Change of variables in a convolution.. I'm in trouble with change of variables in a convolution: 
Definition: The convolution of two $2\pi$-periodic functions $f$ and $g$ is defined as $$(f*g)(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}
f(y)g(x-y)\ dy.$$
I wan't to show $f*(g*h)=(f*g)*h$. Well, $$[f*(g*h)](x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\tau)(g*h)(x-\tau)\ d\tau\\ =\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\tau)\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}g(\rho)h(x-\tau -\rho)\ d\rho\right)\ d\tau.$$ Now I wan't to make the change of variables $\sigma=\tau+\rho$ but I don't know how to proceed with $d\rho$ and with the extremals $-\pi$, $\pi$. Can anyone explain me that? How can I know what is fixed in $h(x-\tau-\rho)$ inside the integral?
 A: First, use change of variables and the periodicity of $f$ and $g$ to deduce four useful facts about the convolution integral.


*

*$(f\star g)(x)$ is a $2\pi$-periodic function of $x$.

*$\displaystyle (f\star g)(x)=\frac{1}{2\pi}\int_{-\pi+a}^{\pi+a} 
f(y)g(x-y)\,\mathrm dy$ for each choice of real number $a$.

*$\displaystyle (f\star g)(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}
f(y)g(x-y)\,\mathrm dy = \frac{1}{2\pi}\int_{-\pi}^{\pi}
f(x-y)g(y)\,\mathrm dy.$

*$\displaystyle \frac{1}{2\pi}\int_{-\pi}^{\pi}
f(a+y)g(b-y)\,\mathrm dy = (f\star g)(a+b \bmod 2\pi)$.
So, 
$$\begin{align}
(f \star (g\star h))(x) &= \frac{1}{2\pi}\int_{-\pi}^{\pi}
f(x-y)\left[(g\star h)(y)\right]\ \mathrm dy\\
&= \frac{1}{2\pi}\int_{-\pi}^{\pi}
f(x-y)\left[\frac{1}{2\pi}\int_{-\pi}^{\pi} g(y-t)h(t)\,\mathrm dt\right]\ \mathrm dy\\
&= \frac{1}{2\pi}\int_{-\pi}^{\pi}\left[\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x-y)g(y-t)\,\mathrm dy\right]h(t)\,\mathrm dt\\
&= \frac{1}{2\pi}\int_{-\pi}^{\pi} \left[(f\star g)(x-t)\right]h(t)\,\mathrm dt\\
&= ((f\star g)\star h)(x).
\end{align}$$
