Question Number 35217 by abdo.msup.com last updated on 16/May/18
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}\:{sin}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$
Commented by prof Abdo imad last updated on 20/May/18
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}\:{sin}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$$$\mathrm{2}{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{x}\:{sin}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}=\:{Im}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{xe}^{{i}\mathrm{2}{x}} }{{x}^{\mathrm{2}} \:+\mathrm{4}}\right){dx} \\ $$$${let}\:{consider}\:{the}\:{compolex}\:{finction} \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{{z}\:{e}^{{i}\mathrm{2}{z}} }{{z}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}\:{e}^{{i}\mathrm{2}{z}} }{\left({z}\:−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)}\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$$\mathrm{2}{i}\:{and}\:−\mathrm{2}{i} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\mathrm{2}{i}\right) \\ $$$${Res}\left(\varphi,\mathrm{2}{i}\right)\:={lim}_{{z}\rightarrow\mathrm{2}{i}} \left({z}−\mathrm{2}{i}\right)\varphi\left({z}\right) \\ $$$$=\:\frac{\left(\mathrm{2}{i}\right)\:{e}^{\mathrm{2}{i}\left(\mathrm{2}{i}\right)} }{\mathrm{4}{i}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−\mathrm{4}} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−\mathrm{4}} }{\mathrm{2}}\:=\:{i}\:\pi\:{e}^{−\mathrm{4}\:} \:\:\Rightarrow \\ $$$$\mathrm{2}\:{I}\:=\:\pi\:{e}^{−\mathrm{4}} \:\:\Rightarrow\:\:{I}\:\:=\:\frac{\pi}{\mathrm{2}}\:{e}^{−\mathrm{4}} \\ $$$$\bigstar\:{I}\:=\:\frac{\pi}{\mathrm{2}}\:{e}^{−\mathrm{4}} \:\bigstar \\ $$