calculate ∫_(−∞) ^(+∞) ((xdx)/((2x+1+i)^3 )) with i^2 =−1 .


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Question Number 36009 by abdo mathsup 649 cc last updated on 27/May/18

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{x d x}{{(2 x + 1 + i)}^{3}} w i t h i^{2} = - 1 .$
Commented by abdo imad last updated on 31/May/18
$let consider the complex function ϕ(z) = (z/((2z +1+i)^3 )) = (z/(8(z−((−1−i)/2))^3 )) so ϕ have one pole triple z_0 =−(1/(√2))e^(i(π/4)) and I=∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_0 ) but Res(ϕ,z_0 ) =lim_(z→z_0 ) (1/((3−1)!)){(z−z_0 )^3 (z/(8(z−z_0 )^3 ))}^((2)) =lim_(z→z_0 ) (1/(16)){z}^((2)) =0 so I =0$
$l e t c o n s i d e r t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{z}{{(2 z + 1 + i)}^{3}} = \frac{z}{8 {(z - \frac{- 1 - i}{2})}^{3}} s o φ h a v e o n e p o l e$ $t r i p l e z_{0} = - \frac{1}{\sqrt{2}} e^{i \frac{π}{4}} a n d$ $I = \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{0}) b u t$ $R e s (φ, z_{0}) = {l i m}_{z \to z_{0}} \frac{1}{(3 - 1)!} {{(z - z_{0})}^{3} \frac{z}{8 {(z - z_{0})}^{3}}}^{(2)}$ $= {l i m}_{z \to z_{0}} \frac{1}{16} {z}^{(2)} = 0 s o I = 0$
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