Double integral of two periodic functions I'm quite stuck with the following practice question. Suppose $f,g: R \to R$ are continuous $2\pi$-periodic functions. Let $h(s) =\int_{0}^{2\pi} f(s-t)g(t)  \ dt$. Prove that
$$\int_{0}^{2\pi} h(s)  \ ds = \left( \int_{0}^{2\pi} f(t)\ dt \right) \left( \int_{0}^{2\pi} g(t) dt \right)$$
My Attempt
$\begin{equation}
\int_{0}^{2\pi} h(s)  \ ds  = \int_{0}^{2\pi}\int_{0}^{2\pi} f(s-t)g(t) \ dt \ ds  \\
= \int_{0}^{2\pi}g(t) [\int_{0}^{2\pi} f(s-t)\ ds] \ dt
\end{equation}$
Could someone point out the right direction to go from here?
 A: Consider changing the bounds of integration on the inner integral, in order to shift the variable under consideration (your goal is to have something like $f(r)$ on the inside of the integral). When the variable shifts, the bounds must shift the other way in order to reflect that change.
For example, $\int_{x=1}^3 x+y\ dx = \int_{r=y+1}^{y+3}r\ dr$.
When dealing with the inner integral, remember that you can think of $t$ as if it's a constant (just imagine $t=1$ or something like that). After that, it should be possible to use the periodic nature of the functions.
A: \begin{align}
&\Rightarrow\int_{0}^{2\pi} h(s)  \ ds  \\
&= \int_{0}^{2\pi}\int_{0}^{2\pi} f(s-t)g(t) \ dt \ ds\\
&= \int_{0}^{2\pi}\int_{0}^{2\pi} f(s-t)g(t)\ ds \ dt&(\text{Changing the order of integration})\\
&= \int_{0}^{2\pi}g(t) \left(\int_{0}^{2\pi} f(s-t)\ ds \right)\ dt&(\text{Separating the independent integrand})\\
&= \int_{0}^{2\pi}g(t) \left(\int_{0-t}^{2\pi-t} f(s)\ ds\right) \ dt&(\text{Shifting property})\\
&= \int_{0}^{2\pi}g(t) \left(\int_{0}^{2\pi} f(s)\ ds\right) \ dt&(\text{Periodicity})\\
&= \left(\int_{0}^{2\pi} f(s)\ ds\right) \left(\int_{0}^{2\pi}g(t) \ dt\right)&(\text{Separating independent integrals})\\
&= \left(\int_{0}^{2\pi} f(t)\ dt\right) \left(\int_{0}^{2\pi}g(t) \ dt\right)&(\text{Changing the variable name})
\end{align}
