Question-193847

Question Number 193847 by Mingma last updated on 21/Jun/23

Answered by witcher3 last updated on 21/Jun/23

=∫_0 ^1 ((ln(1−x))/x)dx=−Li_2 (1)=−(π^2 /6) ((1�)/(1−x))=Σx^k =∫_0 ^1 ln(x)Σ_(k≥0) x^k dx=Σ_(k=0) ^∞ ∫_0 ^1 ln(x)x^k dx =−(1/((k+1)^2 )) I=−Σ_(k≥0) (1/((k+1)^2 ))=−ζ(2)=−(π^2 /6)

$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}=−\mathrm{Li}_{\mathrm{2}} \left(\mathrm{1}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}=\Sigma\mathrm{x}^{\mathrm{k}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{x}\right)\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\mathrm{x}^{\mathrm{k}} \mathrm{dx}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{x}\right)\mathrm{x}^{\mathrm{k}} \mathrm{dx} \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{I}=−\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} }=−\zeta\left(\mathrm{2}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$ \\ $$

Commented by Mingma last updated on 21/Jun/23

Perfect ��

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Question-193847

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