1)calculste A=∫_(−∞) ^(+∞) (dx/((x^2 −ix +1)^2 )) 2) extract Re(A) and Im(A) and determines its values


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Question Number 112449 by mathmax by abdo last updated on 07/Sep/20

$1) calculste A = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} - ix + 1)}^{2}}$ $2) extract Re (A) and Im (A) and determines its values$
Commented by MJS_new last updated on 08/Sep/20

$I get A = \frac{4 π \sqrt{5}}{25} can this be true ?$
Commented by MJS_new last updated on 08/Sep/20
Here's a better place to answer you. I'm well and I hope you're well, too. Nice to be in here again.
Commented by abdomsup last updated on 08/Sep/20

$y o u a r e w e l c o m e s i r w i s h y o u$ $h a p p y r e t u r n . . i h a v e n t t h e v a l u e$ $o f t h i s i n t e g r a l b u t w a i t m y a n s w e r$ $s o o n . . . .$
	Answered by mathmax by abdo last updated on 08/Sep/20
	$1) A =∫_(−∞) ^(+∞) (dx/((x^2 −ix +1)^2 )) let ϕ(z) =(1/((z^2 −iz +1)^2 )) poles of ϕ? z^2 −iz +1 =0 →Δ =(−i)^2 −4 =−5 ⇒z_1 =((i+i(√5))/2) and z_2 =((i−i(√5))/2) ⇒ϕ(z) =(1/((z−z_1 )^2 (z−z_2 )^2 )) residus tbeorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_1 ) and Res(ϕ,z_1 ) =lim_(z→z_1 ) (1/((2−1)!)){(z−z_1 )^2 ϕ(z)}^((1)) =lim_(z→z_1 ) {(1/((z−z_2 )^2 ))}^((1)) =−lim_(z→z_1 ) ((2(z−z_2 ))/((z−z_2 )^4 )) =lim_(z→z_1 ) ((−2)/((z−z_2 )^3 )) =((−2)/((z_1 −z_2 )^3 )) =((−2)/((i(√5))^3 )) =((−2)/(−i5(√5))) =(2/(5i(√5))) ⇒ ∫_(−∞) ^(+∞) ϕ(z) dz =2iπ×(2/(5i(√5))) =((4π)/(5(√5))) ⇒★ A =((4π)/(5(√5)))★$
	$1) A = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} - ix + 1)}^{2}} let φ (z) = \frac{1}{{(z^{2} - iz + 1)}^{2}} poles of φ ?$ $z^{2} - iz + 1 = 0 \to Δ = {(- i)}^{2} - 4 = - 5 \Rightarrow z_{1} = \frac{i + i \sqrt{5}}{2} and z_{2} = \frac{i - i \sqrt{5}}{2}$ $\Rightarrow φ (z) = \frac{1}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}} residus tbeorem give$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, z_{1}) and$ $Res (φ, z_{1}) = \lim_{z \to z_{1}} \frac{1}{(2 - 1)!} {{(z - z_{1})}^{2} φ (z)}^{(1)}$ $= \lim_{z \to z_{1}} {\frac{1}{{(z - z_{2})}^{2}}}^{(1)} = - \lim_{z \to z_{1}} \frac{2 (z - z_{2})}{{(z - z_{2})}^{4}}$ $= \lim_{z \to z_{1}} \frac{- 2}{{(z - z_{2})}^{3}} = \frac{- 2}{{(z_{1} - z_{2})}^{3}} = \frac{- 2}{{(i \sqrt{5})}^{3}} = \frac{- 2}{- i 5 \sqrt{5}} = \frac{2}{5 i \sqrt{5}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π \times \frac{2}{5 i \sqrt{5}} = \frac{4 π}{5 \sqrt{5}} \Rightarrow ★ A = \frac{4 π}{5 \sqrt{5}} ★$
	Answered by MJS_new last updated on 08/Sep/20

	$\frac{1}{{(x^{2} - i x + 1)}^{2}} =$ $= \frac{x^{4} + x^{2} + 1}{{(x^{4} + 3 x^{2} + 1)}^{2}} - \frac{2 x (x^{2} + 1)}{{(x^{4} + 3 x^{2} + 1)}^{2}} i$ $\Rightarrow the imaginary part is uneven$ $\Rightarrow \underset{- a}{\int^{+ a}} \frac{d x}{{(x^{2} - i x + 1)}^{2}} = \underset{- a}{\int^{+ a}} \frac{x^{4} + x^{2} + 1}{{(x^{4} + 3 x^{2} + 1)}^{2}} d x$ $\int \frac{x^{4} + x^{2} + 1}{{(x^{4} + 3 x^{2} + 1)}^{2}} d x =$ $[Ostrogradski]$ $= \frac{x (2 x^{2} + 3)}{5 (x^{4} + 3 x^{2} + 1)} + \frac{2}{5} \int \frac{x^{2} + 1}{x^{4} + 3 x^{2} + 1} d x$ $\int \frac{x^{2} + 1}{x^{4} + 3 x^{2} + 1} d x =$ $= \frac{5 + \sqrt{5}}{10} \int \frac{d x}{x^{2} + \frac{3 + \sqrt{5}}{2}} + \frac{5 - \sqrt{5}}{10} \int \frac{d x}{x^{2} + \frac{3 - \sqrt{5}}{2}} =$ $= \frac{\sqrt{5}}{5} (- \arctan \frac{(1 - \sqrt{5}) x}{2} + \arctan \frac{(1 + \sqrt{5}) x}{2})$ $\Rightarrow$ $\int \frac{x^{4} + x^{2} + 1}{{(x^{4} + 3 x^{2} + 1)}^{2}} d x =$ $= \frac{x (2 x^{2} + 3)}{5 (x^{4} + 3 x^{2} + 1)} + \frac{2 \sqrt{5}}{25} (\arctan \frac{(1 + \sqrt{5}) x}{2} - \arctan \frac{(1 - \sqrt{5}) x}{2}) + C$ $\Rightarrow$ $\underset{- \infty}{\int^{+ \infty}} \frac{d x}{{(x^{2} - i x + 1)}^{2}} = \underset{- \infty}{\int^{+ \infty}} \frac{x^{4} + x^{2} + 1}{{(x^{4} + 3 x^{2} + 1)}^{2}} d x =$ $= \frac{4 π \sqrt{5}}{25}$
	Commented by mathmax by abdo last updated on 08/Sep/20

	$thank you sir M^{j} S$
	Answered by mathmax by abdo last updated on 08/Sep/20
	$A =∫_(−∞) ^(+∞) (dx/((x^2 −ix +1)^2 )) =∫_(−∞) ^(+∞) (((x^2 +1+ix)^2 )/((x^2 +1−ix)^2 (x^2 +1+ix)^2 ))dx =∫_(−∞) ^(+∞) (((x^2 +1)^2 +2ix(x^2 +1)−x^2 )/({(x^2 +1)^2 +x^2 }^2 ))dx =∫_(−∞) ^(+∞) ((x^4 +2x^2 +1 +2ix^3 +2ix −x^2 )/((x^4 +2x^2 +1 +x^2 )^2 ))dx =∫_(−∞) ^(+∞) ((x^4 +x^2 +1 +i(2x^3 +2x))/((x^4 +3x^2 +1)^2 ))dx ⇒ Re(A) =∫_(−∞) ^(+∞) ((x^4 +x^2 +1)/((x^4 +3x^2 +1)^2 )) dx and Im(A) =∫_(−∞) ^(+∞) ((2x^3 +2x)/((x^4 +3x^2 +1)^2 ))dx we conclude that Res(A)=((4π)/(5(√5))) and Im(A)=0$
	$A = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} - ix + 1)}^{2}} = \int_{- \infty}^{+ \infty} \frac{{(x^{2} + 1 + ix)}^{2}}{{(x^{2} + 1 - ix)}^{2} {(x^{2} + 1 + ix)}^{2}} dx$ $= \int_{- \infty}^{+ \infty} \frac{{(x^{2} + 1)}^{2} + 2 ix (x^{2} + 1) - x^{2}}{{{(x^{2} + 1)}^{2} + x^{2}}^{2}} dx$ $= \int_{- \infty}^{+ \infty} \frac{x^{4} + {2 x}^{2} + 1 + {2 ix}^{3} + 2 ix - x^{2}}{{(x^{4} + {2 x}^{2} + 1 + x^{2})}^{2}} dx$ $= \int_{- \infty}^{+ \infty} \frac{x^{4} + x^{2} + 1 + i ({2 x}^{3} + 2 x)}{{(x^{4} + {3 x}^{2} + 1)}^{2}} dx \Rightarrow$ $Re (A) = \int_{- \infty}^{+ \infty} \frac{x^{4} + x^{2} + 1}{{(x^{4} + {3 x}^{2} + 1)}^{2}} dx and Im (A) = \int_{- \infty}^{+ \infty} \frac{{2 x}^{3} + 2 x}{{(x^{4} + {3 x}^{2} + 1)}^{2}} dx$ $we conclude that Res (A) = \frac{4 π}{5 \sqrt{5}} and Im (A) = 0$
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