formula for differentiable function

Question

Prove the formula $$ f(x+\pi)-f(x) = -\frac{4}{\pi}\sum_{m=0}^{\infty}\int_0^{\pi} f(t+x)\cos((2m+1)t)\: dt $$ assuming $f\colon[x,\,x+\pi]\to\mathbb{R}$ is differentiable in $(x,\,x+\pi)$.

Answer 1

Hint: The Fourier series $$ g(t)=\sum_{m\geq 1}\frac{\sin((2m+1)t)}{2m+1} $$ is constantly equal to $\frac{\pi}{4}$ for $t\in(0,\pi)$, since $g(t)=\text{Im}\text{ arctanh}(e^{it})$. Now use integration by parts.