let-A-dx-x-2-j-with-j-e-i-2pi-3-extract-ReA-and-Im-A-and-calculste-its-values-

Question Number 34421 by abdo mathsup 649 cc last updated on 06/May/18

l e t A = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - j} w i t h j = e^{i \frac{2 π}{3}}

e x t r a c t R e A a n d I m (A) a n d c a l c u l s t e i t s v a l u e s .

Commented by abdo mathsup 649 cc last updated on 07/May/18

w e h a v e A = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - (- \frac{1}{2} + i \frac{\sqrt{3}}{2})}

= \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + \frac{1}{2} - i \frac{\sqrt{3}}{2}}

= \int_{- \infty}^{+ \infty} \frac{x^{2} + \frac{1}{2} + i \frac{\sqrt{3}}{2}}{{(x^{2} + \frac{1}{2})}^{2} + \frac{3}{4}} d x \Rightarrow

R e (A) = \int_{- \infty}^{+ \infty} \frac{x^{2} + \frac{1}{2}}{{(x^{2} + \frac{1}{2})}^{2} + \frac{3}{4}} d x a n d

I m (A) = \frac{\sqrt{3}}{2} \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + \frac{1}{2})}^{2} + \frac{3}{4}} l e t i n t r o d u c e

t h e c o m p l e x f u n c t i o n φ (z) = \frac{1}{z^{2} - j}

φ (z) = \frac{1}{(z - \sqrt{j}) (z + \sqrt{j})} = \frac{1}{(z - e^{i \frac{π}{3}}) (z + e^{i \frac{π}{3}})}

t h e p o l e s o f φ a r e e^{i \frac{π}{3}}, - e^{i \frac{π}{3}}

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

R e s (φ, e^{i \frac{π}{3}}) = \frac{1}{2 e^{i \frac{π}{3}}} = \frac{1}{2} e^{- i \frac{π}{3}}

= \frac{1}{2} (c o s (- \frac{π}{3}) + i s i n (- \frac{π}{3}))

= \frac{1}{2} (\frac{1}{2} - i \frac{\sqrt{3}}{2})

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

= i \frac{π}{2} + π \frac{\sqrt{3}}{2} \Rightarrow R e (A) = π \frac{\sqrt{3}}{2}

a n d I m (A) = \frac{π}{2} .