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Question Number 148570 by mathmax by abdo last updated on 29/Jul/21

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{logx}}{{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}$$

Answered by Ar Brandon last updated on 29/Jul/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\log x}{x^2+x+1}dx$$$$={\int }_0^1\frac{\log x}{x^2+x+1}dx+{\int }_1^{\mathrm{\infty }}\frac{\log x}{x^2+x+1}dx$$$$={\int }_0^1\frac{\log x}{x^2+x+1}dx-{\int }_0^1\frac{\log x}{x^2+x+1}dx=0$$

Answered by mathmax by abdo last updated on 29/Jul/21

$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{logx}}{{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}=-\frac{1}{2}\mathrm{Re}(\mathrm{\Sigma }\mathrm{Res}(\mathrm{f}\left(\mathrm{z}\right){\log }^2\mathrm{z},{\mathrm{a}}_{\mathrm{i}})$$$$\mathrm{f}\left(\mathrm{z}\right)=\frac{1}{{\mathrm{z}}^2+\mathrm{z}+1}\mathrm{poles}\mathrm{of}\mathrm{f}!$$$$\mathrm{\Delta }=1-4=-3\Rightarrow {\mathrm{z}}_1=\frac{-1+\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}}\mathrm{and}{\mathrm{z}}_2=\frac{-1-\mathrm{i}\sqrt{3}}{2}$$$$\phi \left(\mathrm{z}\right)=\mathrm{f}\left(\mathrm{z}\right){\log }^2\mathrm{z}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\log }^2\mathrm{z}}{(\mathrm{z}-{\mathrm{z}}_1)(\mathrm{z}-{\mathrm{z}}_2)}$$$$\mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}})=\frac{{\left(\frac{2\mathrm{i}\pi }{3}\right)}^2}{\mathrm{i}\sqrt{3}}=-\frac{4{\pi }^2}{9\left(\mathrm{i}\sqrt{3}\right)}=\frac{4\mathrm{i}{\pi }^2}{9\sqrt{3}}$$$$\mathrm{Res}(\phi ,{\mathrm{e}}^{-\frac{2\mathrm{i}\pi }{3}})=\frac{{(-\frac{2\mathrm{i}\pi }{3})}^2}{-\mathrm{i}\sqrt{3}}=\frac{4{\pi }^2}{9\left(\mathrm{i}\sqrt{3}\right)}=\frac{-4\mathrm{i}{\pi }^2}{9\sqrt{3}}\Rightarrow$$$$\mathrm{\Sigma }\mathrm{Res}(\phi ..)=0\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{\mathrm{logx}}{{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}=0$$

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