Formula for tangent derivatives, how to prove? How to prove?
$$(\tan x)^{(s-1)}=\pi^{-s}\Gamma(s)\left(\zeta\left(s, \frac12-\frac x\pi\right)+(-1)^s\zeta\left(s, \frac12+\frac x\pi\right)\right) $$
 A: We start with the following claim (I hope it has been proven on this web site, but I couldn't find a link):
$$\sum_{k=-\infty}^\infty\frac{1}{(k-\frac{x}{\pi})^2}=\frac{\pi^2}{\sin^2(x)}$$
Because
$$\frac{d}{dx}\tan(x)=\frac{1}{\cos^2(x)}=\frac{1}{\sin^2(x-\frac{\pi}{2})}$$
it follows that
$$\frac{d}{dx}\tan(x)=\frac{1}{\pi^2}\sum_{k=-\infty}^\infty\frac{1}{(k+\frac{1}{2}-\frac{x}{\pi})^2}$$
Differentiating $s-1$ times gives
$$\tan(x)^{(s-1)}=\pi^{-s}\Gamma(s)\sum_{k=-\infty}^\infty\frac{1}{(k+\frac{1}{2}-\frac{x}{\pi})^{s}}$$
The Hurwitz zeta function is the following:
$$\zeta(s,q)=\sum_{k=0}^\infty\frac{1}{(k+q)^s}$$
Rewriting the summation above:
\begin{align}
\sum_{k=-\infty}^\infty\frac{1}{(k+\frac{1}{2}-\frac{x}{\pi})^{s}}&=\sum_{k=1}^\infty\frac{1}{(-k+\frac{1}{2}-\frac{x}{\pi})^{s}}+\sum_{k=0}^\infty\frac{1}{(k+\frac{1}{2}-\frac{x}{\pi})^{s}}\\
&=\sum_{k=0}^\infty\frac{1}{(-k-\frac{1}{2}-\frac{x}{\pi})^{s}}+\sum_{k=0}^\infty\frac{1}{(k+\frac{1}{2}-\frac{x}{\pi})^{s}}\\
&=\sum_{k=0}^\infty\frac{1}{(-1)^s(k+\frac{1}{2}+\frac{x}{\pi})^{s}}+\sum_{k=0}^\infty\frac{1}{(k+\frac{1}{2}-\frac{x}{\pi})^{s}}\\
&=(-1)^s\zeta\left(s,\frac{1}{2}+\frac{x}{\pi}\right)+\zeta\left(s,\frac{1}{2}-\frac{x}{\pi}\right)
\end{align}
Therefore we conclude:
$$\tan(x)^{(s-1)}=\pi^{-s}\Gamma(s)\left[(-1)^s\zeta\left(s,\frac{1}{2}+\frac{x}{\pi}\right)+\zeta\left(s,\frac{1}{2}-\frac{x}{\pi}\right)\right]$$
Inspired by page 22 of this: http://www.staff.science.uu.nl/~ban00101/funcr2014/fr_opgaven_2014.pdf
