I am attempting to understand the behavior of Hurwitz zeta functions and for what $a$ do they have an analytic continuation. Is it possible to write any Hurwitz zeta as an Euler product or are there conditions on $a$?
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2 Answers
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As Hurwitz proved, his zeta function has meromorphic continuation to $\mathbb{C}$. It has a simple pole at $s=1$ and no other pole. Euler products define Dirichlet series with leading term $1$, so the Hurwitz zeta function only has an Euler product decomposition when $a=1$ in which case it equals $\zeta(s)$.
answered Dec 14, 2023 at 13:37
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1$\begingroup$ Do you mind sending a link of the proof of the continuation to $C$? $\endgroup$ Commented Dec 14, 2023 at 15:08
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$\begingroup$ @SamayVarjangbhay See en.wikipedia.org/wiki/Hurwitz_zeta_function and the references therein. $\endgroup$ Commented Dec 14, 2023 at 18:32
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To add a little to the excellent answer above - it is known that for $0<a<1, a \ne 1/2$ the Riemann Hurwitz function has a lot of zeroes in the strip $1 < \sigma < 1+a$, so, in particular, there cannot be a simple product representation for $\Re s >1$ of Euler kind, while for $a=1/2$ there is, of course, a product representation for $\Re s >1$ though it has the extra term $(1-2^{-s})$ so it's not quite an Euler product; as for continuation and functional equation (in terms of the Lerch function at $1-s$ though), Apostol's book Introduction to Analytic Number Theory has it done very nicely.
