find the value of ∫_(−∞) ^(+∞) ((x sin(2x))/((1+4x^2 )^2 )) dx .


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Question Number 33232 by prof Abdo imad last updated on 13/Apr/18

$f i n d t h e v a l u e o f \int_{- \infty}^{+ \infty} \frac{x s i n (2 x)}{{(1 + 4 x^{2})}^{2}} d x .$
Commented by prof Abdo imad last updated on 15/Apr/18

$l e t p u t I = \int_{- \infty}^{+ \infty} \frac{x s i n x}{{(1 + 4 x^{2})}^{2}} d x$ $I = - \frac{1}{8} \int_{- \infty}^{+ \infty} (\frac{- 8 x}{{(1 + 4 x^{2})}^{2}}) s i n (2 x) d x b y p s r t s$ $u^{^{'}} = \frac{- 8 x}{(1 + 4 x^{2})} a n d v (x) = s i n (2 x) \Rightarrow$ $- 8 I = {[\frac{1}{(1 + 4 x^{2})} s i n (2 x)]}_{- \infty}^{+ \infty} - \int_{- \infty}^{+ \infty} \frac{1}{1 + 4 x^{2}} 2 c o s (2 x) d x$ $= - 2 \int_{- \infty}^{+ \infty} \frac{c o s (2 x)}{1 + 4 x^{2}} d x \Rightarrow I = \frac{1}{4} \int_{- \infty}^{+ \infty} \frac{c o s (2 x)}{1 + 4 x^{2}} d x$ $\int_{- \infty}^{+ \infty} \frac{c o s (2 x)}{1 + 4 x^{2}} d x = R e (\int_{- \infty}^{+ \infty} \frac{e^{i 2 x}}{1 + 4 x^{2}} d x)$ $l e t i n t r o d u c e t h e c o m p l e x f i n c t i o n$ $φ (z) = \frac{e^{2 i z}}{1 + 4 z^{2}} w e h a v e φ (z) = \frac{e^{2 i z}}{(2 z - i) (2 z + i)}$ $= \frac{e^{2 i z}}{4 (z - \frac{i}{2}) (z + \frac{i}{2})} s o t h e p o l e s o f φ a r e \frac{i}{2} a n d$ $- \frac{i}{2} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, \frac{i}{2})$ $R e s (φ, \frac{i}{2}) = {l i m}_{z \to \frac{i}{2}} (z - \frac{i}{2}) φ (z)$ $= \frac{e^{2 i \frac{i}{2}}}{4 (2 \frac{i}{2})} = \frac{e^{- 1}}{4 i} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π . \frac{e^{- 1}}{4 i}$ $= \frac{π}{2 e} \Rightarrow I = \frac{1}{4} \frac{π}{2 e} \Rightarrow I = \frac{π}{8 e} .$
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