Question Number 37306 by math khazana by abdo last updated on 11/Jun/18
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:{e}^{{ix}} \:\:\:\frac{{x}−{i}}{\left({x}+{i}\right)\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)}\:{dx}\:. \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 15/Jun/18
$${let}\:\:\varphi\left({z}\right)\:=\:{e}^{{iz}} \:\:\:\frac{{z}−{i}}{\left({z}+{i}\right)\left({z}^{\mathrm{2}} \:+\mathrm{3}\right)}\:\:{the}\:{poles}\:{of} \\ $$$$\varphi\:{are}\:\:−{i}\:,{i}\sqrt{\mathrm{3}},−{i}\sqrt{\mathrm{3}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\right) \\ $$$${we}\:{have}\:\varphi\left({z}\right)=\frac{\left({z}−{i}\right){e}^{{iz}} }{\left({z}+{i}\right)\left({z}−{i}\sqrt{\mathrm{3}}\right)\left({z}+{i}\sqrt{\mathrm{3}}\right)} \\ $$$${Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\right)\:=\:\frac{\left({i}\sqrt{\mathrm{3}}−{i}\right){e}^{{i}\left({i}\sqrt{\mathrm{3}}\right)} }{\left({i}\sqrt{\mathrm{3}}\:+{i}\right)\left(\mathrm{2}{i}\sqrt{\mathrm{3}}\right)} \\ $$$$=\:\frac{{i}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right){e}^{−\sqrt{\mathrm{3}}} }{−\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}\:\:=\:\frac{{i}\left(\mathrm{1}−\sqrt{\mathrm{3}}\right){e}^{−\sqrt{\mathrm{3}}} }{\mathrm{2}\sqrt{\mathrm{3}}\:\:+\mathrm{6}}\:\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\:\mathrm{2}{i}\pi\:\:\frac{{i}\left(\mathrm{1}−\sqrt{\mathrm{3}}\right){e}^{−\sqrt{\mathrm{3}}} }{\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{6}} \\ $$$$=\frac{\pi\left(\sqrt{\mathrm{3}}−\mathrm{1}\right){e}^{−\sqrt{\mathrm{3}}} }{\mathrm{3}+\sqrt{\mathrm{3}}}\:\:\Rightarrow \\ $$$${I}\:\:=\:\frac{\pi\left(\sqrt{\mathrm{3}}−\mathrm{1}\right){e}^{−\sqrt{\mathrm{3}}} }{\mathrm{3}+\sqrt{\mathrm{3}}}\:\:. \\ $$$$ \\ $$