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This paper in the formula F.3.6 (page 271) gives the following formula for the derivative of Hurwitz Zeta function:

$$\frac d{dz}\zeta(z,a)=-\frac12\ln(a)a^{-z}-\frac{\ln(a)a^{1-z}}{(z-1)^2}-\int_0^\infty\frac{\ln(a^2+y^2)\sin (z \arctan \frac ya)}{(a^2+y^2)^{z/2}(e^{2\pi y}-1)}dy+2\int_0^\infty\frac{\arctan \frac ya \cos(z \arctan \frac ya)}{(a^2+y^2)^{z/2}(e^{2\pi y}-1)}dy$$

At $z=0$ this reduces to the closed form

$$\zeta'(0,a)=\ln\left(\frac{\Gamma(a)}{\sqrt{2\pi}}\right)$$

I wonder whether the integrals can be reduced to closed form in general case?

This is a follow-up question of this question.

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