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

Answered by witcher3 last updated on 21/Jun/23

$$={\int }_0^1\frac{\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}=-{\mathrm{Li}}_2\left(1\right)=-\frac{{\pi }^2}{6}$$$$\frac{1}{1-\mathrm{x}}=\mathrm{\Sigma }{\mathrm{x}}^{\mathrm{k}}$$$$={\int }_0^1\ln \left(\mathrm{x}\right)\sum _{\mathrm{k}⩾0}{\mathrm{x}}^{\mathrm{k}}\mathrm{dx}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1\ln \left(\mathrm{x}\right){\mathrm{x}}^{\mathrm{k}}\mathrm{dx}$$$$=-\frac{1}{{(\mathrm{k}+1)}^2}$$$$\mathrm{I}=-\sum _{\mathrm{k}⩾0}\frac{1}{{(\mathrm{k}+1)}^2}=-\zeta \left(2\right)=-\frac{{\pi }^2}{6}$$$$$$

Commented by Mingma last updated on 21/Jun/23

Perfect ��

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