find-0-e-t-2-1-t-2-dt-

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

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

= \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 \dots 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 \dots \frac{e^{- 2}}{2} \times \frac{\sqrt{Π}}{2}