Convolution of step functions

Question

So let $h(x)=1$ when $-1<x<1$ and define $$(f*g)(x)=\int_{-\infty}^{\infty}f(x-y)g(y)\ dy$$ I'm having trouble computing $(h*h)$. As of now, I simplified to $$\int_{-1}^1 h(x-y)\ dy$$ but I'm not sure if this correct. Even if this is correct, I can't simplify it further. I'm interested in taking the Fourier transform of $(h*h)$ so I'm pretty lost at the moment.

Answer 1

Hint: \begin{align} h\ast h(x) =&\ \int^\infty_{-\infty} \chi_{[-1, 1]}(x-t)\chi_{[-1, 1]}(t)\ dt = \int^\infty_{-\infty} \chi_{[-1+x, 1+x]}(t)\chi_{[-1, 1]}(t)\ dt\\ =&\ m\{[-1+x, 1+x]\cap [-1, 1]\}. \end{align}

Answer 2

$h(x-y) = 0$ when $x-y<-1$ or $x-y > 1$

$\int_{x-1}^{x+1} h(x)h(x-y)\ dy$

$h*h(x) \begin{cases} \int_{-1}^{x+1} dy=x+2 & x<0\\\int_{x-1}^1 dy=2-x & x\ge 0 \end{cases}$