Computation of integral involving Heaviside function

Question

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.

Answer 1

I don't know how you get the result but it seems wrong to me.

$H(..)H(..)\neq 0$ requires

1. $t-y/c\ge 0 \Rightarrow y \le ct$
2. $t+(1-1/c)y-1\ge 0 \Rightarrow y\ge c(1-t)/(c-1)$

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.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\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}

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

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