**Background:** I'm facing the computation of the zeta regularization of the infinite product given by


$$\prod\_{m=-\infty}^\infty (km+u)$$


for a real positive $k$ and $\Im(u)\neq 0$. From [J. R. Quine, S. H. Heydari and R. Y. Song](https://www.ams.org/journals/tran/1993-338-01/S0002-9947-1993-1100699-1/home.html#Similar) (Example 10) I know that the zeta regularization of $\prod\_{m=-\infty}^\infty (m+u)$ is given by


$$\prod\_{m=-\infty}^\infty (m+u) = \begin{cases}1-e^{2\pi i u} & \Im(u) > 0 \\ 1-e^{-2\pi i u} & \Im(u) < 0\end{cases}$$


so a possible reasoning is as follows:


$$\prod\_{m=-\infty}^\infty (km+u) = \prod\_{m=-\infty}^\infty k(m+uk^{-1})$$


Now, using formula (1) from [J. R. Quine, S. H. Heydari and R. Y. Song](https://www.ams.org/journals/tran/1993-338-01/S0002-9947-1993-1100699-1/home.html#Similar) we have that


$$\prod\_{m=-\infty}^\infty k(m+uk^{-1}) = k^{Z(0)}\prod\_{m=-\infty}^\infty (m+uk^{-1})$$


and we are reduced to computing $Z(0)$ where $Z(s)$ is the analytic prolongation of $Z(s)=\sum\_{m \in \mathbb{Z}}(m+uk^{-1})^{-s}$. We can write $Z(s)$ as


$$Z(s) = \sum\_{m > 0}(m+uk^{-1})^{-s}+(uk^{-1})^{-s}+\sum\_{m <0}(m+uk^{-1})^{-s}$$


By adding and subtracting the $m=0$ term to both of the series and by changing variable from $m$ to $-m$ in the sum indexed by negative integers we get


$$\begin{align}
Z(s) &= \sum\_{m=0}^\infty(m+uk^{-1})^{-s} - (uk^{-1})^{-s} + \sum\_{m=0}^\infty(uk^{-1}-m)^{-s} \\
&= \sum\_{m=0}^\infty(m+uk^{-1})^{-s} - (uk^{-1})^{-s} + e^{\pi i s}\sum\_{m=0}^\infty(m-uk^{-1})^{-s} \\
&= \zeta(s,uk^{-1}) -(uk^{-1})^{-s} + e^{\pi is}\zeta(s,-uk^{-1}) \end{align}$$


where $\zeta(s,a)$ is the [Hurwitz Zeta function](https://dlmf.nist.gov/25.11#E1). So that by using formula 25.11.13 again in <https://dlmf.nist.gov/25.11#E13> we found that


$$Z(0) = \dfrac{1}{2}-uk^{-1}-1+\dfrac{1}{2}+uk^{-1}=0$$


and we end up with


$$\prod\_{m=-\infty}^\infty k(m+uk^{-1}) = k^{Z(0)}\prod\_{m=-\infty}^\infty (m+uk^{-1}) = \prod\_{m=-\infty}^\infty (m+uk^{-1}) = \begin{cases}1-e^{2\pi i uk^{-1}} & \Im(uk^{-1}) > 0 \\ 1-e^{-2\pi i uk^{-1}} & \Im(uk^{-1}) < 0\end{cases}$$


**Question:** I'm not confident enough with these manipulations involving analytic prolongations so I may be wrong in the above derivation of $Z(0)=0$, Is the above result correct? Does it remain valid if one replaces the condition $k$ real and positive with $k$ a nonzero complex number?



The zeta regularization of $\prod_{m=-\infty}^\infty (km+u)$
I think there are no problems but one should be careful with analytic continuation.


Up until this line nothing happens:


$$\sum\_{m=0}^\infty(m+uk^{-1})^{-s} - (uk^{-1})^{-s} + e^{\pi i s}\sum\_{m=0}^\infty(m-uk^{-1})^{-s}$$


Then you claim that to equal:


$$\zeta(s,uk^{-1}) -(uk^{-1})^{-s} + e^{\pi is}\zeta(s,-uk^{-1})$$.


To be more careful one should make sure that the two sums giving the Hurwitz zetas are absolutely convergent in the same open set, then, applying the standard facts of analytic continuation, you get $Z(s)=\zeta(s,uk^{-1}) -(uk^{-1})^{-s} + e^{\pi is}\zeta(s,-uk^{-1})$ everywhere.


Now, according to the website you linked, the series for $\zeta(s,a)$ converges for $\Re s>1$, so indeed both series converge in the same open set.


To conclude, since your $\sum\_{m\in\mathbb{Z}}(m+uk^{-1})^{-s}$ is analytic for $\Re s>1$ and it is the sum of two series (plus that harmless $(uk^{-1})^{-s}$) that can be both meromorphically extended to the whole plane (and are both analytic in $0$), it is true that $Z(s)=\zeta(s,uk^{-1}) -(uk^{-1})^{-s} + e^{\pi is}\zeta(s,-uk^{-1})$ and that $Z(0)=0$.


About the extension of this result to a complex $k$, it is still valid provided that $uk^{-1}\not\in\mathbb{Z}$ since then one of the two $\zeta$ would not be defined (and of course, one of the factors in your product would then be $0$).


