Suppose that $f : \mathbb R \to \mathbb R$ is a periodic function with period $T \gt 0$ (that is, $f(x+T) = f(x)$ for all $x \in \mathbb R$)...

Suppose that $$f : \mathbb R \to \mathbb R$$ is a periodic function with period $$T \gt 0$$ (that is, $$f(x+T) = f(x)$$ for all $$x \in \mathbb R$$) and integrable into $$[0, T]$$.

Show that $$f$$ admits a primitive F which is also periodic with period $$T$$ if and only if $$\int_0^Tf(x)dx=0$$

My attempt

Let $$F$$ be a primitive of $$f$$ then $$\int^{T}_0 f(x)dx = F(T)-F(0)= 0$$

I'm stuck, this is the only thing i could think of. Thanks in advance for any help.

1 Answer

This is not true!

Take $$f:\mathbb{R} \rightarrow \mathbb{R}$$ defined by $$f(x)=\left\{\begin{matrix} 1\;, & x\in \left [ k,k+1 \right ), k \text{ even}\\ -1\;, & x\in \left [ k,k+1 \right ), k \text{ odd } \end{matrix}\right.$$ Then $$f$$ is Riemann integrable on $$\mathbb{R}$$, in particular on $$[0,2]$$ and it is a periodic function with period $$T=2\,.\,$$ Moreover, $$\;\int_0^T f(x)\,\mathrm dx=0\,.\,$$ However, $$f$$ does not admit a primitive! This is because $$f$$ has a simple discontinuity at $$\,x=1$$.

• But $\int_0^2 f(x) \ dx \ne 0$ in your example.
– fwd
Commented Jun 30, 2021 at 19:57
• You may want to adjust the definition of $f$ to maybe switch between $1$ and $-1$.
– fwd
Commented Jun 30, 2021 at 20:06