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

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

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