Periodic antiderivative

Statement

For a continuous $L$ -periodic function $f$ , an $L$ -periodic antiderivative exists exactly when $\int_0^L f(s)\,ds=0$ . Under this condition a primitive is

F(x)=\int_0^x f(s)\,ds,

and all primitives differ by a constant. There is exactly one with zero mean over a period.

Proof and Fourier formula

Periodicity of a primitive forces $0=F(L)-F(0)=\int_0^L f$ . Conversely, periodicity of $f$ gives $F(x+L)-F(x)=\int_x^{x+L}f=0$ . Subtract its average to normalize $F$ . If $f$ is smooth with period one, the normalized primitive has coefficients $\widehat F(m)=\widehat f(m)/(2\pi im)$ for $m\ne0$ , and $\widehat F(0)=0$ .