TITLE: An integrable and periodic function $f(x)$ satisfies $\int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x)dx$.
QUESTION [26 upvotes]: I want to prove:

For an integrable function $f(x)$ and periodic with period $T$, for every $a \in \mathbb{R}$, $$\int_{0}^{T}f(x)\;dx=\int_{a}^{a+T}f(x)\;dx.$$

I tried to change the values and define $y=a+x$ so that $dy=dx$ and the limits of the integrals are as we want, but I'm not sure how to use the fact that $f(x)$ is periodic.
Thanks a lot!

REPLY [2 votes]: $$\begin{align}
\int_{a}^{a+T}f(x)\ dx&= \int_{T}^{a+T}f(x)\ dx +\int_{a}^{T}f(x)\ dx\\&\overset{y=x-T}{=} \color{red}{\int_{0}^{a}f(y+T)\ dx} +\int_{a}^{T}f(x)\ dx\\&\overset{periodic}{=} \color{red}{\int_{0}^{a}f(y)\ dx} +\int_{a}^{T}f(x)\ dx\\&=\int_0^Tf(x)\ dx.
\end{align}$$