Does this sum of two Euler products converge for $s=1$?

Question

The following sum of two Euler products is always real for real $s$:

$$A(s):=\prod_{p=prime} (1 - (i\,p)^{-s})^{-1}+\prod_{p=prime} (1 - (-i\,p)^{-s})^{-1}$$

ADDED:

Some (trivial) closed forms that I found so far:

$A(2) = \frac{4}{5}\zeta(2)$

$A(4) = 2\,\zeta(4)$

$A(6) = \dfrac{2\,\zeta(12)}{\zeta(6)}$

$A(8) = 2\,\zeta(8)$

$A(10) = \dfrac{2\,\zeta(20)}{\zeta(10)}$

$A(12) = 2\,\zeta(12)$

$A(14) = \dfrac{2\,\zeta(28)}{\zeta(14)}$

$A(16) = 2\,\zeta(16)$

$\cdots$

Numerical evidence suggests that this sum also converges at $s=1$ (edit: I used $\lim_{s\to 1^+}$), albeit slowly. Could this be proven/falsified ?

Answer 1

The boring/trivial answer is no

$A: \Bbb{R} \to \Bbb{C}$ doesn't converge at $x = 1$ in its original definition $$A(x):=\prod_{p \, \text{prime}} (1 - (i\,p)^{-x})^{-1}+\prod_{p \, \text{prime}} (1 - (-i\,p)^{-x})^{-1}$$ because the individual products $F(x) := \prod_{p \, \text{prime}} (1 - (i\,p)^{-x})^{-1}$ and $\overline{F(x)} := \prod_{p \, \text{prime}} (1 - (-i\,p)^{-x})^{-1}$ don't converge at $x = 1$. This can pretty easily be checked via taking logs: $$\log F(x) = \sum_{p \, \text{prime}} \sum_{n \geq 1} \frac{i^{nx}}{n p^{nx}} \sim \sum_{p \, \text{prime}} \frac{i}{p} + \mathcal{O}(1) \text{ as } x \to 1, $$ and we know $\sum_{p \, \text{prime}} \frac{i}{p}$ is divergent. Clearly the product $\overline{F(x)}$ converges at $x = 1$ iff $F(x)$ does.

The interesting answer is also no

Since we're on the real axis, $A(x) = F(x) + \overline{F(x)} = 2 \operatorname{Re}(F(x))$. Even though $F$ and $\overline{F}$ diverge as $x \to 1^+$, we might surmise that the divergence happens "mostly" in the imaginary part, and cancelling that out leaves a real part that converges. Indeed (and this will be relevant shortly), we can show that $\lim_{x \to 1^+} |F(x)| = \frac{\pi}{\sqrt{15}}$ despite the fact that $\lim_{x \to 1^+} F(x)$ DNE by an analogous method, because \begin{align*} |F(x)|^2 &= F(x) \overline{F(x)} \\ &= \prod_{p \, \text{prime}} \left(1 - \frac{i^{-x}}{p^x}\right)^{-1} \left(1 - \frac{i^{x}}{p^x}\right)^{-1} \\ &= \prod_{p \, \text{prime}} \left(1 - \frac{2 \cos( \pi x/2)}{p^x} + \frac{1}{p^{2x}}\right)^{-1} \\ &\to \prod_{p \, \text{prime}} \left(1 + \frac{1}{p^{2}}\right)^{-1} = \frac{\pi^2}{15} \text{ as } x \to 1^+. \ \end{align*}

The final step (demonstrating the stated convergence as $x \to 1^+$) requires us to argue that $$\lim_{x \to 1^+} \cos( \pi x/2) P(x) = 0,$$ where $P(x)$ is the prime zeta function. As $x \to 1^+$, $P(x) \sim \ln (x-1)$, while $\cos(\pi x/2) \sim -(x-1)$, and so we can replace these two complicated functions with their simpler asymptotic equivalents and use L'Hopital to demonstrate that this limit is, in fact, zero.

Why is this relevant?

We know that $\lim_{x \to 1^+} |F(x)| = \frac{\pi}{\sqrt{15}}$ despite the fact that $\lim_{x \to 1^+} F(x)$ does not exist.

The OP's question asks if $\lim_{x \to 1^+} A(x) = \lim_{x \to 1^+} 2 \operatorname{Re}(F(x))$ exists.

But if $\lim_{x \to 1^+} |F(x)|$ and $\lim_{x \to 1^+} \operatorname{Re}(F(x))$ both existed, then so would $\lim_{x \to 1^+} F(x)$. Therefore, since $\lim_{x \to 1^+} F(x)$ does not exist, neither does $\lim_{x \to 1^+} A(x)$.

Answer 2

$$\lim_{x\to \infty} \quad\prod_{p\le x} \frac1{1 +i p^{-1}}+\prod_{p\le x} \frac1{1 -i p^{-1}}$$ diverges, this follows immediately from Mertens thoerem $$\sum_{p\le x} p^{-1}=\log \log x+M+o(1)$$ so that $$\sum_{p\le x} \log(1+ip^{-1})=i\log \log x+a+ib+o(1)$$

$$\prod_{p\le x} \frac1{1 +i p^{-1}}+\prod_{p\le x} \frac1{1 -i p^{-1}}=2 e^{-a} \cos(b+\log\log x)+o(1) $$

Then $$\lim_{s\to 1^+} \Re(\prod_p \frac1{1 -e^{-i\pi s/2} p^{-s}})$$ diverges as well, this is because as $s\to 1^+$, $$-\sum_p \log(1 -e^{i\pi s/2} p^{-s})= C+o(1)+e^{-i\pi s/2} \log \zeta(s)= C+o(1)+i\log (s-1) $$ $$\Re(\prod_p \frac1{1 -e^{i\pi s/2} p^{-s}})=e^{\Re(C)} \cos(\Im(C)+\log(s-1))+o(1)$$