=\frac\pi{x^2}-\frac{\pi}{x^2}\int_0^1\cos(2x\xi)\,\mathrm{d}\xi\\ &=\bbox[5px,border:2px solid #F0A000]{\frac{\pi(x-\sin(x)\cos(x))}{x^3}} \end{align} $$