calculate-dx-x-2-x-1-x-2-x-1-

Question Number 67673 by Abdo msup. last updated on 30/Aug/19

c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} - x + 1) (x^{2} + x + 1)}

Commented by Abdo msup. last updated on 30/Aug/19

l e t A = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} - x + 1) (x^{2} + x + 1)}

a n d φ (z) = \frac{1}{(z^{2} - z + 1) (z^{2} + z + 1)} p o l e s o f φ ?

z^{2} - z + 1 = 0 \to Δ = - 3 = {(i \sqrt{3})}^{2} \Rightarrow

z_{1} = \frac{1 + i \sqrt{3}}{2} a n d z_{2} = \frac{1 - i \sqrt{3}}{2}

z^{2} + z + 1 = 0 \to Δ = - 3 = {(i \sqrt{3})}^{2} \Rightarrow

α_{1} = \frac{- 1 + i \sqrt{3}}{2} a n d α_{2} = \frac{- 1 - i \sqrt{3}}{2}

φ (z) = \frac{1}{(x - z_{1}) (x - z_{2}) (x - α_{1}) (x - α_{2})}

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, z_{1}) + R e s (φ, α_{1})}

R e s (φ, z_{1}) = \frac{1}{(z_{1} - z_{2}) (z_{1}^{2} + z_{1} + 1)}

= \frac{1}{i \sqrt{3} (z_{1}^{2} + z_{1} + 1)}

R e s (φ, α_{1}) = \frac{1}{(α_{1} - α_{2}) (α_{1}^{2} - α_{1} + 1)} = \frac{1}{i \sqrt{3} (α_{1}^{2} - α_{1} + 1)}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{1}{i \sqrt{3} (z_{1}^{2} + z_{1} + 1)} + \frac{1}{i \sqrt{3} (α_{1}^{2} - α_{1} + 1)}}

= \frac{2 π}{\sqrt{3}} {\frac{1}{z_{1}^{2} + z_{1} + 1} + \frac{1}{α_{1}^{2} - α_{1} + 1}}