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_(ξ→+∞) I(ξ) / I(ξ)= ∫_0 ^ξ ((artan(2x)−arctanx)/x)dx I(ξ)=∫_0 ^ξ ((arctan(2x))/x)dx −∫_0 ^ξ ((arctanx)/x)dx ch.2x=t give ∫_0 ^ξ ((arctan(2x))/x)dx = ∫_0 ^(2ξ) ((arctan(t))/(t/2)) (dt/2) = ∫_0 ^(2ξ) ((arctant)/t)dt⇒I(ξ)=∫_0 ^(2ξ) ((arctanx)/x)dx −∫_0 ^ξ ((arctanx)/x)dx = ∫_ξ ^(2ξ) ((arctanx)/x)dx but ∃ c∈]ξ,2ξ[ / I(ξ)=artanξ ∫_ξ ^(2ξ) (dt/t) =ln(2) arctanξ⇒ lim_(ξ→+∞) I(ξ)=(π/2)ln2 ⇒ I=(π/2)ln(2).](https://www.tinkutara.com/question/Q31236.png)
$${I}={lim}_{\xi\rightarrow+\infty} \:{I}\left(\xi\right)\:\:/\:{I}\left(\xi\right)=\:\int_{\mathrm{0}} ^{\xi} \:\:\frac{{artan}\left(\mathrm{2}{x}\right)−{arctanx}}{{x}}{dx} \\ $$$${I}\left(\xi\right)=\int_{\mathrm{0}} ^{\xi} \:\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:−\int_{\mathrm{0}} ^{\xi} \:\frac{{arctanx}}{{x}}{dx} \\ $$$${ch}.\mathrm{2}{x}={t}\:{give}\:\int_{\mathrm{0}} ^{\xi} \:\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\xi} \:\frac{{arctan}\left({t}\right)}{\frac{{t}}{\mathrm{2}}}\:\frac{{dt}}{\mathrm{2}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\xi} \:\:\frac{{arctant}}{{t}}{dt}\Rightarrow{I}\left(\xi\right)=\int_{\mathrm{0}} ^{\mathrm{2}\xi} \:\frac{{arctanx}}{{x}}{dx}\:−\int_{\mathrm{0}} ^{\xi} \:\frac{{arctanx}}{{x}}{dx} \\ $$$$\left.=\:\int_{\xi} ^{\mathrm{2}\xi} \:\:\frac{{arctanx}}{{x}}{dx}\:\:{but}\:\exists\:{c}\in\right]\xi,\mathrm{2}\xi\left[\:/\:{I}\left(\xi\right)={artan}\xi\:\int_{\xi} ^{\mathrm{2}\xi} \:\frac{{dt}}{{t}}\right. \\ $$$$={ln}\left(\mathrm{2}\right)\:{arctan}\xi\Rightarrow\:{lim}_{\xi\rightarrow+\infty} {I}\left(\xi\right)=\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}\:\Rightarrow\:{I}=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right). \\ $$