What is the best lower bound known for $$\prod_{p\equiv 3 \pmod 4} p^{v_p(n!)},$$ where the product is taken over all the primes(congruent to $3$ modulo $4$) less than or equal to $n$.
Fleshing out Wojowu's comment: set $K = \big\lfloor \frac{\log n}{\log p} \big\rfloor$. Since \begin{align*} v_p(n!) = \sum_{1\le k\le K} \bigg\lfloor \frac n{p^k} \bigg\rfloor &\ge \sum_{1\le k\le K} \bigg( \frac n{p^k}  1 \bigg) \\ &= \frac n{p1}  \frac n{p^K(p1)}  K = \frac n{p1} + O\bigg( \frac{\log n}{\log p} \bigg) \end{align*} for $p\le n$, we have \begin{align*} \log\prod_{p\equiv 3 \pmod 4} p^{v_p(n!)} &= \sum_{\substack{p\le n \\ p\equiv 3\pmod 4}} \bigg( \frac n{p1} + O\bigg( \frac{\log n}{\log p} \bigg) \bigg) \log p \\ &= n \sum_{\substack{p\le n \\ p\equiv 3\pmod 4}} \frac{\log p}{p1} + O\big( \pi(n;4,3) \log n \big) \\ &= \frac{n \log n}2 + O(n), \end{align*} where the last equality used partial summation: with $\theta(x;4,3) = \sum_{p\le x,\, p\equiv3\pmod 4} \log p \sim x/2$, \begin{align*} \sum_{\substack{p\le n \\ p\equiv 3\pmod 4}} \frac{\log p}{p1} &= \int_2^n \frac1{t1} \,d\theta(t;4,3) \\ &= \frac{\theta(t;4,3)}{t1} \bigg_2^n + \int_2^n \frac{\theta(t;4,3)}{(t1)^2} \,dt \\ &= O(1) + \int_2^n \frac{t+O(t/\log^2t)}{(t1)^2} \,dt \\ &= O(1) + \bigg( \log(t1)  \frac1{t1} \bigg) \bigg_2^n + O(1) \\ &= \log n + O(1). \end{align*}
In hindsight, of course $n(\log n)/2$ should be the main term: we expect the product to be roughly the square root of $n!$, and $\log\sqrt{n!} \sim n(\log n)/2$ by Stirling's formula.
All the steps of this argument can be given with explicit constants in the inequalities if you want (including corresponding upper bounds); the partial summation step can start with an explicit lower bound for $\theta(n;4,3)$ found in this paper for example.

2$\begingroup$ But our product does not exceed $\prod_p p^{\nu_p(n!)}=n!$, so its logarithm should not grow faster than $\log n!=n\log n+O(n)$. I guess your asymptotic formula for $\sum_{p\leqslant n, p=4k+3} \log p/p$ has to be corrected. $\endgroup$ – Fedor Petrov Aug 1 '19 at 10:16

1$\begingroup$ @FedorPetrov excellent observation and excellent diagnosis! I've corrected the error $\endgroup$ – Greg Martin Aug 1 '19 at 16:47

$\begingroup$ @GregMartin could you please explain the partial summation part in more detail? $\endgroup$ – Kristiyan Vasilev Aug 5 '19 at 20:36
p \equiv 3 (\mod 4)
) on its own produces terrible spacing. $p \equiv 3 \pmod 4$ (p \equiv 3 \pmod 4
) isn't necessarily a lot better in subscripts, but it's the intended useage. I have edited accordingly. $\endgroup$ – LSpice Jul 31 '19 at 17:48