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find-0-xsin-2x-x-2-4-3-dx-




Question Number 133119 by abdomsup last updated on 19/Feb/21
find ∫_0 ^∞   ((xsin(2x))/((x^2 +4)^3 ))dx
find0xsin(2x)(x2+4)3dx
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)
letf(a)=0xsin(2x)x2+adx,aR+{0}f(a)=12Imxe2ixx2+adx..Exe2ixx2+a=2iπRes(xe2ixx2+a,x=ia)=2iπ.ia.e2a2ia=iπe2aEf(a)=π2e2af(a)=0xsin(2x)(x2+a)2dxf(a)=2xsin(2x)dx(x2+a)312f(4)=0xsin(2x)(x2+4)2dxourintegralf(a)=π2ae2a,f(a)=π2ae2a+π4aae2af(4)=5π32e4
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=0xsin(2x)(x2+4)3dxΦ=x=2t02tsin(4t)64(t2+1)3(2dt)=1160tsin(4t)(t2+1)3dt=132+tsin(4t)(t2+1)3dt=132Im(+te4it(t2+1)3dt)φ(z)=ze4iz(z2+1)3φ(z)=ze4iz(zi)3(z+i)3Rφ(z)dz=2iπRes(φ,i)andRes(φ,i)=limzi1(31)!{(zi)3φ(z)}(2)=12limzi{ze4iz(z+i)3}(2)2Res(φ,i)=limzi{(e4iz+4ize4iz)(z+i)33(z+i)2ze4iz(z+i)6}(1)limzi{(1+4iz)e4iz(z+i)3ze4iz(z+i)4}(1)=limzi{(1+4iz)(z+i)3z)e4iz(z+i)4}(1)=limzi{(i+4iz23z)e4iz(z+i)4}(1)=limzi(8iz3)e4iz+4i(4iz23z+1)e4iz)(z+i)44(z+i)3(4iz23z+i)e4iz(z+i)8.resttofinishthecalculus

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