calculate-0-x-2-3-x-4-x-2-2-2-dx-

Question Number 65130 by turbo msup by abdo last updated on 25/Jul/19

c a l c u l a t e \int_{0}^{\infty} \frac{x^{2} - 3}{{(x^{4} + x^{2} + 2)}^{2}} d x

Commented by mathmax by abdo last updated on 26/Jul/19

l e t I = \int_{0}^{\infty} \frac{x^{2} - 3}{{(x^{4} + x^{2} + 2)}^{2}} \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{x^{2} - 3}{{(x^{4} + x^{2} + 2)}^{2}} d x l e t

φ (z) = \frac{z^{2} - 3}{{(z^{4} + z^{2} + 2)}^{2}} p o l e s o f φ ?

z^{4} + z^{2} + 2 = 0 \Rightarrow t^{2} + t + 2 = 0 (t = z^{2})

Δ = 1 - 8 = {(i \sqrt{7})}^{2} \Rightarrow t_{1} = \frac{- 1 + i \sqrt{7}}{2} a n d t_{2} = \frac{- 1 - i \sqrt{7}}{2}

∣ t_{1} ∣ = \sqrt{\frac{1}{4} + \frac{7}{4}} = \sqrt{2} \Rightarrow t_{1} = \sqrt{2} {\frac{- 1}{2 \sqrt{2}} + \frac{i \sqrt{7}}{2 \sqrt{2}}} = \sqrt{2} e^{i a r c t a n (- \sqrt{7})}

t_{2} = c o n j (t_{1}) = \sqrt{2} e^{i a r c t a n (\sqrt{7})} \Rightarrow t^{2} + t + 2 = (t - \sqrt{2} e^{- i a r c t a n (\sqrt{7})}) (t - \sqrt{2} e^{i a r c t a n (\sqrt{7})})

= (z^{2} - \sqrt{2} e^{- i a r c t a n (\sqrt{7})}) (z^{2} - \sqrt{2} e^{i a r c t a n (\sqrt{7})})

= (z - \sqrt{\sqrt{2}} e^{- \frac{i}{2} a r c t a n (\sqrt{7})}) (z + \sqrt{\sqrt{2}} e^{- \frac{i}{2} a r c t a n (\sqrt{7})}) (z - \sqrt{\sqrt{2}} e^{\frac{i}{2} a r c t a n (\sqrt{7})}) (z + \sqrt{\sqrt{2}} e^{\frac{i}{2} a r c t a n (\sqrt{7})}) \Rightarrow

φ (z) = \frac{z^{2} - 3}{{(z - {r e}^{- \frac{i}{2} a r c t a n (\sqrt{7})})}^{2} {(z + r e^{- \frac{i}{2} a r c t a n (\sqrt{7})})}^{2} {(z - {r e}^{\frac{i}{2} a r c t a n (\sqrt{7})})}^{2} {(z + r e^{\frac{i}{2} a r c t a n (\sqrt{7})})}^{2}}

t h e p o l e s o f φ a r e \bar{+} r e^{- \frac{i}{2} a r c t a n (\sqrt{7})} a n d \bar{+} r e^{\frac{i}{2} a r c t a n (\sqrt{7})}

w i t h r = \sqrt{\sqrt{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, r e^{\frac{i}{2} a r c t a n (\sqrt{7})}) + R e s (φ, - r e^{- \frac{i}{2} a r c t a n (\sqrt{7})}}

b e c o n t i n u e d \dots .

Commented by mathmax by abdo last updated on 26/Jul/19

R e s (φ, r e^{\frac{i}{2} a r c t a n (\sqrt{7})}) = {l i m}_{z \to r e^{\frac{i}{2} a r c t a n (\sqrt{7})}} \frac{1}{(2 - 1)!} {{(z - r e^{\frac{i}{2} a r c t a n (\sqrt{7})})}^{2} φ (z)}^{(1)}

= {l i m}_{z \to {r e}^{\frac{i}{2} a r c t a n (\sqrt{7})}} {\frac{z^{2} - 3}{{(z + r e^{\frac{i}{2} a r c t a n (\sqrt{7})})}^{2} {(z^{2} - r^{2} e^{- i a r c t a n (\sqrt{7})})}^{2}}}^{(1)}

= {l i m}_{z \to r e^{\frac{i}{2} a r c t a n (\sqrt{7})}} \frac{2 z D - (z^{2} - 3) D^{^{'}}}{D^{2}}

D^{^{'}} = 2 (z + r e^{\frac{i}{2} a r c t a n (\sqrt{7})}) {(z^{2} - r^{2} e^{- i a r c t a n (\sqrt{7})})}^{2}

+ 4 z (z^{2} - r^{2} e^{- i a r c t a n (\sqrt{7})}) {(z + r e^{\frac{i}{2} a r c t a n (\sqrt{7})})}^{2} = \dots .