Question Number 100969 by mathmax by abdo last updated on 29/Jun/20

find ∫_(−∞) ^∞   ((sin(cosx))/((x^2 −x+1)^2 ))dx

Answered by mathmax by abdo last updated on 30/Jun/20

I =∫_(−∞) ^(+∞ )  ((sin(cosx))/((x^2 −x+1)^2 ))dx   ⇒I =Im(∫_(−∞) ^(+∞)  (e^(icosx) /((x^2 −x+1)^2 ))dx) let ϕ(z) =(e^(icosz) /((z^2 −z+1)^2 ))  poles of ϕ?  z^2 −z+1 =0 →Δ =−3 ⇒z_1 =1+i(√3) =2e^((iπ)/3)   z_2 =1−i(√3)=2e^(−((iπ)/3))  ⇒ϕ(z) =(e^(icosz) /((z−2e^((iπ)/3) )^2 (z+2e^(−((iπ)/3)) )^(2  ) ))  the poles of ϕ are  2e^((iπ)/3)  and 2 e^(−((iπ)/3))  (doubles)  residus theorem give   ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ Res(ϕ,2e^((iπ)/3) )  Res(ϕ ,2e^((iπ)/3) ) =lim_(x→2 e^((iπ)/3) )   (1/((2−1)!)){ (z−e^((iπ)/3) )^2  ϕ(z)}^((1))   =lim_(z→2e^((iπ)/3) )   { (e^(icosz) /((z+2e^(−((iπ)/3)) )^2 ))}^((1))    =lim_(z→2e^((iπ)/3) )    {((−isinz e^(icosz) (z+2e^(−((iπ)/3)) )^2 −2(z+2e^(−((iπ)/3)) )e^(icosz) )/((z+2e^(−((iπ)/3)) )^4 ))}  =lim_(z→2e^((iπ)/3) )   ((−isinz e^(icosz) (z+2e^(−((iπ)/3)) )−2 e^(icosz) )/((z+2e^(−((iπ)/3)) )^3 ))  =lim_(z→ 2e^((iπ)/3) )     (({−isinz (z+2e^(−((iπ)/3)) )−2 }e^(icosz) )/((z+2e^(−((iπ)/3)) )^3 ))  =(({−isin(2e^((iπ)/3) )(4cos((π/3))−2}e^(i cos(2e^((iπ)/3) )) )/(8(2cos((π/3)))))  rest to finish the calculus...