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Question Number 140202 by qaz last updated on 05/May/21

$${\int }_0^{\mathrm{\infty }}{\left(\frac{lnx}{x-1}\right)}^{2}dx=\frac{2}{3}{\pi }^2$$

Answered by mathmax by abdo last updated on 05/May/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{{\log }^2\mathrm{x}}{{(1-\mathrm{x})}^2}\mathrm{dx}\Rightarrow \mathrm{\Phi }={\int }_0^1\frac{{\log }^2\mathrm{x}}{{(1-\mathrm{x})}^2}\mathrm{dx}+{\int }_1^{\mathrm{\infty }}\frac{{\log }^2\mathrm{x}}{{(1-\mathrm{x})}^2}\mathrm{dx}(\to \mathrm{x}=\frac{1}{\mathrm{t}})$$$$={\int }_0^1\frac{{\log }^2\mathrm{x}}{{(1-\mathrm{x})}^2}\mathrm{dx}-{\int }_0^1\frac{{\log }^2\mathrm{t}}{{(1-\frac{1}{\mathrm{t}})}^2}(-\frac{\mathrm{dt}}{{\mathrm{t}}^2})=2{\int }_0^1\frac{{\log }^2\mathrm{x}}{{(1-\mathrm{x})}^2}\mathrm{dx}\mathrm{we}\mathrm{have}$$$$\frac{1}{1-\mathrm{x}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}\Rightarrow \frac{1}{{(1-\mathrm{x})}^2}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}-1}\left(\mathrm{by}\mathrm{derivation}\right)\Rightarrow$$$$\frac{1}{{(1-\mathrm{x})}^2}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}-1}\Rightarrow \mathrm{\Phi }=2{\int }_0^1\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}-1}{\log }^2\mathrm{x}\mathrm{dx}$$$$=2\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}{\mathrm{U}}_{\mathrm{n}}\mathrm{with}{\mathrm{U}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}{\log }^2\mathrm{xdx}\mathrm{by}\mathrm{parts}$$$${\mathrm{U}}_{\mathrm{n}}={\left[\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}{\log }^2\mathrm{x}\right]}_0^1-{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\frac{2\mathrm{logx}}{\mathrm{x}}\mathrm{dx}=-\frac{2}{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}\mathrm{logx}\mathrm{dx}$$$$=-\frac{2}{\mathrm{n}}({\left[\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\mathrm{logx}\right]}_0^1-{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\frac{\mathrm{dx}}{\mathrm{x}})=-\frac{2}{\mathrm{n}}(-\frac{1}{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}\mathrm{dx})$$$$=\frac{2}{{\mathrm{n}}^3}\Rightarrow \mathrm{\Phi }=2\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}\times \frac{2}{{\mathrm{n}}^3}=4\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2}=4\times \frac{{\pi }^2}{6}=\frac{2{\pi }^2}{3}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}{\left(\frac{\mathrm{logx}}{\mathrm{x}-1}\right)}^2\mathrm{dx}=\frac{2}{3}{\pi }^2$$

Answered by Ar Brandon last updated on 05/May/21

$$={\int }_0^1\frac{{\ln }^2\mathrm{x}}{{(1-\mathrm{x})}^2}\mathrm{dx}+{\int }_0^1\frac{{\ln }^2\mathrm{x}}{{(1-\mathrm{x})}^2}\mathrm{dx}=2{\int }_0^1\frac{{\ln }^2\mathrm{x}}{{(1-\mathrm{x})}^2}\mathrm{dx}$$$$=2{[-\frac{{\ln }^2\mathrm{x}}{1-\mathrm{x}}+2{\int }_0^1\frac{\mathrm{lnx}}{\mathrm{x}(1-\mathrm{x})}\mathrm{dx}]}_0^1$$$$=\frac{{2\ln }^2\mathrm{x}}{\mathrm{x}-1}+4{\int }_0^1[\frac{\mathrm{lnx}}{\mathrm{x}}+\frac{\mathrm{lnx}}{1-\mathrm{x}}]\mathrm{dx}$$$$=\frac{{2\ln }^2\mathrm{x}}{\mathrm{x}-1}+{2\ln }^2\mathrm{x}+4{\int }_0^1\frac{\mathrm{lnx}}{1-\mathrm{x}}\mathrm{dx}=-4{\psi }^{\prime }\left(1\right)$$$$=4\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(\mathrm{n}+1)}^2}=4\zeta \left(2\right)=4\times \frac{{\pi }^2}{6}=\frac{2{\pi }^2}{3}$$

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