calculate-dx-x-2-ix-1-2-i-1-

Question Number 110447 by mathmax by abdo last updated on 29/Aug/20

calculate \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} - ix + 1)}^{2}} (i = \sqrt{- 1})

Answered by mathmax by abdo last updated on 29/Aug/20

I = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} - ix + 1)}^{2}} let φ (z) = \frac{1}{{(z^{2} - iz + 1)}^{2}} poles of φ ?

z^{2} - iz + 1 = 0 \to Δ = {(- i)}^{2} - 4 = - 5 \Rightarrow z_{1} = i + i \sqrt{5} and z_{2} = i - i \sqrt{5}

\Rightarrow φ (z) = \frac{1}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}} residus tbeorem give

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, z_{1}) and

Res (φ, z_{1}) = \lim_{z \to z_{1}} \frac{1}{(2 - 1)!} {{(z - z_{1})}^{2} φ (z)}^{(1)}

= \lim_{z \to z_{1}} {\frac{1}{{(z - z_{2})}^{2}}}^{(1)} = \lim_{z \to z_{1}} - \frac{2 (z - z_{2})}{{(z - z_{2})}^{4}} = \lim_{z \to z_{1}} \frac{- 2}{{(z - z_{2})}^{3}}

= \frac{- 2}{{(z_{1} - z_{2})}^{3}} = \frac{- 2}{{(2 i \sqrt{5})}^{3}} = \frac{- 2}{2^{3} (- i) (5 \sqrt{5})} = \frac{1}{20 i \sqrt{5}} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) = 2 i π . \frac{1}{20 i \sqrt{5}} = \frac{π}{10 \sqrt{5}} \Rightarrow I = \frac{π}{10 \sqrt{5}}