TITLE: Convolution doubt
QUESTION [9 upvotes]: Can someone explain why the general formula of the convolution is this one:
$$(f*g)(t)=\int_{-\infty}^\infty f(t-\tau)g(\tau) \, d\tau$$
But when both $f(\tau)$ and $g(\tau)$ are equal to zero for negative values of $\tau$, the convolution turns into:
$$(f*g)(t)=\int_0^t f(t-\tau)g(\tau) \, d\tau$$
I always thought that it would be more logical that it would become:
$$(f*g)(t)=\int_0^\infty f(t-\tau)g(\tau) \, d\tau$$

REPLY [4 votes]: Hint: Use indicator functions for these kind of problems.
Write $f(t)$ as $f(t)1_{\{t \geq 0\}}$ and $g(t)$ as $g(t)1_{\{t \geq 0\}}$. Then the integrand becomes
$$ f(t-\tau)g(\tau)1_{\{\tau \leq t,\tau\geq 0\} } = f(t-\tau)g(\tau)1_{\{0 \leq \tau \leq t\} }$$
$\blacksquare$