If $f$ is a periodic function with period $L$, show that the integral $\int_{a}^{a+L}f(x)dx$ is independent of a.

Question

I appreciate your help.

Show the integral over $[a,a+L]$ is the same as over $[a+nL,a+(n+1)L]$ for any $n$. Then choose $n$ so that $0\le a+nL < L$ and show that the integral over $[0,a+nL]$ is the same as over $[L,a+(n+1)L]$ and so the integral over $[a+nL,a+(n+1)L]$ is the same as $[0,L]$. — copper.hat, Jun 07 '20 at 04:41

score 0 · Answer 1 · edited Dec 08 '21 at 17:13

0

$$I(a)=\int_{a}^{a+L} f(x) dx \implies I'(a)=f(a+L)= f(a)=0$$ Here we have used Leibnitz Rule for the differentiation of the integral w.r.t the limit variable. Next we have used the periodicity: $f(a+L)=f(a) ~~\forall a \in R$. Finally $I'(a)=0$ means that the integral does not depend on $a$.

edited Dec 08 '21 at 17:13

hazelnut_116

1,749

answered Jun 07 '20 at 06:10

Z Ahmed

46,319

If $f$ is a periodic function with period $L$, show that the integral $\int_{a}^{a+L}f(x)dx$ is independent of a.

1 Answers1