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Question Number 142773 by mnjuly1970 last updated on 05/Jun/21

$$$$$$.......nice......integral......$$$$\mathit{\varphi }:={\int }_0^1\frac{{li}_2(1-x)}{2-x}dx=??$$$$.......m.n...$$

Answered by mnjuly1970 last updated on 05/Jun/21

$$\mathit{\varphi }:={\int }_0^1\frac{{li}_2\left(x\right)}{1+x}dx={[{li}_2\left(x\right).ln(1+x)]}_0^1+{\int }_0^1\frac{ln(1-x).ln(1+x)}{x}dx$$$$=\frac{{\pi }^2}{6}ln\left(2\right)-\frac{5}{8}\zeta \left(3\right)...$$$$derivedearlier:$$$${\int }_0^1\frac{ln(1-x)ln(1+x)}{x}dx=\frac{-5}{8}\zeta \left(3\right)$$$$$$$$$$

Answered by qaz last updated on 05/Jun/21

$$\varphi ={\int }_0^1\frac{{\mathrm{Li}}_2(1-\mathrm{x})}{2-\mathrm{x}}\mathrm{dx}={\int }_0^1\frac{{\mathrm{Li}}_2\left(\mathrm{x}\right)}{1+\mathrm{x}}\mathrm{dx}=\frac{{\pi }^2}{6}\ln 2+{\int }_0^1\frac{\ln (1+\mathrm{x})\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$${\int }_0^1\frac{\ln (1+\mathrm{x})\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^1\frac{{\ln }^2(1-{\mathrm{x}}^2)-{\ln }^2(1+\mathrm{x})-{\ln }^2(1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^1\frac{{\ln }^2(1-{\mathrm{x}}^2)}{\mathrm{x}}\mathrm{dx}-\frac{1}{2}{\int }_0^1\frac{{\ln }^2(1+\mathrm{x})}{\mathrm{x}}-\frac{1}{2}{\int }_0^1\frac{{\ln }^2(1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$=-{\int }_0^1\frac{\ln (1-{\mathrm{x}}^2)\mathrm{lnx}}{1-{\mathrm{x}}^2}(-2\mathrm{x})\mathrm{dx}-\frac{1}{2}\cdot \frac{\zeta \left(3\right)}{4}+{\int }_0^1\frac{\ln (1-\mathrm{x})\mathrm{lnx}}{1-\mathrm{x}}(-1)\mathrm{dx}$$$$=2{\int }_0^1\frac{\ln (1-{\mathrm{x}}^2)\mathrm{lnx}}{1-{\mathrm{x}}^2}\mathrm{xdx}-\frac{\zeta \left(3\right)}{8}-{\int }_0^1\frac{\ln (1-\mathrm{x})\mathrm{lnx}}{1-\mathrm{x}}\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^1\frac{\ln (1-\mathrm{u})\mathrm{lnu}}{1-\mathrm{u}}\mathrm{du}-\frac{\zeta \left(3\right)}{8}-{\int }_0^1\frac{\ln (1-\mathrm{x})\mathrm{lnx}}{1-\mathrm{x}}\mathrm{dx}$$$$=-\frac{1}{2}{\int }_0^1\frac{\ln (1-\mathrm{x})\mathrm{lnx}}{\mathrm{x}}\mathrm{dx}-\frac{\zeta \left(3\right)}{8}$$$$=-\frac{1}{2}{\int }_0^1\frac{{\mathrm{Li}}_2\left(\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}-\frac{\zeta \left(3\right)}{8}$$$$=-\frac{1}{2}\zeta \left(3\right)-\frac{\zeta \left(3\right)}{8}$$$$=-\frac{5}{8}\zeta \left(3\right)$$$$\Rightarrow \varphi =\frac{{\pi }^2}{6}\ln 2-\frac{5}{8}\zeta \left(3\right)$$$$---------------------$$$${\int }_0^1\frac{\ln (1-\mathrm{x})\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}\ln (1-\mathrm{x})\mathrm{dx}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}(-\frac{{\mathrm{H}}_{\mathrm{n}}}{\mathrm{n}})$$$$=-\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}-1}{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^2}$$$$=-\frac{5}{8}\zeta \left(3\right)$$

Commented by mnjuly1970 last updated on 05/Jun/21

$$godkeepyou$$$$perfectsolution...$$

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