find-the-value-of-0-e-2x-2-3-x-2-2-dx-

Question Number 27802 by abdo imad last updated on 15/Jan/18

f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{e^{- 2 x^{2}}}{{(3 + x^{2})}^{2}} d x .

Commented by abdo imad last updated on 17/Jan/18

l e t p u t I = \int_{0}^{\infty} \frac{e^{- 2 x^{2}}}{{(3 + x^{2})}^{2}} d x

I = \frac{1}{2} \int_{- \propto}^{+ \propto} \frac{e^{- 2 x^{2}}}{{(3 + x^{2})}^{2}} d x l e t i n t r o d u c e t h e c o m p l e x f u n c t i o n

f (z) = \frac{e^{- 2 z^{2}}}{{(3 + z^{2})}^{2}} . p o l e s o f f ?

f (z) = \frac{e^{- 2 z^{2}}}{{(z - i \sqrt{3})}^{2} {(z + i \sqrt{3})}^{2}} s o f h a v e a d o u b l e p o l e s i \sqrt{3}

a n d - i \sqrt{3}

\int_{- \propto}^{+ \propto} f (z) d z = 2 i π R e s (f, i \sqrt{3})

R e s (f, i \sqrt{3)} = {l i m}_{z \to i \sqrt{3}} \frac{1}{(2 - 1)!} {({(z - i \sqrt{3})}^{2} f (z))}^{^{'}}

{l i m}_{z \to i \sqrt{3}} {(e^{- 2 z^{2}} . {(z + i \sqrt{3})}^{- 2})}^{,}

= {l i m}_{z \to i \sqrt{3}} (- 4 z e^{- z^{2}} {(z + i \sqrt{3})}^{- 2} - 2 e^{- 2 z^{2}} {(z + i \sqrt{3})}^{- 3})

= - 4 (i \sqrt{3}) e^{3} {(2 i \sqrt{3})}^{- 2} - 2 e^{6} {(2 i \sqrt{3})}^{- 3}

= \frac{- 4 i \sqrt{3} e^{3}}{- 12} - \frac{2 e^{6}}{- 24 i \sqrt{3}} = \frac{i \sqrt{3} e^{3}}{3} + \frac{e^{6}}{12 i \sqrt{3}} .

\int_{- \propto}^{+ \propto} f (z) d z = 2 i π (\frac{i \sqrt{3} e^{3}}{3} + \frac{e^{6}}{12 i \sqrt{3}})

= \frac{- 2 \sqrt{3} e^{3}}{3} + \frac{e^{6}}{6 \sqrt{3}} f i n a l l y

I = \frac{1}{2} \int_{R} f (z) d z = \frac{- e^{3} \sqrt{3}}{3} + \frac{e^{6}}{12 \sqrt{3}} .