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

$$\frac{1}{6}\underset{p}{\sum ^{\mathrm{\infty }}}\frac{logp}{p^2-1}=\frac{log1}{{\pi }^2}+\frac{log2}{4{\pi }^2}+\frac{log3}{9{\pi }^2}+...(p=prime)$$

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

$$\zeta \left(s\right)=\underset{p}{\prod ^{\mathrm{\infty }}}{(1-\frac{1}{p^s})}^{-1}$$$$log\left(\zeta \left(s\right)\right)=-\underset{p}{\sum ^{\mathrm{\infty }}}log(1-\frac{1}{p^s})$$$$\frac{{\zeta }^{\prime }\left(s\right)}{\zeta \left(s\right)}=-\sum ^{\mathrm{\infty }}\frac{p^{-s}log\left(p\right)}{1-\frac{1}{p^s}}\Rightarrow {\zeta }^{\prime }\left(s\right)=\zeta \left(s\right)\underset{p}{\sum ^{\mathrm{\infty }}}\frac{logp}{1-p^s}$$$${\zeta }^{\prime }\left(s\right)=-\sum ^{\mathrm{\infty }}\frac{logn}{n^s}$$$$\Rightarrow \frac{log1}{1}+\frac{log2}{2^2}+\frac{log3}{3^2}+...=\frac{{\pi }^2}{6}\underset{p}{\sum ^{\mathrm{\infty }}}\frac{logp}{p^2-1}$$$$\Rightarrow \frac{1}{6}\underset{p}{\sum ^{\mathrm{\infty }}}\frac{logp}{p^2-1}=\frac{log1}{{\pi }^2}+\frac{log2}{4{\pi }^2}+...$$

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