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Question Number 117088 by mnjuly1970 last updated on 09/Oct/20

$. . . a d v a n c e d c a l c u l u s . . .$ $e v s l u a t e ::$ $I = \int_{0}^{\infty} \frac{x s i n (2 x)}{x^{2} + 4} d x = ?$ $h i n t :$ $Φ = \int_{0}^{\infty} \frac{c o s (p x)}{x^{2} + 1} d x = \frac{π}{2} e^{- p} (p > 0)$ $. m . n 1970$
	Answered by mnjuly1970 last updated on 09/Oct/20

	$Φ = \int_{0}^{\infty} \frac{c o s (p x)}{x^{2} + 1} \underset{r e s p e c t t o p}{\overset{(d i f f b o t h s i d e s)}{=}} \frac{π}{2} e^{- p}$ $= - \int_{0}^{\infty} \frac{x s i n (p x)}{x^{2} + 1} d x = - \frac{π}{2} e^{- p}$ $= \int_{0}^{\infty} \frac{x s i n (p x)}{x^{2} + 1} d x = \frac{π}{2} e^{p}$ $I \overset{x = 2 t}{=} \int_{0}^{\infty} \frac{t s i n (4 t)}{t^{2} + 1} d t \overset{Φ}{=} \frac{π}{2} e^{- 4}$ $. . m . n . 1970 . . .$
	Answered by Bird last updated on 10/Oct/20
	$2I =∫_(−∞) ^(+∞ ) ((xsin(2x))/(x^(2 ) +4))dx =Im(∫_(−∞) ^(+∞) ((xe^(2ix) )/(x^2 +4))dx) let ϕ(z) =((z e^(2iz) )/(z^2 +4)) ⇒ϕ(z) =((ze^(2iz) )/((z−2i)(z+2i))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{Res(ϕ,2i)} =2iπ ×((2i e^(−4) )/(4i)) =((−4π e^(−4) )/(4i)) =iπ e^(−4) ⇒2I =π e^(−4) ⇒I =(π/2)e^(−4)$
	$2 I = \int_{- \infty}^{+ \infty} \frac{x s i n (2 x)}{x^{2} + 4} d x$ $= I m (\int_{- \infty}^{+ \infty} \frac{{x e}^{2 i x}}{x^{2} + 4} d x) l e t$ $φ (z) = \frac{z e^{2 i z}}{z^{2} + 4} \Rightarrow φ (z) = \frac{{z e}^{2 i z}}{(z - 2 i) (z + 2 i)}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, 2 i)}$ $= 2 i π \times \frac{2 i e^{- 4}}{4 i} = \frac{- 4 π e^{- 4}}{4 i} = i π e^{- 4}$ $\Rightarrow 2 I = π e^{- 4} \Rightarrow I = \frac{π}{2} e^{- 4}$
	Commented by mnjuly1970 last updated on 10/Oct/20

	$g r a t e f u l m r b i r d$ $v e r y n i c e a s a l w a y s . .$
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