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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 $$ F(x) + c $$

Nzewi Ernest Kenechukwu
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RUTAN GALENO
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    Can you consider whether $F$ has a limit at $+\infty$? Can you show that such a limit, if it exists, must be $\pm\infty$? – Stefan Lafon Jul 27 '19 at 15:42
  • If $f$ is $T$-periodic with mean value $0$ then $F(x)=\int_a^x f(y)dy$ is $T$-periodic, in general $F(x) - m T x$ is $T$-periodic with $m$ the mean value – reuns Jul 27 '19 at 21:56

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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.

celtschk
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  • You proof somehow implicitely assumes that the periodic function is not the zero function. The true statement would be rather any periodic function with periodic antiferivtive is the zero function. – lalala Jul 27 '19 at 16:02
  • @lalala: It nowhere assumes that. Clearly also for the zero function, we have $F(a)-F(0)=0$. In case you meant $f(x)$ is the zero function, then that still works: The zero function has a periodic antiderivative (namely its antiderivatives are exactly the constant functions), and of course also for constant functions we have $F(a)-F(0)=0$. – celtschk Jul 27 '19 at 16:15
  • Your proif cannot work for f beeing the zero function (the conclusion doesnt hold) – lalala Jul 27 '19 at 17:07
  • @lalala: How does the conclusion not hold? If $f$ is the zero function, then $g$ is a non-zero constant function. And a non-zero constant function does not have a periodic antiderivative, does it? – celtschk Jul 27 '19 at 17:13
  • Ah, ok now i see what you are doing. – lalala Jul 27 '19 at 18:01
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Pick any strictly positive periodic function $f$. Then, any antiderivative of $f$ is strictly increasing and so not periodic.

Travis Willse
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