Heaviside function in an integral I having problems understanding how this integral is evaluated. Let $c,\beta$ be constants.
$\begin{align*}u(x,t)&=\frac{c}{\beta}\int_{0}^{t}H(s-x/c)f(t-s)ds \\
&=\frac{c}{\beta}H(t-x/c)\int_{0}^{t-x/c}f(\tau)d\tau \ \ \ \  \text{using the substitution $\tau=t-s$}
\end{align*}$.
I have tried to understand but I cannot see how the second inequality works. If anyone able to add a step(s) between the two equalities.
Thanks

This is not necessary, but it helps to see:
$$
    \int_{0}^{t}H(s-x/c)f(t-s)\,ds=\int_{0}^{t}H(-(t-s)+t-x/c)f(t-s)\,ds.
$$
Now let $\tau=t-s$. Then $d\tau=-ds$, and the new limits of integration are $\tau=t-0$ and $\tau=t-t=0$. So this gives $\int_{t}^{0}\cdots (-d\tau)=\int_{0}^{t}\cdots d\tau$, which leads to the equivalent expression
$$
           \int_{0}^{t}H(t-x/c-\tau)f(\tau)\,d\tau=\int_{0}^{t-x/c}H(t-x/c-\tau)f(\tau)\,d\tau+\int_{t-x/c}^{t}H(t-x/c-\tau)f(\tau)\,d\tau\\
  = H(t-x/c-\tau)|_{\tau=0}\int_{0}^{t-x/c}f(\tau)\,d\tau
   +H(t-x/c-\tau)|_{\tau=t}\int_{t-x/c}^{t}f(\tau)\,d\tau.
$$
The last equality holds because because $t-x/c-\tau$ does not change sign on $[0,t-x/c)$, and does not change sign on $(t-x/c,t]$. So,
$$
   \int_{0}^{t}H(t-x/c-\tau)f(\tau)\,d\tau=H(t-x/c)\int_{0}^{t-x/c}f(\tau)\,d\tau
     + H(-x/c)\int_{t-x/c}^{t}f(\tau)\,d\tau.
$$
You have not given enough information to go further. If $x/c > 0$, then the second term is $0$, and you get what you want.
