...nice mathematics.. please evaluate... Ω =∫_(−∞) ^( +∞) (((x^2 −4)/(x^2 +4))∗ ((sin(2x))/x)) dx =??? m.n.1970


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Question Number 117121 by mnjuly1970 last updated on 09/Oct/20

$. . . n i c e m a t h e m a t i c s . .$ $p l e a s e e v a l u a t e . . .$ $Ω = \int_{- \infty}^{+ \infty} (\frac{x^{2} - 4}{x^{2} + 4} * \frac{s i n (2 x)}{x}) d x = ? ? ?$ $m . n . 1970$
	Answered by Bird last updated on 10/Oct/20
	$I =∫_(−∞) ^(+∞) (((x^2 −4)sin(2x))/(x(x^2 +4)))dx ⇒ I=Im(∫_(−∞) ^(+∞) (((x^2 −4)e^(2ix) )/(x(x^2 +4))) dx) let consider the complex function ϕ(z) =(((z^2 −4)e^(2iz) )/(z(z^2 +4))) ⇒ ϕ(z) =(((z^2 −4)e^(2iz) )/(z(z−2i)(z+2i))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,o)+Res(ϕ,2i)} Res(ϕ,o) =lim_(z→0) zϕ(z) =lim_(z→0) ((−4)/4)=−1 Res(ϕ,2i) =lim_(z→2i) (z−2i)ϕ(z) =lim_(z→2i) (((z^2 −4)e^(2iz) )/(z(z+2i))) =((−8 e^(−4) )/((2i)(4i))) =e^(−4) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{−1 +e^(−4) } ⇒ I =2π{e^(−4) −1}$
	$I = \int_{- \infty}^{+ \infty} \frac{(x^{2} - 4) s i n (2 x)}{x (x^{2} + 4)} d x \Rightarrow$ $I = I m (\int_{- \infty}^{+ \infty} \frac{(x^{2} - 4) e^{2 i x}}{x (x^{2} + 4)} d x)$ $l e t c o n s i d e r t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{(z^{2} - 4) e^{2 i z}}{z (z^{2} + 4)} \Rightarrow$ $φ (z) = \frac{(z^{2} - 4) e^{2 i z}}{z (z - 2 i) (z + 2 i)} r e s i d u s$ $t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, o) + R e s (φ, 2 i)}$ $R e s (φ, o) = {l i m}_{z \to 0} z φ (z)$ $= {l i m}_{z \to 0} \frac{- 4}{4} = - 1$ $R e s (φ, 2 i) = {l i m}_{z \to 2 i} (z - 2 i) φ (z)$ $= {l i m}_{z \to 2 i} \frac{(z^{2} - 4) e^{2 i z}}{z (z + 2 i)}$ $= \frac{- 8 e^{- 4}}{(2 i) (4 i)} = e^{- 4} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {- 1 + e^{- 4}} \Rightarrow$ $I = 2 π {e^{- 4} - 1}$
	Commented by AbduraufKodiriy last updated on 10/Oct/20

	$T h i s s o l u t i o n i s w r o n g . B e c a u s e, a c c o r d i n g t o$ ${C a u c h y}^{'} s f o r m u l a, i f s i n g u l a r i t y p o i n t s$ $o f t h e f u n c t i o n a r e n o t o n t h e b o u n d a r y$ $o f a n y s e t D, t h e f o r m u l a t h a t y o u w r o t e$ $i s c o r r e c t .$ $S o r r y f o r m y m i s t a k e s i n E n g l i s h .$
	Answered by AbduraufKodiriy last updated on 09/Oct/20

	$Ω = \int_{- \infty}^{\infty} \frac{x^{2} - 4}{x^{2} + 4} \cdot \frac{s i n 2 x}{x} d x = 2 \int_{0}^{\infty} (1 - \frac{8}{x^{2} + 4}) \frac{s i n 2 x}{x} d x =$ $= 2 \int_{0}^{\infty} \frac{s i n 2 x}{x} d x - 16 \int_{0}^{\infty} \frac{s i n 2 x}{x (x^{2} + 4)} d x = π - 16 \int_{0}^{\infty} \frac{s i n 2 x}{x (x^{2} + 4)} d x;$ $ρ = \int_{0}^{\infty} \frac{s i n 2 x}{x (x^{2} + 4)} d x; ρ (t) = \int_{0}^{\infty} \frac{s i n (t x)}{x (x^{2} + 4)} d x;$ $L (ρ (t)) = L (\int_{0}^{\infty} \frac{s i n (t x)}{x (x^{2} + 4)} d x) = \int_{0}^{\infty} e^{- s t} \int_{0}^{\infty} \frac{s i n (t x)}{x (x^{2} + 4)} d x d t =$ $= \int_{0}^{\infty} \frac{1}{x (x^{2} + 4)} \int_{0}^{\infty} e^{- s t} s i n (t x) d t d x;$ $I = \int_{0}^{\infty} e^{- s t} s i n (t x) d t = - \frac{1}{s} e^{- s t} s i n (t x) ∣_{0}^{\infty} - \frac{x}{s^{2}} e^{- s t} c o s (t x) ∣_{0}^{\infty} - \frac{x^{2}}{s^{2}} I \Rightarrow$ $\Rightarrow (1 + \frac{x^{2}}{s^{2}}) I = \frac{x}{s^{2}} \Rightarrow I = \frac{x}{s^{2} + x^{2}}$ $L (ρ (t)) = \int_{0}^{\infty} \frac{1}{(x^{2} + 4) (x^{2} + s^{2})} d x = \frac{1}{s^{2} - 4} \int (\frac{1}{x^{2} + 4} - \frac{1}{x^{2} + s^{2}}) d x =$ $= \frac{1}{s^{2} - 4} (\frac{1}{2} a r c t a n (\frac{x}{2}) - \frac{1}{s} a r c t a n (\frac{x}{s})) ∣_{0}^{\infty} =$ $= \frac{1}{s^{2} - 4} \cdot (\frac{1}{2} - \frac{1}{s}) \frac{π}{2} = \frac{π}{4} \cdot \frac{1}{s (s + 2)} = \frac{π}{8} (\frac{1}{s} - \frac{1}{s + 2})$ $L^{- 1} (L (ρ (t))) = L^{- 1} (\frac{π}{8} \cdot (\frac{1}{s} - \frac{1}{s + 2})) \Rightarrow$ $\Rightarrow ρ (t) = \frac{π}{8} (1 - e^{- 2 t}); ρ = ρ (t = 2) \Rightarrow ρ = \frac{π}{8} (1 - e^{- 4})$ $Ω = π - 16 ρ = π - 2 π (1 - e^{- 4}) = - π (1 - 2 e^{- 4})$
	Commented by mnjuly1970 last updated on 09/Oct/20

	$b r a v o b r a v o m r$ $a b d u r a u f . . t h a n k y o u$
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