Convolution - Heaviside

Question

I'm having a hard time seeing how $\int_0^t f(u)H(t-u-1)du$ where H is the Heaviside function, is equal to $0$ for $t<1$ and $\int_0^{t-1}f(u)du$ for $t>1$. I know of course that $H(x)$ is generally zero for $x<0$ and $1$ for $x>0$ but I don't see what happened here. Thank you!

Answer 1

Let's put the question in a form where $H(x)$ is considered explictly without compounding it with the linear map $u\mapsto t-u-1$. Consider the convolution integral and apply the change of variable $t-u-1\mapsto y$: then $$ \begin{split} \int\limits_0^t f(u)H(t-u-1)\mathrm{d}u &= \int\limits^{t-1}_{-1} f(t-y-1)H(y)\mathrm{d}y\\ & = \int\limits^{t-1}_{-1} f(t-y-1)H(y)\mathrm{d}y \\ & = \begin{cases} 0 & t\le 1\\ \\ \displaystyle\int\limits^{t-1}_{0} f(t-y-1)\mathrm{d}y & t >1 \end{cases} = \begin{cases} 0 & t\le 1\\ \\ \displaystyle\int\limits^{t-1}_{0} f(u)\mathrm{d}u & t >1 \end{cases} \end{split} $$

Answer 2

If $t<1$ then $t-u-1<0$ for all $u$ between $0$ and $t$ so $H(t-u-1)=0$ for all $u$ and hence the integral is $0$.

If $t>1$ then $H(t-u-1)=0$ for $u >t-1$ and $H(t-u-1)=1$ for $u <t-1$ so the integral is effectively from $0$ to $t-1$.