calculate-xdx-2x-1-i-3-with-i-2-1-

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

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