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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
(1/6)Σ_p ^∞ ((logp)/(p^2 −1))=((log1)/π^2 )+((log2)/(4π^2 ))+((log3)/(9π^2 ))+...   (p=prime)
$$\frac{\mathrm{1}}{\mathrm{6}}\underset{{p}} {\overset{\infty} {\sum}}\frac{{logp}}{{p}^{\mathrm{2}} −\mathrm{1}}=\frac{{log}\mathrm{1}}{\pi^{\mathrm{2}} }+\frac{{log}\mathrm{2}}{\mathrm{4}\pi^{\mathrm{2}} }+\frac{{log}\mathrm{3}}{\mathrm{9}\pi^{\mathrm{2}} }+…\:\:\:\left({p}={prime}\right) \\ $$
Commented by Dwaipayan Shikari last updated on 05/Jan/21
ζ(s)=Π_p ^∞ (1−(1/p^s ))^(−1)   log(ζ(s))=−Σ_p ^∞ log(1−(1/p^s ))  ((ζ′(s))/(ζ(s)))=−Σ^∞ ((p^(−s) log(p))/(1−(1/p^s )))⇒ζ′(s)=ζ(s)Σ_p ^∞ ((logp)/(1−p^s ))  ζ′(s)=−Σ^∞ ((logn)/n^s )  ⇒((log1)/1)+((log2)/2^2 )+((log3)/3^2 )+...=(π^2 /6)Σ_p ^∞ ((logp)/(p^2 −1))  ⇒(1/6)Σ_p ^∞ ((logp)/(p^2 −1))=((log1)/π^2 )+((log2)/(4π^2 ))+...
$$\zeta\left({s}\right)=\underset{{p}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{{p}^{{s}} }\right)^{−\mathrm{1}} \\ $$$${log}\left(\zeta\left({s}\right)\right)=−\underset{{p}} {\overset{\infty} {\sum}}{log}\left(\mathrm{1}−\frac{\mathrm{1}}{{p}^{{s}} }\right) \\ $$$$\frac{\zeta'\left({s}\right)}{\zeta\left({s}\right)}=−\overset{\infty} {\sum}\frac{{p}^{−{s}} {log}\left({p}\right)}{\mathrm{1}−\frac{\mathrm{1}}{{p}^{{s}} }}\Rightarrow\zeta'\left({s}\right)=\zeta\left({s}\right)\underset{{p}} {\overset{\infty} {\sum}}\frac{{logp}}{\mathrm{1}−{p}^{{s}} } \\ $$$$\zeta'\left({s}\right)=−\overset{\infty} {\sum}\frac{{logn}}{{n}^{{s}} } \\ $$$$\Rightarrow\frac{{log}\mathrm{1}}{\mathrm{1}}+\frac{{log}\mathrm{2}}{\mathrm{2}^{\mathrm{2}} }+\frac{{log}\mathrm{3}}{\mathrm{3}^{\mathrm{2}} }+…=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\underset{{p}} {\overset{\infty} {\sum}}\frac{{logp}}{{p}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{6}}\underset{{p}} {\overset{\infty} {\sum}}\frac{{logp}}{{p}^{\mathrm{2}} −\mathrm{1}}=\frac{{log}\mathrm{1}}{\pi^{\mathrm{2}} }+\frac{{log}\mathrm{2}}{\mathrm{4}\pi^{\mathrm{2}} }+… \\ $$

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