1)calculate I = ∫_0 ^∞ (dx/(x^2 −i)) and J = ∫_0 ^∞ (dx/(x^2 +i)) 2) find the value of ∫


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Question Number 43676 by maxmathsup by imad last updated on 13/Sep/18

$1) c a l c u l a t e I = \int_{0}^{\infty} \frac{d x}{x^{2} - i} a n d J = \int_{0}^{\infty} \frac{d x}{x^{2} + i}$ $2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{d x}{x^{4} + 1}$
Commented by maxmathsup by imad last updated on 13/Sep/18

$i^{2} = - 1$
Commented by maxmathsup by imad last updated on 15/Sep/18
$we have 2I = ∫_(−∞) ^(+∞) (dx/(x^2 −i)) let ϕ(z) = (1/(z^2 −i)) we have ϕ(z) = (1/((z−(√i))(z+(√i)))) = (1/((z−e^((iπ)/4) )(z +e^((iπ)/4) ))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,e^((iπ)/4) ) = 2iπ .(1/(2 e^((iπ)/4) )) =iπ e^(−((iπ)/4)) ⇒ I =((iπ)/2) e^(−((iπ)/4)) also we have 2J =∫_(−∞) ^(+∞) (dx/(x^2 +i)) let ψ(z) =(1/(z^2 +i)) ψ(z) =(1/((z−(√(−i)))(z+(√(−i))))) =(1/((z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) and ∫_(−∞) ^(+∞) ψ(z)dz =2iπ Res(ψ,−e^(−((iπ)/4)) ) =2iπ .(1/(−2 e^(−((iπ)/4)) )) =−iπ e^((iπ)/4) ⇒J=−((iπ)/2)e^((iπ)/4) another way we see that J =I^− ⇒ J = −((iπ)/2) e^((iπ)/4) . 2) we have I −J = ∫_0 ^∞ {(1/(x^2 −i)) −(1/(x^2 +i))}dx =∫_0 ^∞ ((2i)/(x^4 +1))dx ⇒ ∫_0 ^∞ (dx/(x^4 +1)) =(1/(2i)){ I −J} =(1/(2i)){((iπ)/2) e^(−((iπ)/4)) +((iπ)/2) e^((iπ)/4) } = (π/4){ e^((iπ)/4) +e^(−((iπ)/4)) } =(π/4) (2cos((π/4))}=(π/2) .(1/(√2)) =(π/(2(√2))) .$
$w e h a v e 2 I = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - i} l e t φ (z) = \frac{1}{z^{2} - i} w e h a v e$ $φ (z) = \frac{1}{(z - \sqrt{i}) (z + \sqrt{i})} = \frac{1}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}})} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, e^{\frac{i π}{4}}) = 2 i π . \frac{1}{2 e^{\frac{i π}{4}}} = i π e^{- \frac{i π}{4}} \Rightarrow$ $I = \frac{i π}{2} e^{- \frac{i π}{4}} a l s o w e h a v e 2 J = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + i} l e t ψ (z) = \frac{1}{z^{2} + i}$ $ψ (z) = \frac{1}{(z - \sqrt{- i}) (z + \sqrt{- i})} = \frac{1}{(z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})} a n d$ $\int_{- \infty}^{+ \infty} ψ (z) d z = 2 i π R e s (ψ, - e^{- \frac{i π}{4}}) = 2 i π . \frac{1}{- 2 e^{- \frac{i π}{4}}} = - i π e^{\frac{i π}{4}} \Rightarrow J = - \frac{i π}{2} e^{\frac{i π}{4}}$ $a n o t h e r w a y w e s e e t h a t J = \bar{I} \Rightarrow J = - \frac{i π}{2} e^{\frac{i π}{4}} .$ $2) w e h a v e I - J = \int_{0}^{\infty} {\frac{1}{x^{2} - i} - \frac{1}{x^{2} + i}} d x$ $= \int_{0}^{\infty} \frac{2 i}{x^{4} + 1} d x \Rightarrow \int_{0}^{\infty} \frac{d x}{x^{4} + 1} = \frac{1}{2 i} {I - J}$ $= \frac{1}{2 i} {\frac{i π}{2} e^{- \frac{i π}{4}} + \frac{i π}{2} e^{\frac{i π}{4}}} = \frac{π}{4} {e^{\frac{i π}{4}} + e^{- \frac{i π}{4}}} = \frac{π}{4} (2 c o s (\frac{π}{4})} = \frac{π}{2} . \frac{1}{\sqrt{2}} = \frac{π}{2 \sqrt{2}} .$
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