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

Question Number 27619 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{c o s (2 x)}{{(1 + x^{2})}^{2}} d x .

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

l e t p u t I = \int_{0}^{\infty} \frac{c o s (2 x)}{{(1 + x^{2})}^{2}} d x

I = \frac{1}{2} \int_{R} \frac{c o s (2 x)}{{(1 + x^{2})}^{2}} d x = \frac{1}{2} R e (\int_{R} \frac{e^{i 2 x}}{{(1 + 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^{i (2 z)}}{{(1 + z^{2})}^{2}} w e h a v e f (z) = \frac{e^{i (2 z)}}{{(z - i)}^{2} {(z + i)}^{2}} s o t h e p o l e s

o f f a r e i a n d - i (d o u b l e p o l e s)

\int_{R} f (z) d z = 2 i π R e s (f, i) (w e d o n t t a k e - i b e c a u s e I m (- i) < 0)

R e (f, i) = {l i m}_{z - > i} \frac{1}{(2 - 1)!} \frac{d}{d z} ({(z - i)}^{2} f (z))

= {l i m}_{z - > i} \frac{d}{d z} (\frac{e^{i (2 z)}}{{(z + i)}^{2}}) =

= {l i m}_{z - > i} \frac{2 i e^{i (2 z)} {(z + i)}^{2} - 2 (z + i) e^{i (2 z)}}{{(z + i)}^{4}}

= {l i m}_{z - > i} \frac{2 i e^{i (2 z)} (z + i) - 2 e^{i (2 z)}}{{(z + i)}^{3}}

= \frac{2 i e^{- 2} (2 i) - 2 e^{- 2}}{{(2 i)}^{3}} = \frac{- 4 e^{- 2} - 2 e^{- 2}}{- 8 i} = \frac{6 e^{- 2}}{8 i} = \frac{3 e^{- 2}}{4 i}

\int_{R} f (z) d z = 2 i π . \frac{3 e^{- 2}}{4 i} = \frac{3 π}{2} e^{- 2}

I = \frac{1}{2} \int_{R}^{} f (z) d z = \frac{3 π}{4} e^{- 2} = \frac{3 π}{4 e^{2}} .