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Question Number 34717 by abdo mathsup 649 cc last updated on 10/May/18
let I_n = ∫∫_([(1/n),n]^2 ) (((√(xy)) dxdy)/(2 +x^2 +y^2 )) find lim I_n when n→+∞.
$${let}\:{I}_{{n}} =\:\int\int_{\left[\frac{\mathrm{1}}{{n}},{n}\right]^{\mathrm{2}} } \:\:\:\:\:\frac{\sqrt{{xy}}\:{dxdy}}{\mathrm{2}\:+{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} } \\ $$$${find}\:{lim}\:{I}_{{n}} \:{when}\:{n}\rightarrow+\infty. \\ $$
Commented by prof Abdo imad last updated on 11/May/18
let consider the diffeomorphism (r,θ)→(x,y) =(r cosθ,r sinθ) we have (1/n)≤x^ ≤n ,(1/n)≤ y≤n ⇒ (2/n^2 )≤x^2 +y^2 ≤ 2n^2 ⇒ ((√2)/n)≤r≤(√2) n I_n = ∫∫_(((√2)/n)≤r≤(√2) n , 0≤θ≤(π/2)) ((r(√(cosθsinθ)))/(2+r^2 )) r dr dθ = ∫_(((√2)/n) ) ^((√2) n) (r^2 /(2+r^2 )) dr ∫_0 ^(π/2) (√(cosθ sinθ)) dθ but ∫_((√2)/n) ^(n(√2)) (r^2 /(2+r^2 ))dr = ∫_((√2)/n) ^(n(√2)) ((2+r^2 −2)/(2+r^2 )) dr =n(√2) −((√2)/n) −2 ∫_((√2)/n) ^(n(√2)) (dr/(2+r^2 )) ( r=u(√2)) =n(√2) −((√2)/n) − ∫_(1/n) ^n (((√2) du)/(2(1+u^2 ))) =n(√2) −((√2)/n) −((√2)/2) (arctan(n) −arctan((1/n))) =n(√2) −((√2)/n) −((√2)/2)( 2arctan(n) −(π/2)) ∫_0 ^(π/2) (√(cosθsinθ)) dθ = (1/( (√2))) ∫_0 ^(π/2) (√(sin(2θ))) dθ (2θ=t) =(1/( (√2))) ∫_0 ^π (√(sint)) (dt/2) = (1/(2(√2))) ∫_0 ^π (√(sint)) dt ch. (√(sint_ )) =x give sint =x^2 ⇒ t =arcsin(x^2 ) ∫_0 ^π (√(sint)) dt = ∫_0 ^(π/2) (√(sint)) dt + ∫_(π/2) ^π (√(sint))dt = ∫_0 ^1 x ((2x)/( (√(1−x^4 )))) dx = ∫_0 ^1 ((2x^2 )/( (√(1−x^4 )))) dx.... but n(√2) −((√2)/n) −((√2)/2){ 2arctan(n)−(π/2)} ∼n(√2) −((√2)/2) (π/2) ∼ n(√2) ⇒ lim_(n→+∞) I_n =+∞
$${let}\:{consider}\:{the}\:{diffeomorphism} \\ $$$$\left({r},\theta\right)\rightarrow\left({x},{y}\right)\:=\left({r}\:{cos}\theta,{r}\:{sin}\theta\right)\:\:{we}\:{have} \\ $$$$\frac{\mathrm{1}}{{n}}\leqslant{x}^{} \leqslant{n}\:\:,\frac{\mathrm{1}}{{n}}\leqslant\:{y}\leqslant{n}\:\Rightarrow\:\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\:\mathrm{2}{n}^{\mathrm{2}} \:\Rightarrow \\ $$$$\frac{\sqrt{\mathrm{2}}}{{n}}\leqslant{r}\leqslant\sqrt{\mathrm{2}}\:{n} \\ $$$${I}_{{n}} \:=\:\int\int_{\frac{\sqrt{\mathrm{2}}}{{n}}\leqslant{r}\leqslant\sqrt{\mathrm{2}}\:{n}\:\:\:,\:\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{r}\sqrt{{cos}\theta{sin}\theta}}{\mathrm{2}+{r}^{\mathrm{2}} }\:{r}\:{dr}\:{d}\theta \\ $$$$=\:\int_{\frac{\sqrt{\mathrm{2}}}{{n}}\:} ^{\sqrt{\mathrm{2}}\:{n}} \:\:\:\frac{{r}^{\mathrm{2}} }{\mathrm{2}+{r}^{\mathrm{2}} }\:{dr}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\sqrt{{cos}\theta\:{sin}\theta}\:{d}\theta\:\:{but} \\ $$$$\int_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:\:\frac{{r}^{\mathrm{2}} }{\mathrm{2}+{r}^{\mathrm{2}} }{dr}\:=\:\int_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:\:\frac{\mathrm{2}+{r}^{\mathrm{2}} \:−\mathrm{2}}{\mathrm{2}+{r}^{\mathrm{2}} }\:{dr} \\ $$$$={n}\sqrt{\mathrm{2}}\:\:−\frac{\sqrt{\mathrm{2}}}{{n}}\:\:\:\:−\mathrm{2}\:\int_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:\:\:\:\frac{{dr}}{\mathrm{2}+{r}^{\mathrm{2}} }\:\:\:\:\left(\:{r}={u}\sqrt{\mathrm{2}}\right) \\ $$$$={n}\sqrt{\mathrm{2}}\:\:−\frac{\sqrt{\mathrm{2}}}{{n}}\:−\:\:\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:\:\:\:\frac{\sqrt{\mathrm{2}}\:\:{du}}{\mathrm{2}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$$={n}\sqrt{\mathrm{2}}\:\:−\frac{\sqrt{\mathrm{2}}}{{n}}\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\left({arctan}\left({n}\right)\:−{arctan}\left(\frac{\mathrm{1}}{{n}}\right)\right) \\ $$$$={n}\sqrt{\mathrm{2}}\:\:−\frac{\sqrt{\mathrm{2}}}{{n}}\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\:\mathrm{2}{arctan}\left({n}\right)\:−\frac{\pi}{\mathrm{2}}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{cos}\theta{sin}\theta}\:{d}\theta\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\sqrt{{sin}\left(\mathrm{2}\theta\right)}\:{d}\theta\:\:\left(\mathrm{2}\theta={t}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\pi} \:\:\sqrt{{sint}}\:\:\:\frac{{dt}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\pi} \:\sqrt{{sint}}\:\:\:{dt} \\ $$$${ch}.\:\sqrt{{sint}_{\:} }\:={x}\:\:{give}\:{sint}\:={x}^{\mathrm{2}} \:\Rightarrow\:{t}\:={arcsin}\left({x}^{\mathrm{2}} \right) \\ $$$$\int_{\mathrm{0}} ^{\pi} \sqrt{{sint}}\:{dt}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\sqrt{{sint}}\:{dt}\:\:+\:\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:\sqrt{{sint}}{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{x}\:\:\:\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:{dx}\:=\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:{dx}…. \\ $$$${but}\:{n}\sqrt{\mathrm{2}}\:−\frac{\sqrt{\mathrm{2}}}{{n}}\:\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left\{\:\mathrm{2}{arctan}\left({n}\right)−\frac{\pi}{\mathrm{2}}\right\} \\ $$$$\sim{n}\sqrt{\mathrm{2}}\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\frac{\pi}{\mathrm{2}}\:\sim\:{n}\sqrt{\mathrm{2}}\:\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{I}_{{n}} =+\infty \\ $$

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