Is $\prod_{n=1}^\infty P_{2n-1}$ regularizable?

Assume that $P_n$ denotes the $n$'th prime for this entire question.

Inspriation: I was dumbfounded by the fact that: $$\hat\prod_\limits{n=1}^\infty P_{n}=4\pi^2$$ After further investigation, I learned of many other properties of zeta-regulation, as well as their proofs (to a reasonable extent). I realized nothing had been done of this question: $$2*5*11\cdots=\hat\prod_\limits{n=1}^\infty P_{2n-1}=\kappa$$ and solve for $\kappa$.

Issues: I am fairly competent in the usage of zeta-regularization, but am lost here, because it seems that zeta regularization doesn't work with $2n-1$ used instead of $n$. Unfortunately, I couldn't really figure out how to apply that on to this here. I was driven, from Resource 1, that a potential to use bounds was created, but my inability to logically understand this problem made it impossible to determine if $4\pi^2$ or $\sqrt{2\pi}$ would be upper or lower.

Questions:

a) Can $\kappa$ be zeta-regularized?

b) If it can, could you please assist me in a calculation of $\kappa$?

Side notes: I have had this question on my mind almost forever, and would really love an answer. Although I would most appreciate a proof, really anything will help me here. I am also somewhat uncertain with my tag choices, so please consider editing before immediately downvoting. The following were helpful in the construction of this problem.

1) When is an infinite product of natural numbers regularizable?

2) http://mathworld.wolfram.com/Zeta-RegularizedProduct.html

(I realize the former has been unanswered, but the problem had helpful comments as well as the idea to find a bound. )

Solution 1:

Given a set of indices $K\subseteq \mathbb{N}$, we can define the generalized prime zeta function $$ {\cal P}_{K}(s)=\sum_{k\in K}\frac{1}{p_k^s}, $$ where $p_k$ is the $k$-th prime. Following the thread of the proof that $\prod p = 4\pi^2$, we note that $$ \begin{eqnarray} e^{{\cal P}_K (s)} &=& \prod_{k\in K}\exp\left(p_k^{-s}\right) \\ &=& \prod_{k \in K} \prod_{n=1}^{\infty} \left(1 - p_k^{-ns}\right)^{-\mu(n)/n} \\ &=& \prod_{n=1}^{\infty} \prod_{k \in K} \left(1 - p_k^{-ns}\right)^{-\mu(n)/n} \\ &=& \prod_{n=1}^{\infty}\zeta_K(ns)^{\mu(n)/n}, \end{eqnarray} $$ where $\mu(n)$ is the Moebius function, and $\zeta_K$ is a suitably generalized Riemann zeta function: $\zeta_K(s) = \sum_{m}m^{-s}$, where the sum is over natural numbers $m$ all of whose prime factors are in $\{ p_{k\in K} \}$. Unless $K$ is a finite set, this series definition for $\zeta_K(s)$ converges only for $\Re(s)>1$; it needs to be defined by analytic continuation elsewhere. The logarithmic derivative is $$ {\cal P}^{\prime}_{K}(s)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\frac{n\zeta^{\prime}_K(ns)}{\zeta_K(ns)}=\sum_{n=1}^{\infty}\mu(n)\frac{\zeta^{\prime}_K(ns)}{\zeta_K(ns)}, $$ and so $$ {\cal P}^{\prime}_{K}(0) = \left(\sum_{n=1}^{\infty}\mu(n)\right)\frac{\zeta^{\prime}_K(0)}{\zeta_K(0)}=\frac{1}{\zeta(0)}\frac{\zeta^{\prime}_K(0)}{\zeta_K(0)}=-2\frac{\zeta^{\prime}_K(0)}{\zeta_K(0)}.$$ We conclude that $$ \prod_{k\in K}p_k=\exp\left(-{\cal P}^{\prime}_K(0)\right)=\exp\left(2\frac{\zeta^{\prime}_K(0)}{\zeta_K(0)}\right). $$ This puts your original question in terms of a hopefully easier-to-answer result, which depends only on the analytic continuation of a particular infinite sum.