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Question Number 140200 by qaz last updated on 05/May/21

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

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

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}{\left(\frac{\mathrm{lnx}}{\mathrm{x}-1}\right)}^3\mathrm{dx}={\int }_0^1{\left(\frac{\mathrm{lnx}}{\mathrm{x}-1}\right)}^3\mathrm{dx}+{\int }_1^{\mathrm{\infty }}{\left(\frac{\mathrm{lnx}}{\mathrm{x}-1}\right)}^3\mathrm{dx}$$$$={\int }_0^1\frac{{\ln }^3\mathrm{x}}{{(\mathrm{x}-1)}^3}\mathrm{dx}+{\int }_0^1\frac{{\mathrm{xln}}^3\mathrm{x}}{{(\mathrm{x}-1)}^3}\mathrm{dx}$$$$={\int }_0^1\frac{{\ln }^3\mathrm{x}}{{(\mathrm{x}-1)}^2}\mathrm{dx}+2{\int }_0^1\frac{{\ln }^3\mathrm{x}}{{(\mathrm{x}-1)}^3}\mathrm{dx}$$$$={[-\frac{{\ln }^3\mathrm{x}}{\mathrm{x}-1}+3\int \frac{{\ln }^2\mathrm{x}}{\mathrm{x}(\mathrm{x}-1)}\mathrm{dx}]}_0^1+2{[-\frac{{\ln }^3\mathrm{x}}{2{(\mathrm{x}-1)}^2}+\frac{3}{2}\int \frac{{\ln }^2\mathrm{x}}{\mathrm{x}{(\mathrm{x}-1)}^2}\mathrm{dx}]}_0^1$$$$={[-\frac{{\ln }^3\mathrm{x}}{\mathrm{x}-1}-\frac{{\ln }^3\mathrm{x}}{{(\mathrm{x}-1)}^2}]}_0^1-3{[\int \frac{{\ln }^2\mathrm{x}}{\mathrm{x}}\mathrm{dx}-\int \frac{{\ln }^2\mathrm{x}}{\mathrm{x}-1}\mathrm{dx}]}_0^1-3{[\int \frac{{\ln }^2\mathrm{x}}{\mathrm{x}(\mathrm{x}-1)}\mathrm{dx}-\int \frac{{\ln }^2\mathrm{x}}{{(\mathrm{x}-1)}^2}\mathrm{dx}]}_0^1$$$$=-{\ln }^3\mathrm{x}+3{\psi }^{″}\left(1\right)+3[\int \frac{{\ln }^2\mathrm{x}}{\mathrm{x}}\mathrm{dx}-\int \frac{{\ln }^2\mathrm{x}}{\mathrm{x}-1}\mathrm{dx}]-3[-\frac{{\ln }^2\mathrm{x}}{\mathrm{x}-1}+2\int \frac{\mathrm{lnx}}{\mathrm{x}(\mathrm{x}-1)}\mathrm{dx}]$$$$=-{\ln }^3\mathrm{x}+3{\psi }^{″}\left(1\right)+{\ln }^3\mathrm{x}-3{\psi }^{″}\left(1\right)+\frac{{3\ln }^2\mathrm{x}}{\mathrm{x}-1}-6[\int \frac{\mathrm{lnx}}{\mathrm{x}}\mathrm{dx}+\int \frac{\mathrm{lnx}}{\mathrm{x}-1}\mathrm{dx}]$$$$=\frac{{3\ln }^2\mathrm{x}}{\mathrm{x}-1}-{3\ln }^2\mathrm{x}-6{\psi }^{\prime }\left(1\right)=6\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(\mathrm{n}+1)}^2}=6\zeta \left(2\right)={\pi }^2$$

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

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\log }^3\mathrm{x}}{{(1-\mathrm{x})}^3}\mathrm{dx}\Rightarrow \mathrm{I}=-{\int }_0^1\frac{{\log }^3\mathrm{x}}{{(1-\mathrm{x})}^3}\mathrm{dx}-{\int }_1^{\mathrm{\infty }}\frac{{\log }^3\mathrm{x}}{{(1-\mathrm{x})}^3}\mathrm{dx}(\to \mathrm{x}=\frac{1}{\mathrm{t}})$$$$=-{\int }_0^1\frac{{\log }^3\mathrm{x}}{{(1-\mathrm{x})}^3}+{\int }_0^1\frac{-{\log }^3\mathrm{t}}{{(1-\frac{1}{\mathrm{t}})}^3}(-\frac{\mathrm{dt}}{{\mathrm{t}}^2})$$$$=-{\int }_0^1\frac{{\log }^3\mathrm{x}}{{(1-\mathrm{x})}^3}\mathrm{dx}-{\int }_0^1\frac{{\mathrm{tlog}}^3\mathrm{t}}{{(1-\mathrm{t})}^3}\mathrm{dt}$$$$=-{\int }_0^1\frac{(1+\mathrm{x}){\log }^3\mathrm{x}}{{(1-\mathrm{x})}^3}\mathrm{dx}\mathrm{we}\mathrm{have}\frac{1}{1-\mathrm{x}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}\Rightarrow \frac{1}{{(1-\mathrm{x})}^2}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}-1}$$$$\Rightarrow \frac{2(1-\mathrm{x})}{{(1-\mathrm{x})}^4}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\mathrm{n}(\mathrm{n}-1){\mathrm{x}}^{\mathrm{n}-2}\Rightarrow \frac{2}{{(1-\mathrm{x})}^3}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\mathrm{n}(\mathrm{n}-1){\mathrm{x}}^{\mathrm{n}-2}\Rightarrow$$$$\mathrm{I}=-\frac{1}{2}{\int }_0^1\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\mathrm{n}(\mathrm{n}-1){\mathrm{x}}^{\mathrm{n}-2}(1+\mathrm{x}){\log }^3\mathrm{x}\mathrm{dx}$$$$=-\frac{1}{2}\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\mathrm{n}(\mathrm{n}-1){\int }_0^1({\mathrm{x}}^{\mathrm{n}-2}+{\mathrm{x}}^{\mathrm{n}-1}){\log }^3\mathrm{xdx}$$$$=-\frac{1}{2}\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\mathrm{n}(\mathrm{n}-1){\mathrm{X}}_{\mathrm{n}}$$$${\mathrm{X}}_{\mathrm{n}}={\int }_0^1({\mathrm{x}}^{\mathrm{n}-2}+{\mathrm{x}}^{\mathrm{n}-1}){\log }^3\mathrm{xdx}$$$${=[\frac{{\mathrm{x}}^{\mathrm{n}-1}}{\mathrm{n}-1}+\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}){\log }^3\mathrm{x}]}_0^1-{\int }_0^1(\frac{{\mathrm{x}}^{\mathrm{n}-1}}{\mathrm{n}-1}+\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}){3\log }^2\mathrm{x}\frac{\mathrm{dx}}{\mathrm{x}}$$$$=-3{\int }_0^1(\frac{{\mathrm{x}}^{\mathrm{n}-2}}{\mathrm{n}-1}+\frac{{\mathrm{x}}^{\mathrm{n}-1}}{\mathrm{n}}){\log }^2\mathrm{x}\mathrm{dx}$$$$=-3{\left\{[\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{(\mathrm{n}-1)}^2}+\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^2}){\log }^2\mathrm{x}\right]}_0^1-{\int }_0^1(\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{(\mathrm{n}-1)}^2}+\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^2})\frac{2\mathrm{logx}}{\mathrm{x}}\mathrm{dx}$$$$=6{\int }_0^1(\frac{{\mathrm{x}}^{\mathrm{n}-2}}{{(\mathrm{n}-1)}^2}+\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{n}}^2})\log \mathrm{xdx}$$$$=6\{{\left[(\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{(\mathrm{n}-1)}^3}+\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^3})\mathrm{logx}\right]}_0^1-{\int }_0^1(\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{(\mathrm{n}-1)}^3}+\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^3})\frac{\mathrm{dx}}{\mathrm{x}}$$$$=-6{\int }_0^1(\frac{{\mathrm{x}}^{\mathrm{n}-2}}{{(\mathrm{n}-1)}^3}+\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{n}}^3})\mathrm{dx}=-6{[\frac{1}{{(\mathrm{n}-1)}^4}{\mathrm{x}}^{\mathrm{n}-1}+\frac{1}{{\mathrm{n}}^4}{\mathrm{x}}^{\mathrm{n}}]}_0^1$$$$=-6(\frac{1}{{(\mathrm{n}-1)}^4}+\frac{1}{{\mathrm{n}}^4})\Rightarrow$$$$\mathrm{I}=3\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\mathrm{n}(\mathrm{n}-1)(\frac{1}{{(\mathrm{n}-1)}^4}+\frac{1}{{\mathrm{n}}^4})$$$$=3\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{\mathrm{n}}{{(\mathrm{n}-1)}^3}+3\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{\mathrm{n}-1}{{\mathrm{n}}^3}$$$$=3\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\mathrm{n}+1}{{\mathrm{n}}^3}+3\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\mathrm{n}-1}{{\mathrm{n}}^3}$$$$=6\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2}=6.\frac{{\pi }^2}{6}={\pi }^2\Rightarrow {\int }_0^{\mathrm{\infty }}{\left(\frac{\mathrm{logx}}{\mathrm{x}-1}\right)}^3={\pi }^2$$

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