find-the-value-of-0-1-x-2-3-x-2-dx-

Question Number 27620 by abdo imad last updated on 11/Jan/18

f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{{(- 1)}^{x^{2}}}{3 + x^{2}} d x .

Commented by abdo imad last updated on 12/Jan/18

l e t p u t I = \int_{0}^{\infty} \frac{{(- 1)}^{x^{2}}}{3 + x^{2}} d x d u e t o (- 1) = e^{i π}

I = \int_{0}^{\infty} \frac{e^{i π x^{2}}}{3 + x^{2}} d x = \frac{1}{2} \int_{R} \frac{e^{i π x^{2}}}{3 + x^{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^{i π z^{2}}}{z^{2} + 3} w e h a v e

f (z) = \frac{e^{i π z^{2}}}{(z - i \sqrt{3}) (z + i \sqrt{3)}} t h e p o l e s o f f a r e z_{1} = i \sqrt{3} a n d

z_{2} = - i \sqrt{3} t h e r e s i d u s t h e o r e m g i v e

\int_{R} f (z) d z = 2 i π R e s (f, i \sqrt{3}) b u t

R e s (f, i \sqrt{3}) = {l i m}_{z - > i \sqrt{3}} (z - i \sqrt{3}) f (z) = {l i m}_{z - > i \sqrt{3}} \frac{e^{i π z^{2}}}{z + i \sqrt{3}}

= \frac{e^{i π (- 3)}}{2 i \sqrt{3}} = \frac{{(- 1)}^{- 3}}{2 i \sqrt{3}} = \frac{- 1}{2 i \sqrt{3}}

\int_{R} f (z) d z = 2 i π . \frac{- 1}{2 i \sqrt{3}} = \frac{- π}{\sqrt{3}} a n d I = \frac{- π}{2 \sqrt{3}} .