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

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

$${find}\:\:\int_{−\infty} ^{+\infty} \:\frac{{sin}\left(\mathrm{2}{cosx}\right)}{\left({x}^{\mathrm{2}} \:−{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$

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.

$$\Phi=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{sin}\left(\mathrm{2cosx}\right)}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow\Phi\:=\mathrm{Im}\left(\int_{−\infty} ^{\infty} \:\frac{\mathrm{e}^{\mathrm{2icosx}} }{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\right) \\ $$$$\mathrm{let}\:\Psi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{2icosz}} }{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{z}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\mathrm{poles}\:\mathrm{of}\:\Psi? \\ $$$$\mathrm{z}^{\mathrm{2}} −\mathrm{z}+\mathrm{1}\:=\mathrm{0}\:\rightarrow\Delta=−\mathrm{3}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\Rightarrow \\ $$$$\Psi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{2icosz}} }{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} } \\ $$$$\int_{−\infty} ^{+\infty} \:\Psi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\Psi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:\mathrm{and}\: \\ $$$$\mathrm{Res}\left(\Psi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} } \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \Psi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} } \:\:\:\left\{\frac{\mathrm{e}^{\mathrm{2icosz}} }{\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \: \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} } \:\:\:\frac{\left(−\mathrm{2isinz}\:\mathrm{e}^{\mathrm{2icosz}} \right)\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\mathrm{e}^{\mathrm{2icosz}} }{\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{4}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} } \:\:\frac{\left.\left(−\mathrm{2isinz}\right)\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)−\mathrm{2}\right)\mathrm{e}^{\mathrm{2icosz}} }{\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\left\{\left(\mathrm{2isin}\left(\frac{\pi}{\mathrm{3}}\right)\right)\left(−\mathrm{2isin}\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)−\mathrm{2}\right\}\mathrm{e}^{\mathrm{2icos}\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)} \right.}{\left(\mathrm{2isin}\left(\frac{\pi}{\mathrm{3}}\right)\right)^{\mathrm{3}} } \\ $$$$=\:\frac{\left\{\left(\mathrm{i}\sqrt{\mathrm{3}}\right)\left(−\mathrm{2i}\:\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right)−\mathrm{2}\right\}\mathrm{e}^{\mathrm{2icos}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} }{−\mathrm{8i}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left\{\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\mathrm{2}\right\}\mathrm{e}^{\mathrm{2icos}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} }{−\mathrm{3}\sqrt{\mathrm{3}}\mathrm{i}}\:\mathrm{we}\:\mathrm{know} \\ $$$$\mathrm{sinz}\:=\frac{\mathrm{e}^{\mathrm{iz}} −\mathrm{e}^{−\mathrm{iz}} }{\mathrm{2i}}\:\Rightarrow\mathrm{sin}\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:=\frac{\mathrm{e}^{\mathrm{ie}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} } −\mathrm{e}^{−\mathrm{i}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} } }{\mathrm{2i}} \\ $$$$=\frac{\mathrm{e}^{\mathrm{i}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} −\mathrm{e}^{−\mathrm{i}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} }{\mathrm{2i}}\:=\frac{\mathrm{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \left(\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{isin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \left(\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{isin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\right.}{\mathrm{2i}} \\ $$$$\mathrm{rest}\:\mathrm{to}\:\mathrm{finich}\:\mathrm{the}\:\mathrm{calculus}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com