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Question Number 212470 by Spillover last updated on 14/Oct/24

Answered by Ar Brandon last updated on 14/Oct/24

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{(1-x)\ln x}{1-x^6}dx$$$$={\int }_0^1\frac{1-x}{1-x^6}\ln xdx+{\int }_1^{\mathrm{\infty }}\frac{1-x}{1-x^6}\ln xdx$$$$={\int }_0^1\frac{1-x}{1-x^6}\ln xdx-{\int }_0^1\frac{1-\frac{1}{x}}{x^2-\frac{1}{x^4}}\ln xdx$$$$={\int }_0^1\frac{1-x-x^3+x^4}{1-x^6}\ln xdx,{\psi }^{\prime }\left(t\right)=-{\int }_0^1\frac{x^{t-1}\ln x}{1-x}dx$$$$=\frac{1}{36}{\int }_0^1\frac{x^{-\frac{5}{6}}-x^{-\frac{2}{3}}-x^{-\frac{1}{3}}+x^{-\frac{1}{6}}}{1-x}\ln \left(x\right)dx$$$$=-\frac{1}{36}({\psi }^{\prime }\left(\frac{1}{6}\right)-{\psi }^{\prime }\left(\frac{1}{3}\right)-{\psi }^{\prime }\left(\frac{2}{3}\right)+{\psi }^{\prime }\left(\frac{5}{6}\right))$$$$=-\frac{1}{36}(({\psi }^{\prime }\left(\frac{1}{6}\right)+{\psi }^{\prime }\left(\frac{5}{6}\right))-({\psi }^{\prime }\left(\frac{1}{3}\right)+{\psi }^{\prime }\left(\frac{2}{3}\right))$$$$[{\psi }^{\prime }\left(x\right)+{\psi }^{\prime }(1-x)={\pi }^2{\mathrm{cosec}}^2\left(\pi x\right)]$$$$=-\frac{1}{36}({\pi }^2{\mathrm{cosec}}^2\left(\frac{\pi }{6}\right)-{\pi }^2{\mathrm{cosec}}^2\left(\frac{\pi }{3}\right))$$$$=-\frac{{\pi }^2}{36}(2^2-{\left(\frac{2}{\sqrt{3}}\right)}^2)=-\frac{{\pi }^2}{36}\left(\frac{12-4}{3}\right)=-\frac{2{\pi }^2}{27}$$

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