Gamma function arises when we consecutively differentiate an [Abel sequence][1]. An example of Abel polynomials are Bernoulli polynomials. When we differentiate it, the factors combine with themselves: $$B_n'(x)=nB_{n-1}(x)$$ $$B_n''(x)=n(n-1)B_{n-2}(x)$$ $$B_n'''(x)=n(n-1)(n-2)B_{n-3}(x)$$ They are just another name for Hurwitz Zeta function: $$B_n(x) = -n \zeta(1-n,x)$$ Thus, $$\frac\partial{\partial q}f(s,q)= s f(s+1,q)$$ $$\frac{\partial^2}{\partial q^2}f(s,q)= s(s+1) f(s+2,q)$$ $$\frac{\partial^3}{\partial q^3}f(s,q)= s(s+1)(s+2) f(s+3,q)$$ Since Reihmann zeta is Hurwitz zeta evaluated at $q=1$, the expression you give is apparently consecutive derivative of Hurwitz Zeta, with factor $\pi^{-s}$ appearing if we normalize Hurwitz Zeta by stretching it horizontally by factor of pi. Consecutive derivatives of Hurwitz Zeta in turn are nothing more than just polygamma function. ---------- For instance, here is the function $-1/x$: ![enter image description here][2] If we add infinitely many similar functions with a shift of pi/2 each in both directions, we get $\tan x$. But if we do the same only in one direction, we get "incomplete tangent": ![http://storage7.static.itmages.ru/i/14/0910/h_1410326921_7988832_91f3fd7d7d.png][3] The yellow one is $\operatorname{pg}(x)=\frac 1\pi \psi (\frac x\pi)$, the blue one is $\operatorname{cpg}(x)=-\frac 1\pi \psi (1-\frac x\pi)$. They obey $\operatorname{cpg}(x)+\operatorname{pg}(x)=-\cot(x)$. Now if we differentiate cpg(x) we get: $$(\operatorname{cpg}(x))^{(s-1)}=\pi^{-s}\Gamma(s)\zeta(s,1-\frac x\pi)$$ Compare it with yours formula: $$\xi(2s) = \pi^{-s}\Gamma\left(s\right)\zeta(2s)$$ [1]: http://en.wikipedia.org/wiki/Abel_polynomials [2]: http://storage6.static.itmages.ru/i/14/0910/h_1410327935_9959844_7e87ae4078.png [3]: https://i.sstatic.net/XR4Is.png