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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.

Lorenz H Menke
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1 Answers1

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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 $$ S = -\frac{\zeta'(2)}{\zeta(2)} - \frac{\log2}3 = -\frac{6\zeta'(2)}{\pi^2} - \frac{\log2}3 \approx 0.33891193290788436993. $$

Greg Martin
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  • using this, we can re-write this as $$S=\log\frac{A^{12}}{2^{4/3}\pi}-\gamma,$$ where $A$ is the Glaisher constant and $\gamma$ is the Euler constant – clathratus Sep 29 '23 at 18:50