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Consider $J(s)$ a Dirichlet series defined by its Euler product as follows \begin{align*} J(s)=\prod_{p \in \mathbb{P}}\left(1+\sum_{k=1}^{\infty} \frac{2}{p^{k^{2} s}}\right) \end{align*} After some formal manipulations and following the clues of certain patterns, I was able to assemble the above "identity". Now, I'm not completely sure of its correctness, so I'm asking for a confirmation or else a refutation.


This comes from $$ \mathcal{\zeta}(s)=\prod_{p \in \mathbb{P}}\left(1+\sum_{k=1}^{\infty} \frac{2}{p^{k^{2} s}}\right) \times \frac{\zeta(2 s)}{\zeta(s)} \times \underbrace{ \prod_{p \in \mathbb{P}} \prod_{k=1}^{\infty} \frac{\left(1+p^{-2 k s}\right)\left(1-p^{-(2 k+1) s}\right)}{\left(1-p^{-2 k s}\right)\left(1+p^{-(2 k+1) s}\right)}}_{B(s)} $$ and then observing that $B(s)=B(-s)$.

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    $\begingroup$ Could you edit in your reasoning? $\endgroup$
    – J.G.
    Aug 27 '20 at 12:07
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    $\begingroup$ I think the product defining $J(s)$ only converges for $\Re(s)>0$, so $J(-s)$ is undefined. May be the analytic continuation can be used to define $J$ in a bigger region, but then the formula of $J$ can't be used for the values in the extended domain. $\endgroup$
    – jjagmath
    Aug 27 '20 at 12:17
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The product for $J(s)$ converges if $\Re s>1$ [@jjagmath is a bit off], and the Jacobi triple product gives \begin{align*} J(s)&=\prod_{p\in\mathbb{P}}\sum_{k\in\mathbb{Z}}p^{-k^2 s} \\&=\prod_{p\in\mathbb{P}}\prod_{n\geqslant 1}(1-p^{-2ns})(1+p^{-(2n-1)s})^2 \\&=\prod_{n\geqslant 1}\prod_{p\in\mathbb{P}}\frac{(1-p^{-2ns})(1-p^{-(4n-2)s})^2}{(1-p^{-(2n-1)s})^2} \\&=\prod_{n\geqslant 1}\frac{\zeta^2\big((2n-1)s\big)}{\zeta(2ns)\zeta^2\big((4n-2)s\big)}. \end{align*} With $\zeta$ analytically continued, the last product converges locally uniformly for $\Re s>0$ and $s$ not a singularity of any term, thus it defines the analytic continuation of $J(s)$ for these values of $s$. The poles of $J(s)$ are nontrivial zeros of $\zeta(ks)$ for even $k$, and we know that any neighborhood of any $s$ with $\Re s=0$ contains infinitely many of the poles. Thus, $\Re s=0$ is the natural boundary of $J(s)$, and the notation "$J(-s)$" makes no sense for $\Re s\geqslant 0$.

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  • $\begingroup$ The poles generated by $\zeta(2ns)$ are all at horizontal lines passing through the nontrivial zeros of $\zeta(s)$, so there are neighborhoods of many $\Re s=0$ with no poles. $\endgroup$
    – Neves
    Aug 29 '20 at 10:21
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    $\begingroup$ @Neves: Not horizontal but vertical. And, due to the "$2n$", with denser and denser points as $n\to\infty$. $\endgroup$
    – metamorphy
    Aug 29 '20 at 10:29

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