Question Number 31099 by abdo imad last updated on 02/Mar/18
$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left(\mathrm{2}{x}\right)\:−{arctanx}}{{x}}{dx}. \\ $$
Commented by abdo imad last updated on 04/Mar/18
$${I}={lim}_{\xi\rightarrow+\infty} \:{I}\left(\xi\right)\:\:/\:{I}\left(\xi\right)=\:\int_{\mathrm{0}} ^{\xi} \:\:\frac{{artan}\left(\mathrm{2}{x}\right)−{arctanx}}{{x}}{dx} \\ $$$${I}\left(\xi\right)=\int_{\mathrm{0}} ^{\xi} \:\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:−\int_{\mathrm{0}} ^{\xi} \:\frac{{arctanx}}{{x}}{dx} \\ $$$${ch}.\mathrm{2}{x}={t}\:{give}\:\int_{\mathrm{0}} ^{\xi} \:\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\xi} \:\frac{{arctan}\left({t}\right)}{\frac{{t}}{\mathrm{2}}}\:\frac{{dt}}{\mathrm{2}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\xi} \:\:\frac{{arctant}}{{t}}{dt}\Rightarrow{I}\left(\xi\right)=\int_{\mathrm{0}} ^{\mathrm{2}\xi} \:\frac{{arctanx}}{{x}}{dx}\:−\int_{\mathrm{0}} ^{\xi} \:\frac{{arctanx}}{{x}}{dx} \\ $$$$\left.=\:\int_{\xi} ^{\mathrm{2}\xi} \:\:\frac{{arctanx}}{{x}}{dx}\:\:{but}\:\exists\:{c}\in\right]\xi,\mathrm{2}\xi\left[\:/\:{I}\left(\xi\right)={artan}\xi\:\int_{\xi} ^{\mathrm{2}\xi} \:\frac{{dt}}{{t}}\right. \\ $$$$={ln}\left(\mathrm{2}\right)\:{arctan}\xi\Rightarrow\:{lim}_{\xi\rightarrow+\infty} {I}\left(\xi\right)=\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}\:\Rightarrow\:{I}=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right). \\ $$