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Question Number 97235 by bobhans last updated on 07/Jun/20

$$\underset{0}{\stackrel{1}{\int }}\ln \left(\mathrm{x}\right)\ln (1-\mathrm{x})\mathrm{dx}?$$

Answered by abdomathmax last updated on 07/Jun/20

$$\mathrm{I}={\int }_0^1\mathrm{lnxln}(1-\mathrm{x})\mathrm{dx}\mathrm{we}\mathrm{have}\mathrm{for}\mid \mathrm{x}\mid <1$$$$\frac{\mathrm{d}}{\mathrm{dx}}\ln (1-\mathrm{x})=\frac{-1}{1-\mathrm{x}}=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}\Rightarrow$$$$\ln (1-\mathrm{x})=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\Rightarrow$$$$\mathrm{I}=-{\int }_0^1\ln \left(\mathrm{x}\right)\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\right)=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}}\ln \left(\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{by}\mathrm{parts}{\mathrm{A}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{\mathrm{n}}\ln \left(\mathrm{x}\right)\mathrm{dx}$$$$={\left[\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}\ln \left(\mathrm{x}\right)\right]}_0^1-{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}\frac{\mathrm{dx}}{\mathrm{x}}=-\frac{1}{\mathrm{n}+1}{\int }_0^1{\mathrm{x}}^{\mathrm{n}}\mathrm{dx}$$$$=-\frac{1}{{(\mathrm{n}+1)}^2}\Rightarrow \mathrm{I}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{\mathrm{n}{(\mathrm{n}+1)}^2}$$$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}{(\mathrm{x}+1)}^2}\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{x}+1}+\frac{\mathrm{c}}{{(\mathrm{x}+1)}^2}$$$$\mathrm{a}=1,\mathrm{c}=-1$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{xF}\left(\mathrm{x}\right)=0=\mathrm{a}+\mathrm{b}\Rightarrow \mathrm{b}=-1\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+1}-\frac{1}{{(\mathrm{x}+1)}^2}\Rightarrow$$$$\mathrm{I}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1})-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+1)}^2}$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1})={lim}_{n\to \mathrm{\infty }}\sum _{\mathrm{k}=1}^{\mathrm{n}}(\frac{1}{\mathrm{k}}-\frac{1}{\mathrm{k}+1})$$$${\lim }_{\mathrm{n}\to \mathrm{\infty }}(1-\frac{1}{\mathrm{n}+1})=1$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+1)}^2}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2}=\frac{{\pi }^2}{6}-1\Rightarrow$$$$\mathrm{I}=1-(\frac{{\pi }^2}{6}-1)=2-\frac{{\pi }^2}{6}$$$$$$$$$$

Commented by bobhans last updated on 07/Jun/20

$$\mathrm{thank}\mathrm{you}.\mathrm{great}$$

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