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If $\zeta(s)$ is the Riemann Zeta function, then $\zeta(n)^z$, with $z \in \mathbb{C}$, $\Re(s)>1$, can be represented as

$$\zeta(s)^z=\sum_{n=1}^\infty \frac{d_z(n)}{n^{-s}}$$

where $d_z(n)$ can be represented as

$$d_z(n) = \prod_{p^a \mid n} (-1)^a \binom{-z}{a}$$

(see pg. 421 of Ivic's "The Riemann-Function: Theory and Practice"). There are a few more representations of $d_z(n)$ in Chapter 2 of Bateman and Diamond's "Analytic Number Theory: An Introductory Course" as well.

MY QUESTION: Can this be generalized for the Hurwitz Zeta function $\zeta(s,q) = \sum_{n=0}^{\infty}\frac{1}{(q+n)^s}$ with the added restriction that $q \in \mathbb{N}$? Which is to say, is there some $f_{q,z}(n)$ that satisfies

$$\zeta(s,q)^z=\sum_{n=1}^\infty \frac{f_{q,z}(n)}{n^{-s}}$$

? And if so, how can it be expressed?

It's trivial to see that $f_{q,1}(n) = 1$ if $n >= q$ and 0 otherwise for $\zeta(n,q)$.

It's not much more difficult to see that

$$f_{q,k}(n) = \sum_{a \cdot b = n}f_{q,1}(a) \cdot f_{q,k-1}(b)$$

for $\zeta(n,q)^k$, with $k \in \mathbb{N}$, through the standard mechanics of Dirichlet convolution.

But I'm stumped for the more general case of $z \in \mathbb{C}$, despite my firm conviction that the Hurwitz zeta function can indeed be raised to complex powers.

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  • $\begingroup$ The problem you are running into is due to the fact that the Hurwitz zeta-function does not, in general, have an Euler product representation. $\endgroup$ Dec 21, 2014 at 15:53
  • $\begingroup$ Does this prove, absolutely, that there can be no such $f_{q,z}(n)$, specifically for $q \in \mathbb{N}$? Or just that the usual methods can't be used to find such a function, and that it won't resemble the tidy definition for $d_z(n)$? $\endgroup$ Dec 21, 2014 at 19:17

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