find ∫_0 ^(+∞) e^(−(t^2 +(1/t^2 ))) dt


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Question Number 37634 by math khazana by abdo last updated on 16/Jun/18

$f i n d \int_{0}^{+ \infty} e^{- (t^{2} + \frac{1}{t^{2}})} d t$
Commented by prof Abdo imad last updated on 17/Jun/18
$let I = ∫_0 ^∞ e^(−(t^2 +(1/t^2 ))) dt 2I = ∫_(−∞) ^∞ e^(−{ ( t−(1/t))^2 +2}) dt = e^(−2) ∫_(−∞) ^(+∞) e^(−(t−(1/t))^2 ) dt changement t−(1/t)=x give t^2 −1=xt ⇒t^2 −xt−1=0 Δ =x^2 +4 ⇒ t_1 =((x +(√(x^2 +4)))/2) and t_2 = ((x−(√(x^2 +4)))/2) let take t =((x +(√(x^2 +4)))/2) ⇒ dt =(1/2)( 1+ (x/(√(x^2 +4)))) dx⇒ 2I = (e^(−2) /2)∫_(−∞) ^(+∞) e^(−x^2 ) ( 1+(x/(√(x^2 +4))))dx =(e^(−2) /2)∫_(−∞) ^(+∞) e^(−x^2 ) dx +(e^(−2) /2) ∫_(−∞) ^(+∞) ((x e^(−x^2 ) )/(√(x^2 +4)))dx =(((√π) e^(−2) )/2) +0 because the function x→ ((x e^(−x^2 ) )/(√(x^2 +4))) is odd ⇒ I = ((e^(−2) (√π))/4) .$
$l e t I = \int_{0}^{\infty} e^{- (t^{2} + \frac{1}{t^{2}})} d t$ $2 I = \int_{- \infty}^{\infty} e^{- {{(t - \frac{1}{t})}^{2} + 2}} d t$ $= e^{- 2} \int_{- \infty}^{+ \infty} e^{- {(t - \frac{1}{t})}^{2}} d t c h a n g e m e n t$ $t - \frac{1}{t} = x g i v e t^{2} - 1 = x t \Rightarrow t^{2} - x t - 1 = 0$ $Δ = x^{2} + 4 \Rightarrow t_{1} = \frac{x + \sqrt{x^{2} + 4}}{2} a n d$ $t_{2} = \frac{x - \sqrt{x^{2} + 4}}{2} l e t t a k e t = \frac{x + \sqrt{x^{2} + 4}}{2} \Rightarrow$ $d t = \frac{1}{2} (1 + \frac{x}{\sqrt{x^{2} + 4}}) d x \Rightarrow$ $2 I = \frac{e^{- 2}}{2} \int_{- \infty}^{+ \infty} e^{- x^{2}} (1 + \frac{x}{\sqrt{x^{2} + 4}}) d x$ $= \frac{e^{- 2}}{2} \int_{- \infty}^{+ \infty} e^{- x^{2}} d x + \frac{e^{- 2}}{2} \int_{- \infty}^{+ \infty} \frac{x e^{- x^{2}}}{\sqrt{x^{2} + 4}} d x$ $= \frac{\sqrt{π} e^{- 2}}{2} + 0 b e c a u s e t h e f u n c t i o n$ $x \to \frac{x e^{- x^{2}}}{\sqrt{x^{2} + 4}} i s o d d \Rightarrow$ $I = \frac{e^{- 2} \sqrt{π}}{4} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jun/18
	$=(1/2)∫_0 ^∞ (1−(1/t^2 )+1+(1/(t^2 )))e^(−(t^2 +(1/t^2 ))) dt =(1/2)∫_0 ^∞ (1−(1/t^2 ))e^(−{(t+(1/t))^2 −2}) dt+ (1/2)∫_0 ^∞ (1+(1/t^2 ))e^(−{(t−(1/t))^2 +2}) dt =(1/2)∫_0 ^∞ (1−(1/t^2 ))e^(−{(t+(1/t))^2 }) ×e^2 dt+ (1/2)∫_0 ^∞ (1+(1/t^2 ))e^(−{(t−(1/t))^2 }) ×e^(−2) dt =(e^2 /2)∫_∞ ^∞ e^(−k_1 ^2 ) dk_1 +(e^(−2) /2)∫_(−∞) ^∞ e^(−k_2 ^2 ) dk_(2 ) so first intregsl value=0 2nd intregal=(e^(−2) /2)×2∫_0 ^∞ e^(−k_2 ^2 ) dk_2 =e^(−2) ×(((√Π) )/2) formula∫_0 ^∞ e^(−x^2 ) dx=(((√Π) )/2) i have done a small error marking with red and later corrected...this red marked 2 should not be there..correct answer...(e^(−2) /2)×((√Π)/2)$
	$= \frac{1}{2} \int_{0}^{\infty} (1 - \frac{1}{t^{2}} + 1 + \frac{1}{t^{2}}) e^{- (t^{2} + \frac{1}{t^{2}})} d t$ $= \frac{1}{2} \int_{0}^{\infty} (1 - \frac{1}{t^{2}}) e^{- {{(t + \frac{1}{t})}^{2} - 2}} d t +$ $\frac{1}{2} \int_{0}^{\infty} (1 + \frac{1}{t^{2}}) e^{- {{(t - \frac{1}{t})}^{2} + 2}} d t$ $= \frac{1}{2} \int_{0}^{\infty} (1 - \frac{1}{t^{2}}) e^{- {{(t + \frac{1}{t})}^{2}}} \times e^{2} d t +$ $\frac{1}{2} \int_{0}^{\infty} (1 + \frac{1}{t^{2}}) e^{- {{(t - \frac{1}{t})}^{2}}} \times e^{- 2} d t$ $= \frac{e^{2}}{2} \int_{\infty}^{\infty} e^{- k_{1}^{2}} {d k}_{1} + \frac{e^{- 2}}{2} \int_{- \infty}^{\infty} e^{- k_{2}^{2}} {d k}_{2}$ $s o f i r s t i n t r e g s l v a l u e = 0$ $2 n d i n t r e g a l = \frac{e^{- 2}}{2} \times 2 \int_{0}^{\infty} e^{- k_{2}^{2}} {d k}_{2}$ $= e^{- 2} \times \frac{\sqrt{Π}}{2} f o r m u l a \int_{0}^{\infty} e^{- x^{2}} d x = \frac{\sqrt{Π}}{2}$ $i h a v e d o n e a s m a l l e r r o r m a r k i n g w i t h r e d$ $a n d l a t e r c o r r e c t e d . . . t h i s r e d m a r k e d 2 s h o u l d$ $n o t b e t h e r e . . c o r r e c t a n s w e r . . . \frac{e^{- 2}}{2} \times \frac{\sqrt{Π}}{2}$
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