1-6-p-logp-p-2-1-log1-pi-2-log2-4pi-2-log3-9pi-2-p-prime-

Question Number 128236 by Dwaipayan Shikari last updated on 05/Jan/21

\frac{1}{6} \underset{p}{\sum^{\infty}} \frac{l o g p}{p^{2} - 1} = \frac{l o g 1}{π^{2}} + \frac{l o g 2}{4 π^{2}} + \frac{l o g 3}{9 π^{2}} + \dots (p = p r i m e)

Commented by Dwaipayan Shikari last updated on 05/Jan/21

ζ (s) = \underset{p}{\prod^{\infty}} {(1 - \frac{1}{p^{s}})}^{- 1}

l o g (ζ (s)) = - \underset{p}{\sum^{\infty}} l o g (1 - \frac{1}{p^{s}})

\frac{ζ^{'} (s)}{ζ (s)} = - \sum^{\infty} \frac{p^{- s} l o g (p)}{1 - \frac{1}{p^{s}}} \Rightarrow ζ^{'} (s) = ζ (s) \underset{p}{\sum^{\infty}} \frac{l o g p}{1 - p^{s}}

ζ^{'} (s) = - \sum^{\infty} \frac{l o g n}{n^{s}}

\Rightarrow \frac{l o g 1}{1} + \frac{l o g 2}{2^{2}} + \frac{l o g 3}{3^{2}} + \dots = \frac{π^{2}}{6} \underset{p}{\sum^{\infty}} \frac{l o g p}{p^{2} - 1}

\Rightarrow \frac{1}{6} \underset{p}{\sum^{\infty}} \frac{l o g p}{p^{2} - 1} = \frac{l o g 1}{π^{2}} + \frac{l o g 2}{4 π^{2}} + \dots