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Question Number 138970 by mathmax by abdo last updated on 20/Apr/21
calculate ∫_(−∞) ^∞ ((xsin(2x))/((x^2 +x+1)^2 ))dx
$$\mathrm{calculate}\:\int_{−\infty} ^{\infty} \:\:\frac{\mathrm{xsin}\left(\mathrm{2x}\right)}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$
Answered by mathmax by abdo last updated on 23/Apr/21
Φ=∫_(−∞) ^(+∞) ((xsin(2x))/((x^2 +x+1)^2 ))dx ⇒Φ =Im(∫_(−∞) ^(+∞) ((xe^(2ix) )/((x^2 +x+1)^2 ))dx) let Ψ(z)=((ze^(2iz) )/((z^2 +z+1)^2 )) poles of Ψ? z^2 +z+1=0→Δ=−3 ⇒z_1 =((−1+i(√3))/2)=e^((2iπ)/3) and z_2 =e^(−((2iπ)/3)) ⇒Ψ(z)=((ze^(2iz) )/((z−e^((2iπ)/3) )^2 (z−e^(−((2iπ)/3)) )^2 )) and ∫_(−∞) ^(+∞) Ψ(z)dz=2iπRes(Ψ,e^((2iπ)/3) ) Res(Ψ,e^((2iπ)/3) ) =lim_(z→e^((2iπ)/3) ) (1/((2−1)!)){(z−e^((2iπ)/3) )^2 Ψ(z)}^((1)) =lim_(z→e^((2iπ)/3) ) {((ze^(2iz) )/((z−e^(−((2iπ)/z)) )^2 ))}^((1)) =lim_(z→e^((2iπ)/3) ) (((e^(2iz ) +2ize^(2iz) )(z−e^(−((2iπ)/z)) )^2 −2(z−e^(−((2iπ)/z)) )ze^(2iz) )/((z−e^(−((2iπ)/z)) )^4 )) =lim_(z→e^((2iπ)/3) ) (((2iz+1)e^(2iz) (z−e^(−((2iπ)/3)) )−2ze^(2iz) )/((z−e^(−((2iπ)/3)) )^3 )) =(((2ie^((2iπ)/3) +1)e^(2ie^((2iπ)/3) ) (2isin(((2π)/3))−2e^((2iπ)/3) e^(2ie^((2iπ)/3) ) )/((2isin(((2π)/3))^3 )) =((2i.((√3)/2)(2i(−(1/2)+i((√3)/2))e^(2i(−(1/2)+((i(√3))/2))) −2(−(1/2)+i((√3)/2))e^(2i(−(1/2)+((i(√3))/2))) )/(−8i(((√3)/2))^3 )) =((i(√3)(−i−(√3))e^(−i−(√3)) +(1−i(√3))e^(−i−(√3)) )/(−i(3(√3)))) =((((√3)−3i+1−i(√3))e^(−(√3)) {cos(1)−isin(1)})/(−3i(√3))) =((((√3)+1−(3+(√3)i)e^(−(√3)) {cos1−isin1})/(−3i(√3))) ....rest to extract Im(of this quantity...)
$$\Phi=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{xsin}\left(\mathrm{2x}\right)}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow\Phi\:=\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{xe}^{\mathrm{2ix}} }{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\right) \\ $$$$\mathrm{let}\:\Psi\left(\mathrm{z}\right)=\frac{\mathrm{ze}^{\mathrm{2iz}} }{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{z}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\mathrm{poles}\:\mathrm{of}\:\Psi? \\ $$$$\mathrm{z}^{\mathrm{2}} \:+\mathrm{z}+\mathrm{1}=\mathrm{0}\rightarrow\Delta=−\mathrm{3}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \:\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \\ $$$$\Rightarrow\Psi\left(\mathrm{z}\right)=\frac{\mathrm{ze}^{\mathrm{2iz}} }{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} }\:\mathrm{and}\:\int_{−\infty} ^{+\infty} \:\Psi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\mathrm{Res}\left(\Psi,\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right) \\ $$$$\mathrm{Res}\left(\Psi,\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} } \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \Psi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} } \:\:\:\left\{\frac{\mathrm{ze}^{\mathrm{2iz}} }{\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{z}}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} } \:\:\:\frac{\left(\mathrm{e}^{\mathrm{2iz}\:} +\mathrm{2ize}^{\mathrm{2iz}} \right)\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{z}}} \right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{z}}} \right)\mathrm{ze}^{\mathrm{2iz}} }{\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{z}}} \right)^{\mathrm{4}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} } \:\:\:\frac{\left(\mathrm{2iz}+\mathrm{1}\right)\mathrm{e}^{\mathrm{2iz}} \left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)−\mathrm{2ze}^{\mathrm{2iz}} }{\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\left(\mathrm{2ie}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} +\mathrm{1}\right)\mathrm{e}^{\mathrm{2ie}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} } \left(\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)−\mathrm{2e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \:\mathrm{e}^{\mathrm{2ie}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} } \right.}{\left(\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)^{\mathrm{3}} \right.} \\ $$$$=\frac{\mathrm{2i}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{2i}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\mathrm{e}^{\mathrm{2i}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} −\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\mathrm{e}^{\mathrm{2i}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} \right.}{−\mathrm{8i}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{i}\sqrt{\mathrm{3}}\left(−\mathrm{i}−\sqrt{\mathrm{3}}\right)\mathrm{e}^{−\mathrm{i}−\sqrt{\mathrm{3}}} +\left(\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}\right)\mathrm{e}^{−\mathrm{i}−\sqrt{\mathrm{3}}} }{−\mathrm{i}\left(\mathrm{3}\sqrt{\mathrm{3}}\right)} \\ $$$$=\frac{\left(\sqrt{\mathrm{3}}−\mathrm{3i}+\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}\right)\mathrm{e}^{−\sqrt{\mathrm{3}}} \left\{\mathrm{cos}\left(\mathrm{1}\right)−\mathrm{isin}\left(\mathrm{1}\right)\right\}}{−\mathrm{3i}\sqrt{\mathrm{3}}} \\ $$$$=\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}−\left(\mathrm{3}+\sqrt{\mathrm{3}}\mathrm{i}\right)\mathrm{e}^{−\sqrt{\mathrm{3}}} \left\{\mathrm{cos1}−\mathrm{isin1}\right\}\right.}{−\mathrm{3i}\sqrt{\mathrm{3}}} \\ $$$$….\mathrm{rest}\:\mathrm{to}\:\mathrm{extract}\:\mathrm{Im}\left(\mathrm{of}\:\mathrm{this}\:\mathrm{quantity}…\right) \\ $$
Answered by qaz last updated on 24/Apr/21
I=∫_(−∞) ^(+∞) ((xsin 2x)/((x^2 +x+1)^2 ))dx f(z)=−(e^(i2z) /(z^2 +az+1))=−(e^(i2z) /((z−((−a+(√(a^2 −4)))/2))(z−((−a−(√(a^2 −4)))/2))))=−(e^(i2z) /((z−z_1 )(z−z_2 ))) Res(f(z),z_1 )=−(e^(i2z_1 ) /(z_1 −z_2 ))=−(e^(i(−a+(√(a^2 −4)))) /( (√(a^2 −4)))) I=ℑ(d/da)∣_(a=1) {−((2πie^(i(−a+(√(a^2 −4)))) )/( (√(a^2 −4))))} =−ℑ{2πi∙{−a(a^2 −4)^(−3/2) e^(i(−a+(√(a^2 −4)))) +(a^2 −4)^(−1/2) e^(i(−a+(√(a^2 −4)))) [−i+((ia)/( (√(a^2 −4))))]}}_(a=1) =−ℑ{2πi∙{(e^(−(√3)) /(i3(√3)))cos (1)−(e^(−(√3)) /(3(√3)))sin (1)+((−e^(−(√3)) )/( (√3)))cos (1)+(e^(−(√3)) /(i3))cos (1)+((ie^(−(√3)) )/( (√3)))sin (1)−(e^(−(√3)) /3)sin (1)}} =((2π)/(3(√3)))e^(−(√3)) [(1+(√3))sin (1)+3cos (1)] ....
$${I}=\int_{−\infty} ^{+\infty} \frac{{x}\mathrm{sin}\:\mathrm{2}{x}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$${f}\left({z}\right)=−\frac{{e}^{{i}\mathrm{2}{z}} }{{z}^{\mathrm{2}} +{az}+\mathrm{1}}=−\frac{{e}^{{i}\mathrm{2}{z}} }{\left({z}−\frac{−{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\right)\left({z}−\frac{−{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\right)}=−\frac{{e}^{{i}\mathrm{2}{z}} }{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$$${Res}\left({f}\left({z}\right),{z}_{\mathrm{1}} \right)=−\frac{{e}^{{i}\mathrm{2}{z}_{\mathrm{1}} } }{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }=−\frac{{e}^{{i}\left(−{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)} }{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}} \\ $$$${I}=\Im\frac{{d}}{{da}}\mid_{{a}=\mathrm{1}} \left\{−\frac{\mathrm{2}\pi{ie}^{{i}\left(−{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)} }{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}\right\} \\ $$$$=−\Im\left\{\mathrm{2}\pi{i}\centerdot\left\{−{a}\left({a}^{\mathrm{2}} −\mathrm{4}\right)^{−\mathrm{3}/\mathrm{2}} {e}^{{i}\left(−{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)} +\left({a}^{\mathrm{2}} −\mathrm{4}\right)^{−\mathrm{1}/\mathrm{2}} {e}^{{i}\left(−{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)} \left[−{i}+\frac{{ia}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}\right]\right\}\right\}_{{a}=\mathrm{1}} \\ $$$$=−\Im\left\{\mathrm{2}\pi{i}\centerdot\left\{\frac{{e}^{−\sqrt{\mathrm{3}}} }{{i}\mathrm{3}\sqrt{\mathrm{3}}}\mathrm{cos}\:\left(\mathrm{1}\right)−\frac{{e}^{−\sqrt{\mathrm{3}}} }{\mathrm{3}\sqrt{\mathrm{3}}}\mathrm{sin}\:\left(\mathrm{1}\right)+\frac{−{e}^{−\sqrt{\mathrm{3}}} }{\:\sqrt{\mathrm{3}}}\mathrm{cos}\:\left(\mathrm{1}\right)+\frac{{e}^{−\sqrt{\mathrm{3}}} }{{i}\mathrm{3}}\mathrm{cos}\:\left(\mathrm{1}\right)+\frac{{ie}^{−\sqrt{\mathrm{3}}} }{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\left(\mathrm{1}\right)−\frac{{e}^{−\sqrt{\mathrm{3}}} }{\mathrm{3}}\mathrm{sin}\:\left(\mathrm{1}\right)\right\}\right\} \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}{e}^{−\sqrt{\mathrm{3}}} \left[\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\mathrm{sin}\:\left(\mathrm{1}\right)+\mathrm{3cos}\:\left(\mathrm{1}\right)\right] \\ $$$$…. \\ $$

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