The original question is:
Show that $$f(x)=\sum_{k=-\infty}^\infty c_k e^{-i kx}=\frac{2\sinh(\pi)}{\pi} \sum_{k=1}^\infty \frac{(-1)^{k-1}k}{1+k^2}\sin(kx)$$ where $\displaystyle c_k=\frac{1}{4\pi} \int_{-\pi}^\pi e^{ikx}(e^x-e^{-x})\,dx$.
I managed to show that
$$c_k=\frac{1}{4\pi} \left[\frac{1}{i k+1} e^{(ik+1)x}-\frac{1}{ik-1} e^{(ik-1)x}\right]_{-\pi}^{\pi}$$
then got stuck at expanding it in terms of trigonometric function $\sinh(\pi)$.
The answer given says:
$$c_k = -\frac{i k (-1)^k}{\pi (1+k^2)} \sinh(\pi),\hspace{.5cm} f(x)=-\frac{\sinh(\pi)}{\pi}\sum_{k=-\infty}^\infty \frac{i k (-1)^k}{(1+k^2)} e^{-i k x}$$ Hence the result.
Thanks in advance,
Consider first $$A_k(x)=\frac{1}{i k+1} e^{(ik+1)x}-\frac{1}{ik-1} e^{(ik-1)x}$$ $$A_k(x)=\frac{ik-1}{(i k+1)(ik-1)} e^{(ik+1)x}-\frac{ik+1}{(ik-1)(ik+1)} e^{(ik-1)x}$$ $$A_k(x)=\frac{1-ik}{1+k^2} e^{(ik+1)x}+\frac{1+ik}{1+k^2} e^{(ik-1)x}=\frac{e^{ikx}}{1+k^2}\Big((1-ik) e^{x}+(1+ik) e^{-x} \Big)$$ Now, replace $e^{\pm x}=\cosh(x)\pm\sinh(x)$ to get $$A_k(x)=\frac{2 e^{i k x} (\cosh (x)-i k \sinh (x))}{1+k^2}$$ Now, using the integration bounds $$A_k(\pi)-A_k(-\pi)=\frac{4 i (\cosh (\pi ) \sin (\pi k)-k \sinh (\pi ) \cos (\pi k))}{1+k^2}$$ and, since $k$ is an integer, $$A_k(\pi)-A_k(-\pi)=-\frac{4 i k \sinh (\pi ) \cos (\pi k)}{1+k^2}=-\frac{4 i k \sinh (\pi ) (-1)^k}{1+k^2}$$ and hence $$c_k = -\frac{i k (-1)^k}{\pi (1+k^2)} \sinh(\pi)$$