How to prove the anti-derivative of a periodic function?

Question

How to prove that the anti-derivative of a periodic function is not necessarily periodic? (Without considering random examples)

I know that integral of $$ f(x) = f(a+x) $$ is $$ F(x) + c $$

Answer 1

A counterexample would be the simplest way. But since you don't want this, here's an alternative proof:

Assume that the periodic function $f$ with period $a$ has a periodic antiderivative $F$. Then we have $$\int_0^a f(x)\,\mathrm dx = F(a)-F(0) = 0$$ where the latter equality is due to the periodicity of $F$. However, if $f(x)$ is periodic, then so is $g(x):=f(x)+c$ with $c\ne 0$. But then, $$\int_0^a g(x)\,\mathrm dx = \int_0^a f(x)\,\mathrm dx +\int_0^a c\,\mathrm dx = ac \ne 0.$$ Therefore $g(x)$ does not have a periodic antiderivative.

Answer 2

Pick any strictly positive periodic function $f$. Then, any antiderivative of $f$ is strictly increasing and so not periodic.