Solve-using-Residue-Theorem-I-x-2-x-4-16-dx-

Question Number 35960 by Joel579 last updated on 26/May/18

Solve using Residue Theorem

I = \int_{- \infty}^{+ \infty} \frac{x^{2}}{x^{4} + 16} d x

Commented by abdo mathsup 649 cc last updated on 26/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^{4} + 16}

φ (z) = \frac{z^{2}}{(z^{2} - 4 i) (z^{2} + 4 i)}

= \frac{z^{2}}{(z - 2 \sqrt{i}) (z + 2 \sqrt{i}) (z - 2 \sqrt{- i}) (z + 2 \sqrt{- i})}

φ (z) = \frac{z^{2}}{(z - 2 e^{i \frac{π}{4}}) (z + 2 e^{i \frac{π}{4}}) (z - 2 e^{- i \frac{π}{4}}) (z + 2 e^{- i \frac{π}{4}})}

t h e p o l e s o f φ a r e 2 e^{i \frac{π}{4}}, - 2 e^{i \frac{π}{4}}, 2 e^{- i \frac{π}{4}}, - 2 e^{- i \frac{π}{4}}

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

R e s (φ, 2 e^{i \frac{π}{4}}) = {l i m}_{z \to 2 e^{i \frac{π}{4}}} (z - 2 e^{i \frac{π}{4}}) φ (z)

= \frac{4 e^{i \frac{π}{2}}}{4 e^{i \frac{π}{4}} (4 i + 4 i)} = \frac{4 i}{4 e^{i \frac{π}{4}} (8 i)} = \frac{1}{8} e^{- i \frac{π}{4}}

R e s (φ, - 2 e^{- i \frac{π}{4}}) = {l i m}_{z \to - 2 e^{- i \frac{π}{4}}} (z + e^{- i \frac{π}{4}}) φ (z)

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

= \frac{- 4 i}{- 4 e^{- i \frac{π}{4}} (- 8 i)} = - \frac{1}{8} e^{i \frac{π}{4}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{1}{8} (e^{- i \frac{π}{4}} - e^{i \frac{π}{4}})}

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

= \frac{π}{2} \frac{\sqrt{2}}{2} = \frac{π \sqrt{2}}{4} s o I = \frac{π \sqrt{2}}{4} .

Commented by Joel579 last updated on 26/May/18

t h a n k y o u v e r y m u c h

Commented by abdo mathsup 649 cc last updated on 26/May/18

n e v e r m i n d s i r j o e l .