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Question Number 222424 by wewji12 last updated on 26/Jun/25
∫_2 ^( ∞) (dz/(ln(z)))−Σ_(l=2) ^∞ (1/(ln(l)))=??
$$\int_{\mathrm{2}} ^{\:\infty} \:\:\:\:\frac{\mathrm{d}{z}}{\mathrm{ln}\left({z}\right)}−\underset{{l}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{ln}\left({l}\right)}=?? \\ $$
Answered by MrGaster last updated on 26/Jun/25
I=lim_(n→∞) (∫_z ^n (dz/(ln z))−Σ_(l=2) ^n (1/(ln l))) f(x)=(1/(ln x)),f′(x)=−(1/(x ln^2 x)) Σ_(l=2) ^n f(t)=∫_2 ^n f(x)dx+((f(2)+f(n))/2)+(B_2 /(2!))(f′(n)−f′(2))+… B_2 =6,(B_2 /(2!))=(1/(12)) f(2)=(1/(ln 2)),f′(2)=(1/(2 ln^2 2)) ∫_2 ^n f(x)dx−Σ_(l=2) ^n f(l)=−((f(2)+f(n))/2)−(B_2 /(2!))(f′(n)−f′(2))+… lim_(n→∞) f(n)=0,lim_(n→∞) f′(n)=0 lim_(n→∞) (∫_2 ^n f(x)dx−Σ_(l=2) ^n f(l))=−((f(2))/2)+(B_2 /(2!))f′(2) =−(1/2)∙(1/(ln 2))+(1/(12))∙(−(1/(2 ln^2 2))) =−(1/(2 ln 2))−(1/(24 ln^2 2))
$${I}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\int_{{z}} ^{{n}} \frac{{dz}}{\mathrm{ln}\:{z}}−\underset{{l}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{ln}\:{l}}\right) \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{ln}\:{x}},{f}'\left({x}\right)=−\frac{\mathrm{1}}{{x}\:\mathrm{ln}^{\mathrm{2}} {x}} \\ $$$$\underset{{l}=\mathrm{2}} {\overset{{n}} {\sum}}{f}\left({t}\right)=\int_{\mathrm{2}} ^{{n}} {f}\left({x}\right){dx}+\frac{{f}\left(\mathrm{2}\right)+{f}\left({n}\right)}{\mathrm{2}}+\frac{{B}_{\mathrm{2}} }{\mathrm{2}!}\left({f}'\left({n}\right)−{f}'\left(\mathrm{2}\right)\right)+\ldots \\ $$$${B}_{\mathrm{2}} =\mathrm{6},\frac{{B}_{\mathrm{2}} }{\mathrm{2}!}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$${f}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}},{f}'\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{ln}^{\mathrm{2}} \mathrm{2}} \\ $$$$\int_{\mathrm{2}} ^{{n}} {f}\left({x}\right){dx}−\underset{{l}=\mathrm{2}} {\overset{{n}} {\sum}}{f}\left({l}\right)=−\frac{{f}\left(\mathrm{2}\right)+{f}\left({n}\right)}{\mathrm{2}}−\frac{{B}_{\mathrm{2}} }{\mathrm{2}!}\left({f}'\left({n}\right)−{f}'\left(\mathrm{2}\right)\right)+\ldots \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left({n}\right)=\mathrm{0},\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}'\left({n}\right)=\mathrm{0} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\int_{\mathrm{2}} ^{{n}} {f}\left({x}\right){dx}−\underset{{l}=\mathrm{2}} {\overset{{n}} {\sum}}{f}\left({l}\right)\right)=−\frac{{f}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{{B}_{\mathrm{2}} }{\mathrm{2}!}{f}'\left(\mathrm{2}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{12}}\centerdot\left(−\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{ln}^{\mathrm{2}} \mathrm{2}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{ln}\:\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{24}\:\mathrm{ln}^{\mathrm{2}} \mathrm{2}} \\ $$

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