Concerning Hurwitz Zeta function, how to prove the following identity? It is claimed that
$$\zeta'(0,s)=\ln\left(\frac{\Gamma(s)}{\sqrt{2\pi}}\right)$$
where the derivative is meant by the first argument (as usual with Hurwitz Zeta).
How to prove this?
Wolfram Alpha confirms this: http://www.wolframalpha.com/input/?i=D[HurwitzZeta[s%2C+p]%2C+s]+at+s%3D0 but I have no other clue.
Is there a more general form of the identity (with a variable instead of 0)?
 A: Ok let's see:
There is a famous integral representation for the Hurwitz function due to Hermite which reads as follows:
$$
\zeta(s,a)= \frac{a^{-s}}{2}+\frac{a^{1-s}}{1-s}+2\int_0^{\infty}dt \frac{\sin(s\arctan(t/a))}{(e^{2 \pi t}-1)(t^2+a^2)^{s/2}} \quad (1)
$$
You may check for yourself that it is allowed to switch differentiation w.r.t to $s$ with integration. 
So the next task is differentiating and taking the limit afterwards, because this is straightforward but tedious i will omit the details here and just state that
$$
\zeta'(0,a)= \frac{-\log(a)}{2}+ a\log(a)-a +2\int_0^{\infty}dt \frac{\arctan(t/a)}{e^{2 \pi t}-1}
$$
Because we have our lucky day, the remaining integral is well known:
$$
2\int_0^{\infty}dt \frac{\arctan(t/a)}{e^{2 \pi t}-1}= \log(\Gamma(a))- \log(a)(1/2-a)+a-\log(2\pi)/2 \quad (2)
$$
Putting everything together we obtain the desired result:
$$
\zeta'(0,a)=\log\left(\frac{\Gamma(a)}{\sqrt{2\pi}}\right)
$$
Edit: One way to proof (1) is by the aid of the famous Abel-Plana formula using the fact that $\frac{1}{(\pm i t+a)^s}=\frac{e^{ \pm i s\arctan(t/a)}}{( t^2+a^2)^{s/2}}$
Edit2: A route for proving (2) may be found here, differentiating w.r.t to $1/a$ and then use @M.N.C.E's beautiful solution, the integrating back w.r.t to  $1/a$.
