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


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 140635 by Mathspace last updated on 10/May/21

$f i n d \int_{- \infty}^{+ \infty} \frac{s i n (2 c o s x)}{{(x^{2} - x + 1)}^{2}} d x$
	Answered by mathmax by abdo last updated on 12/May/21
	$Φ=∫_(−∞) ^(+∞) ((sin(2cosx))/((x^2 −x+1)^2 ))dx ⇒Φ =Im(∫_(−∞) ^∞ (e^(2icosx) /((x^2 −x+1)^2 ))dx) let Ψ(z)=(e^(2icosz) /((z^2 −z+1)^2 )) poles of Ψ? z^2 −z+1 =0 →Δ=−3 ⇒z_1 =((1+i(√3))/2) =e^((iπ)/3) and z_2 =((1−i(√3))/2)=e^(−((iπ)/3)) ⇒ Ψ(z) =(e^(2icosz) /((z−e^((iπ)/3) )^2 (z−e^(−((iπ)/3)) )^2 )) ∫_(−∞) ^(+∞) Ψ(z)dz =2iπ Res(Ψ,e^((iπ)/3) ) and Res(Ψ,e^((iπ)/3) ) =lim_(z→e^((iπ)/3) ) (1/((2−1)!)){(z−e^((iπ)/3) )^2 Ψ(z)}^((1)) =lim_(z→e^((iπ)/3) ) {(e^(2icosz) /((z−e^(−((iπ)/3)) )^2 ))}^((1)) =lim_(z→e^((iπ)/3) ) (((−2isinz e^(2icosz) )(z−e^(−((iπ)/3)) )^2 −2(z−e^(−((iπ)/3)) )e^(2icosz) )/((z−e^(−((iπ)/3)) )^4 )) =lim_(z→e^((iπ)/3) ) (((−2isinz)(z−e^(−((iπ)/3)) )−2)e^(2icosz) )/((z−e^(−((iπ)/3)) )^3 )) =(({(2isin((π/3)))(−2isin(e^((iπ)/3) )−2}e^(2icos(e^((iπ)/3) )) )/((2isin((π/3)))^3 )) = (({(i(√3))(−2i sin((1/2)+i((√3)/2)))−2}e^(2icos((1/2)+((i(√3))/2))) )/(−8i(((√3)/2))^3 )) =(({2(√3)sin((1/2)+i((√3)/2))−2}e^(2icos((1/2)+((i(√3))/2))) )/(−3(√3)i)) we know sinz =((e^(iz) −e^(−iz) )/(2i)) ⇒sin(e^((iπ)/3) ) =((e^(ie^((iπ)/3) ) −e^(−i e^((iπ)/3) ) )/(2i)) =((e^(i((1/2)+((i(√3))/2))) −e^(−i((1/2)+((i(√3))/2))) )/(2i)) =((e^(−((√3)/2)) (cos((1/2))+isin((1/2))−e^((√3)/2) (cos((1/2))−isin((1/2))))/(2i)) rest to finich the calculus.$
	$Φ = \int_{- \infty}^{+ \infty} \frac{\sin (2 cosx)}{{(x^{2} - x + 1)}^{2}} dx \Rightarrow Φ = Im (\int_{- \infty}^{\infty} \frac{e^{2 icosx}}{{(x^{2} - x + 1)}^{2}} dx)$ $let Ψ (z) = \frac{e^{2 icosz}}{{(z^{2} - z + 1)}^{2}} poles of Ψ ?$ $z^{2} - z + 1 = 0 \to Δ = - 3 \Rightarrow z_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}} and z_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{3}} \Rightarrow$ $Ψ (z) = \frac{e^{2 icosz}}{{(z - e^{\frac{i π}{3}})}^{2} {(z - e^{- \frac{i π}{3}})}^{2}}$ $\int_{- \infty}^{+ \infty} Ψ (z) dz = 2 i π Res (Ψ, e^{\frac{i π}{3}}) and$ $Res (Ψ, e^{\frac{i π}{3}}) = \lim_{z \to e^{\frac{i π}{3}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i π}{3}})}^{2} Ψ (z)}^{(1)}$ $= \lim_{z \to e^{\frac{i π}{3}}} {\frac{e^{2 icosz}}{{(z - e^{- \frac{i π}{3}})}^{2}}}^{(1)}$ $= \lim_{z \to e^{\frac{i π}{3}}} \frac{(- 2 isinz e^{2 icosz}) {(z - e^{- \frac{i π}{3}})}^{2} - 2 (z - e^{- \frac{i π}{3}}) e^{2 icosz}}{{(z - e^{- \frac{i π}{3}})}^{4}}$ $= \lim_{z \to e^{\frac{i π}{3}}} \frac{(- 2 isinz) (z - e^{- \frac{i π}{3}}) - 2) e^{2 icosz}}{{(z - e^{- \frac{i π}{3}})}^{3}}$ $= \frac{{(2 isin (\frac{π}{3})) (- 2 isin (e^{\frac{i π}{3}}) - 2} e^{2 icos (e^{\frac{i π}{3}})}}{{(2 isin (\frac{π}{3}))}^{3}}$ $= \frac{{(i \sqrt{3}) (- 2 i \sin (\frac{1}{2} + i \frac{\sqrt{3}}{2})) - 2} e^{2 icos (\frac{1}{2} + \frac{i \sqrt{3}}{2})}}{- 8 i {(\frac{\sqrt{3}}{2})}^{3}}$ $= \frac{{2 \sqrt{3} \sin (\frac{1}{2} + i \frac{\sqrt{3}}{2}) - 2} e^{2 icos (\frac{1}{2} + \frac{i \sqrt{3}}{2})}}{- 3 \sqrt{3} i} we know$ $sinz = \frac{e^{iz} - e^{- iz}}{2 i} \Rightarrow \sin (e^{\frac{i π}{3}}) = \frac{e^{{ie}^{\frac{i π}{3}}} - e^{- i e^{\frac{i π}{3}}}}{2 i}$ $= \frac{e^{i (\frac{1}{2} + \frac{i \sqrt{3}}{2})} - e^{- i (\frac{1}{2} + \frac{i \sqrt{3}}{2})}}{2 i} = \frac{e^{- \frac{\sqrt{3}}{2}} (\cos (\frac{1}{2}) + isin (\frac{1}{2}) - e^{\frac{\sqrt{3}}{2}} (\cos (\frac{1}{2}) - isin (\frac{1}{2}))}{2 i}$ $rest to finich the calculus .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum