let I_n = ∫∫_([(1/n),n]^2 ) (((√(xy)) dxdy)/(2 +x^2 +y^2 )) find lim I


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Question Number 34717 by abdo mathsup 649 cc last updated on 10/May/18

$l e t I_{n} = \int \int_{{[\frac{1}{n}, n]}^{2}} \frac{\sqrt{x y} d x d y}{2 + x^{2} + y^{2}}$ $f i n d l i m I_{n} w h e n n \to + \infty .$
Commented by prof Abdo imad last updated on 11/May/18
$let consider the diffeomorphism (r,θ)→(x,y) =(r cosθ,r sinθ) we have (1/n)≤x^ ≤n ,(1/n)≤ y≤n ⇒ (2/n^2 )≤x^2 +y^2 ≤ 2n^2 ⇒ ((√2)/n)≤r≤(√2) n I_n = ∫∫_(((√2)/n)≤r≤(√2) n , 0≤θ≤(π/2)) ((r(√(cosθsinθ)))/(2+r^2 )) r dr dθ = ∫_(((√2)/n) ) ^((√2) n) (r^2 /(2+r^2 )) dr ∫_0 ^(π/2) (√(cosθ sinθ)) dθ but ∫_((√2)/n) ^(n(√2)) (r^2 /(2+r^2 ))dr = ∫_((√2)/n) ^(n(√2)) ((2+r^2 −2)/(2+r^2 )) dr =n(√2) −((√2)/n) −2 ∫_((√2)/n) ^(n(√2)) (dr/(2+r^2 )) ( r=u(√2)) =n(√2) −((√2)/n) − ∫_(1/n) ^n (((√2) du)/(2(1+u^2 ))) =n(√2) −((√2)/n) −((√2)/2) (arctan(n) −arctan((1/n))) =n(√2) −((√2)/n) −((√2)/2)( 2arctan(n) −(π/2)) ∫_0 ^(π/2) (√(cosθsinθ)) dθ = (1/(√2)) ∫_0 ^(π/2) (√(sin(2θ))) dθ (2θ=t) =(1/(√2)) ∫_0 ^π (√(sint)) (dt/2) = (1/(2(√2))) ∫_0 ^π (√(sint)) dt ch. (√(sint_ )) =x give sint =x^2 ⇒ t =arcsin(x^2 ) ∫_0 ^π (√(sint)) dt = ∫_0 ^(π/2) (√(sint)) dt + ∫_(π/2) ^π (√(sint))dt = ∫_0 ^1 x ((2x)/(√(1−x^4 ))) dx = ∫_0 ^1 ((2x^2 )/(√(1−x^4 ))) dx.... but n(√2) −((√2)/n) −((√2)/2){ 2arctan(n)−(π/2)} ∼n(√2) −((√2)/2) (π/2) ∼ n(√2) ⇒ lim_(n→+∞) I_n =+∞$
$l e t c o n s i d e r t h e d i f f e o m o r p h i s m$ $(r, θ) \to (x, y) = (r c o s θ, r s i n θ) w e h a v e$ $\frac{1}{n} ⩽ x^{} ⩽ n, \frac{1}{n} ⩽ y ⩽ n \Rightarrow \frac{2}{n^{2}} ⩽ x^{2} + y^{2} ⩽ 2 n^{2} \Rightarrow$ $\frac{\sqrt{2}}{n} ⩽ r ⩽ \sqrt{2} n$ $I_{n} = \int \int_{\frac{\sqrt{2}}{n} ⩽ r ⩽ \sqrt{2} n, 0 ⩽ θ ⩽ \frac{π}{2}} \frac{r \sqrt{c o s θ s i n θ}}{2 + r^{2}} r d r d θ$ $= \int_{\frac{\sqrt{2}}{n}}^{\sqrt{2} n} \frac{r^{2}}{2 + r^{2}} d r \int_{0}^{\frac{π}{2}} \sqrt{c o s θ s i n θ} d θ b u t$ $\int_{\frac{\sqrt{2}}{n}}^{n \sqrt{2}} \frac{r^{2}}{2 + r^{2}} d r = \int_{\frac{\sqrt{2}}{n}}^{n \sqrt{2}} \frac{2 + r^{2} - 2}{2 + r^{2}} d r$ $= n \sqrt{2} - \frac{\sqrt{2}}{n} - 2 \int_{\frac{\sqrt{2}}{n}}^{n \sqrt{2}} \frac{d r}{2 + r^{2}} (r = u \sqrt{2})$ $= n \sqrt{2} - \frac{\sqrt{2}}{n} - \int_{\frac{1}{n}}^{n} \frac{\sqrt{2} d u}{2 (1 + u^{2})}$ $= n \sqrt{2} - \frac{\sqrt{2}}{n} - \frac{\sqrt{2}}{2} (a r c t a n (n) - a r c t a n (\frac{1}{n}))$ $= n \sqrt{2} - \frac{\sqrt{2}}{n} - \frac{\sqrt{2}}{2} (2 a r c t a n (n) - \frac{π}{2})$ $\int_{0}^{\frac{π}{2}} \sqrt{c o s θ s i n θ} d θ = \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \sqrt{s i n (2 θ)} d θ (2 θ = t)$ $= \frac{1}{\sqrt{2}} \int_{0}^{π} \sqrt{s i n t} \frac{d t}{2} = \frac{1}{2 \sqrt{2}} \int_{0}^{π} \sqrt{s i n t} d t$ $c h . \sqrt{{s i n t}_{}} = x g i v e s i n t = x^{2} \Rightarrow t = a r c s i n (x^{2})$ $\int_{0}^{π} \sqrt{s i n t} d t = \int_{0}^{\frac{π}{2}} \sqrt{s i n t} d t + \int_{\frac{π}{2}}^{π} \sqrt{s i n t} d t$ $= \int_{0}^{1} x \frac{2 x}{\sqrt{1 - x^{4}}} d x = \int_{0}^{1} \frac{2 x^{2}}{\sqrt{1 - x^{4}}} d x . . . .$ $b u t n \sqrt{2} - \frac{\sqrt{2}}{n} - \frac{\sqrt{2}}{2} {2 a r c t a n (n) - \frac{π}{2}}$ $\sim n \sqrt{2} - \frac{\sqrt{2}}{2} \frac{π}{2} \sim n \sqrt{2} \Rightarrow {l i m}_{n \to + \infty} I_{n} = + \infty$
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