Question Number 29551 by abdo imad last updated on 09/Feb/18
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{arctan}\left(\mathrm{2}{x}\right)−{arctanx}}{{x}}{dx}. \\ $$
Commented by prof Abdo imad last updated on 11/Feb/18
![let put I=∫_0 ^∞ ((arctan(2x)−arctanx)/x)dx we have I=lim _(ξ→+∞) I(ξ) /I(ξ)= ∫_0 ^ξ ((arctan(2x)−arctanx)/x)dx = ∫_0 ^ξ ((arctan(2x))/x)dx − ∫_0 ^ξ ((srctanx)/x)dx = ∫_0 ^(2ξ) ((arctant)/(t/2)) (dt/2) −∫_0 ^ξ ((arctanx)/x)dx = ∫_ξ ^(2ξ) ((arctanx)/x)dx but ∃ c ∈]ξ,2ξ[ / I(ξ)= arctanξ ∫_ξ ^(2ξ) (dx/x)=ln(2)arctan(ξ) and lim_(ξ→+∞) I(ξ)= (π/2)ln(2) . ⇒I= (π/2)ln(2).](Q29676.png)
$${let}\:{put}\:{I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{x}\right)−{arctanx}}{{x}}{dx}\:{we}\:{have} \\ $$$${I}={lim}\:_{\xi\rightarrow+\infty} \:{I}\left(\xi\right)\:\:/{I}\left(\xi\right)=\:\int_{\mathrm{0}} ^{\xi} \:\:\frac{{arctan}\left(\mathrm{2}{x}\right)−{arctanx}}{{x}}{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\xi} \:\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:−\:\int_{\mathrm{0}} ^{\xi} \:\:\frac{{srctanx}}{{x}}{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\xi} \:\:\:\:\:\frac{{arctant}}{\frac{{t}}{\mathrm{2}}}\:\frac{{dt}}{\mathrm{2}}\:−\int_{\mathrm{0}} ^{\xi} \:\:\frac{{arctanx}}{{x}}{dx} \\ $$$$\left.=\:\int_{\xi} ^{\mathrm{2}\xi} \:\:\:\:\frac{{arctanx}}{{x}}{dx}\:\:\:{but}\:\exists\:{c}\:\in\right]\xi,\mathrm{2}\xi\left[\:/\right. \\ $$$${I}\left(\xi\right)=\:{arctan}\xi\:\int_{\xi} ^{\mathrm{2}\xi} \:\frac{{dx}}{{x}}={ln}\left(\mathrm{2}\right){arctan}\left(\xi\right)\:{and} \\ $$$${lim}_{\xi\rightarrow+\infty} \:\:{I}\left(\xi\right)=\:\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:.\:\:\Rightarrow{I}=\:\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right). \\ $$