Very simple question. Assume we are given a continuously differentiable $T$-periodic function $f: \Bbb R \to \Bbb R$, $f(x)=f(x+T)$. What are the necessary and sufficient conditions for the integral \begin{equation*} I = \int_0^Tf(x) dx \end{equation*}
to be identically zero? Clearly $f$ being odd is sufficient, but what are the necessary conditions if $f \neq 0$?

  • $\begingroup$ Being odd is not necessary. $\sin(x)$ is odd, but if you horizontally translate it slightly it is not (odd implies some symmetry about the origin) but still has the property that integrating over a period gives you zero. I mean that (for small, positive $\epsilon$), $\sin(x+\epsilon)$ is not odd, but $\int_{0}^{2\pi} \sin(x+\epsilon) dx=0$. $\endgroup$
    – TravisJ
    Aug 23, 2016 at 15:49
  • $\begingroup$ I doubt there are any "simple" necessary conditions for the integral of a non-zero, periodic function (over one period) to be zero. $\endgroup$
    – TravisJ
    Aug 23, 2016 at 15:53
  • $\begingroup$ Your $I$ is a constant, how can it be identically $0$ ? Don't you mean $\int_x^{x+T}$ ? $\endgroup$
    – user65203
    Aug 23, 2016 at 16:24

1 Answer 1


$f$ being odd is not necessary, as pointed out by TravisJ in the comments.

A simple sufficient condition for the integral $\begin{equation*} \int_0^Tf(x) dx \end{equation*}$ to be zero is : $f$ is the derivative of a $\mathcal{C}^1$ T-periodic function.

Let $g$ be a $\mathcal{C}^1$ periodic function such that $f=g'$, then :

$$\int_0^Tf(x) dx= \int_0^Tg'(x) dx=g(T)-g(0)=0.$$

Since you want $f$ to be $\mathcal{C}^1(\mathbb{R})$, then a necessary and sufficient condition is $f$ to be the derivative of a $\mathcal{C}^2$ $T$-periodic function.

Suppose $f$ a $\mathcal{C}^1(\mathbb{R})$ $T$-periodic function with : $\int_0^Tf(x) dx=0.$ Then $g(t)=\int_0^t f(x) dx$ exists, is $\mathcal{C}^2(\mathbb{R})$ and $T$-periodic.

  • $\begingroup$ Thank you for the partial answer (I have changed the question slightly!), can you please confirm precisely what is meant by "this is also a necessary condition for $g \mathcal{C}^2$"? I do not follow the last line. $\endgroup$
    – Harry
    Aug 23, 2016 at 16:51
  • $\begingroup$ Edited. Is it better ? $\endgroup$
    – nicomezi
    Aug 23, 2016 at 16:58
  • $\begingroup$ I think you mean $g(t)$ in the last equality. $\endgroup$
    – Harry
    Aug 23, 2016 at 17:02
  • $\begingroup$ Afraid I still don't follow why this condition is necessary? Or do you mean necessary under the assumption $f=g'$? $\endgroup$
    – Harry
    Aug 23, 2016 at 17:03
  • $\begingroup$ If you search a function $f$, $T$-periodic and of mean $0$ over a period, then $f$ must be the derivative of a $T$-periodic function. If $f$ is the derivative of a function which is not $T$-periodic, then the mean over a period $T$ will not be $0$. $\endgroup$
    – nicomezi
    Aug 23, 2016 at 17:21

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