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-

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) 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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