For $t\geq0$, let $g_\beta(t)=e^{-t}\sin(\beta t)$, where $\beta$ is a real number, and for $t<0$, $g_\beta(t)=0$. Find $h*g_\beta(t)$ for all $t\geq0$, where $h(t)=\begin{cases}1/d&t\in[0,d]\\0&\text{otherwise}\end{cases}$.
I know there are two cases I have to consider: $0\leq t<d$ and $d\leq t$. When $0\leq t<d$, I found that $$h*g_\beta=\frac{1}{d}\int_0^te^{-\tau}\sin(\beta\tau)d\tau,$$ but I don't see what the bounds would be when $d\leq t$. Would it be $\int_{t-d}^d$? Or $\int_{t-d}^t$? $h=0$ when $t>d$ so I think it should be $\int_{t-d}^d$ but my intuition on these is more often wrong than right.
The convolution is $\int_0^\infty h(t-\tau)\,g_\beta(\tau)\,d\tau$. Now, $h(t-\tau)\ne0$ if and only if $0\le t-\tau\le d$, that is, $t-d\le\tau\le t$. If $t\ge d$, the interval of integration is $[t-d,t]$.