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Question Number 36009 by abdo mathsup 649 cc last updated on 27/May/18

calculate ∫_(−∞) ^(+∞) ((xdx)/((2x+1+i)^3 )) with i^2 =−1 .

$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{xdx}}{\left(\mathrm{2}{x}+\mathrm{1}+{i}\right)^{\mathrm{3}} }\:\:{with}\:{i}^{\mathrm{2}} \:=−\mathrm{1}\:. \\ $$

Commented by abdo imad last updated on 31/May/18

let consider the complex function ϕ(z) = (z/((2z +1+i)^3 )) = (z/(8(z−((−1−i)/2))^3 )) so ϕ have one pole triple z_0 =−(1/(√2))e^(i(π/4)) and I=∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_0 ) but Res(ϕ,z_0 ) =lim_(z→z_0 ) (1/((3−1)!)){(z−z_0 )^3 (z/(8(z−z_0 )^3 ))}^((2)) =lim_(z→z_0 ) (1/(16)){z}^((2)) =0 so I =0

$${let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}}{\left(\mathrm{2}{z}\:+\mathrm{1}+{i}\right)^{\mathrm{3}} }\:=\:\frac{{z}}{\mathrm{8}\left({z}−\frac{−\mathrm{1}−{i}}{\mathrm{2}}\right)^{\mathrm{3}} }\:{so}\:\varphi\:{have}\:{one}\:{pole} \\ $$$${triple}\:{z}_{\mathrm{0}} =−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\:\:{and} \\ $$$${I}=\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{0}} \right)\:{but} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{0}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{0}} } \:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left({z}−{z}_{\mathrm{0}} \right)^{\mathrm{3}} \:\frac{{z}}{\mathrm{8}\left({z}−{z}_{\mathrm{0}} \right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{0}} } \:\:\frac{\mathrm{1}}{\mathrm{16}}\left\{{z}\right\}^{\left(\mathrm{2}\right)} \:=\mathrm{0}\:\:{so}\:\:{I}\:=\mathrm{0} \\ $$

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