Primitive of Periodic Real Function

Theorem
Let $f: \mathbb \R \to \mathbb \R$ be a real function.

Let $F$ be a primitive of $f$ that is bounded on all of $\R$.

Then if $f$ is periodic with period $L$, then $F$ is also periodic with period $L$.

Proof
Let $f$ be periodic with period $L$.

Let $f$ have a primitive $F$ that is bounded on all of $\R$.

By definition of a periodic function, it is seen that $f \left({x}\right) = f \left({x+L}\right)$.

Then:


 * $\displaystyle \int f \left({x}\right) \, \mathrm d x \quad$ and $\quad \displaystyle \int f \left({x+L}\right) \, \mathrm d x$

are both primitives to the same function, so by Primitives which Differ by Constant:

Suppose that $C \ne 0$. Then it is to be shown that for all $k \in \N_{\gt 0}$:
 * $F \left({x}\right) + kC = F \left({x+kL}\right)$

The case for $k=1$ has already been proven, so this will be proven by induction. Suppose for some $n \in \N_{\gt 0}$ we have
 * $F \left({x}\right) + nC = F \left({x+nL}\right)$

Then

Holding $x$ fixed yields
 * $\displaystyle \lim_{k \to \infty} \lvert F \left({x+kL}\right) \rvert = \lim_{k \to \infty} \lvert F \left({x}\right) + kC \rvert = \infty$

And so $F \left({x}\right)$ is unbounded.

But we had previously established that $F \left({x}\right)$ was bounded.

This is a contradiction, therefore $C=0$ and $F \left({x}\right) = F \left({x+L}\right)$.

Hence the result.

Also see

 * Derivative of Periodic Function