Let $f(t)$ be a signal that is $0$ when $t<0$ or $t>1$. Show that, for the Butterworth filter, one has $$Ae^{-\alpha t}\int_{0}^{\min\{t,1\}}e^{\alpha\tau}f(\tau)d\tau$$
My attempt:
\begin{equation*} \begin{aligned} L(f) =& \int_{-\infty}^\infty Ae^{-\alpha\tau}f(t-\tau)d\tau \\ =& \int_{-\infty}^{\infty}Ae^{-\alpha(t-\tau)}f(\tau)d\tau \\ =& \int_{-\infty}^\infty Ae^{\alpha(\tau-t)}f(\tau)d\tau \\ =& \int_{-\infty}^\infty Ae^{-\alpha t}e^{\alpha\tau}f(\tau)d\tau \\ =& Ae^{-\alpha t}\int_{-\infty}^\infty e^{-\alpha\tau}f(\tau)d\tau \end{aligned} \end{equation*} $f(\tau)\neq0$ if $\tau\in[0,1]$, and $Ae^{-\alpha(t-\tau)}\neq0$ if $t-\tau>0$. Also, $f(t-\tau)\neq0$ if $0<t-\tau<1\Rightarrow t-1\leq0<\tau<t$. But since $\tau<1$ and $\tau<1$, $\tau<\min\{t,1\}$. $$\therefore L(f)=Ae^{-\alpha t}\int_0^{\min\{1,t\}}e^{\alpha\tau}f(\tau)d\tau,$$ $t>0$.
Is this correct? I find determining the bounds for convolutions of piecewise functions very confusing.
Here is what I eventually got:
$f(t)=0,t\notin[0,1]$, and $h(t)=Ae^{-\alpha t},t\geq0$.
\begin{equation*} \begin{aligned} L(f) =& \int_{-\infty}^\infty Ae^{-\alpha\tau}f(t-\tau)d\tau \\ =& \int_{-\infty}^{\infty}Ae^{-\alpha(t-\tau)}f(\tau)d\tau \\ =& \int_{-\infty}^\infty Ae^{\alpha(\tau-t)}f(\tau)d\tau \\ =& \int_{-\infty}^\infty Ae^{-\alpha t}e^{\alpha\tau}f(\tau)d\tau \\ =& Ae^{-\alpha t}\int_{-\infty}^\infty e^{-\alpha\tau}f(\tau)d\tau \\ \end{aligned} \end{equation*} $f(\tau)\neq0$ if $\tau\in[0,1]$, and $Ae^{-\alpha(t-\tau)}\neq0$ if $t-\tau>0$. Also, $f(t-\tau)\neq0$ if $0<t-\tau<1\Rightarrow t-1\leq0<\tau<t$. But since $\tau<1$ and $\tau<1$, $\tau<\min\{t,1\}$. $$\therefore L(f)=Ae^{-\alpha t}\int_0^{\min\{1,t\}}e^{\alpha\tau}f(\tau)d\tau,$$ $t>0$.