Question Number 36009 by abdo mathsup 649 cc last updated on 27/May/18
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{xdx}}{\left(\mathrm{2}{x}+\mathrm{1}+{i}\right)^{\mathrm{3}} }\:\:{with}\:{i}^{\mathrm{2}} \:=−\mathrm{1}\:. \\ $$
Commented by abdo imad last updated on 31/May/18
$${let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}}{\left(\mathrm{2}{z}\:+\mathrm{1}+{i}\right)^{\mathrm{3}} }\:=\:\frac{{z}}{\mathrm{8}\left({z}−\frac{−\mathrm{1}−{i}}{\mathrm{2}}\right)^{\mathrm{3}} }\:{so}\:\varphi\:{have}\:{one}\:{pole} \\ $$$${triple}\:{z}_{\mathrm{0}} =−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\:\:{and} \\ $$$${I}=\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{0}} \right)\:{but} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{0}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{0}} } \:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left({z}−{z}_{\mathrm{0}} \right)^{\mathrm{3}} \:\frac{{z}}{\mathrm{8}\left({z}−{z}_{\mathrm{0}} \right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{0}} } \:\:\frac{\mathrm{1}}{\mathrm{16}}\left\{{z}\right\}^{\left(\mathrm{2}\right)} \:=\mathrm{0}\:\:{so}\:\:{I}\:=\mathrm{0} \\ $$