let A =∫_(−∞) ^(+∞) ((x+1)/((x^2 +x+1)( x^2 −2i)))dx 1) calculate A 2) extract Re(A) and Im(A) and determine its values (i^2 =−1)


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Question Number 61921 by maxmathsup by imad last updated on 11/Jun/19

$l e t A = \int_{- \infty}^{+ \infty} \frac{x + 1}{(x^{2} + x + 1) (x^{2} - 2 i)} d x$ $1) c a l c u l a t e A$ $2) e x t r a c t R e (A) a n d I m (A) a n d d e t e r m i n e i t s v a l u e s (i^{2} = - 1)$
Commented by maxmathsup by imad last updated on 12/Jun/19
$1) residus method let w(z) =((z+1)/((z^2 +z +1)(z^2 −2i))) poles of w? z^2 +z +1 =0 →Δ=1−4 =−3 =(i(√3))^2 ⇒z_1 =((−1+i(√3))/2) =e^(i((2π)/3)) z_2 =((−1−i(√3))/2) =e^(−i((2π)/3)) also z^2 −2i =z^2 −((√2)(√i))^2 =(z−(√2)e^((iπ)/4) )(z+(√2)e^((iπ)/4) ) ⇒ w(z) =((z+1)/((z−e^(i((2π)/3)) )(z+e^((i2π)/3) )(z−(√2)e^((iπ)/4) )(z+(√2)e^((iπ)/4) ))) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ{ Res(w,e^(i((2π)/3)) ) +Res(w,(√2)e^((iπ)/4) )} the poles are simples ⇒ Res(w,e^((i2π)/3) ) =lim_(z→e^((i2π)/3) ) (z−e^((i2π)/3) )w(z) =((1+e^((i2π)/3) )/(2e^((i2π)/3) ( e^(i((4π)/3)) −2i))) =((e^(−((i2π)/3)) +1)/(2( e^(−((i2π)/3)) −2i))) =−((1+e^(−((i2π)/3)) )/(2i(ie^(−((i2π)/3)) +2))) Res(w,(√2)e^((iπ)/4) ) =lim_(z→(√2)e^((iπ)/4) ) (z−(√2)e^((iπ)/4) ) w(z) =((1+(√2)e^((iπ)/4) )/(2(√2)e^((iπ)/4) ( 2 e^((iπ)/2) +(√2)e^((iπ)/4) +1))) =((e^(−((iπ)/4)) +(√2))/(2(√2)(2i +(√2)e^((iπ)/4) +1))) =((e^(−((iπ)/4)) +(√2))/(2(√2)i(2 −i(√2)e^((iπ)/4) −i))) ⇒ ∫_(−∞) ^(+∞) w(z)dz =2iπ{ −((1+e^(−((i2π)/3)) )/(2i(i e^(−((i2π)/3)) +2))) +((e^(−((iπ)/4)) +(√2))/(2(√2)i(2−i−i(√2)e^((iπ)/4) ))} =−π ((1+e^(−i((2π)/3)) )/(2+ie^(−((i2π)/3)) )) +(π/(√2)) (((√2)+e^(−((iπ)/4)) )/(2−i −i(√2)e^((iπ)/4) )) =A rest to find A at form α+iλ$
$1) r e s i d u s m e t h o d l e t w (z) = \frac{z + 1}{(z^{2} + z + 1) (z^{2} - 2 i)} p o l e s o f w ?$ $z^{2} + z + 1 = 0 \to Δ = 1 - 4 = - 3 = {(i \sqrt{3})}^{2} \Rightarrow z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{i \frac{2 π}{3}}$ $z_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- i \frac{2 π}{3}} a l s o z^{2} - 2 i = z^{2} - {(\sqrt{2} \sqrt{i})}^{2} = (z - \sqrt{2} e^{\frac{i π}{4}}) (z + \sqrt{2} e^{\frac{i π}{4}}) \Rightarrow$ $w (z) = \frac{z + 1}{(z - e^{i \frac{2 π}{3}}) (z + e^{\frac{i 2 π}{3}}) (z - \sqrt{2} e^{\frac{i π}{4}}) (z + \sqrt{2} e^{\frac{i π}{4}})} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} w (z) d z = 2 i π {R e s (w, e^{i \frac{2 π}{3}}) + R e s (w, \sqrt{2} e^{\frac{i π}{4}})} t h e p o l e s a r e s i m p l e s \Rightarrow$ $R e s (w, e^{\frac{i 2 π}{3}}) = {l i m}_{z \to e^{\frac{i 2 π}{3}}} (z - e^{\frac{i 2 π}{3}}) w (z) = \frac{1 + e^{\frac{i 2 π}{3}}}{2 e^{\frac{i 2 π}{3}} (e^{i \frac{4 π}{3}} - 2 i)}$ $= \frac{e^{- \frac{i 2 π}{3}} + 1}{2 (e^{- \frac{i 2 π}{3}} - 2 i)} = - \frac{1 + e^{- \frac{i 2 π}{3}}}{2 i ({i e}^{- \frac{i 2 π}{3}} + 2)}$ $R e s (w, \sqrt{2} e^{\frac{i π}{4}}) = {l i m}_{z \to \sqrt{2} e^{\frac{i π}{4}}} (z - \sqrt{2} e^{\frac{i π}{4}}) w (z) = \frac{1 + \sqrt{2} e^{\frac{i π}{4}}}{2 \sqrt{2} e^{\frac{i π}{4}} (2 e^{\frac{i π}{2}} + \sqrt{2} e^{\frac{i π}{4}} + 1)}$ $= \frac{e^{- \frac{i π}{4}} + \sqrt{2}}{2 \sqrt{2} (2 i + \sqrt{2} e^{\frac{i π}{4}} + 1)} = \frac{e^{- \frac{i π}{4}} + \sqrt{2}}{2 \sqrt{2} i (2 - i \sqrt{2} e^{\frac{i π}{4}} - i)} \Rightarrow$ $\int_{- \infty}^{+ \infty} w (z) d z = 2 i π {- \frac{1 + e^{- \frac{i 2 π}{3}}}{2 i (i e^{- \frac{i 2 π}{3}} + 2)} + \frac{e^{- \frac{i π}{4}} + \sqrt{2}}{2 \sqrt{2} i (2 - i - i \sqrt{2} e^{\frac{i π}{4}}}}$ $= - π \frac{1 + e^{- i \frac{2 π}{3}}}{2 + {i e}^{- \frac{i 2 π}{3}}} + \frac{π}{\sqrt{2}} \frac{\sqrt{2} + e^{- \frac{i π}{4}}}{2 - i - i \sqrt{2} e^{\frac{i π}{4}}} = A r e s t t o f i n d A a t f o r m α + i λ$
	Answered by MJS last updated on 12/Jun/19

	$\frac{x + 1}{(x^{2} + x + 1) (x^{2} - 2 i)} = \frac{x^{2} (x + 1)}{(x^{2} + x + 1) (x^{4} + 4)} + \frac{2 (x + 1)}{(x^{2} + x + 1) (x^{4} + 4)} i$ $A = \underset{- \infty}{\int^{+ \infty}} \frac{x^{2} (x + 1)}{(x^{2} + x + 1) (x^{4} + 4)} d x + i \underset{- \infty}{\int^{+ \infty}} \frac{2 (x + 1)}{(x^{2} + x + 1) (x^{4} + 4)} d x$ $\int \frac{x^{2} (x + 1)}{(x^{2} + x + 1) (x^{4} + 4)} d x = \int \frac{x^{2} (x + 1)}{(x^{2} + x + 1) (x^{2} - 2 x + 2) (x^{2} + 2 x + 2)} d x =$ $= - \frac{1}{13} \int \frac{4 x + 1}{x^{2} + x + 1} d x + \frac{1}{52} \int \frac{3 x + 8}{x^{2} - 2 x + 2} d x + \frac{1}{4} \int \frac{x}{x^{2} + 2 x + 2} d x =$ $[all are easy to solve I hope so I allow myself$ $to not type the paths]$ $= - \frac{2}{13} \ln (x^{2} + x + 1) + \frac{2 \sqrt{3}}{39} \arctan \frac{\sqrt{3} (2 x + 1)}{3} +$ $+ \frac{3}{104} \ln (x^{2} - 2 x + 2) + \frac{11}{52} \arctan (x - 1) +$ $+ \frac{1}{8} \ln (x^{2} + 2 x + 2) - \frac{1}{4} \arctan (x + 1) + C$ $\Rightarrow real (A) = \frac{π}{78} (- 3 + 4 \sqrt{3})$ $\int \frac{2 (x + 1)}{(x^{2} + x + 1) (x^{4} + 4)} d x = \int \frac{2 (x + 1)}{(x^{2} + x + 1) (x^{2} - 2 x + 2) (x^{2} + 2 x + 2)} d x =$ $= \frac{2}{13} \int \frac{3 x + 4}{x^{2} + x + 1} d x - \frac{1}{52} \int \frac{11 x - 14}{x^{2} - 2 x + 2} d x - \frac{1}{4} \int \frac{x + 2}{x^{2} + 2 x + 2} d x =$ $= \frac{3}{12} \ln (x^{2} + x + 1) + \frac{10 \sqrt{3}}{39} \arctan \frac{\sqrt{3} (2 x + 1)}{3} -$ $- \frac{11}{104} \ln (x^{2} - 2 x + 2) + \frac{3}{52} \arctan (x - 1) -$ $- \frac{1}{8} \ln (x^{2} + 2 x + 2) - \frac{1}{4} \arctan (x + 1) + C$ $\Rightarrow imag (A) = \frac{5 π}{78} (- 3 + 4 \sqrt{3})$ $A = \frac{π}{78} (- 3 + 4 \sqrt{3}) (1 + 5 i)$
	Commented by maxmathsup by imad last updated on 12/Jun/19

	$t h a n k y o u s i r m j s .$
	Commented by MJS last updated on 12/Jun/19

	${you}^{'} re welcome$
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