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Question Number 197821 by mnjuly1970 last updated on 30/Sep/23

$$$$$$findthevalueof:$$$$$$$$\mathrm{\Omega }={\int }_0^1\frac{\ln (1+\frac{1}{x^2})}{2+x^2}dx=?$$$$$$

Answered by witcher3 last updated on 05/Oct/23

$$\mathrm{x}\to \frac{1}{\mathrm{x}}$$$$\mathrm{\Omega }={\int }_1^{\mathrm{\infty }}\frac{\ln (1+{\mathrm{x}}^2)}{({2\mathrm{x}}^2+1)}\mathrm{dx}$$$$\mathrm{\Omega }\left(\mathrm{a}\right)={\int }_1^{\mathrm{\infty }}\frac{\ln (1+{\mathrm{ax}}^2)}{({2\mathrm{x}}^2+1)},\mathrm{\Omega }\left(\mathrm{a}\right)=0$$$${\mathrm{\Omega }}^{\prime }\left(\mathrm{a}\right)={\int }_1^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{(1+{\mathrm{ax}}^2)({2\mathrm{x}}^2+1)}$$$$={\int }_1^{\mathrm{\infty }}\frac{{2\mathrm{x}}^2+1-(1+{\mathrm{ax}}^2)}{(2-\mathrm{a})({2\mathrm{x}}^2+1)(1+{\mathrm{ax}}^2)}$$$$=\frac{1}{2-\mathrm{a}}\{{\int }_1^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{ax}}^2}-{\int }_1^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{2\mathrm{x}}^2}\}$$$$=\frac{1}{(2-\mathrm{a})\sqrt{\mathrm{a}}}(\frac{\pi }{2}-{\tan }^{-1}\left(\sqrt{\mathrm{a}}\right))-\frac{1}{(2-\mathrm{a})\sqrt{2}}(\frac{\pi }{2}-{\tan }^{-1}\left(\sqrt{2}\right))$$$$-{\int }_0^1\frac{{\tan }^{-1}\left(\sqrt{\mathrm{a}}\right)}{\sqrt{\mathrm{a}}(2-\mathrm{a})}\mathrm{da}$$$$=-2{\int }_0^1\frac{{\tan }^{-1}\left(\mathrm{a}\right)}{2-{\mathrm{a}}^2}..\mathrm{can}\mathrm{solved}{\mathrm{Li}}_2..$$$$\mathrm{but}\mathrm{non}\mathrm{nice}\mathrm{close}\mathrm{formez}$$$$$$

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