# Convolution - prove commutative

1. Sep 30, 2006

### benndamann33

anyone know how to prove that it is commutative...

as if f *g = g*f

2. Sep 30, 2006

### shmoe

Try a change of variables.

3. Oct 1, 2006

### HallsofIvy

Staff Emeritus
You might want to start with the definition: the convolution of f and g is
$f*g(x)= \int_0^\infty f(x-t)g(t)dt$ and, of course, $g*f(x)= \int_0^\infty g(x-u)f(u)du$ (I have intentionally used a different variable of integration here). Hmm, in one you have f(x-t) and in the other f(u). Does that substitution benndamann33 mentioned leap to mind?