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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?
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    $\begingroup$ You should be coming up with $-\int_\pi^{-\pi}f(z)g(x-z)dz$. :-) $\endgroup$ – Robin Chapman Sep 11 '10 at 22:08
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    $\begingroup$ @Robin: are kids these days still saying "duh"? $\endgroup$ – Tom Stephens Sep 11 '10 at 22:15
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You need to check the bounds on your integral. since $y$ ranges from $-\pi$ to $\pi$, you'll have $z = x-y$ ranging from $x+\pi$ to $x-\pi$. Therefore: $$ \int_{-\pi}^{\pi}f(x-y)g(y)dy = -\int_{x+\pi}^{x-\pi}f(z)g(x-z)dz = \int_{x-\pi}^{x+\pi}f(z)g(x-z)dz = \int_{-\pi}^{\pi}f(z)g(x-z)dz $$ in the second-last step, I swapped the two bounds on the integral (this changes the sign). In the final step, I shifted both bounds on the integral by $-x$, which does not change the value because we are integrating over an interval of length $2\pi$ and the function is $2\pi$-periodic.

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    $\begingroup$ ouch, I'm not going to lie - this one hurts a bit! $\endgroup$ – Tom Stephens Sep 11 '10 at 22:19
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\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*}

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Here is something I've sometimes wondered about. If $f,g$ are both nonnegative proving commutativity of convolution can be done without a tedious change of variable.

Indeed, let $X$ be a random variable with density $f$ and let $Y$ be a random variable with density $g$. Its easy to see that $f$ convolved with $g$ is the density of $X+Y$ (or in your case $X+Y ~{\rm mod} ~2 \pi$). By commutativity of addition, the density of $X+Y$ is the same as the density of $Y+X$ and we are done!

I wonder if an argument of this sort can work in the general case.

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