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Suppose $f(x)$ is periodic with period $l$ and integrable. Prove that, for any $a$ $$ \int_a^{a+l} f(x)\,dx=\int_0^l f(x)\,dx.$$.

i was thinkikng of using definate integral properties such as, $\int_0^l f(x)\,dx =F(l)-F(0) = F(l)$. and $\int_a^{a+l} f(x) = F(a+l) -F(a)$

now the idea is to break $F(a+l)$ to get $F(a) + F(l)$ is this even true.

any idea of how to approach it

Travis Willse
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user146269
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2 Answers2

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Your edit contains a very good idea!

Let $F(x) = \int_x^{x + l}f(t)\,dt$. Notice that by the FTC, $F'(x) = f(x + l) - f(x) = 0$ by the assumption that $f$ is $l$-periodic. This proves that $F$ is constant. Then in particular $F(0) = F(a)$, which is exactly the desired equality, indeed $$\int_0^lf(t)\,dt = F(0) = F(a) = \int_a^{a + l}f(t)\,dt.$$

Giovanni
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  • I am thinking that but i don't see how that is one of the reason i post it – user146269 Oct 20 '15 at 19:44
  • @user146269: I have edited to include all the details. – Giovanni Oct 20 '15 at 19:48
  • that makes perfect sense thanks Giovanni – user146269 Oct 20 '15 at 19:57
  • @user146269: I am glad I could help. I have noticed that you asked 82 questions but you have not accepted a single answer. Please read this (http://stackoverflow.com/help/someone-answers) and start accept all the good answers you have received so far :) – Giovanni Oct 20 '15 at 20:15
  • @user146269: you have not accepted my answer, you probably upvoted it, it is a different thing. There should be a check mark you can click right below the down vote button, that's what you have to click to accept an answer. Also, you get reputation by doing so! it should be +2 rep for each accepted answer. – Giovanni Oct 20 '15 at 21:47
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    oh i see thanks again my friend – user146269 Oct 21 '15 at 17:20
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We have: $\int_0^{a+l}f(x)dx=\int_0^{a}f(x)dx+\int_a^{a+l}f(x)dx.$

Now, using the periodicity of $f$ and a change of variable we can re-write the second member of the equality:

$\int_0^af(x)dx=\int_0^af(x+l)dx=\int_l^{a+l}f(s)ds$.

So, we get

$\int_0^{a+l}f(x)dx=\int_l^{a+l}f(s)ds+\int_a^{a+l}f(x)dx.$

and then

$\int_a^{a+l}f(x)dx=\int_0^{a+l}f(x)dx-\int_l^{a+l}f(x)dx=\int_0^lf(x)dx$

Kamala
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