Question Number 48171 by Abdo msup. last updated on 20/Nov/18
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({cosx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx} \\ $$
Commented by Abdo msup. last updated on 21/Nov/18
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({cosx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{sin}\left({cosx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx} \\ $$$$={Im}\left(\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{icosx}} }{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\right)\:{let}\:{consider}\:{the}\:{complexfunction} \\ $$$$\varphi\left({z}\right)=\frac{{e}^{{icosz}} }{{z}^{\mathrm{2}} \:+\mathrm{3}}\:\:{we}\:{have}\:\varphi\left({z}\right)=\frac{{e}^{{icosz}} }{\left({z}−{i}\sqrt{\mathrm{3}}\right)\left({z}+{i}\sqrt{\mathrm{3}}\right)}?{so}\:{the}\:{poles} \\ $$$${of}\:\varphi\:{are}\:\overset{−} {+}{i}\sqrt{\mathrm{3}}\:\:\:{residus}\:{theorem}?{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\right) \\ $$$${Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\:\:\left({z}−{i}\sqrt{\mathrm{3}}\right)\varphi\left({z}\right) \\ $$$$=\:\frac{{e}^{{icos}\left({i}\sqrt{\mathrm{3}}\right)} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:{but}\:\:{cos}\left({i}\sqrt{\mathrm{3}}\right)=\frac{{e}^{{i}\left({i}\sqrt{\mathrm{3}}\right)} +{e}^{−{i}\left({i}\sqrt{\mathrm{3}}\right)} }{\mathrm{2}}\:=\frac{{e}^{−\sqrt{\mathrm{3}}} +{e}^{\sqrt{\mathrm{3}}} }{\mathrm{2}} \\ $$$$={ch}\left(\sqrt{\mathrm{3}}\right)\:\Rightarrow{Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\right)=\frac{{e}^{{ich}\left(\sqrt{\mathrm{3}}\right)} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi.\frac{{e}^{{ich}\left(\sqrt{\mathrm{3}}\right)} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:=\frac{\pi}{\:\sqrt{\mathrm{3}}}\:\left\{{cos}\left({ch}\left(\sqrt{\mathrm{3}}\right)+{isin}\left({ch}\left(\sqrt{\mathrm{3}}\right)\right)\right\}\Rightarrow\right. \\ $$$$\mathrm{2}{I}\:=\frac{\pi}{\:\sqrt{\mathrm{3}}}{sin}\left({ch}\left(\sqrt{\mathrm{3}}\right)\right)\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\:{sin}\left(\frac{{e}^{\sqrt{\mathrm{3}}} +{e}^{−\sqrt{\mathrm{3}}} }{\mathrm{2}}\right). \\ $$