calculate ∫_0 ^∞ (dx/((x^2 +2i)( x^2 +4j))) with i=e^((iπ)/2) and j=e^(i((2π)/3))


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Question Number 66465 by mathmax by abdo last updated on 15/Aug/19

$c a l c u l a t e \int_{0}^{\infty} \frac{d x}{(x^{2} + 2 i) (x^{2} + 4 j)} w i t h i = e^{\frac{i π}{2}} a n d j = e^{i \frac{2 π}{3}}$
Commented by mathmax by abdo last updated on 16/Aug/19
$let A =∫_0 ^∞ (dx/((x^2 +2i)(x^2 +4j))) ⇒2A =∫_(−∞) ^(+∞) (dx/((x^2 +2i)(x^2 +4j))) let ϕ(z) =(1/((z^2 +2i)(z^2 +4j))) poles of ϕ? we have ϕ(z) =(1/((z^2 −(−2i)(z^2 −(−4j)))) =(1/((z−(√(−2i)))(z+(√(−2i)))(z−2(√(−j)))(z+2(√(−j))))) −j =e^(iπ) .e^(i((2π)/3)) =e^(i(π+((2π)/3))) =e^(i(((5π)/3))) =e^(−((iπ)/3)) ⇒(√(−j))=e^(−((iπ)/6)) ⇒ ϕ(z) =(1/((z−(√2)e^(−((iπ)/4)) )(z+(√2)e^(−((iπ)/4)) )(z−2e^(−((iπ)/6)) )(z+2 e^(−((iπ)/6)) ))) the poles of ϕ are +^− (√2)e^(−((iπ)/4)) and +^− 2e^(−((iπ)/6)) residus tbeorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,−(√2)e^(−((iπ)/4)) )+Res(ϕ,−2e^(−((iπ)/6)) )} Res(ϕ,−(√2)e^(−((iπ)/4)) ) =(1/(−2(√2)e^(−((iπ)/4)) (2 e^(−((iπ)/2)) +4j))) =−(e^((iπ)/4) /(4(√2)(−i +4j))) Res(ϕ,−2e^(−((iπ)/6)) ) =(1/(−4e^(−((iπ)/6)) (e^(−((iπ)/3)) +2i))) =−(e^(i(π/6)) /(4(2i +e^(−((iπ)/3)) ))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =−((iπ)/2){ (e^((iπ)/4) /((√2)(−i+4j))) +(e^((iπ)/6) /(2i+e^(−((iπ)/3)) ))} =2A$
$l e t A = \int_{0}^{\infty} \frac{d x}{(x^{2} + 2 i) (x^{2} + 4 j)} \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 2 i) (x^{2} + 4 j)}$ $l e t φ (z) = \frac{1}{(z^{2} + 2 i) (z^{2} + 4 j)} p o l e s o f φ ?$ $w e h a v e φ (z) = \frac{1}{(z^{2} - (- 2 i) (z^{2} - (- 4 j))}$ $= \frac{1}{(z - \sqrt{- 2 i}) (z + \sqrt{- 2 i}) (z - 2 \sqrt{- j}) (z + 2 \sqrt{- j})}$ $- j = e^{i π} . e^{i \frac{2 π}{3}} = e^{i (π + \frac{2 π}{3})} = e^{i (\frac{5 π}{3})} = e^{- \frac{i π}{3}} \Rightarrow \sqrt{- j} = e^{- \frac{i π}{6}} \Rightarrow$ $φ (z) = \frac{1}{(z - \sqrt{2} e^{- \frac{i π}{4}}) (z + \sqrt{2} e^{- \frac{i π}{4}}) (z - 2 e^{- \frac{i π}{6}}) (z + 2 e^{- \frac{i π}{6}})}$ $t h e p o l e s o f φ a r e \bar{+} \sqrt{2} e^{- \frac{i π}{4}} a n d \bar{+} 2 e^{- \frac{i π}{6}} r e s i d u s t b e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, - \sqrt{2} e^{- \frac{i π}{4}}) + R e s (φ, - 2 e^{- \frac{i π}{6}})}$ $R e s (φ, - \sqrt{2} e^{- \frac{i π}{4}}) = \frac{1}{- 2 \sqrt{2} e^{- \frac{i π}{4}} (2 e^{- \frac{i π}{2}} + 4 j)} = - \frac{e^{\frac{i π}{4}}}{4 \sqrt{2} (- i + 4 j)}$ $R e s (φ, - 2 e^{- \frac{i π}{6}}) = \frac{1}{- 4 e^{- \frac{i π}{6}} (e^{- \frac{i π}{3}} + 2 i)} = - \frac{e^{i \frac{π}{6}}}{4 (2 i + e^{- \frac{i π}{3}})} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = - \frac{i π}{2} {\frac{e^{\frac{i π}{4}}}{\sqrt{2} (- i + 4 j)} + \frac{e^{\frac{i π}{6}}}{2 i + e^{- \frac{i π}{3}}}} = 2 A$
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