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I have this question. I would like to help me with this problem please . If $f'(x)$ is a periodic function, with period $a$, prove that $f(x)$ is a periodic function, if and only if $f(a)=f(0)$. I appreciate your help.

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  • $\begingroup$ It is given that $f'(x)$ is periodic. $\endgroup$ – Doug M Jul 7 '16 at 22:23
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    $\begingroup$ if $f(x)$ is $a$-periodic then $f(x)+C$ is $a$-periodic, so the integration constant isn't the problem. An example of the problem is $f'(x) = 1$, to be compared with $f'(x) = \cos(2\pi x / a)$ $\endgroup$ – reuns Jul 7 '16 at 22:25
  • $\begingroup$ and I prefer a more intuitive and explicit statement : if $g(x)$ is $a$-periodic, then there is a unique constant $\bar{g}$ such that $f(x) = \int_{x_0}^x (g(x)-\bar{g}) dx$ is $a$-periodic $\endgroup$ – reuns Jul 7 '16 at 22:29
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if $f(x)$ is periodic with period $a$ then $f(x) = f(x+a)$ for all x. $f(0) = f(a)$ and $f'(x) = \lim_\limits{x\to 0} \frac {f(x+h) + f(x)}{h} = \lim_\limits{x\to 0} \frac {f(x+a + h) + f(x+a)}{h} = f'(x+a)$

To go the other direction.

It is a necessary condition that $f(0) = f(a)$ for $f(x)$ to be periodic.

but is $f'(x)$ periodic and $f(0) = f(a)$ sufficient?

$f(x+a) - f(x) = \int_x^{x+a} f'(x) dx$

Given that $f'(x)$ is periodic $\int_x^{x+a} f'(x) dx$ is constant.

$f(0) - f(a) \implies\int_0^{a} f'(x) dx = 0\\ f(x+a) = f(x)$

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\begin{equation} \int_0^xf^\prime(t)\,dt=\int_a^{a+x}f^\prime(t)\,dt \end{equation}

\begin{equation} f(x)-f(0)=f(a+x)-f(a) \end{equation}

So

\begin{equation} f(x)=f(x+a)+f(0)-f(a) \end{equation}

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