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...