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Question Number 157469 by mnjuly1970 last updated on 23/Oct/21

$$$$$$calculate:$$$$\mathrm{\Omega }:={\int }_0^1\frac{ln(1+x).ln(1-x)}{1+x}dx=?$$$$$$

Answered by qaz last updated on 23/Oct/21

$${\int }_0^1\frac{\ln (1+\mathrm{x})\ln (1-\mathrm{x})}{1+\mathrm{x}}\mathrm{dx}$$$$={\int }_1^2\frac{\mathrm{lnxln}(2-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$={\int }_{1/2}^1\frac{(\ln 2+\mathrm{lnx})(\ln 2+\ln (1-\mathrm{x}))}{\mathrm{x}}\mathrm{dx}$$$$={\int }_{1/2}^1\frac{{\ln }^{2}2+\ln 2(\mathrm{lnx}+\ln (1-\mathrm{x}))+\mathrm{lnxln}(1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$={\ln }^{3}2+\ln 2\cdot (-\frac{1}{2}{\ln }^{2}2+{\int }_{1/2}^1\frac{\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx})+{\int }_{1/2}^1\frac{\mathrm{lnxln}(1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$=\frac{1}{2}{\ln }^{3}2+\ln 2({\mathrm{Li}}_2\left(\frac{1}{2}\right)-{\mathrm{Li}}_2\left(1\right))-\mathrm{lnx}\cdot {\mathrm{Li}}_2\left(\mathrm{x}\right){\mid }_{1/2}^1+{\mathrm{Li}}_3\left(1\right)-{\mathrm{Li}}_3\left(\frac{1}{2}\right)$$$$=\frac{1}{2}{\ln }^{3}2-\frac{{\pi }^2}{6}\ln 2-\ln 2\cdot {\mathrm{Li}}_2\left(\frac{1}{2}\right)-\ln 2\cdot {\mathrm{Li}}_2\left(\frac{1}{2}\right)+\zeta \left(3\right)-{\mathrm{Li}}_3\left(\frac{1}{2}\right)$$$$=-\frac{{\pi }^2}{4}\ln 2+\frac{4}{3}{\ln }^{3}2+\frac{1}{8}\zeta \left(3\right)$$$$-----------------$$$${\mathrm{Li}}_2\left(\frac{1}{2}\right)=-\frac{1}{2}{\ln }^{2}2+\frac{{\pi }^2}{12}$$$${\mathrm{Li}}_3\left(\frac{1}{2}\right)=-\frac{{\pi }^2}{12}\ln 2+\frac{1}{6}{\ln }^{3}2+\frac{7}{8}\zeta \left(3\right)$$

Commented by mnjuly1970 last updated on 23/Oct/21

$$thanksalot...$$

Commented by mnjuly1970 last updated on 23/Oct/21

$$pleaserechecktheanswer$$

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