# Computation of integral involving Heaviside function

Let $H : \mathbb{R} \to \mathbb{R}$ denote the Heaviside function:

$$H(y) = \begin{cases} 0 & y < 0, \\ 1 & y \ge 0. \end{cases}$$

Suppose that $c > 1$ is fixed with $t$ ranging over $\mathbb{R}$. Prove that: $$\int_0^\infty H(t + (1- \frac{1}{c})y - 1)H(t - \frac{y}{c})dy = \begin{cases} 0 & t < \frac{c}{2c-1}, \\ \frac{c}{c-1}((2c-1)t-c) & \frac{c}{2c-1} \le t \le 1, \\ ct & t > 1. \end{cases}$$

This integral appears in a research paper (on inverse problems) that I've been studying. I've been trying to work out all the cases for awhile.

The case $t >1$ is easy enough to work out. In that situation, $t + (1 - \frac{1}{c})y - 1 > (1 - \frac{1}{c})y \ge 0$, all $y \ge 0$. And $t - \frac{y}{c} \ge 0 \iff y \le ct$ Hence, $H(t + (1- \frac{1}{c})y - 1)H(t - \frac{y}{c})$ = 1 for $y \in [0,ct]$ and $0$ otherwise (this gives the correct answer for $t > 1$).

With the other cases, my main stumbling point is determining why the function should behave differently as we cross $t = \frac{c}{2c-1}$.

Hints or solutions are greatly appreciated.

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