Convolution
Background
We define the following operation called convolution.
\[f(t) * g(t) = \int_{-\infty}^{\infty} f(t-\tau) g(\tau) \ d\tau\]Convolution is a bit of a funny operation, but it has a lot of great applications in signal processing (which is also a main aplication of Laplace Transforms). To understand what this operation is going, fix two functions $f(\tau)$ and $g(\tau)$. Horizontally flip $f$, i.e. $f(-\tau)$. Then put $f$ all the way on the left and $g$ all the way on the right. Now, move $f$ and $g$ towards and through each other. This “movement” is parameterized by the variables $t$.
The yellow region in the figure above is the convolution of $f$ and $g$ at $t$, i.e. the intersecting area between $f(t - \tau)$ and $g(\tau)$
If you are still unclear with the interpretation of convolution, Wikipidia’s page on convolution has excellent visualizations. Also 3Blue1Brown’s video has far better visualizations than I could produce here.
Restrictions on f and g
For our application, we are going to assume that $f(t)$ and $g(t)$ are $0$ for all $t < 0$. This is not an unreasonable assumption since $t$ typically represents time. What this means is that we can simplify our convolution definition.
\[f(t) * g(t) = \int_{0}^{t} f(t-\tau) g(\tau) \ d\tau\]Why is this the same as the previous definition? At $\tau < 0$, $g(\tau) = 0$. Likewise, at $\tau > t$, $f(t - \tau) = 0$.
The Laplace Transform Proof
Given functions $f(t)$ and $g(t)$ with Laplace Transforms $F(s)$ and $G(s)$, respectively.
\[\begin{align} \mathcal{L}\{ f(t) * g(t) \} &= \int_{0}^{\infty} [f(t) * g(t)] e^{-st} \ dt \\[10pt] &= \int_{0}^{\infty} \left [ \int_{0}^{t} f(t-\tau) g(\tau) \ d\tau \right ] e^{-st} \ dt \\[10pt] &= \int_{0}^{\infty} \int_{0}^{t} f(t-\tau) g(\tau) e^{-st} \ d\tau \ dt \\[10pt] \end{align}\]We have to pause here because what we want to do is exchange the order of integration. To do this, lets consider the region that we are integrating over.
Thus, we can construct the region by integrating $\tau$ from $0$ to $\infty$. Then $t$ will range from $\tau$ to $\infty$.
\[\begin{align} \mathcal{L}\{ f(t) * g(t) \} &= \int_{0}^{\infty} \int_{\tau}^{\infty} f(t-\tau) g(\tau) e^{-st} \ dt \ d\tau \\[10pt] &= \int_{0}^{\infty} \left [ \int_{\tau}^{\infty} f(t-\tau) e^{-st} \ dt \right ] g(\tau) \ d\tau \\[10pt] &\text{let } u = t-\tau \implies du = dt \\[10pt] &= \int_{0}^{\infty} \left [ \int_{0}^{\infty} f(u) e^{-s(u+\tau)} \ du \right ] g(\tau) \ d\tau \\[10pt] &= \int_{0}^{\infty} \left [ \int_{0}^{\infty} f(u) e^{-su} \ du \right ] g(\tau)e^{-s\tau} \ d\tau \\[10pt] &= \left [ \int_{0}^{\infty} f(u) e^{-su} \ du \right ] \cdot \left [ \int_{0}^{\infty} g(\tau)e^{-s\tau} \ d\tau \right ]\\[10pt] &= F(s) \cdot G(s) \end{align}\]