changing argument of integrand of a periodic function inside an integral

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a $2\pi$-periodic function that is integrable on $[0,2\pi]$. Then for all $x \in \mathbb{R}$: \begin{align}\int^{x+2\pi}_x f(y) \ dy =\int^{2\pi}_0 f(y) \ dy = \int^{2\pi}_0 f(y+x) \ dy. \end{align} I get the first equation, because $F(x) = \int^{x+2\pi}_x f(y) \ dy$ is a constant function. However, can someone show me why the second equation is true ? I thought that $f(y) = f(y+x)$ is only true when $x = 2\pi$ , since it's a $2\pi$-periodic function ?

Answer 1

We can set: $u=x+y$ and it follows $\frac{\textrm{d}u}{\textrm{d}y}=1 \rightarrow \textrm{d}u=\textrm{d}y$.

The new integration borders are [$x+0$ und $x+2\pi$].

After you put everything in your integral you get what you want: \begin{align}\int^{x+2\pi}_x f(y) \ dy =\int^{2\pi}_0 f(y) \ dy = \int^{2\pi}_0 f(y+x) \ dy= \int^{x+2\pi}_x f(u) \ du \end{align}