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Question Number 123034 by benjo_mathlover last updated on 21/Nov/20

Answered by mathmax by abdo last updated on 21/Nov/20

$$\chi ={\int }_0^{\mathrm{\infty }}\frac{\ln \left(\mathrm{x}\right)}{1+{\mathrm{x}}^2}\mathrm{dx}\Rightarrow \chi {=}_{\mathrm{x}=\frac{1}{\mathrm{t}}}-{\int }_0^{\mathrm{\infty }}\frac{-\mathrm{lnt}}{1+\frac{1}{{\mathrm{t}}^2}}(-\frac{\mathrm{dt}}{{\mathrm{t}}^2})$$$$=-{\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnt}}{1+{\mathrm{t}}^2}\mathrm{dt}=-\chi \Rightarrow 2\chi =0\Rightarrow \chi =0$$$$\mathrm{another}\mathrm{put}\mathrm{x}=\tan \theta \Rightarrow \chi ={\int }_0^{\mathrm{\infty }}\frac{\ln \left(\tan \theta \right)}{1+{\tan }^2\theta }(1+{\tan }^2\theta )\mathrm{d}\theta$$$$={\int }_0^{\mathrm{\infty }}\ln \left(\sin \theta \right)\mathrm{d}\theta -{\int }_0^{\mathrm{\infty }}\ln \left(\cos \theta \right)\mathrm{d}\theta =0\left(\mathrm{those}\mathrm{integrals}\mathrm{are}\mathrm{equals}\right)$$

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