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.
Best 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 $\,x=1$.