find-0-dt-1-t-2-4-

Question Number 28883 by abdo imad last updated on 31/Jan/18

f i n d \int_{0}^{\infty} \frac{d t}{{(1 + t^{2})}^{4}}

Commented by abdo imad last updated on 02/Feb/18

I = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d t}{{(1 + t^{2})}^{4}} l e t p u t f (z) = \frac{1}{{(1 + z^{2})}^{4}}

f (z) = \frac{1}{{(z - i)}^{4} {(z + i)}^{4}} t h e p o l e s o f f a r e i a n d - i w i t h o r d r 4

\int_{- \infty}^{+ \infty} f (z) d z = 2 i π R e s (f, i) b u t

R e s (f, i) = {l i m}_{z \to i} \frac{1}{(4 - 1)!} {({(z - i)}^{4} f (z))}^{(3)}

= {l i m}_{z \to i} \frac{1}{6} {({(z + i)}^{- 4})}^{(3)} a n d w e h a v e

{{(z + i)}^{- 4})}^{(1)} = - 4 {(z + i)}^{- 5}

{({(z + i)}^{- 4})}^{(2)} = 20 {(z + i)}^{- 6}

{({(z + i)}^{- 4})}^{(3)} = - 120 {(z + i)}^{- 7}

R e s (f, i) = \frac{1}{6} (- 120) {(2 i)}^{- 7} = \frac{- 20}{2^{7} i^{7}} b u t i^{7} = \frac{i^{8}}{i} = \frac{1}{i}

R e s (f, i) = \frac{- 20 i}{2^{7}} = - \frac{2^{2} . 5 i}{2^{7}} = - i \frac{5}{32} a n d

\int_{- \infty}^{+ \infty} f (z) d z = 2 i π (- i \frac{5}{32}) = \frac{10 π}{32} = \frac{5 π}{16} . f i n a l l y

I = \frac{1}{2} \int_{- \infty}^{+ \infty} f (z) d z = \frac{5 π}{32} .