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


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 56629 by maxmathsup by imad last updated on 19/Mar/19

$1) c a l c u l a t e I = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - i} a n d J = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - i}$ $2) f i n d t h e v a l u e o f \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + 1}$
Commented by maxmathsup by imad last updated on 19/Mar/19

$J = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + i}$
Commented by maxmathsup by imad last updated on 20/Mar/19
$1) let ϕ(z)=(1/(z^2 −i)) ⇒ ϕ(z)= (1/((z−(√i))(z+(√i)))) =(1/((z−e^((iπ)/4) )(z+e^((iπ)/4) ))) so the poles of ϕ are +^− e^((iπ)/4) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz=2iπRes(ϕ,e^((iπ)/4) ) Res(ϕ, e^((iπ)/4) )=lim_(z→e^((iπ)/4) ) (z−e^((iπ)/4) )ϕ(z)=(1/(2 e^((iπ)/4) )) =(1/2) e^(−((iπ)/4)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ(1/2) e^(−((iπ)/4)) =iπ e^(−((iπ)/4)) ⇒ I =iπ e^(−((iπ)/4)) let calculate J =∫_(−∞) ^(+∞) (dx/(x^2 +i)) let W(z)=(1/(z^2 +i)) ⇒W(z)=(1/((z−e^(−((iπ)/4)) )(z +e^(−((iπ)/4)) ))) so the poles of ϕ are +^− e^(−((iπ)/4)) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,−e^(−((iπ)/4)) ) Res(W,−e^(−((iπ)/4)) ) =lim_(z→−e^(−((iπ)/4)) ) (z+e^(−((iπ)/4)) )W(z) =(1/(−2 e^(−((iπ)/4)) )) =−(1/2) e^((iπ)/4) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ (−(1/2) e^((iπ)/4) ) =−iπ e^((iπ)/4) =J also we can use that J =conj(I) =conj(iπ e^(−((iπ)/4)) )=−iπ e^((iπ)/4) 2) ∫_(−∞) ^(+∞) (dx/(x^4 +1)) =∫_(−∞) ^(+∞) (dx/((x^2 −i)(x^2 +i))) =(1/(2i))∫_(−∞) ^(+∞) {(1/(x^2 −i)) −(1/(x^2 +i))}dx =(1/(2i)){ I−J} =(1/(2i)){iπ e^(−((iπ)/4)) +i π e^((iπ)/4) } =(π/2) (2cos((π/4)))=π (1/(√2)) =(π/(√2)) .$
$1) l e t φ (z) = \frac{1}{z^{2} - i} \Rightarrow φ (z) = \frac{1}{(z - \sqrt{i}) (z + \sqrt{i})} = \frac{1}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}})} s o t h e p o l e s o f$ $φ a r e \bar{+} e^{\frac{i π}{4}} r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, e^{\frac{i π}{4}})$ $R e s (φ, e^{\frac{i π}{4}}) = {l i m}_{z \to e^{\frac{i π}{4}}} (z - e^{\frac{i π}{4}}) φ (z) = \frac{1}{2 e^{\frac{i π}{4}}} = \frac{1}{2} e^{- \frac{i π}{4}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{1}{2} e^{- \frac{i π}{4}}$ $= i π e^{- \frac{i π}{4}} \Rightarrow I = i π e^{- \frac{i π}{4}} l e t c a l c u l a t e J = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + i}$ $l e t W (z) = \frac{1}{z^{2} + i} \Rightarrow W (z) = \frac{1}{(z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})} s o t h e p o l e s o f φ a r e \bar{+} e^{- \frac{i π}{4}}$ $r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, - e^{- \frac{i π}{4}})$ $R e s (W, - e^{- \frac{i π}{4}}) = {l i m}_{z \to - e^{- \frac{i π}{4}}} (z + e^{- \frac{i π}{4}}) W (z) = \frac{1}{- 2 e^{- \frac{i π}{4}}} = - \frac{1}{2} e^{\frac{i π}{4}} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π (- \frac{1}{2} e^{\frac{i π}{4}}) = - i π e^{\frac{i π}{4}} = J a l s o w e c a n u s e t h a t$ $J = c o n j (I) = c o n j (i π e^{- \frac{i π}{4}}) = - i π e^{\frac{i π}{4}}$ $2) \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + 1} = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} - i) (x^{2} + i)} = \frac{1}{2 i} \int_{- \infty}^{+ \infty} {\frac{1}{x^{2} - i} - \frac{1}{x^{2} + i}} d x$ $= \frac{1}{2 i} {I - J} = \frac{1}{2 i} {i π e^{- \frac{i π}{4}} + i π e^{\frac{i π}{4}}} = \frac{π}{2} (2 c o s (\frac{π}{4})) = π \frac{1}{\sqrt{2}} = \frac{π}{\sqrt{2}} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum