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Question Number 51994 by maxmathsup by imad last updated on 01/Jan/19

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

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} .