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Question Number 158272 by HongKing last updated on 01/Nov/21

Answered by qaz last updated on 04/Nov/21

$$\mathrm{\Omega }=4{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2\mathrm{lnx}}{{\mathrm{x}}^4-{\mathrm{x}}^2+1}\mathrm{dx}$$$$=4{\int }_0^1\frac{{\mathrm{x}}^2\mathrm{lnx}}{{\mathrm{x}}^4-{\mathrm{x}}^2+1}\mathrm{dx}+4{\int }_1^{\mathrm{\infty }}\frac{{\mathrm{x}}^2\mathrm{lnx}}{{\mathrm{x}}^4-{\mathrm{x}}^2+1}\mathrm{dx}$$$$=4{\int }_0^1\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^4-{\mathrm{x}}^2+1}\mathrm{lnxdx}$$$$=4{\int }_0^1\frac{\mathrm{lnx}}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-3}\mathrm{d}(\mathrm{x}+\frac{1}{\mathrm{x}})$$$$=\frac{2}{\sqrt{3}}\ln \frac{\mathrm{x}+\frac{1}{\mathrm{x}}-\sqrt{3}}{\mathrm{x}+\frac{1}{\mathrm{x}}+\sqrt{3}}\cdot \mathrm{lnx}{\mid }_0^1-\frac{2}{\sqrt{3}}{\int }_0^1\ln \frac{\mathrm{x}+\frac{1}{\mathrm{x}}-\sqrt{3}}{\mathrm{x}+\frac{1}{\mathrm{x}}+\sqrt{3}}\frac{\mathrm{dx}}{\mathrm{x}}$$$$=-\frac{2}{\sqrt{3}}{\int }_0^1(\frac{\ln ({\mathrm{x}}^2-\sqrt{3}\mathrm{x}+1)}{\mathrm{x}}-\frac{\ln ({\mathrm{x}}^2+\sqrt{3}\mathrm{x}+1)}{\mathrm{x}})\mathrm{dx}$$$$\mathrm{I}\left(\mathrm{a}\right)={\int }_0^1\frac{\ln ({\mathrm{x}}^2+\mathrm{ax}+1)}{\mathrm{x}}\mathrm{dx}$$$$={\int }_0^{\mathrm{a}}\mathrm{dy}{\int }_0^1\frac{\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{yx}+1}+{\int }_0^1\frac{\ln ({\mathrm{x}}^2+1)}{\mathrm{x}}\mathrm{dx}$$$$={\int }_0^{\mathrm{a}}\frac{2}{\sqrt{4-{\mathrm{y}}^2}}\arctan \sqrt{\frac{2-\mathrm{y}}{2+\mathrm{y}}}\mathrm{dy}+\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{2\mathrm{n}-1}\mathrm{dx}$$$$=\frac{\pi }{2}\arcsin \frac{\mathrm{a}}{2}-\frac{1}{2}{\arcsin }^2\frac{\mathrm{a}}{2}+\frac{{\pi }^2}{24}$$$$\mathrm{\Omega }=-\frac{2}{\sqrt{3}}(\mathrm{I}(-\sqrt{3})-\mathrm{I}\left(\sqrt{3}\right))=\frac{2{\pi }^2}{3\sqrt{3}}$$

Commented by HongKing last updated on 09/Nov/21

$$\mathrm{cool}\mathrm{my}\mathrm{dear}\mathrm{Ser}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

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