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Question Number 88424 by abdomathmax last updated on 10/Apr/20
$${calculate}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({n}^{\mathrm{2}} {x}\right)−{arctan}\left({nx}\right)}{{x}}{dx} \\ $$$${and}\:{xetermine}\:{nature}\:{of}\:{the}\:{serie}\:\Sigma\:{U}_{{n}} \\ $$
Commented by mathmax by abdo last updated on 11/Apr/20
![let I(ξ) =∫_0 ^ξ ((arctan(ax)−arctan(bx))/x)dx ⇒ I(ξ) =∫_0 ^ξ ((arctan(ax))/x)dx−∫_0 ^ξ ((arctan(bx))/x)dx but ∫_0 ^ξ ((arctan(ax))/x)dx =_(ax=t) ∫_0 ^(aξ) ((arctan(t))/(t/a))×(dt/a) =∫_0 ^(aξ) ((arctan(t))/t)dt also ∫_0 ^ξ ((arctan(bx))/x)dx =∫_0 ^(bξ) ((arctan(t))/t)dt ⇒ I(ξ) =∫_0 ^(aξ) ((arctant)/t)dt+∫_(bξ) ^0 ((arctan(t))/t)dt =∫_(bξ) ^(aξ) ((arctan(t))/t)dt ∃c ∈]bξ,aξ[ / I(ξ) =arctan(ξ)∫_(bξ) ^(aξ) (dt/t) =arctan(ξ)ln∣(a/b)∣ ⇒ lim_(ξ→+∞) I(ξ) =∫_0 ^∞ ((arctan(ax)−arctan(bx))/x)dx=(π/2)ln∣(a/b)∣ ⇒U_n =(π/2)arctan((n^2 /n)) =(π/2)arctan(n) we have lim_(n→+∞) U_n =(π^2 /4) ≠0 ⇒Σ U_n diverges](Q88576.png)
$${let}\:{I}\left(\xi\right)\:=\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left({ax}\right)−{arctan}\left({bx}\right)}{{x}}{dx}\:\Rightarrow \\ $$$${I}\left(\xi\right)\:=\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left({ax}\right)}{{x}}{dx}−\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left({bx}\right)}{{x}}{dx}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left({ax}\right)}{{x}}{dx}\:=_{{ax}={t}} \:\:\:\int_{\mathrm{0}} ^{{a}\xi} \:\frac{{arctan}\left({t}\right)}{\frac{{t}}{{a}}}×\frac{{dt}}{{a}}\:=\int_{\mathrm{0}} ^{{a}\xi} \:\frac{{arctan}\left({t}\right)}{{t}}{dt} \\ $$$${also}\:\int_{\mathrm{0}} ^{\xi} \:\frac{{arctan}\left({bx}\right)}{{x}}{dx}\:=\int_{\mathrm{0}} ^{{b}\xi} \:\frac{{arctan}\left({t}\right)}{{t}}{dt}\:\Rightarrow \\ $$$${I}\left(\xi\right)\:=\int_{\mathrm{0}} ^{{a}\xi} \:\frac{{arctant}}{{t}}{dt}+\int_{{b}\xi} ^{\mathrm{0}} \:\frac{{arctan}\left({t}\right)}{{t}}{dt}\:=\int_{{b}\xi} ^{{a}\xi} \:\frac{{arctan}\left({t}\right)}{{t}}{dt} \\ $$$$\left.\exists{c}\:\:\in\right]{b}\xi,{a}\xi\left[\:\:/\:{I}\left(\xi\right)\:={arctan}\left(\xi\right)\int_{{b}\xi} ^{{a}\xi} \:\frac{{dt}}{{t}}\:={arctan}\left(\xi\right){ln}\mid\frac{{a}}{{b}}\mid\:\Rightarrow\right. \\ $$$${lim}_{\xi\rightarrow+\infty} \:{I}\left(\xi\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({ax}\right)−{arctan}\left({bx}\right)}{{x}}{dx}=\frac{\pi}{\mathrm{2}}{ln}\mid\frac{{a}}{{b}}\mid \\ $$$$\Rightarrow{U}_{{n}} =\frac{\pi}{\mathrm{2}}{arctan}\left(\frac{{n}^{\mathrm{2}} }{{n}}\right)\:=\frac{\pi}{\mathrm{2}}{arctan}\left({n}\right) \\ $$$${we}\:{have}\:{lim}_{{n}\rightarrow+\infty} \:{U}_{{n}} =\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\:\neq\mathrm{0}\:\Rightarrow\Sigma\:{U}_{{n}} \:{diverges} \\ $$