calculate-e-ix-x-i-x-i-x-2-3-dx-

Question Number 37306 by math khazana by abdo last updated on 11/Jun/18

c a l c u l a t e \int_{- \infty}^{+ \infty} e^{i x} \frac{x - i}{(x + i) (x^{2} + 3)} d x .

Commented by prof Abdo imad last updated on 15/Jun/18

l e t φ (z) = e^{i z} \frac{z - i}{(z + i) (z^{2} + 3)} t h e p o l e s o f

φ a r e - i, i \sqrt{3}, - i \sqrt{3}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i \sqrt{3})

w e h a v e φ (z) = \frac{(z - i) e^{i z}}{(z + i) (z - i \sqrt{3}) (z + i \sqrt{3})}

R e s (φ, i \sqrt{3}) = \frac{(i \sqrt{3} - i) e^{i (i \sqrt{3})}}{(i \sqrt{3} + i) (2 i \sqrt{3})}

= \frac{i (\sqrt{3} - 1) e^{- \sqrt{3}}}{- 2 \sqrt{3} (1 + \sqrt{3})} = \frac{i (1 - \sqrt{3}) e^{- \sqrt{3}}}{2 \sqrt{3} + 6} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{i (1 - \sqrt{3}) e^{- \sqrt{3}}}{2 \sqrt{3} + 6}

= \frac{π (\sqrt{3} - 1) e^{- \sqrt{3}}}{3 + \sqrt{3}} \Rightarrow

I = \frac{π (\sqrt{3} - 1) e^{- \sqrt{3}}}{3 + \sqrt{3}} .