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

Question Number 39119 by math khazana by abdo last updated on 02/Jul/18

c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{x^{2} c o s (4 x)}{{(x^{2} + 1)}^{2}} d x

Commented by math khazana by abdo last updated on 03/Jul/18

l e t I = \int_{- \infty}^{+ \infty} \frac{x^{2} c o s (4 x)}{{(x^{2} + 1)}^{2}} d x

I = R e (\int_{- \infty}^{+ \infty} \frac{x^{2} e^{i 4 x}}{{(x^{2} + 1)}^{2}}) l e t φ (z) = \frac{z^{2} e^{i 4 z}}{{(z^{2} + 1)}^{2}}

w e h a v e φ (z) = \frac{z^{2} e^{i 4 z}}{{(z - i)}^{2} {(z + i)}^{2}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i) b u t

R e s (φ, i) = {l i m}_{z \to i} {{(z - i)}^{2} φ (z)}^{(1)}

= {l i m}_{z \to i} {\frac{z^{2} e^{i 4 z}}{{(z + i)}^{2}}}^{(1)}

= {l i m}_{z \to i} \frac{(2 z e^{i 4 z} + 4 {i z}^{2} e^{i 4 z}) {(z + i)}^{2} - 2 (z + i) z^{2} e^{i 4 z}}{{(z + i)}^{4}}

= {l i m}_{z \to i} \frac{(2 z + 4 {i z}^{2}) e^{i 4 z} (z + i) - 2 z^{2} e^{i 4 z}}{{(z + i)}^{3}}

= \frac{(2 i - 4 i) e^{- 4} (2 i) + 2 e^{- 4}}{{(2 i)}^{3}} = \frac{4 e^{- 4} + 2 e^{- 4}}{- 8 i}

= \frac{6 e^{- 4}}{- 8 i} = i \frac{3}{4} e^{- 4} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{3 i}{4} e^{- 4}

= - \frac{3 π}{2} e^{- 4} \Rightarrow I = - \frac{3 π}{2} e^{- 4} .