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Question Number 110447 by mathmax by abdo last updated on 29/Aug/20

calculate ∫_(−∞) ^(+∞) (dx/((x^2 −ix +1)^2 )) (i=(√(−1)))

$$\mathrm{calculate}\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{ix}\:+\mathrm{1}\right)^{\mathrm{2}} }\:\:\left(\mathrm{i}=\sqrt{−\mathrm{1}}\right) \\ $$

Answered by mathmax by abdo last updated on 29/Aug/20

I =∫_(−∞) ^(+∞) (dx/((x^2 −ix +1)^2 )) let ϕ(z) =(1/((z^2 −iz +1)^2 )) poles of ϕ? z^2 −iz +1 =0→Δ =(−i)^2 −4 =−5 ⇒z_1 =i+i(√5) and z_2 =i−i(√5) ⇒ϕ(z) =(1/((z−z_1 )^2 (z−z_2 )^2 )) residus tbeorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_1 ) and Res(ϕ,z_1 ) =lim_(z→z_1 ) (1/((2−1)!)){(z−z_1 )^2 ϕ(z)}^((1)) =lim_(z→z_1 ) {(1/((z−z_2 )^2 ))}^((1)) =lim_(z→z_1 ) −((2(z−z_2 ))/((z−z_2 )^4 )) =lim_(z→z_1 ) ((−2)/((z−z_2 )^3 )) =((−2)/((z_1 −z_2 )^3 )) =((−2)/((2i(√5))^3 )) =((−2)/(2^3 (−i)(5(√5)))) =(1/(20i(√5))) ⇒ ∫_(−∞) ^(+∞) ϕ(z) =2iπ.(1/(20i(√5))) =(π/(10(√5))) ⇒ I =(π/(10(√5)))

$$\mathrm{I}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{ix}\:+\mathrm{1}\right)^{\mathrm{2}} }\:\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{iz}\:+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{poles}\:\mathrm{of}\:\varphi? \\ $$$$\mathrm{z}^{\mathrm{2}} −\mathrm{iz}\:+\mathrm{1}\:=\mathrm{0}\rightarrow\Delta\:=\left(−\mathrm{i}\right)^{\mathrm{2}} −\mathrm{4}\:=−\mathrm{5}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\mathrm{i}+\mathrm{i}\sqrt{\mathrm{5}}\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =\mathrm{i}−\mathrm{i}\sqrt{\mathrm{5}} \\ $$$$\Rightarrow\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)^{\mathrm{2}} \left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\:\:\mathrm{residus}\:\mathrm{tbeorem}\:\mathrm{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{1}} \right)\:\mathrm{and}\: \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{1}} \right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{z}_{\mathrm{1}} \:} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)^{\mathrm{2}} \varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{z}_{\mathrm{1}} } \:\:\:\:\left\{\frac{\mathrm{1}}{\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{z}_{\mathrm{1}} } \:\:−\frac{\mathrm{2}\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)}{\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)^{\mathrm{4}} }\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{z}_{\mathrm{1}} } \:\frac{−\mathrm{2}}{\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\frac{−\mathrm{2}}{\left(\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \right)^{\mathrm{3}} }\:=\frac{−\mathrm{2}}{\left(\mathrm{2i}\sqrt{\mathrm{5}}\right)^{\mathrm{3}} }\:=\frac{−\mathrm{2}}{\mathrm{2}^{\mathrm{3}} \left(−\boldsymbol{\mathrm{i}}\right)\left(\mathrm{5}\sqrt{\mathrm{5}}\right)}\:=\frac{\mathrm{1}}{\mathrm{20}\boldsymbol{\mathrm{i}}\sqrt{\mathrm{5}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\:=\mathrm{2i}\pi.\frac{\mathrm{1}}{\mathrm{20i}\sqrt{\mathrm{5}}}\:=\frac{\pi}{\mathrm{10}\sqrt{\mathrm{5}}}\:\Rightarrow\:\mathrm{I}\:=\frac{\pi}{\mathrm{10}\sqrt{\mathrm{5}}} \\ $$

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