calculate-dx-x-2-x-1-3-

Question Number 146197 by mathmax by abdo last updated on 11/Jul/21

calculate \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} - x + 1)}^{3}}

Answered by Olaf_Thorendsen last updated on 12/Jul/21

Ω = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} - x + 1)}^{3}}

f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - x + a}, a > \frac{1}{4} (1)

f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{{(x - \frac{1}{2})}^{2} + a - \frac{1}{4}}

f (a) = \frac{1}{\sqrt{a - \frac{1}{4}}} {[\arctan (\frac{x - \frac{1}{2}}{\sqrt{a - \frac{1}{4}}})]}_{- \infty}^{+ \infty}

f (a) = \frac{π}{\sqrt{a - \frac{1}{4}}} = π {(a - \frac{1}{4})}^{- 1 / 2} (2)

(1) : f^{'} (a) = - \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} - x + a)}^{2}}

f^{″} (a) = 2 \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} - x + a)}^{3}}

\Rightarrow Ω = \frac{1}{2} f^{″} (1) (3)

(2) : f^{'} (a) = - \frac{π}{2} {(a - \frac{1}{4})}^{- 3 / 2}

f^{″} (a) = \frac{3 π}{4} {(a - \frac{1}{4})}^{- 5 / 2}

(3) : Ω = \frac{1}{2} \times \frac{3 π}{4} {(1 - \frac{1}{4})}^{- 5 / 2}

Ω = \frac{4 π}{3 \sqrt{3}}

Answered by mathmax by abdo last updated on 12/Jul/21

Υ = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} - x + 1)}^{3}} we considere φ (z) = \frac{1}{{(z^{2} - z + 1)}^{3}} poles of φ

z^{2} - z + 1 = 0 \to Δ = - 3 \Rightarrow z_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}} and z_{2} = e^{- \frac{i π}{3}}

\Rightarrow φ (z) = \frac{1}{{(z - e^{\frac{i π}{3}})}^{3} {(z - e^{- \frac{i π}{3}})}^{3}} residus theorem

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, e^{\frac{i π}{3}})

Res (φ, e^{\frac{i π}{3}}) = \lim_{z \to e^{\frac{i π}{3}}} \frac{1}{(3 - 1)!} {{(z - e^{\frac{i π}{3}})}^{3} φ (z)}^{(2)}

= \frac{1}{2} \lim_{z \to e^{\frac{i π}{3}}} {{(z - e^{- \frac{i π}{3}})}^{- 3}}^{(2)}

= \frac{1}{2} \lim_{z \to e^{\frac{i π}{3}}} {- 3 {(z - e^{- \frac{i π}{3}})}^{- 4}}^{(1)}

= \frac{1}{2} \lim_{z \to e^{\frac{i π}{3}}} 12 {(z - e^{- \frac{i π}{3}})}^{- 5}

= 6 {(e^{\frac{i π}{3}} - e^{- \frac{i π}{3}})}^{- 5} = 6 {(2 i \times \frac{\sqrt{3}}{2})}^{- 5} = \frac{6}{i^{5} {(\sqrt{3})}^{5}} = \frac{6}{i 9 \sqrt{3}} \frac{2}{3 i \sqrt{3}}

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π \times \frac{2}{3 i \sqrt{3}} = \frac{4 π}{3 \sqrt{3}} \Rightarrow Υ = \frac{4 π}{3 \sqrt{3}}