Convolution theorem for the Laplace transform

Statement

Let $f, g : [0, \infty) \to \mathbb{C}$ be piecewise continuous, and let $M, a$ be real constants with $|f(t)| \le M e^{a t}$ and $|g(t)| \le M e^{a t}$ for all $t \ge 0$. Let the convolution $f * g : [0, \infty) \to \mathbb{C}$ be $$(f * g)(t) = \int_0^t f(\tau)\, g(t - \tau)\, d\tau.$$ Let $F$ and $G$ be the Laplace transforms of $f$ and $g$. Then for each $s \in \mathbb{C}$ with $\operatorname{Re} s > a$, the Laplace transform of $f * g$ converges at $s$, and $\mathcal{L}\{f * g\}(s) = F(s)\, G(s)$.

Remarks

The proof writes $F(s) G(s)$ as a double integral over the quadrant $\tau, u \ge 0$, and changes variables to $t = \tau + u$. The absolute convergence for $\operatorname{Re} s > a$ justifies the change of the order of integration. The convolution is commutative: $f * g = g * f$. Example: $\mathcal{L}^{-1}\left\{\frac{1}{s^2 (s - 1)}\right\}$ is the convolution of $t$ and $e^t$, that is $e^t - 1 - t$. In solving initial value problems, the solution with zero initial values is the convolution of the forcing term with the impulse response.

Explanations

L1 human

Multiplying two transforms corresponds to a special kind of product of the functions, the convolution. The convolution mixes the past values of one function with the other, as a memory of earlier inputs.