let D_n = {(x,y)∈R^2 /(x,y)∈[(1/n) ,n[ } 1) find the value of ∫∫_D_n e^(−x^2 −y^2 ) dxdy 2) calculate ∫


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Question Number 51994 by maxmathsup by imad last updated on 01/Jan/19
$let D_n = {(x,y)∈R^2 /(x,y)∈[(1/n) ,n[ } 1) find the value of ∫∫_D_n e^(−x^2 −y^2 ) dxdy 2) calculate ∫_0 ^(+∞) e^(−x^2 ) dx .$
$l e t D_{n} = {(x, y) \in R^{2} / (x, y) \in [\frac{1}{n}, n [}$ $1) f i n d t h e v a l u e o f \int \int_{D_{n}} e^{- x^{2} - y^{2}} d x d y$ $2) c a l c u l a t e \int_{0}^{+ \infty} e^{- x^{2}} d x .$
Commented by Abdo msup. last updated on 03/Jan/19
$changement x=r cosθ and y=r sinθ give x^2 +y^2 =r^2 but (1/n^2 )≤x^2 ≤n^2 and (1/n^2 )≤y^2 ≤n^2 ⇒ (2/n^2 )≤x^2 +y^2 ≤2n^2 ⇒(2/n^2 ) ≤r^2 ≤2n^2 ((√2)/n) ≤r≤(√2)n also x≥0 and y≥0 ⇒0≤θ≤(π/2) ∫∫_(D_n ) e^(−x^2 −y^2 ) dxdy =∫∫_(((√2)/n)≤r<(√2)n and 0≤θ≤(π/2)) e^(−r^2 ) rdr dθ =(π/2) (−(1/2))[ e^(−r^2 ) ]_((√2)/n) ^((√2)n) =(π/4){ −e^(−2n^2 ) +e^(−(2/n^2 )) } 2) we have ∫∫_D_n e^(−x^2 −y^2 ) dxdy =∫_(1/n) ^n e^(−x^2 ) dx.∫_(1/n) ^n e^(−y^2 ) dy =(∫_(1/n) ^n e^(−x^2 ) dx)^2 ⇒ lim_(n→+∞) (∫_(1/n) ^n e^(−x^2 ) dx)^2 =lim_(n→+∞) D_n =(π/4) ⇒ (∫_0 ^(+∞) e^(−x^2 ) dx)^2 =(π/4) ⇒∫_0 ^∞ e^(−x^2 ) dx =((√π)/2) .$
$c h a n g e m e n t x = r c o s θ a n d y = r s i n θ g i v e$ $x^{2} + y^{2} = r^{2} b u t \frac{1}{n^{2}} ⩽ x^{2} ⩽ n^{2} a n d \frac{1}{n^{2}} ⩽ y^{2} ⩽ n^{2} \Rightarrow$ $\frac{2}{n^{2}} ⩽ x^{2} + y^{2} ⩽ 2 n^{2} \Rightarrow \frac{2}{n^{2}} ⩽ r^{2} ⩽ 2 n^{2}$ $\frac{\sqrt{2}}{n} ⩽ r ⩽ \sqrt{2} n a l s o x ⩾ 0 a n d y ⩾ 0 \Rightarrow 0 ⩽ θ ⩽ \frac{π}{2}$ $\int \int_{D_{n}} e^{- x^{2} - y^{2}} d x d y = \int \int_{\frac{\sqrt{2}}{n} ⩽ r < \sqrt{2} n a n d 0 ⩽ θ ⩽ \frac{π}{2}} e^{- r^{2}} r d r d θ$ $= \frac{π}{2} (- \frac{1}{2}) {[e^{- r^{2}}]}_{\frac{\sqrt{2}}{n}}^{\sqrt{2} n} = \frac{π}{4} {- e^{- 2 n^{2}} + e^{- \frac{2}{n^{2}}}}$ $2) w e h a v e \int \int_{D_{n}} e^{- x^{2} - y^{2}} d x d y$ $= \int_{\frac{1}{n}}^{n} e^{- x^{2}} d x . \int_{\frac{1}{n}}^{n} e^{- y^{2}} d y = {(\int_{\frac{1}{n}}^{n} e^{- x^{2}} d x)}^{2} \Rightarrow$ ${l i m}_{n \to + \infty} {(\int_{\frac{1}{n}}^{n} e^{- x^{2}} d x)}^{2} = {l i m}_{n \to + \infty} D_{n} = \frac{π}{4} \Rightarrow$ ${(\int_{0}^{+ \infty} e^{- x^{2}} d x)}^{2} = \frac{π}{4} \Rightarrow \int_{0}^{\infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{2} .$
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