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Question Number 166437 by mnjuly1970 last updated on 20/Feb/22

$$$$$$calculate$$$$\mathrm{\Omega }={\int }_0^1\frac{ln(1-x).ln(1+x)}{x^2}dx=?$$$$-------$$

Answered by qaz last updated on 21/Feb/22

$${\int }_0^1\frac{\ln (1-\mathrm{x})\ln (1+\mathrm{x})}{{\mathrm{x}}^2}\mathrm{dx}$$$$={\int }_0^1(\frac{1}{\mathrm{x}}-1)(\frac{\ln (1+\mathrm{x})}{\mathrm{x}-1}+\frac{\ln (1-\mathrm{x})}{1+\mathrm{x}})\mathrm{dx}$$$$=-{\int }_0^1\frac{\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}+{\int }_0^1\frac{\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}-2{\int }_0^1\frac{\ln (1-\mathrm{x})}{1+\mathrm{x}}\mathrm{dx}$$$$=-\frac{1}{12}{\pi }^2+{\int }_0^1\frac{\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}-2{\int }_1^2\frac{\ln (2-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$=-\frac{1}{12}{\pi }^2+{\int }_0^1\frac{\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}-2{\int }_{1/2}^1\frac{\ln 2+\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$=-\frac{1}{12}{\pi }^2-{2\ln }^{2}2+{\mathrm{Li}}_2\left(1\right)-{2\mathrm{Li}}_2\left(\frac{1}{2}\right)$$$$=-\frac{1}{12}{\pi }^2-{3\ln }^{2}2$$

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