= Translation invariance of an integral over one period
{title2=$\int_a^{a+P}f(x)\,dx=\int_0^Pf(x)\,dx$}
For an <integrable> <periodic function> of period $P>0$, its <integral> over an interval of length $P$ is independent of the starting point. Write $a=mP+r$ with $m$ an <integer> and $0\leq r<P$. By a <change of variables> and the defining identity of a <periodic function>,
$$
\int_a^{a+P}f=\int_r^{r+P}f=\int_r^Pf+\int_P^{P+r}f=\int_r^Pf+\int_0^rf=\int_0^Pf.
$$
No <differentiability> is required. This also allows the <Fourier coefficients> of a <periodic function> to be computed on any interval of one full <period>.
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