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Question Number 68877 by mathmax by abdo last updated on 16/Sep/19
calculate ∫_(−∞) ^(+∞) ((xdx)/((x^2 −x+i)^2 )) with i^2 =−1
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{xdx}}{\left({x}^{\mathrm{2}} −{x}+{i}\right)^{\mathrm{2}} }\:\:\:\:\:{with}\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 18/Sep/19
let A =∫_(−∞) ^(+∞) ((xdx)/((x^2 −x +i)^2 )) let ϕ(z) =(z/((z^2 −z +i)^2 )) poles of ϕ? z^2 −z +i =0 ⇒Δ =1−4i =(√(17))e^(iarctan(−4)) =(√(17))e^(−i arctan(4)) z_1 =((1+17^(1/4) e^(−(i/2)arctan(4)) )/2) and z_2 =((1−17^(1/4) e^(−(i/2)arctan(4)) )/2) ϕ(z) =(z/((z−z_1 )^2 (z−z_2 )^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_2 ) Res(ϕ,z_2 ) =lim_(z→z_2 ) (1/((2−1)!)){ (z−z_2 )^2 ϕ(z)}^((1)) =lim_(z→z_2 ) {(z/((z−z_1 )^2 ))}^((1)) =lim_(z→z_1 ) (((z−z_1 )^2 −2(z−z_1 )z)/((z−z_1 )^4 )) =lim_(z→z_2 ) ((z−z_1 −2z)/((z−z_1 )^3 )) =lim_(z→z_2 ) ((−z−z_1 )/((z−z_1 )^3 )) =((−z_2 −z_1 )/((z_2 −z_1 )^3 )) =−(1/((−17^(1/4) e^(−(i/2)arctan(4)) )^3 )) =(1/(17^(3/4) e^(−((3i)/2) arctan(4)) )) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ×(1/(17^(3/4) e^(−((3i)/2) arctan(4)) )) =((2iπ e^(((3i)/2) arctan(4)) )/(17^(3/4) )) = A
$${let}\:{A}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{xdx}}{\left({x}^{\mathrm{2}} −{x}\:+{i}\right)^{\mathrm{2}} }\:\:{let}\:\:\varphi\left({z}\right)\:=\frac{{z}}{\left({z}^{\mathrm{2}} −{z}\:+{i}\right)^{\mathrm{2}} }\:{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{2}} −{z}\:+{i}\:=\mathrm{0}\:\Rightarrow\Delta\:=\mathrm{1}−\mathrm{4}{i}\:=\sqrt{\mathrm{17}}{e}^{{iarctan}\left(−\mathrm{4}\right)} \:=\sqrt{\mathrm{17}}{e}^{−{i}\:{arctan}\left(\mathrm{4}\right)} \\ $$$${z}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{17}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\mathrm{4}\right)} }{\mathrm{2}}\:\:{and}\:{z}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{17}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\mathrm{4}\right)} }{\mathrm{2}} \\ $$$$\varphi\left({z}\right)\:=\frac{{z}}{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{2}} \right) \\ $$$${Res}\left(\varphi,{z}_{\mathrm{2}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{2}} } \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{2}} } \:\:\left\{\frac{{z}}{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\frac{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} −\mathrm{2}\left({z}−{z}_{\mathrm{1}} \right){z}}{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{2}} \:\:\:} \:\:\:\:\:\frac{{z}−{z}_{\mathrm{1}} −\mathrm{2}{z}}{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{3}} }\:={lim}_{{z}\rightarrow{z}_{\mathrm{2}} } \:\:\:\frac{−{z}−{z}_{\mathrm{1}} }{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{3}} }\:=\frac{−{z}_{\mathrm{2}} −{z}_{\mathrm{1}} }{\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)^{\mathrm{3}} } \\ $$$$=−\frac{\mathrm{1}}{\left(−\mathrm{17}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\mathrm{4}\right)} \right)^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{17}^{\frac{\mathrm{3}}{\mathrm{4}}} \:{e}^{−\frac{\mathrm{3}{i}}{\mathrm{2}}\:{arctan}\left(\mathrm{4}\right)} }\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:×\frac{\mathrm{1}}{\mathrm{17}^{\frac{\mathrm{3}}{\mathrm{4}}} \:{e}^{−\frac{\mathrm{3}{i}}{\mathrm{2}}\:{arctan}\left(\mathrm{4}\right)} }\:=\frac{\mathrm{2}{i}\pi\:{e}^{\frac{\mathrm{3}{i}}{\mathrm{2}}\:{arctan}\left(\mathrm{4}\right)} }{\mathrm{17}^{\frac{\mathrm{3}}{\mathrm{4}}} }\:=\:{A}\: \\ $$

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