Asymptotics for $\prod(1-\frac{1}{p})$ over all primes $p\leq x$ with $p \equiv 3 \bmod 4$

Question

Let us define the following functions:

\begin{equation*} \small A(x)=\prod_{\substack{p\leq x\\ p\equiv 3 \bmod 4}} \Big(1-\frac{1}{p}\Big), \mbox{ } \mbox{ } B(x)=\prod_{\substack{p\leq x\\ p\equiv 1 \bmod 4}} \Big(1-\frac{1}{p}\Big), \mbox{ } \mbox{ } C(x)=\prod_{\substack{p\leq x\\ p\equiv 3 \bmod 4}} \Big(1+\frac{1}{p}\Big) \end{equation*} The following is known: $$A(x) \sim \frac{\alpha}{\sqrt{\log x}}, \mbox{ } \mbox{ } B(x)\sim \frac{\beta}{\sqrt{\log x}}, \mbox{ } \mbox{ } C(x)\sim \gamma \cdot \sqrt{\log x}$$ where $\alpha,\beta,\gamma$ are positive constants with

$$\alpha\beta =2e^{-\gamma}, \mbox{ } \mbox{ } \alpha\gamma = \frac{1}{2K^2}, \mbox{ } \mbox{ } \alpha\neq\beta.$$

The fact that $\alpha\gamma=1/(2K^2)$ where $K$ is the Landau–Ramanujan constant, can easily be derived from equations (2.1) and (2.2) in this article. The fact that $\alpha\beta=2e^{-\gamma}$ where $\gamma$ is the Euler–Mascheroni constant, can be derived from the formula $A(x)B(x)\sim 2e^{-\gamma}/\log x$, used for instance here in my previous MO question. It is confirmed when you perform the computations numerically. Products such as $A(x)$ are related to Gaussian primes and the prime density in sums of two square integers. See also this article about quadractic polynomials with very high density of primes, and the related Hardy and Littlewood's Conjecture F (see here). My interest in $A(x)$ in particular is for the same reason: it is the main asymptotic function in some other quadratic polynomials with very high prime density, and the constant $\alpha$ is related to the highest possible prime density.

My question

What is the exact value of $\alpha$? I feel I am getting close, but I am stuck with a system of two equations and three variables $\alpha,\beta,\gamma$, unable to get the exact value of any of them. If you can find the exact value for one of these variables, then you automatically get the exact value for the three of them.

Answer 1

I assume that you meant to write product and not sum. Defining \begin{equation*} \small D(x)=\prod_{\substack{p\leq x\\ p\equiv 1 \bmod 4}} \Big(1+\frac{1}{p}\Big) \mbox{ } \mbox{ } \end{equation*} We see that $$A(x)\cdot D(x) \sim L(1,\chi) = 1 - \frac{1}{3} + \frac{1}{5} - \cdots = \frac{\pi}{4}$$ Where $\chi$ is the Dirichlet character modulo 4 such that $\chi(-1) = -1$, and $L(1, \chi)$ is the associated $L$-function.

Also, $$A(x) \cdot B(x) \cdot C(x) \cdot D(x) \sim \prod_{p \neq 2} \left(1-\frac{1}{p^2} \right) = \frac{4}{3} \cdot \frac{1}{\zeta(2)} = \frac{8}{\pi ^ 2}$$

Dividing these two equations shows that $\beta \gamma = \frac{32}{\pi ^ 3}$, and so we see that

$$\alpha = \sqrt{\frac{\alpha \beta \cdot \alpha \gamma}{\beta \gamma}} = \sqrt{e ^ {-\gamma} \cdot \pi ^ 3 \cdot2^{-5} \cdot K^{-2}} = \frac{e ^ {-\frac{\gamma}{2}} \pi ^ {\frac{3}{2}} \sqrt{2}}{8K}$$

$\beta$ and $\gamma$ are of course easily recoverable as well.