# Convolution integral involving two Heaviside functions

I am having trouble solving the following integral involving two Heaviside functions, obtained from a Laplace transform convolution:

$\Large \int_0^t \frac{\tau}{\sqrt{\tau^2 - \alpha^2}} H(\tau - \alpha) H(t - \tau) \text{d}\tau$

where $H(\cdot)$ is the Heaviside function.

It is my understanding that for $\tau < \alpha$, the integrand cancels, but I'm quite confused about how this is affected by the second Heaviside function.

Can anyone help me? Any help will be appreciated.

Another way to approach it in this case: we are guaranteed that $0 < \tau < t$ at all times, just due to the integration bounds. This means that $t - \tau > 0$ for the whole integration, so $H(t - \tau) = 1$. This reduces it to $$\int_0^t \frac{\tau}{\sqrt{\tau^2-\alpha^2}} H(\tau-\alpha) d\tau$$
At this point, as you stated -- if $\tau < a$, then the integrand vanishes. And for $\tau > \alpha$, the $H(\tau - \alpha)$ becomes 1. So we can break it up into these two parts. (As I said, case-splitting!)
$$\int_0^t \frac{\tau}{\sqrt{\tau^2-\alpha^2}} H(\tau-\alpha) d\tau =$$ $$\int_0^\alpha \frac{\tau}{\sqrt{\tau^2-\alpha^2}} H(\tau-\alpha) d\tau + \int_\alpha^t \frac{\tau}{\sqrt{\tau^2-\alpha^2}} H(\tau-\alpha) d\tau =$$ $$\int_\alpha^t \frac{\tau}{\sqrt{\tau^2-\alpha^2}} d\tau$$
which should work out fine. (Keep in mind that this final form is only true if $\alpha>0$, though! )