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

Question Number 102165 by mathmax by abdo last updated on 07/Jul/20

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

Answered by MAB last updated on 07/Jul/20

f (z) = \frac{z^{2}}{{(z^{2} - z + 1)}^{3}} h a s 2 t r i p l e p o l e s :

z_{1} = e^{i \frac{π}{3}}, z_{2} = e^{- i \frac{π}{3}}

I m (z_{2}) < 0 a n d I m (z_{1}) > 0

\int_{- \infty}^{\infty} \frac{x^{2} dx}{{(x^{2} - x + 1)}^{3}} = 2 π i . R e s (f (z), z_{1})

= 2 π i . \frac{1}{(3 - 1)!} {(\frac{z^{2}}{{(z - z_{2})}^{3}})}_{z = z_{1}}^{(2)}

= \frac{2 π}{3 \sqrt{3}}

f i n a l l y

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

Answered by mathmax by abdo last updated on 07/Jul/20

I = \int_{- \infty}^{+ \infty} \frac{x^{2} dx}{{(x^{2} - x + 1)}^{3}} let φ (z) = \frac{z^{2}}{{(z^{2} - z + 1)}^{3}} pole 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} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{3}} \Rightarrow φ (z) = \frac{z^{2}}{{(z - z_{1})}^{3} {(z - z_{2})}^{3}}

the poles of φ are e^{\frac{i π}{3}} and e^{- \frac{i π}{3}} (triples) residus theorem give

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

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}}} {\frac{z^{2}}{{(z - e^{- \frac{i π}{3}})}^{3}}}^{(2)}

= \lim_{z \to e^{\frac{i π}{3}}} {\frac{2 z {(z - e^{- \frac{i π}{3}})}^{3} - 3 {(z - e^{- \frac{i π}{3}})}^{2} z^{2}}{{(z - e^{- \frac{i π}{3}})}^{6}}}^{(1)}

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

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

- \lim_{z \to e^{\frac{i π}{3}}} \frac{(2 z + {2 e}^{- \frac{i π}{3}}) {(z - e^{- \frac{i π}{3}})}^{4} - 4 {(z - e^{- \frac{i π}{3}})}^{3} (z^{2} + 2 z e^{- \frac{i π}{3}})}{{(z - e^{- \frac{i π}{3}})}^{8}}

= - \lim_{z \to e^{\frac{i π}{3}}} \frac{(2 z + {2 e}^{- \frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) - 4 (z^{2} + 2 z e^{- \frac{i π}{3}})}{{(z - e^{- \frac{i π}{3}})}^{5}}

= - \frac{4 \cos (\frac{π}{3}) 2 isin (\frac{π}{3}) - 4 {e^{\frac{2 i π}{3}} + 2}}{{(2 isin (\frac{π}{3}))}^{5}}

= - \frac{8 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} i - 8 - {4 e}^{\frac{i 2 π}{3}}}{32 i {(\frac{\sqrt{3}}{2})}^{5}} = \frac{2 \sqrt{3} i - 8 - 4 (- \frac{1}{2} + \frac{i \sqrt{3}}{2})}{32 i {(\frac{\sqrt{3}}{2})}^{5}}

= \frac{6}{32 i {(\sqrt{3})}^{5}} \times \frac{2^{5}}{\sqrt{3}} = \frac{6}{i {(\sqrt{3})}^{6}} = \frac{6}{3^{3} i} = \frac{2}{9 i} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) dz = 2 i π \times \frac{2}{9 i} = \frac{4 π}{9} \Rightarrow

I = \frac{4 π}{9}

Commented by mathmax by abdo last updated on 07/Jul/20

error at final line Res (φ, e^{\frac{i π}{3}}) = \frac{1}{2 i} \times \frac{6}{32 {(\frac{\sqrt{3}}{2})}^{5} i} = \frac{3 \times 2^{5}}{32 i \times {(\sqrt{3})}^{5}} = \frac{1}{i {(\sqrt{3})}^{3}} = \frac{1}{3 \sqrt{3} i}

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

Commented by MAB last updated on 07/Jul/20

e x a c t s i r

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