I'm trying to tackle the following question:
Let $\displaystyle g_a(x)=\begin{cases}1-\frac{|x|}{a},&x<0\\0,|x|\ge0\end{cases}$. Find $g_a\ast g_a$.
So, I tried to compute it by definition $\displaystyle g_a\ast g_a=\int\limits_{-\infty}^{\infty}g_a(t)g_a(x-t)\text{d}t$.
I found that $-a< t< a$ and $-a+x<t<a+x$, so I think I could write that $$g_a\ast g_a=\begin{cases}\displaystyle \int\limits_{-a+x}^{a}\left(1-\frac{|t|}{a}\right)\left(1-\frac{|x-t|}{a}\right),&x\ge0 \\ \displaystyle\int\limits_{-a+x}^{a}\left(1-\frac{|t|}{a}\right)\left(1-\frac{|x-t|}{a}\right),&x\le 0\end{cases}$$
Is it correct? How should I continue? Is there an easier way (maybe using Fourier transform?)?
Please help, thank you!
if $h(t) = 0$ and $g(t) = 0$ for $t < 0$ : $$h \ast g (t) = \int_{-\infty}^\infty h(t-\tau) g(\tau) d\tau = \int_0^\tau h(t-\tau) g(\tau) d\tau$$