Question Number 68240 by mathmax by abdo last updated on 07/Sep/19
$${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{3}{x}\right)−{arctan}\left(\mathrm{2}{x}\right)}{{x}}{dx} \\ $$
Commented by mathmax by abdo last updated on 08/Sep/19
![let I =∫_0 ^∞ ((arctan(3x)−arctan(2x))/x)dx and I(ξ) =∫_0 ^ξ ((arctan(3x)−arctan(2x))/x)dx we have lim_(ξ→+∞) I(ξ)=I I(ξ) =∫_0 ^ξ ((arctan(3x))/x)dx−∫_0 ^ξ ((arctan(2x))/x)dx but ∫_0 ^ξ ((arctan(3x))/x)dx =_(3x=t) ∫_0 ^(3ξ) ((arctan(t))/(t/3))(dt/3) =∫_0 ^(3ξ) ((arctan(t))/t)dt ∫_0 ^ξ ((arctan(2x))/x)dx =_(2x=t) ∫_0 ^(2ξ) ((arctan(t))/(t/2))(dt/2) =∫_0 ^(2ξ) ((arctan(t))/t)dt ⇒ I(ξ) =∫_0 ^(3ξ) ((arctan(t))/t)dt −∫_0 ^(2ξ) ((arctant)/t)dt =∫_(2ξ) ^(3ξ) ((arctan(t))/t)dt ∃ c ∈]2ξ,3ξ[ / I(ξ) =arctan(c)∫_(2ξ) ^(3ξ) (dt/(t )) =arctan(c)ln(((3ξ)/(2ξ))) ⇒ lim_(ξ→+∞) I(ξ) =(π/2)ln((3/2)) ⇒ I =(π/2)ln((3/2)).](https://www.tinkutara.com/question/Q68302.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{3}{x}\right)−{arctan}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:{and} \\ $$$${I}\left(\xi\right)\:=\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left(\mathrm{3}{x}\right)−{arctan}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:{we}\:{have}\:{lim}_{\xi\rightarrow+\infty} {I}\left(\xi\right)={I} \\ $$$${I}\left(\xi\right)\:=\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left(\mathrm{3}{x}\right)}{{x}}{dx}−\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left(\mathrm{3}{x}\right)}{{x}}{dx}\:=_{\mathrm{3}{x}={t}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{3}\xi} \:\frac{{arctan}\left({t}\right)}{\frac{{t}}{\mathrm{3}}}\frac{{dt}}{\mathrm{3}}\:=\int_{\mathrm{0}} ^{\mathrm{3}\xi} \:\frac{{arctan}\left({t}\right)}{{t}}{dt} \\ $$$$\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:=_{\mathrm{2}{x}={t}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\xi} \:\frac{{arctan}\left({t}\right)}{\frac{{t}}{\mathrm{2}}}\frac{{dt}}{\mathrm{2}}\:=\int_{\mathrm{0}} ^{\mathrm{2}\xi} \frac{{arctan}\left({t}\right)}{{t}}{dt}\:\Rightarrow \\ $$$${I}\left(\xi\right)\:=\int_{\mathrm{0}} ^{\mathrm{3}\xi} \:\frac{{arctan}\left({t}\right)}{{t}}{dt}\:−\int_{\mathrm{0}} ^{\mathrm{2}\xi} \:\frac{{arctant}}{{t}}{dt}\:=\int_{\mathrm{2}\xi} ^{\mathrm{3}\xi} \:\:\frac{{arctan}\left({t}\right)}{{t}}{dt} \\ $$$$\left.\exists\:{c}\:\in\right]\mathrm{2}\xi,\mathrm{3}\xi\left[\:/\:{I}\left(\xi\right)\:={arctan}\left({c}\right)\int_{\mathrm{2}\xi} ^{\mathrm{3}\xi} \:\frac{{dt}}{{t}\:}\:={arctan}\left({c}\right){ln}\left(\frac{\mathrm{3}\xi}{\mathrm{2}\xi}\right)\:\Rightarrow\right. \\ $$$${lim}_{\xi\rightarrow+\infty} \:{I}\left(\xi\right)\:=\frac{\pi}{\mathrm{2}}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{2}}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right). \\ $$