How to prove that $\sqrt{3}\pi/6=\prod_{p \equiv 1 \pmod{6}} \frac{p}{p-1}\prod_{p \equiv 5 \pmod{6}} \frac{p}{p+1}$ with $p \in \mathbb{P}$?

Question

I would like to prove the formula $$\frac{\sqrt{3}\pi}{6}=\left(\prod_{\substack{p \equiv 1 \pmod{6} \\ p \in \mathbb{P}}} \frac{p}{p-1}\right) \cdot \left(\prod_{\substack{p \equiv 5 \pmod{6} \\ p \in \mathbb{P}}} \frac{p}{p+1}\right)=\frac{5}{6} \cdot \frac{7}{6} \cdot \frac{11}{12} \cdot \frac{13}{12} \cdot \frac{17}{18} \cdots ,$$ it appears listed on the wikipedia article List of formulae involving π#Infinite products, I put a citation needed tag on it and left a message on the talk page of the user who added the formula but this was 5 months ago and is unlikely to get an answer as this formula is that user only edit ever so it looks more like throw away account.

Numerically the formula seems to be correct so my guess is that it was taken from some paper or that it is not that hard to prove, I don't much about this stuff my only idea was try something to similar to the dirichlet beta function $\beta(s) = \prod_{p \equiv 1 \ \mathrm{mod} \ 4} \frac{1}{1 - p^{-s}} \prod_{p \equiv 3 \ \mathrm{mod} \ 4} \frac{1}{1 + p^{-s}}$ since $\beta(1)=\pi/4$ like this answer but couldn't make it work mod 6 so I got nowhere. Also looked for the formula with approach0.xyz/searchonmath.com but couldn't find it either.

The code "Product[Piecewise[{{p/(p - 1), Element[p, Primes] && Mod[p, 6] == 1}}, 1], {p, 2, 1000}] Product[Piecewise[{{p/(p + 1), Element[p, Primes] && Mod[p, 6] == 5}}, 1], {p, 2, 1000}]" can be used on wolfram online to evaluate the series (Thanks to @J.M.can'tdealwithit who provided me with code over the mathematica stackexchange chat)

Any help would be appreciated.

Answer 1

Knowing $L(s,\chi)$${}=\prod_{p\in\mathbb{P}}(1-\chi(p)p^{-s})^{-1}=\sum_{n>0}\chi(n)n^{-s}$ for $\Re s>1$, where $$\chi(n):=\begin{cases}1,&n\equiv1\pmod{6}\\-1,&n\equiv-1\pmod{6}\\0,&\text{otherwise}\end{cases}$$ the claim follows from the ability to take $s\to1^+$ (a non-trivial result!), since $$\sum_{n>0}\frac{\chi(n)}{n}=\sum_{n=0}^\infty\int_0^1(x^{6n}-x^{6n+4})\,dx=\int_0^1\frac{1-x^4}{1-x^6}\,dx=\frac{\pi\sqrt{3}}{6}$$ (an elementary way, or using $\int_0^1\frac{t^{a-1}-t^{-a}}{1-t}\,dt=\pi\cot a\pi$ at $a=1/6$ after $t=x^6$).