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find-the-value-of-cos-ax-1-x-x-2-dx-




Question Number 33204 by prof Abdo imad last updated on 12/Apr/18
find the value of ∫_(−∞) ^(+∞)     ((cos(ax))/(1+x+x^2 )) dx.
findthevalueof+cos(ax)1+x+x2dx.
Commented by abdo imad last updated on 13/Apr/18
let put f(a) =∫_(−∞) ^(+∞)   ((cos(ax))/(1+x+x^2 ))dx  f(a) =Re (∫_(−∞) ^(+∞)   (e^(iax) /(1+x+x^2 ))dx) let consider the complex  function ϕ(z) = (e^(iaz) /(1+z+z^2 )) the poles of ϕ are j and j^−   j =e^(i((2π)/3))   and ϕ(z) =(e^(iaz) /((z−j)(z−j^− )))  ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ Res(ϕ,j)  Res(ϕ,j) = (e^(iaj) /(j−j^− )) =(e^(ia(−(1/2)+i((√3)/2))) /(2iIm(j))) = ((e^(−((ia)/2))  .e^(−a((√3)/2)) )/(2i((√3)/2)))  =(e^(−a((√3)/2)) /(i(√3))) ( cos((a/2)) −isin((a/2))) ⇒  ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ (e^(−a((√3)/2)) /(i(√3)))( cos((a/2))−isin((a/2)))⇒  f(a) =((2π)/( (√3))) cos((a/2)).e^(−a((√3)/2))  .
letputf(a)=+cos(ax)1+x+x2dxf(a)=Re(+eiax1+x+x2dx)letconsiderthecomplexfunctionφ(z)=eiaz1+z+z2thepolesofφarejandjj=ei2π3andφ(z)=eiaz(zj)(zj)+φ(z)dz=2iπRes(φ,j)Res(φ,j)=eiajjj=eia(12+i32)2iIm(j)=eia2.ea322i32=ea32i3(cos(a2)isin(a2))+φ(z)dz=2iπea32i3(cos(a2)isin(a2))f(a)=2π3cos(a2).ea32.

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