Can the derivative of Hurwitz Zeta function by the first argument be expressed in terms of Hurwitz Zeta and elementary fuctions?

There is a formula which expresses Hurwitz Zeta through its derivative:

$$\zeta '\left(z,\frac{q}{2}\right)-2^z \zeta '(z,q)+\zeta '\left(z,\frac{q+1}{2}\right)=\zeta(z,q)2^{z}\ln 2$$

So I wonder whether the opposite can be done?

  • $\begingroup$ In any case this relation will not be equally straightforward to obtain. The identity that you mention is an easy consequence of the multiplication formula for $\zeta(z,q)$ (just differentiate the second identity here). $\endgroup$ Dec 3, 2014 at 17:40
  • $\begingroup$ @O.L. I would be OK even if it deals with a particular $q$. $\endgroup$
    – Anixx
    Dec 3, 2014 at 20:52

1 Answer 1


Note that the derivative of the Hurwitz Zeta Function can be calculated as following (similiar to calculating the derivative of the Riemann Zeta):

\begin{align} \frac{d}{ds} \zeta(s,q) &= \frac{d}{ds} \sum_{n=0}^{\infty} (n+q)^{-s} \\ &= \sum_{n=0}^{\infty} \frac{d}{ds} (n+q)^{-s} \\ & = \sum_{n=0}^{\infty} - \ln(n+q) (n+q)^{-s} \\ &= -\sum_{n=0}^{\infty} \frac{\ln(n+q)}{(n+q)^s} \end{align}

Because $\ln(n+q)-\ln(n+q-1) = \ln\left(\frac{n+q}{n+q-1}\right)$, we actually have:

$$\zeta'(s,q) = - \ln(q) \zeta(s,q) + \sum_{n=1}^{\infty} \ln\left(\frac{n+q-1}{n+q}\right) \zeta(s, n+q)$$

I doubt a nicer relation exists.


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