# Proving the "identity" $\frac{\zeta^{2}(s)}{\zeta(2 s) J(s)}=\frac{\zeta^{2}(-s)}{\zeta(-2 s) J(-s)}$

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 $$B(s)=B(-s)$$.

• Could you edit in your reasoning?
– J.G.
Aug 27 '20 at 12:07
• 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. Aug 27 '20 at 12:17

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 $$\Re s\geqslant 0$$.
• 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. Aug 29 '20 at 10:21
• @Neves: Not horizontal but vertical. And, due to the "$2n$", with denser and denser points as $n\to\infty$. Aug 29 '20 at 10:29