∫_(−∞) ^∞ e^(−x^2 ) dx=?


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Question Number 99707 by student work last updated on 22/Jun/20

$\int_{- \infty}^{\infty} e^{- x^{2}} dx = ?$
Commented by student work last updated on 22/Jun/20

$help me$
	Answered by smridha last updated on 22/Jun/20

	$2 \int_{0}^{\infty} e^{- x^{2}} x^{2 . \frac{1}{2} - 1} d x$ $= Γ (\frac{1}{2}) = \sqrt{π}$
	Answered by smridha last updated on 22/Jun/20

	$\int_{- \infty}^{\infty} e^{- x^{2}} dx = ?$ ${l e t}^{'} s d o i t a n o t h e r w a y$ $I = \int_{- \infty}^{+ \infty} e^{- x^{2}} d x$ $s i m i l a r l y I = \int_{- \infty}^{+ \infty} e^{- y^{2}} d y$ $s o n o w I^{2} = \int_{- \infty}^{+ \infty} \int_{- \infty}^{+ \infty} e^{- (x^{2} + y^{2})} d x d y$ $n o w b y t r a n s f o r m i n g t o t h i s$ $i n t e g r a l c a r t e s i a n t o p l a n e - p o l a r$ $c o - o r d i n a t e w e g e t$ $I^{2} = \int_{0}^{2 π} d θ \int_{0}^{\infty} e^{- r^{2}} r d r = 2 π . (- \frac{1}{2}) {[e^{- r^{2}}]}_{0}^{\infty}$ $= π$ $s o I = \sqrt{π} (c o n s i d e r p o s i t i v e r o o t) .$ $t h i s i s o n e o f t h e v a l u a b l e i n t e g r a l .$
	Answered by mathmax by abdo last updated on 23/Jun/20
	$A =∫_(−∞) ^(+∞) e^(−x^2 ) dx ⇒A =2 ∫_0 ^∞ e^(−x^2 ) dx let A_n =∫∫_([(1/n),n[^2 ) e^(−x^2 −y^2 ) dxdy we have A_n =∫_((1 )/n) ^(n ) e^(−x^2 ) dx.∫_(1/n) ^n e^(−y^2 ) dy ⇒lim_(n→+∞) A_n =(∫_0 ^∞ e^(−x^2 ) dx)^2 let considere the diffeomorphisme { ((x = r cosθ)),((y =rsinθ)) :} (1/n)≤x <n and (1/n)≤y<n ⇒(2/n^2 )≤x^2 +y^2 <2n^2 ⇒(2/n^2 )≤r^2 <2n^2 ⇒((√2)/n)≤r<n(√2) ⇒A_n =∫∫_(((√2)/n)≤r<n(√2)and 0≤θ≤(π/2)) e^(−r^2 ) rdrdθ =∫_((√2)/n) ^(n(√2)) re^(−r^2 ) dr ∫_0 ^(π/2) dθ =(π/2)[−(1/2)e^(−r^2 ) ]_((√2)/n) ^(n(√2)) =−(π/4){ e^(−2n^2 ) −e^(−(2/n^2 )) } ⇒ lim_(n→+∞) A_n =(π/4) =(∫_0 ^∞ e^(−x^2 ) dx)^2 but ∫_0 ^∞ e^(−x^2 ) dx >0 ⇒ ∫_0 ^∞ e^(−x^2 ) dx =((√π)/2) ⇒★ A =(√π)★$
	$A = \int_{- \infty}^{+ \infty} e^{- x^{2}} dx \Rightarrow A = 2 \int_{0}^{\infty} e^{- x^{2}} dx let A_{n} = \int \int_{[\frac{1}{n}, n [^{2}} e^{- x^{2} - y^{2}} dxdy$ $we have A_{n} = \int_{\frac{1}{n}}^{n} e^{- x^{2}} dx . \int_{\frac{1}{n}}^{n} e^{- y^{2}} dy \Rightarrow \lim_{n \to + \infty} A_{n} = {(\int_{0}^{\infty} e^{- x^{2}} dx)}^{2}$ $let considere the diffeomorphisme {\begin{cases} x = r \cos θ \\ y = rsin θ \end{cases}$ $\frac{1}{n} ⩽ x < n and \frac{1}{n} ⩽ y < n \Rightarrow \frac{2}{n^{2}} ⩽ x^{2} + y^{2} < {2 n}^{2} \Rightarrow \frac{2}{n^{2}} ⩽ r^{2} < {2 n}^{2} \Rightarrow \frac{\sqrt{2}}{n} ⩽ r < n \sqrt{2}$ $\Rightarrow A_{n} = \int \int_{\frac{\sqrt{2}}{n} ⩽ r < n \sqrt{2} and 0 ⩽ θ ⩽ \frac{π}{2}} e^{- r^{2}} rdrd θ$ $= \int_{\frac{\sqrt{2}}{n}}^{n \sqrt{2}} {re}^{- r^{2}} dr \int_{0}^{\frac{π}{2}} d θ = \frac{π}{2} {[- \frac{1}{2} e^{- r^{2}}]}_{\frac{\sqrt{2}}{n}}^{n \sqrt{2}} = - \frac{π}{4} {e^{- {2 n}^{2}} - e^{- \frac{2}{n^{2}}}} \Rightarrow$ $\lim_{n \to + \infty} A_{n} = \frac{π}{4} = {(\int_{0}^{\infty} e^{- x^{2}} dx)}^{2} but \int_{0}^{\infty} e^{- x^{2}} dx > 0 \Rightarrow$ $\int_{0}^{\infty} e^{- x^{2}} dx = \frac{\sqrt{π}}{2} \Rightarrow ★ A = \sqrt{π} ★$
	Answered by mathmax by abdo last updated on 23/Jun/20

	$let use Γ function we have \int_{- \infty}^{+ \infty} e^{- x^{2}} dx = 2 \int_{0}^{\infty} e^{- x^{2}} dx changement x^{2} = t$ $give \int_{0}^{\infty} e^{- x^{2}} dx = \int_{0}^{\infty} e^{- t} \frac{dt}{2 \sqrt{t}} = \frac{1}{2} \int_{0}^{\infty} e^{- t} t^{- \frac{1}{2}} dt$ $we know Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} dt (x > 0) \Rightarrow \int_{0}^{\infty} e^{- x^{2}} dx = \frac{1}{2} \int_{0}^{\infty} t^{\frac{1}{2} - 1} e^{- t} dt$ $= \frac{1}{2} Γ (\frac{1}{2}) = \frac{\sqrt{π}}{2} \Rightarrow \int_{- \infty}^{+ \infty} e^{- x^{2}} dx = \sqrt{π}$
	Commented by student work last updated on 25/Jun/20

	$how i can Γ (\frac{1}{2}) calculate ?$
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