let-I-n-1-n-n-2-xy-dxdy-2-x-2-y-2-find-lim-I-n-when-n-

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

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 \dots .

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