# Talk:Primitive of Periodic Function

Is this even true?

Take $f \left({x}\right) = 1 + \cos x$, which is periodic.

Then $\displaystyle \int f \left({x}\right) \ \mathrm d x = x + \sin x$ which actually is not periodic. The $x$ term makes it so that $f \left({x + 2 \pi}\right) = 2 \pi + f \left({x}\right) \ne f \left({x}\right)$. --prime mover (talk) 20:24, 30 August 2015 (UTC)

I think this becomes true if we add the condition that the primitive is bounded. --Wanfactory (talk) 12:12, 16 December 2016 (EST)

Another mistake comes at the very end of the proof when he (or she) invokes Periodic Function plus Constant. That theorem only works for when the constant is on both sides of the equation, but for

$F \left({x}\right) = F \left({x + L}\right) + C$

The constant is only on one side of the equation, so unless $C = 0$, there's no way that $F$ is periodic. --HumblePi (talk) 10:14, 17 December 2016 (EST)

if $F$ is bounded, then $C$ becomes zero Wanfactory (talk) 00:19, 19 December 2016 (EST)
I'll make the appropriate changes. --HumblePi (talk) 10:44, 19 December 2016 (EST)
I may be missing something obvious, but have we actually proved that $f \left({x}\right) = f \left({x + L}\right) \implies \int f \left({x}\right) \, \mathrm d x + C = \int f \left({x + L}\right) \, \mathrm d x$? We seem to be taking it for granted, but is it actually a valid statement? Unfortunately I can't think my way through this at the moment, it's all making my eyes glaze over. --prime mover (talk) 11:52, 19 December 2016 (EST)
Yes, it is valid. $f \left({x}\right) = f \left({x+L}\right)$ so from Primitives which Differ by Constant, $\int f \left({x}\right) \mathrm d x$ and $\int f \left({x+L}\right) \mathrm d x$ are both primitives of the same function. And therefore they only differ by a constant. This is true for every closed interval of $\R$, which essentially makes it true for all $x \in \R$ --HumblePi (talk) 12:42, 19 December 2016 (EST)
Good job, but all what you just typed above is on the wrong page. Needs to be on the main page. --prime mover (talk) 13:36, 19 December 2016 (EST)

Right, I'll get on that. --HumblePi (talk) 13:55, 19 December 2016 (EST)

This looks like it covers the case of real-number functions, but i'm not too sure about the complex case. I seem to recall that a bounded differentiable complex function degenerates into a constant so maybe we need to weaken the conditions to something like 'bounded in the direction of L'. For example exp(x) is bounded in the direction of its period 2 pi i Wanfactory (talk) 15:01, 19 December 2016 (EST)
Complex Analysis isn't my strongest subject, so I'm not too sure about the complex case either. The thing that you recalled is called Liouville's Theorem, which says that every bounded entire function is a constant. That would make $F \left({x}\right) = C$ and $f \left({x}\right) = 0$ in the complex case. But there exist entire functions that are periodic in a sense (like your example), so I'll just make it a theorem for real functions. --HumblePi (talk) 15:19, 19 December 2016 (EST)
sounds good, my complex analysis is pretty rusty as well. I think now we're in the same boat as Derivative of Periodic Function with still needing to show that L is the smallest number over which the function repeats. However, I think that the 'primary' part of each of these proofs can be used to prove the 'secondary' part of the other proof. This might be an opportunity to complete both proofs. Wanfactory (talk) 20:16, 19 December 2016 (EST)

Is anyone able to source this proof and its differential-calculus counterpart? This can't be the first time someone has come up with it. --prime mover (talk) 01:24, 20 December 2016 (EST)