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Question Number 206592 by NasaSara last updated on 19/Apr/24

Answered by namphamduc last updated on 20/Apr/24

$${\int }_0^1\frac{\ln \left(x\right)}{1-x}dx=\frac{4}{3}{\int }_0^1\frac{\ln \left(x\right)}{1-x}dx-\frac{1}{3}{\int }_0^1\frac{\ln \left(x\right)}{1-x}dx$$$$x\to x^2\Rightarrow {\int }_0^1\frac{\ln \left(x\right)}{1-x}dx=\frac{4}{3}{\int }_0^1\frac{\ln \left(x\right)}{1-x}dx-\frac{4}{3}{\int }_0^1\frac{x\ln \left(x\right)}{1-x^2}dx=\frac{4}{3}{\int }_0^1\frac{\ln \left(x\right)}{1-x^2}dx=\frac{2}{3}{\int }_0^{\mathrm{\infty }}\frac{\ln \left(x\right)}{1-x^2}$$$$\mathrm{Let}:I\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{\ln (1-a^2+a^2{x}^2)}{1-x^2}dx,I\left(0\right)=0,I\left(1\right)=2{\int }_0^{\mathrm{\infty }}\frac{\ln \left(x\right)}{1-x^2}dx$$$$I^{\prime }\left(a\right)=-{\int }_0^{\mathrm{\infty }}\frac{2a}{a^2{x}^2+1-a^2}dx=-\frac{\pi }{\sqrt{1-a^2}}$$$$\Rightarrow I\left(1\right)=-\pi {\int }_0^1\frac{1}{\sqrt{1-a^2}}da=-\pi {\sin }^{-1}\left(a\right){\mid }_0^1=-\pi .\frac{\pi }{2}=-\frac{{\pi }^2}{2}$$$$\Rightarrow {\int }_0^1\frac{\ln \left(x\right)}{1-x}dx=\frac{1}{3}I\left(1\right)=-\frac{{\pi }^2}{2}.\frac{1}{3}=-\frac{{\pi }^2}{6}$$$$\Rightarrow {\int }_0^1\frac{\ln \left(x\right)}{x-1}dx=\frac{{\pi }^2}{6}$$

Commented by NasaSara last updated on 20/Apr/24

$$thankyou$$

Answered by Berbere last updated on 20/Apr/24

$${\int }_0^1\frac{ln\left(x\right)}{1-x}dx=\frac{\partial }{\partial s}{\int }_0^1\frac{x^s-1}{x-1}dx{\mid }_{s=0}=\frac{\partial }{\partial s}.-{\int }_0^1\frac{x^s-1}{1-x}dx$$$$=\frac{\partial }{\partial s}\mathrm{\Psi }(1+s)={\mathrm{\Psi }}^{\prime }(1+s){\mid }_{s=0}={\mathrm{\Psi }}^{\left(1\right)}\left(1\right)=\zeta \left(2\right)=\frac{{\pi }^2}{6}$$

Commented by NasaSara last updated on 20/Apr/24

$$thanks$$

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