In my research I cam across this sum
$$S = \sum_{p\ge 3} \frac{\log \left({p}\right)}{{p}^{2} - 1}$$
where $p$ is a prime number. Numerical evaluation of the first $10^8$ primes yields $S = 0.338911932\cdots$. So far I have not found anything concerning this sum in the literature. I am looking for an analytical expression (sum over Möbius function?) and its asymptotic expansion if possible.
It turns out this constant has an exact expression in terms of the Riemann zeta function $$ \zeta(s) = \sum_{n=1}^\infty n^{-s} = \prod_p ( 1-p^{-s} )^{-1}. \quad(s>1) $$ We have \begin{align*} -\frac{\zeta'(s)}{\zeta(s)} &= -\frac d{ds} \log\zeta(s) \\ &= -\frac d{ds} \sum _p \log \bigl( ( 1-p^{-s} )^{-1} \bigr) \\ &= \sum_p \frac{\log p}{p^s-1}. \end{align*} Therefore $$ -\frac{\zeta'(2)}{\zeta(2)} = \sum_p \frac{\log p}{p^2-1} = S + \frac{\log2}{2^2-1}, $$ and therefore $$ S = -\frac{\zeta'(2)}{\zeta(2)} - \frac{\log2}3 = -\frac{6\zeta'(2)}{\pi^2} - \frac{\log2}3 \approx 0.33891193290788436993. $$