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Question Number 185517 by Noorzai last updated on 23/Jan/23

Answered by Ar Brandon last updated on 23/Jan/23

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\ln x}{x^6-1}dx={\int }_0^1\frac{\ln x}{x^6-1}dx+{\int }_1^{\mathrm{\infty }}\frac{\ln x}{x^6-1}dx$$$$={\int }_0^1\frac{\ln x}{x^6-1}dx-{\int }_0^1\frac{x^4\ln x}{1-x^6}dx=-\frac{1}{36}{\int }_0^1\frac{x^{-\frac{5}{6}}+x^{-\frac{1}{6}}}{1-x}\ln xdx$$$$=\frac{1}{36}({\psi }^{\left(1\right)}\left(\frac{1}{6}\right)+{\psi }^{\left(1\right)}\left(\frac{5}{6}\right))=\frac{{\pi }^2}{36}{\mathrm{cosec}}^2\left(\frac{\pi }{6}\right)=\frac{{\pi }^2}{9}$$

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