calculate ∫_(−∞) ^(+∞) (dx/((x^2 −ix +1)^2 )) (i=(√(−1)))


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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
	$I =∫_(−∞) ^(+∞) (dx/((x^2 −ix +1)^2 )) let ϕ(z) =(1/((z^2 −iz +1)^2 )) poles of ϕ? z^2 −iz +1 =0→Δ =(−i)^2 −4 =−5 ⇒z_1 =i+i(√5) and z_2 =i−i(√5) ⇒ϕ(z) =(1/((z−z_1 )^2 (z−z_2 )^2 )) residus tbeorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_1 ) and Res(ϕ,z_1 ) =lim_(z→z_1 ) (1/((2−1)!)){(z−z_1 )^2 ϕ(z)}^((1)) =lim_(z→z_1 ) {(1/((z−z_2 )^2 ))}^((1)) =lim_(z→z_1 ) −((2(z−z_2 ))/((z−z_2 )^4 )) =lim_(z→z_1 ) ((−2)/((z−z_2 )^3 )) =((−2)/((z_1 −z_2 )^3 )) =((−2)/((2i(√5))^3 )) =((−2)/(2^3 (−i)(5(√5)))) =(1/(20i(√5))) ⇒ ∫_(−∞) ^(+∞) ϕ(z) =2iπ.(1/(20i(√5))) =(π/(10(√5))) ⇒ I =(π/(10(√5)))$
	$I = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} - ix + 1)}^{2}} let φ (z) = \frac{1}{{(z^{2} - iz + 1)}^{2}} poles of φ ?$ $z^{2} - iz + 1 = 0 \to Δ = {(- i)}^{2} - 4 = - 5 \Rightarrow z_{1} = i + i \sqrt{5} and z_{2} = i - i \sqrt{5}$ $\Rightarrow φ (z) = \frac{1}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}} residus tbeorem give$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, z_{1}) and$ $Res (φ, z_{1}) = \lim_{z \to z_{1}} \frac{1}{(2 - 1)!} {{(z - z_{1})}^{2} φ (z)}^{(1)}$ $= \lim_{z \to z_{1}} {\frac{1}{{(z - z_{2})}^{2}}}^{(1)} = \lim_{z \to z_{1}} - \frac{2 (z - z_{2})}{{(z - z_{2})}^{4}} = \lim_{z \to z_{1}} \frac{- 2}{{(z - z_{2})}^{3}}$ $= \frac{- 2}{{(z_{1} - z_{2})}^{3}} = \frac{- 2}{{(2 i \sqrt{5})}^{3}} = \frac{- 2}{2^{3} (- i) (5 \sqrt{5})} = \frac{1}{20 i \sqrt{5}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) = 2 i π . \frac{1}{20 i \sqrt{5}} = \frac{π}{10 \sqrt{5}} \Rightarrow I = \frac{π}{10 \sqrt{5}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum