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Question Number 38724 by maxmathsup by imad last updated on 28/Jun/18
calculate  ∫_0 ^∞    ((x^2 cos(πx))/((x^2  +4)^2 ))dx
calculate0x2cos(πx)(x2+4)2dx
Commented by maxmathsup by imad last updated on 30/Jun/18
let I =∫_0 ^∞  ((x^2 cos(πx))/((x^2  +4)^2 ))dx  2I = ∫_(−∞) ^(+∞)   ((x^2 cos(πx))/((x^2  +4)^2 ))dx  =Re( ∫_(−∞) ^(+∞)  ((x^2  e^(iπx) )/((x^2  +4)^2 ))) let  ϕ(z)=((z^2  e^(iπz) )/((z^2 +4)^2 ))  we have ϕ(z) =((z^2  e^(iπz) )/((z−2i)^2 (z+2i)^2 )) so the poles of ϕ are 2i and −2i  ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ Res(ϕ,2i) but  Res(ϕ,2i) =lim_(z→2i)    (1/((2−1)!)) { (z−2i)^2 ϕ(z)}^((1))   =lim_(z→2i)   { ((z^2 e^(iπz) )/((z+2i)^2 ))}^((1)) =lim_(z→2i)  { (((2z e^(iπz)  +iπz^2  e^(iπz) )(z+2i)^2  −2(z+2i)z^2 e^(iπz) )/((z+2i)^4 ))}  lim_(z→2i)   (((2z+iπz^2 )e^(iπz) (z+2i)−2z^2  e^(iπz) )/((z+2i)^3 ))  = (((4i −4iπ)e^(−2π) (4i) +8 e^(−2π) )/((4i)^3 )) =(((−16 +16 π +8)e^(−2π) )/(−64i)) =(((8−16π)e^(−2π) )/(64i))  ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ (((8−16π)e^(−2π) )/(64i)) =π (((8−16π)e^(−2π) )/(32))  =(π/(32)) 8(1−2π)e^(−2π)  =(π/4)(1−2π)e^(−2π)   ⇒
letI=0x2cos(πx)(x2+4)2dx2I=+x2cos(πx)(x2+4)2dx=Re(+x2eiπx(x2+4)2)letφ(z)=z2eiπz(z2+4)2wehaveφ(z)=z2eiπz(z2i)2(z+2i)2sothepolesofφare2iand2i+φ(z)dz=2iπRes(φ,2i)butRes(φ,2i)=limz2i1(21)!{(z2i)2φ(z)}(1)=limz2i{z2eiπz(z+2i)2}(1)=limz2i{(2zeiπz+iπz2eiπz)(z+2i)22(z+2i)z2eiπz(z+2i)4}limz2i(2z+iπz2)eiπz(z+2i)2z2eiπz(z+2i)3=(4i4iπ)e2π(4i)+8e2π(4i)3=(16+16π+8)e2π64i=(816π)e2π64i+φ(z)dz=2iπ(816π)e2π64i=π(816π)e2π32=π328(12π)e2π=π4(12π)e2π
Commented by maxmathsup by imad last updated on 30/Jun/18
2I = (π/4)(1−2π)e^(−2π)  ⇒ I =(π/8)(1−2π)e^(−2π) .  π
2I=π4(12π)e2πI=π8(12π)e2π.π

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