integral over interval of length $2 \pi$

Question

Let $f$ be a $2 \pi$ periodic function. I think the integral of $f$ over $[a- \pi ,\; a+ \pi]$ is same as integral of $f$ over $[ -\pi , \; \pi]$. I.e. I think the integral is same in any interval of length $2\pi$. Am I correct?

Answer 1

Let $$F (x)=\int_0^xf (t)dt $$ and

$$G (a)=\int_{a-\pi}^{a+\pi}f (x)dx $$ $$=F (a+\pi)-F (a-\pi) $$

by FTC, $$G'(a)=F'(a+\pi)-F'(a-\pi) $$ $$=f (a+\pi)-f (a-\pi) $$ $$=f (a+\pi-2\pi)-f (a-\pi)=0$$ thus $G $ is constant and $$G (a)=G (0)=\int_{-\pi}^\pi f $$

Answer 2

Yes, it is correct. You can break your interval at some odd multiple of $\pi$ into two pieces. The integral over the bottom piece of your interval is equal to the integral over the top piece of the same length of $[-\pi,\pi]$ by the periodicity. Similarly, the integral over the top piece of your interval is equal to the integral over the bottom piece of $[-\pi, \pi]$