TITLE: Proving commutativity of convolution $(f \ast g)(x) = (g \ast f)(x)$
QUESTION [19 upvotes]: From any textbook on fourier analysis: 
"It is easily shown that for $f$ and $g$, both $2 \pi$-periodic functions on $[-\pi,\pi]$, we have $$(f \ast g)(x) = \int_{-\pi}^{\pi}f(x-y)g(y)\;dy = \int_{-\pi}^{\pi}f(z)g(x-z)\;dz = (g \ast f)(x),$$ by using the substitution $z = x-y.$"

I don't doubt that this is true, but I cannot figure out what happened to the negative sign coming from $dy = -dz\;$ after the change of variable $z = x - y$.  In particular, after the change of variable $z = x-y,\;$ I am coming up with $ -\int_{-\pi}^{\pi}f(z)g(x-z)\;dz$. What am I missing here?

REPLY [6 votes]: \begin{align*}
\\&\int_0^tf(t-u)g(u)\cdot\text du\\
\\\text{let } v=t-u\\
\text dv=-\text du\\
\\ u=0\rightarrow v=t\\u=t\rightarrow v=0\\
&=\int_t^0f(v)g(t-v)\cdot -\text dv\\
&=\int_0^tf(v)g(t-v)\cdot \text dv\\
\\&\\
\\&\\
\\&\\
\\&\\
\end{align*}
\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\}
&=\int\limits_0^\infty\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\
&=\int\limits_0^\infty e^{-pt}\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot\text dt\\
u=0\rightarrow t=u\\
u=t \rightarrow t=\infty\\
&=\int\limits_0^\infty g(u)\int\limits_u^\infty e^{-pt}f(t-u)\cdot\text dt\cdot\text du\\
\text{let }v=t-u\\
\text dv =\text dt\\
t=u\rightarrow v=0\\
t=\infty \rightarrow v=\infty\\
&=\int\limits_0^\infty g(u)\int\limits_{0}^\infty e^{-p(v+u)}f(v)\cdot\text dv\cdot\text du\\
&=\int\limits_0^\infty g(u)e^{-up}\cdot\text du\times\int\limits_{0}^\infty f(v)e^{-pv}\cdot\text dv\\
\\&=G(p)F(p)
\end{align*}