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


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

$find \int_{- \infty}^{\infty} \frac{\sin (cosx)}{{(x^{2} - x + 1)}^{2}} dx$
	Answered by mathmax by abdo last updated on 30/Jun/20
	$I =∫_(−∞) ^(+∞ ) ((sin(cosx))/((x^2 −x+1)^2 ))dx ⇒I =Im(∫_(−∞) ^(+∞) (e^(icosx) /((x^2 −x+1)^2 ))dx) let ϕ(z) =(e^(icosz) /((z^2 −z+1)^2 )) poles of ϕ? z^2 −z+1 =0 →Δ =−3 ⇒z_1 =1+i(√3) =2e^((iπ)/3) z_2 =1−i(√3)=2e^(−((iπ)/3)) ⇒ϕ(z) =(e^(icosz) /((z−2e^((iπ)/3) )^2 (z+2e^(−((iπ)/3)) )^(2 ) )) the poles of ϕ are 2e^((iπ)/3) and 2 e^(−((iπ)/3)) (doubles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,2e^((iπ)/3) ) Res(ϕ ,2e^((iπ)/3) ) =lim_(x→2 e^((iπ)/3) ) (1/((2−1)!)){ (z−e^((iπ)/3) )^2 ϕ(z)}^((1)) =lim_(z→2e^((iπ)/3) ) { (e^(icosz) /((z+2e^(−((iπ)/3)) )^2 ))}^((1)) =lim_(z→2e^((iπ)/3) ) {((−isinz e^(icosz) (z+2e^(−((iπ)/3)) )^2 −2(z+2e^(−((iπ)/3)) )e^(icosz) )/((z+2e^(−((iπ)/3)) )^4 ))} =lim_(z→2e^((iπ)/3) ) ((−isinz e^(icosz) (z+2e^(−((iπ)/3)) )−2 e^(icosz) )/((z+2e^(−((iπ)/3)) )^3 )) =lim_(z→ 2e^((iπ)/3) ) (({−isinz (z+2e^(−((iπ)/3)) )−2 }e^(icosz) )/((z+2e^(−((iπ)/3)) )^3 )) =(({−isin(2e^((iπ)/3) )(4cos((π/3))−2}e^(i cos(2e^((iπ)/3) )) )/(8(2cos((π/3))))) rest to finish the calculus...$
	$I = \int_{- \infty}^{+ \infty} \frac{\sin (cosx)}{{(x^{2} - x + 1)}^{2}} dx \Rightarrow I = Im (\int_{- \infty}^{+ \infty} \frac{e^{icosx}}{{(x^{2} - x + 1)}^{2}} dx) let φ (z) = \frac{e^{icosz}}{{(z^{2} - z + 1)}^{2}}$ $poles of φ ?$ $z^{2} - z + 1 = 0 \to Δ = - 3 \Rightarrow z_{1} = 1 + i \sqrt{3} = {2 e}^{\frac{i π}{3}}$ $z_{2} = 1 - i \sqrt{3} = {2 e}^{- \frac{i π}{3}} \Rightarrow φ (z) = \frac{e^{icosz}}{{(z - {2 e}^{\frac{i π}{3}})}^{2} {(z + {2 e}^{- \frac{i π}{3}})}^{2}} the poles of φ are$ ${2 e}^{\frac{i π}{3}} and 2 e^{- \frac{i π}{3}} (doubles) residus theorem give$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, {2 e}^{\frac{i π}{3}})$ $Res (φ, {2 e}^{\frac{i π}{3}}) = \lim_{x \to 2 e^{\frac{i π}{3}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i π}{3}})}^{2} φ (z)}^{(1)}$ $= \lim_{z \to {2 e}^{\frac{i π}{3}}} {\frac{e^{icosz}}{{(z + {2 e}^{- \frac{i π}{3}})}^{2}}}^{(1)}$ $= \lim_{z \to {2 e}^{\frac{i π}{3}}} {\frac{- isinz e^{icosz} {(z + {2 e}^{- \frac{i π}{3}})}^{2} - 2 (z + {2 e}^{- \frac{i π}{3}}) e^{icosz}}{{(z + {2 e}^{- \frac{i π}{3}})}^{4}}}$ $= \lim_{z \to {2 e}^{\frac{i π}{3}}} \frac{- isinz e^{icosz} (z + {2 e}^{- \frac{i π}{3}}) - 2 e^{icosz}}{{(z + {2 e}^{- \frac{i π}{3}})}^{3}}$ $= \lim_{z \to {2 e}^{\frac{i π}{3}}} \frac{{- isinz (z + {2 e}^{- \frac{i π}{3}}) - 2} e^{icosz}}{{(z + {2 e}^{- \frac{i π}{3}})}^{3}}$ $= \frac{{- isin ({2 e}^{\frac{i π}{3}}) (4 \cos (\frac{π}{3}) - 2} e^{i \cos ({2 e}^{\frac{i π}{3}})}}{8 (2 \cos (\frac{π}{3}))} rest to finish the calculus . . .$
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