calculate-xsin-2x-1-x-2-2-dx-

Question Number 39033 by maxmathsup by imad last updated on 01/Jul/18

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

Commented by math khazana by abdo last updated on 02/Jul/18

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

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

φ (z) = \frac{z e^{i z}}{{(z - i)}^{2} {(z + i)}^{2}}

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

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

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

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

= {l i m}_{z \to i} \frac{(1 + i z) e^{i x} (z + i) - 2 {z e}^{i z}}{{(z + i)}^{3}}

= \frac{- 2 i e^{- 1}}{{(2 i)}^{3}} = \frac{- 2 i e^{- 1}}{- 8 i} = \frac{e^{- 1}}{4} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{e^{- 1}}{4} = \frac{i π e^{- 1}}{2} \Rightarrow

I = I m (\int_{- \infty}^{+ \infty} φ (z) d z) = \frac{π}{2 e} .

Commented by math khazana by abdo last updated on 02/Jul/18

t h e Q i s f i n d I = \int_{- \infty}^{+ \infty} \frac{x s i n (x)}{{(1 + x^{2})}^{2}} d x