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Given an odd function $f$, defined everywhere, periodic with period $2$, and integrable on every interval. Let $g(x) = \displaystyle\int_{0}^{x}f(t)dt$. I know that $\displaystyle\int_{-b}^{b}f(t)dt=0$ for $b\in\mathbb{R}$ if it is odd function, and $f(t)= f(t+2n)$ where $n$ is integer if it is periodic with period $2$.

Prove that $g(2n)= 0$ for every integer n.

And I do the next:

$g(2n)= \displaystyle\int_{0}^{2n}f(t)dt= \displaystyle\int_{-n}^{n}f(t-n)dt$ if $n=2k$ where $k$ is integer, so $\displaystyle\int_{-n}^{n}f(t)dt=0$

But if $n = 2k+1$ then $f(t+n)=f(t+1)$ then I can be sure that $\displaystyle\int_{-n}^{n}f(t+1)dt=0$, some hint to solve that.


If I take $h(t) = f(t+1)$, we can see that $h(t)=h(t+2)$ (because $f$ is 2-periodic function too) and $h(-t)=-h(t)$ (because $f$ odd function).

$h(t+2)= f((t+2)+1)=f(t+1)= h(t)$

$h(-t)= f(-t+1)=f(-(t-1))=-f(t-1)=-f(t+1)=-h(t)$

Then $\displaystyle\int_{-n}^{n}f(t+1)dt=\displaystyle\int_{-n}^{n}h(t)dt=0$

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You don't need to divide into the cases when $n$ is even or odd: $$ g(2n)=\int_0^{2n}f(t)\,dt= \int_{-n}^n f(u)\,du=0 $$ with the substitution $t=u+n$.

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  • $\begingroup$ But when you use substitution then your interval change . And I dont know how it s gonna work. $\endgroup$ – Elll Jun 4 '15 at 10:48
  • $\begingroup$ @Elll Of course it does; but where's the problem? $\endgroup$ – egreg Jun 4 '15 at 10:58
  • $\begingroup$ ok. I understood, theres no problem, thks. $\endgroup$ – Elll Jun 4 '15 at 11:05
  • $\begingroup$ wait, I´m still lost. if $f(t)$ is odd then $f(t+1)$ is odd too? $\endgroup$ – Elll Jun 4 '15 at 11:11
  • $\begingroup$ more, for your substitution $t = u +n$ $f(t)$ is odd too? If It´s true, There´s no problem. $\endgroup$ – Elll Jun 4 '15 at 11:14
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$$g(2n)=ng(2)=n\int_{0}^2f(t)dt=n\int_0^1f(t)dt+n\int_{-1}^0f(t)dt$$ and $$\int_{-1}^0f(t)dt=-\int_1^0f(-t)dt=\int_1^0f(t)dt=-\int_0^1f(t)dt$$

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  • $\begingroup$ I dont know haw you get to $g(2n)=ng(2)$. $\endgroup$ – Elll Jun 4 '15 at 10:45

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