calculate-dx-x-2-x-1-2-2x-2-5-2-

Question Number 103593 by mathmax by abdo last updated on 16/Jul/20

calculate \int_{- \infty}^{\infty} \frac{dx}{{(x^{2} + x + 1)}^{2} {({2 x}^{2} + 5)}^{2}}

Answered by mathmax by abdo last updated on 18/Jul/20

A = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} + x + 1)}^{2} {({2 x}^{2} + 5)}^{2}} let φ (z) = \frac{1}{{(z^{2} + z + 1)}^{2} {({2 z}^{2} + 5)}^{2}} poles of φ ?

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

{2 z}^{2} + 5 = 0 \Rightarrow z^{2} + \frac{5}{2} = 0 \Rightarrow z^{2} = - \frac{5}{2} \Rightarrow z = \bar{+} i \sqrt{\frac{5}{2}}

\Rightarrow φ (z) = \frac{1}{4 {(z - e^{\frac{i 2 π}{3}})}^{2} {(z + e^{\frac{i 2 π}{3}})}^{2} {(z - i \sqrt{\frac{5}{2}})}^{2} {(z + i \sqrt{\frac{5}{2}})}^{2}}

residus theorem give

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, e^{\frac{i 2 π}{3}}) + Res (φ, i \sqrt{\frac{5}{2}})}

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

= \lim_{z \to e \frac{i 2 π}{3}} {\frac{1}{4 {(z + e^{\frac{i 2 π}{3}})}^{2} {(z^{2} + \frac{5}{2})}^{2}}}^{(1)}

= \lim_{z \to e^{\frac{i 2 π}{3}}} - \frac{1}{4} \times \frac{2 (z + e^{\frac{i 2 π}{3}}) {(z^{2} + \frac{5}{2})}^{2} + 4 z (z^{2} + \frac{5}{2}) {(z + e^{\frac{i 2 π}{3}})}^{2}}{{(z + e^{\frac{i 2 π}{3}})}^{4} {(z^{2} + \frac{5}{2})}^{4}}

= - \frac{1}{4} \lim_{z \to e^{\frac{i 2 π}{3}}} {\frac{2}{{(z + e^{\frac{i 2 π}{3}})}^{3} {(z^{2} + \frac{5}{2})}^{2}} + \frac{4 z}{{(z + e^{\frac{i 2 π}{3}})}^{2} {(z^{2} + \frac{5}{2})}^{3}}}

\dots be continued \dots