find ∫_0 ^∞ ((xsin(2x))/((x^2 +4)^3 ))dx


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Question Number 133119 by abdomsup last updated on 19/Feb/21

$f i n d \int_{0}^{\infty} \frac{x s i n (2 x)}{{(x^{2} + 4)}^{3}} d x$
	Answered by mindispower last updated on 19/Feb/21
	$let f(a)=∫_0 ^∞ ((xsin(2x))/(x^2 +a))dx,∀a∈R_+ −{0} f(a)=(1/2)Im∫_(−∞) ^∞ ((xe^(2ix) )/(x^2 +a))dx.....E ∫_(−∞) ^∞ ((xe^(2ix) )/(x^2 +a))=2iπRes(((xe^(2ix) )/(x^2 +a)),x=i(√a)) =2iπ.((i(√a).e^(−2(√a)) )/(2i(√a)))=iπe^(−2(√a)) E⇒f(a)=(π/2)e^(−2(√a)) f′(a)=∫_0 ^∞ ((−xsin(2x))/((x^2 +a)^2 ))dx f′′(a)=2∫((xsin(2x)dx)/((x^2 +a)^3 )) (1/2)f′′(4)=∫_0 ^∞ ((xsin(2x))/((x^2 +4)^2 ))dx...our integral f′(a)=−(π/(2(√a)))e^(−2(√a)) ,f′′(a)=(π/(2a))e^(−2(√a)) +(π/(4a(√a)))e^(−2(√a)) f′′(4)=((5π)/(32))e^(−4)$
	$l e t$ $f (a) = \int_{0}^{\infty} \frac{x s i n (2 x)}{x^{2} + a} d x, \forall a \in R_{+} - {0}$ $f (a) = \frac{1}{2} I m \int_{- \infty}^{\infty} \frac{{x e}^{2 i x}}{x^{2} + a} d x . . . . . E$ $\int_{- \infty}^{\infty} \frac{{x e}^{2 i x}}{x^{2} + a} = 2 i π R e s (\frac{{x e}^{2 i x}}{x^{2} + a}, x = i \sqrt{a})$ $= 2 i π . \frac{i \sqrt{a} . e^{- 2 \sqrt{a}}}{2 i \sqrt{a}} = i π e^{- 2 \sqrt{a}}$ $E \Rightarrow f (a) = \frac{π}{2} e^{- 2 \sqrt{a}}$ $f^{'} (a) = \int_{0}^{\infty} \frac{- x s i n (2 x)}{{(x^{2} + a)}^{2}} d x$ $f^{″} (a) = 2 \int \frac{x s i n (2 x) d x}{{(x^{2} + a)}^{3}}$ $\frac{1}{2} f^{″} (4) = \int_{0}^{\infty} \frac{x s i n (2 x)}{{(x^{2} + 4)}^{2}} d x . . . o u r i n t e g r a l$ $f^{'} (a) = - \frac{π}{2 \sqrt{a}} e^{- 2 \sqrt{a}}, f^{″} (a) = \frac{π}{2 a} e^{- 2 \sqrt{a}} + \frac{π}{4 a \sqrt{a}} e^{- 2 \sqrt{a}}$ $f^{″} (4) = \frac{5 π}{32} e^{- 4}$
	Answered by mathmax by abdo last updated on 20/Feb/21
	$Φ=∫_0 ^∞ ((xsin(2x))/((x^2 +4)^3 ))dx ⇒Φ=_(x=2t) ∫_0 ^∞ ((2tsin(4t))/(64(t^2 +1)^3 ))(2dt) =(1/(16))∫_0 ^∞ ((tsin(4t))/((t^2 +1)^3 ))dt =(1/(32))∫_(−∞) ^(+∞) ((tsin(4t))/((t^2 +1)^3 ))dt =(1/(32))Im(∫_(−∞) ^(+∞) ((te^(4it) )/((t^2 +1)^3 ))dt) ϕ(z)=((ze^(4iz) )/((z^2 +1)^3 )) ⇒ϕ(z)=((ze^(4iz) )/((z−i)^3 (z+i)^3 )) ∫_R ϕ(z)dz =2iπ Res(ϕ,i) and Res(ϕ,i)=lim_(z→i) (1/((3−1)!)){(z−i)^3 ϕ(z)}^((2)) =(1/2)lim_(z→i) {((ze^(4iz) )/((z+i)^3 ))}^((2)) ⇒2Res(ϕ,i)=lim_(z→i) { (((e^(4iz) +4iz e^(4iz) )(z+i)^3 −3(z+i)^2 ze^(4iz) )/((z+i)^6 ))}^((1)) lim_(z→i) {(((1+4iz)e^(4iz) (z+i)−3ze^(4iz) )/((z+i)^4 ))}^((1)) =lim_(z→i) {(((1+4iz)(z+i)−3z)e^(4iz) )/((z+i)^4 ))}^((1)) =lim_(z→i) {(((i+4iz^2 −3z)e^(4iz) )/((z+i)^4 ))}^((1)) =lim_(z→i) (((8iz−3)e^(4iz) +4i (4iz^2 −3z+1)e^(4iz) )(z+i)^4 −4(z+i)^3 (4iz^2 −3z+i)e^(4iz) )/((z+i)^8 )) ....rest to finish the calculus ...$
	$Φ = \int_{0}^{\infty} \frac{xsin (2 x)}{{(x^{2} + 4)}^{3}} dx \Rightarrow Φ =_{x = 2 t} \int_{0}^{\infty} \frac{2 tsin (4 t)}{64 {(t^{2} + 1)}^{3}} (2 dt)$ $= \frac{1}{16} \int_{0}^{\infty} \frac{tsin (4 t)}{{(t^{2} + 1)}^{3}} dt = \frac{1}{32} \int_{- \infty}^{+ \infty} \frac{tsin (4 t)}{{(t^{2} + 1)}^{3}} dt = \frac{1}{32} Im (\int_{- \infty}^{+ \infty} \frac{{te}^{4 it}}{{(t^{2} + 1)}^{3}} dt)$ $φ (z) = \frac{{ze}^{4 iz}}{{(z^{2} + 1)}^{3}} \Rightarrow φ (z) = \frac{{ze}^{4 iz}}{{(z - i)}^{3} {(z + i)}^{3}}$ $\int_{R} φ (z) dz = 2 i π Res (φ, i) and Res (φ, i) = \lim_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} φ (z)}^{(2)}$ $= \frac{1}{2} \lim_{z \to i} {\frac{{ze}^{4 iz}}{{(z + i)}^{3}}}^{(2)} \Rightarrow 2 Res (φ, i) = \lim_{z \to i} {\frac{(e^{4 iz} + 4 iz e^{4 iz}) {(z + i)}^{3} - 3 {(z + i)}^{2} {ze}^{4 iz}}{{(z + i)}^{6}}}^{(1)}$ $\lim_{z \to i} {\frac{(1 + 4 iz) e^{4 iz} (z + i) - {3 ze}^{4 iz}}{{(z + i)}^{4}}}^{(1)}$ $= \lim_{z \to i} {\frac{(1 + 4 iz) (z + i) - 3 z) e^{4 iz}}{{(z + i)}^{4}}}^{(1)}$ $= \lim_{z \to i} {\frac{(i + {4 iz}^{2} - 3 z) e^{4 iz}}{{(z + i)}^{4}}}^{(1)}$ $= \lim_{z \to i} \frac{(8 iz - 3) e^{4 iz} + 4 i ({4 iz}^{2} - 3 z + 1) e^{4 iz}) {(z + i)}^{4} - 4 {(z + i)}^{3} ({4 iz}^{2} - 3 z + i) e^{4 iz}}{{(z + i)}^{8}}$ $. . . . rest to finish the calculus . . .$
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