calculate-dt-1-it-1-it-2-

Question Number 32715 by caravan msup abdo. last updated on 31/Mar/18

c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d t}{(1 + i t) (1 + {i t}^{2})} .

Commented by abdo imad last updated on 03/Apr/18

l e t p u t φ (z) = \frac{1}{(1 + i z) (1 + {i z}^{2})}

φ (z) = \frac{1}{(i z - i^{2}) ({i z}^{2} - i^{2})} \frac{- 1}{(z - i) (z^{2} - i)}

= \frac{- 1}{(z - i) (z - \sqrt{i}) (z + \sqrt{i})} = \frac{- 1}{(z - i) (z - e^{i \frac{π}{4}}) (z + e^{i \frac{π}{4}})}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (R e s (φ, i) + R e s (φ, e^{i \frac{π}{4}}))

R e s (φ, i) = {l i m}_{z \to i} (z - i) φ (z) = \frac{- 1}{- 1 - i} = \frac{1}{1 + i} = \frac{1 - i}{2}

R e s (φ, e^{i \frac{π}{4}}) = {l i m}_{z \to e^{i \frac{π}{4}}} (z - e^{i \frac{π}{4}}) φ (z)

= \frac{- 1}{(e^{i \frac{π}{4}} - i) (2 e^{i \frac{π}{4}})} = \frac{- 1}{2 i - 2 i e^{i \frac{π}{4}}} = \frac{i}{2 (1 - e^{i \frac{π}{4}})}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (\frac{1 - i}{2} + \frac{i}{2 (1 - e^{i \frac{π}{4}})}) .