---
tag: 0JNMJEDD93
kind: statement
role: theorem
name: Convolution theorem for the Laplace transform
msc: [44A10, 44A35]
uses: [05XT5AVJBK]
sources:
  - url: https://www.jirka.org/diffyqs/html/convolution_section.html
    locator: Section 6.3, Theorem 6.3.1
    note: J. Lebl, Notes on Diffy Qs. The theorem states the result for functions of exponential order.
  - url: https://en.wikipedia.org/wiki/Laplace_transform
    locator: Properties and theorems
---

## 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|05XT5AVJBK]] 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|0K3HFZJK5Q]], 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.