I'm having a misconception.
Here's the definition of convolution from my math book:
Let $f(t)$ and $g(t)$ be continuous functions by parts on $[0, \infty)$. Convolution of these functions is $f(t) * g(t)$ = $(f * g)(t)$ = $\int_0^t f(\tau)g(t-\tau)d\tau$, $t>0$
However, on I came across a question:
- The convolution of functions $f(x)$ and $g(x)$, whose domain is the interval $(-\infty, \infty)$
is equal to the integral $\int_{-\infty}^{\infty}f(t)g(x-t)dt$
We had to choose if this statement was true or false.Judging by the book definition (It's the only one we have), I'd say that it is not true because we haven't ever mentioned that the domain could be negative. However, the answer was that it is true.
Can anyone clarify why? Also, the possibility of the solution being incorrect shouldn't be ruled out.
$f(t)$ and $g(t)$ are continuous functions on $[0,\infty)$. Let $F(x)$ and $G(x)$ be the extensions as piecewise continuous functions to $(-\infty,\infty)$ by defining them to be $0$ for all $x < 0$.
$$ \int_{-\infty}^\infty F(t) G(x-t) dt = \int_{-\infty}^0 F(t) G(x-t) dt + \int_0^x F(t) G(x-t) dt + \int_x^\infty F(t) G(x-t) dt\\ \int_{-\infty}^0 F(t) G(x-t) dt = \int_{-\infty}^0 0 * G(x-t) dt = 0\\ \int_x^\infty F(t) G(x-t) dt = \int_x^\infty F(t)*0 dt = 0\\ \int_{-\infty}^\infty F(t) G(x-t) dt = \int_0^x F(t) G(x-t) dt = \int_0^x f(t) g(x-t) dt\\ $$