calculate-0-cos-2x-x-2-1-2-x-2-4-dx-

Question Number 145745 by mathmax by abdo last updated on 07/Jul/21

calculate \int_{0}^{\infty} \frac{\cos (2 x)}{{(x^{2} + 1)}^{2} (x^{2} + 4)} dx

Answered by mathmax by abdo last updated on 11/Jul/21

Ψ = \int_{0}^{\infty} \frac{\cos (2 x)}{{(x^{2} + 1)}^{2} (x^{2} / + 4)} dx \Rightarrow 2 Ψ = Re (\int_{- \infty}^{+ \infty} \frac{e^{2 ix}}{{(x^{2} + 1)}^{2} (x^{2} + 4)} dx)

φ (z) = \frac{e^{2 iz}}{{(z^{2} + 1)}^{2} (z^{2} + 4)} \Rightarrow φ (z) = \frac{e^{2 iz}}{{(z - i)}^{2} {(z + i)}^{2} (z - 2 i) (z + 2 i)}

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

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

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

= \lim_{z \to i} \frac{{2 ie}^{2 iz} {(z + i)}^{2} (z^{2} + 4) - (2 (z + i) (z^{2} + 4) + 2 z {(z + i)}^{2}) e^{2 iz}}{{(z + i)}^{4} {(z^{2} + 4)}^{2}}

= \frac{{2 ie}^{- 2} (- 4) (3) - (4 i (3) + 2 i (- 4)) e^{- 2}}{{(- 4)}^{2} . 3^{2}}

= \frac{- {24 e}^{- 2} - {4 ie}^{- 2}}{16.9}

Res (φ, 2 i) = \lim_{z \to 2 i} (z - 2 i) φ (z) = \lim_{z \to 2 i} \frac{e^{- 4}}{4 i {(- 3)}^{2}} = \frac{e^{- 4}}{36 i} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {\frac{- {24 e}^{- 2} - {4 ie}^{- 2}}{144} + \frac{e^{- 4}}{36 i}}

= 2 i π (- \frac{24}{144} e^{- 2}) + \frac{8 π}{144} e^{- 2} + \frac{π}{18} e^{- 4}

2 Ψ = Re (\dots .) \Rightarrow 2 Ψ = \frac{4 π}{72} e^{- 2} + \frac{π}{18} e^{- 4} \Rightarrow

Ψ = \frac{π}{36} e^{- 2} + \frac{π}{36} e^{- 4}