How to solve this?
$$ g(t) = t \Theta (t) $$
$$ g * g(t)$$
I had hope to be able to use the $\delta $ function in some way to get eaiser calculations, but I can't see how. Is there any way to divide this into easier functions?
I would think that $(t*t)\Theta(t)$ would be easier to solve, but I can't find any rule that allows me to do that rewrite.
Suppose you have two (sufficiently regular) functions $a,b \colon \mathbb{R} \to \mathbb{R}$, and set $u(t) = a(t)\Theta(t)$, $v(t) = b(t) \Theta(t)$. Then for the convolution $u\ast v$ you get
\begin{align} (u\ast v)(t) &= \int_{\mathbb{R}} u(x)v(t-x)\,dx\\ &= \int_{\mathbb{R}} a(x)\Theta(x)v(t-x)\,dx\\ &= \int_{[0,\infty)} a(x)v(t-x)\,dx\\ &= \int_{[0,\infty)} a(x) b(t-x)\Theta(t-x)\,dx\\ &= \int_{[0,t]} a(x) b(t-x)\,dx, \end{align}
since $\Theta(y) = 0$ for $y < 0$.
Using that with $a = b = \operatorname{id}$, it is easy to compute $(g\ast g)(t)$.