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. 
 A: I don't know how you get the result but it seems wrong to me.
$H(..)H(..)\neq 0$ requires


*

*$t-y/c\ge 0 \Rightarrow y \le ct$

*$t+(1-1/c)y-1\ge 0 \Rightarrow y\ge c(1-t)/(c-1)$


If $1\ge t\ge 1/c$, then $0\le c(1-t)/(c-1)\le ct$. then the integral gives 
$$
\int_0^\infty H(...)H(...)dy = ct-c\frac{1-t}{c-1}=\frac{c}{c-1}(ct-1)
$$
If $t\le 1/c$, then $c(1-t)/(c-1)\ge ct$, thus $\forall y\ge0 ,H(..)H(...)=0$. So the integral gives 0.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
? &\equiv \int_{0}^{\infty}{\rm H}\pars{t + \bracks{1- {1 \over c}}y - 1}
{\rm H}\pars{t - {y \over c}}\,\dd y
=
\int_{0}^{\infty}{\rm H}\pars{{c\bracks{1 - t} \over c - 1} - y}{\rm H}\pars{ct - y}\,\dd y
\\[3mm]&=
\int_{0}^{\infty}{\rm H}\pars{\min\braces{ct,{c\bracks{1 - t} \over c -1}} - y}\,\dd y
\end{align}



*
*$\ds{ct < {c\pars{1 - t} \over c - 1}\quad\imp\quad
             c^{2}t - ct < c - ct\quad\imp\quad t < {1 \over c}}$
    $$
? = \int_{0}^{\infty}{\rm H}\pars{ct - y}\,\dd y
  = {\rm H}\pars{t}\int_{0}^{ct}\dd y = {\rm H}\pars{t}ct
$$
    

*$\ds{t > {1 \over c}}$
    $$
? = \int_{0}^{\infty}{\rm H}\pars{{c\bracks{1 - t} \over c - 1} - y}\,\dd y
  = {\rm H}\pars{1 - t}\int_{0}^{c\pars{1 - t}/\pars{c - 1}}\,\dd y
  = {\rm H}\pars{1 - t}\,{c\pars{1 - t} \over c - 1}
$$




$$
?=\left\lbrace%
\begin{array}{lcl}
0 & \mbox{if} & t < 0\ \wedge\ t > 1
\\
ct & \mbox{if} & 0 \leq t < {1 \over c}
\\
{c \over c - 1}\,\pars{1 - t} & \mbox{if} &  {1 \over c} \leq t < 1
\end{array}\right.
$$

