Formula for tangent derivatives, how to prove?

Question

How to prove?

$$(\tan x)^{(s-1)}=\pi^{-s}\Gamma(s)\left(\zeta\left(s, \frac12-\frac x\pi\right)+(-1)^s\zeta\left(s, \frac12+\frac x\pi\right)\right) $$

Answer 1

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