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Question Number 140310 by liberty last updated on 06/May/21

$$\mathrm{prove}\mathrm{that}{\int }_0^{\mathrm{\infty }}\frac{\ln \mathrm{x}}{{\mathrm{x}}^2+1}\mathrm{dx}=0$$

Answered by qaz last updated on 06/May/21

$${\int }_0^{\mathrm{\infty }}\frac{lnx}{1+x^2}dx={\int }_{\mathrm{\infty }}^0\frac{-lnx}{1+x^2}(-1)dx$$$$=-{\int }_0^{\mathrm{\infty }}\frac{lnx}{1+x^2}dx=0$$

Commented by TheSupreme last updated on 06/May/21

$$t=\frac{1}{x}$$$$dx=-\frac{1}{t^2}dt$$$${\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)}{1+x^2}dx={\int }_{\mathrm{\infty }}^0\frac{ln\left(\frac{1}{t}\right)}{1+\frac{1}{t^2}}(-\frac{1}{t^2})dt={\int }_{\mathrm{\infty }}^0\frac{ln\left(t\right)}{t^2+1}dt$$$$I=-I=0$$$$$$

Answered by mathmax by abdo last updated on 06/May/21

$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{{\mathrm{x}}^2+1}\mathrm{dx}=-\frac{1}{2}\mathrm{Re}\left(\mathrm{\Sigma }\mathrm{Res}\left(\mathrm{f}{\mathrm{a}}_{\mathrm{i}}\right)\right)\mathrm{with}\mathrm{f}\left(\mathrm{z}\right)=\frac{{\log }^2\mathrm{z}}{{\mathrm{z}}^2+1}$$$$\mathrm{f}\left(\mathrm{z}\right)=\frac{{\log }^2\mathrm{z}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})}\Rightarrow \mathrm{Res}(\mathrm{f},\mathrm{i})=\frac{{\log }^2\mathrm{i}}{2\mathrm{i}}=\frac{{\left(\log \left({\mathrm{e}}^{\frac{\mathrm{i}\pi }{2}}\right)\right)}^2}{2\mathrm{i}}=\frac{{\left(\frac{\mathrm{i}\pi }{2}\right)}^2}{2\mathrm{i}}=\frac{-{\pi }^2}{8\mathrm{i}}$$$$\mathrm{Res}(\mathrm{f},-\mathrm{i})=\frac{{\log }^2(-\mathrm{i})}{2\mathrm{i}}=\frac{{\log }^2\left({\mathrm{e}}^{-\frac{\mathrm{i}\pi }{2}}\right)}{2\mathrm{i}}=-\frac{{\pi }^2}{8\mathrm{i}}\Rightarrow$$$$\mathrm{\Sigma }\mathrm{Res}=\frac{\mathrm{i}{\pi }^2}{8}+\frac{\mathrm{i}{\pi }^2}{8}=\frac{\mathrm{i}{\pi }^2}{4}\Rightarrow \mathrm{Re}\left(\mathrm{\Sigma }\mathrm{Res}\right)=0\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{\mathrm{logx}}{{\mathrm{x}}^2+1}\mathrm{dx}=0$$

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