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Question Number 67673 by Abdo msup. last updated on 30/Aug/19
calculate ∫_(−∞) ^(+∞) (dx/((x^2 −x+1)(x^2 +x+1)))
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)} \\ $$
Commented by Abdo msup. last updated on 30/Aug/19
let A =∫_(−∞) ^(+∞) (dx/((x^2 −x+1)(x^2 +x+1))) and ϕ(z) =(1/((z^2 −z+1)(z^2 +z+1))) poles of ϕ? z^2 −z +1=0 →Δ =−3 =(i(√3))^2 ⇒ z_1 =((1+i(√3))/2) and z_2 =((1−i(√3))/2) z^2 +z +1=0 →Δ =−3 =(i(√3))^2 ⇒ α_1 =((−1+i(√3))/2) and α_2 =((−1−i(√3))/2) ϕ(z) = (1/((x−z_1 )(x−z_2 )(x−α_1 )(x−α_2 ))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,z_1 ) +Res(ϕ,α_1 )} Res (ϕ,z_1 ) =(1/((z_1 −z_2 )(z_1 ^2 +z_1 +1))) =(1/(i(√3)(z_1 ^2 +z_1 +1))) Res(ϕ,α_1 ) =(1/((α_1 −α_2 )(α_1 ^2 −α_1 +1))) =(1/(i(√3)(α_1 ^2 −α_1 +1))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(1/(i(√3)(z_1 ^2 +z_1 +1))) +(1/(i(√3)(α_1 ^2 −α_1 +1)))} =((2π)/( (√3))){ (1/(z_1 ^2 +z_1 +1)) +(1/(α_1 ^2 −α_1 +1))}
$${let}\:{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)} \\ $$$${and}\:\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −{z}+\mathrm{1}\right)\left({z}^{\mathrm{2}} \:+{z}+\mathrm{1}\right)}\:\:{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{2}} −{z}\:+\mathrm{1}=\mathrm{0}\:\rightarrow\Delta\:=−\mathrm{3}\:=\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${z}_{\mathrm{1}} =\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:{and}\:{z}_{\mathrm{2}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} +{z}\:+\mathrm{1}=\mathrm{0}\:\rightarrow\Delta\:=−\mathrm{3}\:=\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\alpha_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:{and}\:\alpha_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({x}−{z}_{\mathrm{1}} \right)\left({x}−{z}_{\mathrm{2}} \right)\left({x}−\alpha_{\mathrm{1}} \right)\left({x}−\alpha_{\mathrm{2}} \right)} \\ $$$${residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:+{Res}\left(\varphi,\alpha_{\mathrm{1}} \right)\right\} \\ $$$${Res}\:\left(\varphi,{z}_{\mathrm{1}} \right)\:\:=\frac{\mathrm{1}}{\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)\left({z}_{\mathrm{1}} ^{\mathrm{2}} \:+{z}_{\mathrm{1}} +\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left({z}_{\mathrm{1}} ^{\mathrm{2}} \:+{z}_{\mathrm{1}} +\mathrm{1}\right)} \\ $$$${Res}\left(\varphi,\alpha_{\mathrm{1}} \right)\:=\frac{\mathrm{1}}{\left(\alpha_{\mathrm{1}} −\alpha_{\mathrm{2}} \right)\left(\alpha_{\mathrm{1}} ^{\mathrm{2}} −\alpha_{\mathrm{1}} +\mathrm{1}\right)}\:=\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left(\alpha_{\mathrm{1}} ^{\mathrm{2}} −\alpha_{\mathrm{1}} +\mathrm{1}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left({z}_{\mathrm{1}} ^{\mathrm{2}} \:+{z}_{\mathrm{1}} +\mathrm{1}\right)}\:+\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left(\alpha_{\mathrm{1}} ^{\mathrm{2}} −\alpha_{\mathrm{1}} +\mathrm{1}\right)}\right\} \\ $$$$=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\left\{\:\:\frac{\mathrm{1}}{{z}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{1}} +\mathrm{1}}\:+\frac{\mathrm{1}}{\alpha_{\mathrm{1}} ^{\mathrm{2}} −\alpha_{\mathrm{1}} +\mathrm{1}}\right\} \\ $$$$ \\ $$

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