Convolution is communicative from 2 view

Question

Assume convolution between $f(x)$ and $g(x)$. $f(x)*g(x)=\int_{\infty}^{\infty}f(\tau)g(t-\tau)d_{\tau}$. We know that convolution is communicative, to prove this property, we can change the variable $\tau'=t-\tau$, so $f(x)*g(x) = -\int_{\infty}^{\infty}f(t-\tau')g(\tau')d_{\tau'}$ and to remove minus we change the bounds of integral, but if we define the variable as $\tau' = \tau-t$ , the convolution would be $f(x)*g(x) = \int_{\infty}^{\infty}f(\tau'+t)g(-\tau')d_{\tau'}$, is there any way to prove that the first way and the second one are the same?

Answer 1

By the change of variable $\tau=-\tau'$, $d\tau'=-d\tau$, one gets $$ \int_{-\infty}^{\infty}f(\tau'+t)g(-\tau')\:d{\tau'}=\int_{\infty}^{-\infty}f(-\tau+t)g(\tau)\:(-d{\tau})=\int_{-\infty}^{\infty}f(t-\tau)g(\tau)\:d{\tau} $$ which is your first expression.