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

Question Number 103974 by abdomsup last updated on 18/Jul/20

c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} - x + 1) {(x^{2} + 3)}^{2}}

Answered by OlafThorendsen last updated on 18/Jul/20

R (x) = \frac{1}{(x^{2} - x + 1) {(x^{2} + 3)}^{2}}

R (x) = \frac{x - 2}{7 {(x^{2} + 3)}^{2}} + \frac{5 x - 3}{49 (x^{2} + 3)} - \frac{5 x - 8}{x^{2} - x + 1}

\int_{- \infty}^{+ \infty} R (x) d x =

- \frac{2}{7} \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 3)}^{2}} (I)

- \frac{3}{49} \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + 3} (J)

- \int_{- \infty}^{+ \infty} \frac{5 x - 8}{x^{2} - x + 1} (K)

(we {don}^{'} t keep odd functions)

I = - \frac{2}{7} {[\frac{x}{6 (x^{2} + 3)} + \frac{1}{6 \sqrt{3}} \arctan \frac{x}{\sqrt{3}}]}_{- \infty}^{+ \infty}

I = - \frac{π}{21 \sqrt{3}}

J = - \frac{3}{49} {[\frac{1}{\sqrt{3}} \arctan \frac{x}{\sqrt{3}}]}_{- \infty}^{+ \infty}

J = - \frac{3 π}{49 \sqrt{3}}

K = - {[\frac{11}{\sqrt{3}} \arctan (\frac{1}{\sqrt{3}} - \frac{2 x}{\sqrt{3}}) + \frac{5}{2} \ln (x^{2} - x + 1)]}_{- \infty}^{+ \infty}

K = \frac{11 π}{\sqrt{3}}

\int_{- \infty}^{+ \infty} R (x) d x = I + J + K

= - \frac{π}{21 \sqrt{3}} - \frac{3 π}{49 \sqrt{3}} + \frac{11 π}{\sqrt{3}}

= \frac{1601 π}{147 \sqrt{3}}

sorry, i^{'} m not sure of my result .

Commented by mathmax by abdo last updated on 19/Jul/20

nevermind you do a work thanks sir .

Answered by mathmax by abdo last updated on 19/Jul/20

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

\Rightarrow φ (z) = \frac{1}{(z - e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) {(z - i \sqrt{3})}^{2} {(z + i \sqrt{3})}^{2}} rwsidus theorem give

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

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

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

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

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

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

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

= - {\frac{2 i \sqrt{3} - 1}{{(2 + i \sqrt{3})}^{2} (- 12)} + \frac{2}{(- 2 - i \sqrt{3}) (- 24 i)}} \dots . be continued \dots