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Question Number 124360 by Lordose last updated on 02/Dec/20

$${\int }_0^{\mathrm{\infty }}\frac{\ln \left(\mathrm{x}\right)}{{\mathrm{x}}^4+{\mathrm{x}}^2+1}\mathrm{dx}$$

Answered by mathmax by abdo last updated on 02/Dec/20

$$\mathrm{we}\mathrm{use}{\int }_0^{\mathrm{\infty }}\mathrm{q}\left(\mathrm{x}\right)\ln \left(\mathrm{x}\right)\mathrm{dx}=-\frac{1}{2}\mathrm{Re}(\mathrm{\Sigma }\mathrm{Res}(\mathrm{q}\left(\mathrm{z}\right){\ln }^2\mathrm{z},{\mathrm{a}}_{\mathrm{i}})$$$$\mathrm{q}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^4+{\mathrm{x}}^2+1}\Rightarrow \mathrm{q}\left(\mathrm{z}\right){\ln }^2\left(\mathrm{z}\right)=\frac{{\ln }^2\mathrm{z}}{{\mathrm{z}}^4+{\mathrm{x}}^2+1}=\mathrm{w}\left(\mathrm{z}\right)$$$${\mathrm{z}}^4+{\mathrm{z}}^2+1=0\Rightarrow {\mathrm{u}}^2+\mathrm{u}+1(\mathrm{u}={\mathrm{z}}^2)$$$$\mathrm{\Delta }=1-4=-3\Rightarrow {\mathrm{u}}_1=\frac{-1+\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{\frac{\mathrm{i}2\pi }{3}}\mathrm{and}{\mathrm{u}}_2=\frac{-1-\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{-\frac{\mathrm{i}2\pi }{3}}$$$$\Rightarrow \mathrm{w}\left(\mathrm{z}\right)=\frac{{\ln }^2\mathrm{z}}{({\mathrm{z}}^2-{\mathrm{e}}^{\frac{\mathrm{i}2\pi }{3}})({\mathrm{z}}^2-{\mathrm{e}}^{-\frac{\mathrm{i}2\pi }{3}})}=\frac{{\ln }^2\mathrm{z}}{(\mathrm{z}-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}})(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}})(\mathrm{z}-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})(\mathrm{z}+{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})}$$$$\mathrm{\Sigma }\mathrm{Res}(\mathrm{w}.)=\mathrm{Res}(\mathrm{w},{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}})+\mathrm{Res}(\mathrm{w},-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}})+\mathrm{Res}(\mathrm{w},{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})+\mathrm{Res}(\mathrm{w},-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})$$$$\mathrm{we}\mathrm{have}\mathrm{Res}(\mathrm{w},{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}})=\frac{{\left(\frac{\mathrm{i}\pi }{3}\right)}^2}{{2\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}\left(2\mathrm{isin}\left(\frac{2\pi }{3}\right)\right)}=-\frac{{\pi }^2}{36}\frac{{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}}{\frac{\sqrt{3}}{2}}=-\frac{{\pi }^2}{18\mathrm{i}\sqrt{3}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}$$$$\mathrm{Res}(\mathrm{w},-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}})=\frac{{(\mathrm{i}\pi +\frac{\mathrm{i}\pi }{3})}^2}{-{2\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}\left(2\mathrm{isin}\left(\frac{2\pi }{3}\right)\right)}=-\frac{{\left(\frac{4\pi }{3}\right)}^2}{-4\mathrm{i}\times \frac{\sqrt{3}}{2}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}=\frac{16{\pi }^2}{18\mathrm{i}\sqrt{3}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}=\frac{8{\pi }^2}{9\mathrm{i}\sqrt{3}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}$$$$\mathrm{Res}(\mathrm{w},{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})=\frac{{(-\frac{\mathrm{i}\pi }{3})}^2}{-2\mathrm{isin}\left(\frac{2\pi }{3}\right){2\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}}=\frac{{\pi }^2}{36\mathrm{i}\times \frac{\sqrt{3}}{2}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}$$$$=\frac{{\pi }^2}{18\mathrm{i}\sqrt{3}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}$$$$\mathrm{Res}(\mathrm{w},-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})=\frac{{(\mathrm{i}\pi -\frac{\mathrm{i}\pi }{3})}^2}{-{2\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}(-2\mathrm{isin}\left(\frac{2\pi }{3}\right))}=-\frac{{\left(\frac{2\pi }{3}\right)}^2}{4\mathrm{i}.\frac{\sqrt{3}}{2}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}$$$$=-\frac{4{\pi }^2}{18\mathrm{i}\sqrt{3}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}=-\frac{2{\pi }^2}{9\mathrm{i}\sqrt{3}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}\Rightarrow$$$$\mathrm{\Sigma }\mathrm{Res}(\mathrm{w},{\mathrm{z}}_{\mathrm{i}})=\frac{\mathrm{i}{\pi }^2}{18\sqrt{3}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}-\mathrm{i}\frac{8{\pi }^2}{9\sqrt{3}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}-\mathrm{i}\frac{{\pi }^2}{18\sqrt{3}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}+\mathrm{i}\frac{2{\pi }^2}{9\sqrt{3}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}$$$$=-\frac{\mathrm{i}{\pi }^2}{18\sqrt{3}}\left(2\mathrm{isin}\left(\frac{\pi }{3}\right)\right)+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}(-\frac{\mathrm{i}{\pi }^2}{18\sqrt{3}}+\frac{4\mathrm{i}{\pi }^2}{18\sqrt{3}})$$$$=\frac{{\pi }^2}{9\sqrt{3}}\times \frac{\sqrt{3}}{2}+(\frac{1}{2}+\frac{\mathrm{i}\sqrt{3}}{2})\left(\frac{\mathrm{i}{\pi }^2}{6\sqrt{3}}\right)$$$$=\frac{{\pi }^2}{18}+\frac{\mathrm{i}{\pi }^2}{12\sqrt{3}}-\frac{{\pi }^2}{12}\Rightarrow \mathrm{I}=-\frac{1}{2}(\frac{{\pi }^2}{18}-\frac{{\pi }^2}{12})=\frac{{\pi }^2}{24}-\frac{{\pi }^2}{36}$$

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