calculate ∫_(−∞) ^(+∞) ((xdx)/((x^2 −x+i)^2 )) with i^2 =−1


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Question Number 68877 by mathmax by abdo last updated on 16/Sep/19

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{x d x}{{(x^{2} - x + i)}^{2}} w i t h i^{2} = - 1$
Commented by mathmax by abdo last updated on 18/Sep/19
$let A =∫_(−∞) ^(+∞) ((xdx)/((x^2 −x +i)^2 )) let ϕ(z) =(z/((z^2 −z +i)^2 )) poles of ϕ? z^2 −z +i =0 ⇒Δ =1−4i =(√(17))e^(iarctan(−4)) =(√(17))e^(−i arctan(4)) z_1 =((1+17^(1/4) e^(−(i/2)arctan(4)) )/2) and z_2 =((1−17^(1/4) e^(−(i/2)arctan(4)) )/2) ϕ(z) =(z/((z−z_1 )^2 (z−z_2 )^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_2 ) Res(ϕ,z_2 ) =lim_(z→z_2 ) (1/((2−1)!)){ (z−z_2 )^2 ϕ(z)}^((1)) =lim_(z→z_2 ) {(z/((z−z_1 )^2 ))}^((1)) =lim_(z→z_1 ) (((z−z_1 )^2 −2(z−z_1 )z)/((z−z_1 )^4 )) =lim_(z→z_2 ) ((z−z_1 −2z)/((z−z_1 )^3 )) =lim_(z→z_2 ) ((−z−z_1 )/((z−z_1 )^3 )) =((−z_2 −z_1 )/((z_2 −z_1 )^3 )) =−(1/((−17^(1/4) e^(−(i/2)arctan(4)) )^3 )) =(1/(17^(3/4) e^(−((3i)/2) arctan(4)) )) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ×(1/(17^(3/4) e^(−((3i)/2) arctan(4)) )) =((2iπ e^(((3i)/2) arctan(4)) )/(17^(3/4) )) = A$
$l e t A = \int_{- \infty}^{+ \infty} \frac{x d x}{{(x^{2} - x + i)}^{2}} l e t φ (z) = \frac{z}{{(z^{2} - z + i)}^{2}} p o l e s o f φ ?$ $z^{2} - z + i = 0 \Rightarrow Δ = 1 - 4 i = \sqrt{17} e^{i a r c t a n (- 4)} = \sqrt{17} e^{- i a r c t a n (4)}$ $z_{1} = \frac{1 + 17^{\frac{1}{4}} e^{- \frac{i}{2} a r c t a n (4)}}{2} a n d z_{2} = \frac{1 - 17^{\frac{1}{4}} e^{- \frac{i}{2} a r c t a n (4)}}{2}$ $φ (z) = \frac{z}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{2})$ $R e s (φ, z_{2}) = {l i m}_{z \to z_{2}} \frac{1}{(2 - 1)!} {{(z - z_{2})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to z_{2}} {\frac{z}{{(z - z_{1})}^{2}}}^{(1)} = {l i m}_{z \to z_{1}} \frac{{(z - z_{1})}^{2} - 2 (z - z_{1}) z}{{(z - z_{1})}^{4}}$ $= {l i m}_{z \to z_{2}} \frac{z - z_{1} - 2 z}{{(z - z_{1})}^{3}} = {l i m}_{z \to z_{2}} \frac{- z - z_{1}}{{(z - z_{1})}^{3}} = \frac{- z_{2} - z_{1}}{{(z_{2} - z_{1})}^{3}}$ $= - \frac{1}{{(- 17^{\frac{1}{4}} e^{- \frac{i}{2} a r c t a n (4)})}^{3}} = \frac{1}{17^{\frac{3}{4}} e^{- \frac{3 i}{2} a r c t a n (4)}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \times \frac{1}{17^{\frac{3}{4}} e^{- \frac{3 i}{2} a r c t a n (4)}} = \frac{2 i π e^{\frac{3 i}{2} a r c t a n (4)}}{17^{\frac{3}{4}}} = A$
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