How to compute constants in asymptotic density of numbers divisible by subset of primes I'm interested in the asymptotic density of the set $S$ of natural numbers divisible only by primes $p \equiv 1 \bmod 4$ (and similar subsets of $\mathbb{N}$). I'm aware of results which show that the number of members of $S$ which are at most $n$ is given by the asymptotic density
$$ \pi_S(n) \approx Cn\ln(n)^{-1/2}. $$
What isn't clear to me is how to obtain an expression which allows for computation of the constant $C$. Various sources I've read suggest that $C$ can be expressed in terms of special values of Dirichlet $L$-functions, or as an Euler product in similar fashion to the twin prime constant.
There's a Tauberian theorem which states that if $F(s) = \displaystyle\sum\limits_{n \geq 1}{\frac{a_n}{n^s}}$ is a Dirichlet series with non-negative coefficients, converging for $\text{Re}(s) > 1$ and suppose $F(s)$ extends analytically to $\text{Re}(s) = 1$, except at $s = 1$ where we have $F(s) = H(s)(s-1)^{\alpha-1}$ for some $\alpha \in \mathbb{R}$ and some $H(s)$ holomorphic and non-vanishing for $\text{Re}(s) \geq 1$, then
$$ \sum\limits_{n \leq x}{a_n} \approx Cx\ln(x)^{-\alpha}, \quad \quad C = \frac{H(1)}{\Gamma(1-\alpha)}. $$
It seems that one could use this result by taking $a_n = 1$ if $n \in S$ and $a_n = 0$ otherwise, which gives
$$ F(s) = \displaystyle\prod\limits_{p \equiv 1 \bmod 4}{(1-p^{-s})^{-1}}. $$
However, I've been unsuccessful in expressing $F(s)$ in terms of $L$-functions. Letting $G(s)$ be similar to $F(s)$, but with primes $p \equiv 3 \bmod 4$, and letting $\chi_0, \chi_1$ be the trivial and non-trivial Dirichlet characters with modulus 4 respectively, then $$L(s,\chi_0) = F(s)G(s) \quad \text{and} \quad L(s,\chi_1) = F(s)G(2s)/G(s).$$
My question is can the above theorem be used to find an expression for $C$ by isolating an expression for $F(s)$ in terms of $L$-functions, or otherwise?
 A: As reuns states in their comments, we should consider
$$H(s) = F(s)/\zeta(s)^{1/2} = ((1-2^{-s})L(s,\chi_1)/G(2s))^{1/2}.$$
$L(s,\chi_1)$ is known to be analytic and non-vanishing for $\text{Re}(s) \geq 1$ and as $1/G(2s) = \displaystyle\prod\limits_{p \equiv 3 \bmod 4} (1-p^{-2s})$, this is clearly convergent and non-vanishing for $\text{Re}(s) > 1/2$, in particular on $\text{Re}(s) \geq 1$. Moreover, we have
$$\lim_{s \rightarrow 1} F(s)(s-1)^{1/2}/H(s) = \lim_{s \rightarrow 1} (s-1)^{1/2} \zeta(s)^{1/2} = \left( \lim_{s \rightarrow 1} (s-1) \zeta(s) \right)^{1/2} = 1.$$
So indeed $H(s)$ satisfies the criteria in the theorem given in the question for $\alpha = 1/2$, so the constant $C$ may be written as the Euler product
$$C = H(1)/\Gamma(1/2) = \frac{1}{\sqrt{2\pi}} \displaystyle\prod\limits_{p \equiv 1 \bmod 4} (1-p^{-1})^{-1/2} \displaystyle\prod\limits_{q \equiv 3 \bmod 4} (1-q^{-1})^{1/2}.$$
However, this is a very slowly converging product, with 2 million terms giving just 5 correct decimal places. Using the relations between the $L$-functions, $F(s)$ and $G(s)$, one can produce the recurrence $G(s) = (L(s,\chi_0)G(2s)/L(s,\chi_1))^{1/2}$. As this recurrence is sub-linear and as the $L$-functions tend to $0$ as $s \rightarrow \infty$, this produces a rapidly converging product for $C$, with 20 terms giving over 100000 correct decimal places
$$C = \frac{\sqrt{\text{arcoth}(\sqrt{2})}}{\sqrt{2\pi\sqrt{2}}} \displaystyle\prod\limits_{k=1}^{\infty} \left( \frac{L(s,\chi_1)}{L(s,\chi_0)} \right)^{2^{-(k+1)}} \approx 0.287785482294975806193...$$
