Solve using Residue Theorem I = ∫_(−∞) ^(+∞) (x^2 /(x^4 + 16)) dx


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
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
$let consider the complex functionϕ(z)=(z^2 /(z^4 +16)) ϕ(z)= (z^2 /((z^2 −4i)(z^2 +4i))) = (z^2 /((z −2(√i))(z +2(√i))( z −2(√(−i)))(z+2(√(−i))))) ϕ(z) = (z^2 /((z −2e^(i(π/4)) )(z+2e^(i(π/4)) )(z −2 e^(−i(π/4)) )( z+2 e^(−i(π/4)) ))) the poles of ϕ are 2 e^(i(π/4)) , −2 e^(i(π/4)) , 2 e^(−i(π/4)) , −2 e^(−i(π/4)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res( ϕ, 2 e^(i(π/4)) ) +Res( ϕ,−2e^(−i(π/4)) )} Res(ϕ, 2 e^(i(π/4)) ) = lim_(z→2e^(i(π/4)) ) (z −2e^(i(π/4)) )ϕ(z) = ((4 e^(i(π/2)) )/(4 e^(i(π/4)) ( 4i +4i))) = ((4i)/(4 e^(i(π/4)) (8i))) = (1/8)e^(−i(π/4)) Res(ϕ ,−2 e^(−i(π/4)) ) = lim_(z→ −2e^(−i(π/4)) ) (z+e^(−i(π/4)) )ϕ(z) = ((4 e^(−i(π/2)) )/(−4 e^(−i(π/4)) ( 4 e^(−i(π/4)) −4i))) = ((−4i)/(−4 e^(−i(π/4)) (−8i))) = −(1/8) e^(i(π/4)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (1/8)( e^(−i(π/4)) −e^(i(π/4)) )} =((−iπ)/4){ e^(i(π/4)) −e^(−i(π/4)) } = ((−iπ)/4){2isin((π/4))} = (π/2) ((√2)/2) = ((π(√2))/4) so I = ((π(√2))/4) .$
$l e t c o n s i d e r t h e c o m p l e x f u n c t i o n φ (z) = \frac{z^{2}}{z^{4} + 16}$ $φ (z) = \frac{z^{2}}{(z^{2} - 4 i) (z^{2} + 4 i)}$ $= \frac{z^{2}}{(z - 2 \sqrt{i}) (z + 2 \sqrt{i}) (z - 2 \sqrt{- i}) (z + 2 \sqrt{- i})}$ $φ (z) = \frac{z^{2}}{(z - 2 e^{i \frac{π}{4}}) (z + 2 e^{i \frac{π}{4}}) (z - 2 e^{- i \frac{π}{4}}) (z + 2 e^{- i \frac{π}{4}})}$ $t h e p o l e s o f φ a r e 2 e^{i \frac{π}{4}}, - 2 e^{i \frac{π}{4}}, 2 e^{- i \frac{π}{4}}, - 2 e^{- i \frac{π}{4}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, 2 e^{i \frac{π}{4}}) + R e s (φ, - 2 e^{- i \frac{π}{4}})}$ $R e s (φ, 2 e^{i \frac{π}{4}}) = {l i m}_{z \to 2 e^{i \frac{π}{4}}} (z - 2 e^{i \frac{π}{4}}) φ (z)$ $= \frac{4 e^{i \frac{π}{2}}}{4 e^{i \frac{π}{4}} (4 i + 4 i)} = \frac{4 i}{4 e^{i \frac{π}{4}} (8 i)} = \frac{1}{8} e^{- i \frac{π}{4}}$ $R e s (φ, - 2 e^{- i \frac{π}{4}}) = {l i m}_{z \to - 2 e^{- i \frac{π}{4}}} (z + e^{- i \frac{π}{4}}) φ (z)$ $= \frac{4 e^{- i \frac{π}{2}}}{- 4 e^{- i \frac{π}{4}} (4 e^{- i \frac{π}{4}} - 4 i)}$ $= \frac{- 4 i}{- 4 e^{- i \frac{π}{4}} (- 8 i)} = - \frac{1}{8} e^{i \frac{π}{4}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{1}{8} (e^{- i \frac{π}{4}} - e^{i \frac{π}{4}})}$ $= \frac{- i π}{4} {e^{i \frac{π}{4}} - e^{- i \frac{π}{4}}} = \frac{- i π}{4} {2 i s i n (\frac{π}{4})}$ $= \frac{π}{2} \frac{\sqrt{2}}{2} = \frac{π \sqrt{2}}{4} s o I = \frac{π \sqrt{2}}{4} .$
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

$n e v e r m i n d s i r j o e l .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum