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Question Number 48171 by Abdo msup. last updated on 20/Nov/18

calculate ∫_0 ^∞ ((sin(cosx))/(x^2 +3))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({cosx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx} \\ $$

Commented by Abdo msup. last updated on 21/Nov/18

let I =∫_0 ^∞ ((sin(cosx))/(x^2 +3))dx ⇒2I =∫_(−∞) ^(+∞) ((sin(cosx))/(x^2 +3))dx =Im(∫_(−∞) ^(+∞) (e^(icosx) /(x^2 +3))dx) let consider the complexfunction ϕ(z)=(e^(icosz) /(z^2 +3)) we have ϕ(z)=(e^(icosz) /((z−i(√3))(z+i(√3))))?so the poles of ϕ are +^− i(√3) residus theorem?give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i(√3)) Res(ϕ,i(√3)) =lim_(z→i(√3)) (z−i(√3))ϕ(z) = (e^(icos(i(√3))) /(2i(√3))) but cos(i(√3))=((e^(i(i(√3))) +e^(−i(i(√3))) )/2) =((e^(−(√3)) +e^(√3) )/2) =ch((√3)) ⇒Res(ϕ,i(√3))=(e^(ich((√3))) /(2i(√3))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ.(e^(ich((√3))) /(2i(√3))) =(π/(√3)) {cos(ch((√3))+isin(ch((√3)))}⇒ 2I =(π/(√3))sin(ch((√3))) ⇒ I =(π/(2(√3))) sin(((e^(√3) +e^(−(√3)) )/2)).

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({cosx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{sin}\left({cosx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx} \\ $$$$={Im}\left(\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{icosx}} }{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\right)\:{let}\:{consider}\:{the}\:{complexfunction} \\ $$$$\varphi\left({z}\right)=\frac{{e}^{{icosz}} }{{z}^{\mathrm{2}} \:+\mathrm{3}}\:\:{we}\:{have}\:\varphi\left({z}\right)=\frac{{e}^{{icosz}} }{\left({z}−{i}\sqrt{\mathrm{3}}\right)\left({z}+{i}\sqrt{\mathrm{3}}\right)}?{so}\:{the}\:{poles} \\ $$$${of}\:\varphi\:{are}\:\overset{−} {+}{i}\sqrt{\mathrm{3}}\:\:\:{residus}\:{theorem}?{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\right) \\ $$$${Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\:\:\left({z}−{i}\sqrt{\mathrm{3}}\right)\varphi\left({z}\right) \\ $$$$=\:\frac{{e}^{{icos}\left({i}\sqrt{\mathrm{3}}\right)} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:{but}\:\:{cos}\left({i}\sqrt{\mathrm{3}}\right)=\frac{{e}^{{i}\left({i}\sqrt{\mathrm{3}}\right)} +{e}^{−{i}\left({i}\sqrt{\mathrm{3}}\right)} }{\mathrm{2}}\:=\frac{{e}^{−\sqrt{\mathrm{3}}} +{e}^{\sqrt{\mathrm{3}}} }{\mathrm{2}} \\ $$$$={ch}\left(\sqrt{\mathrm{3}}\right)\:\Rightarrow{Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\right)=\frac{{e}^{{ich}\left(\sqrt{\mathrm{3}}\right)} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi.\frac{{e}^{{ich}\left(\sqrt{\mathrm{3}}\right)} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:=\frac{\pi}{\sqrt{\mathrm{3}}}\:\left\{{cos}\left({ch}\left(\sqrt{\mathrm{3}}\right)+{isin}\left({ch}\left(\sqrt{\mathrm{3}}\right)\right)\right\}\Rightarrow\right. \\ $$$$\mathrm{2}{I}\:=\frac{\pi}{\sqrt{\mathrm{3}}}{sin}\left({ch}\left(\sqrt{\mathrm{3}}\right)\right)\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\:{sin}\left(\frac{{e}^{\sqrt{\mathrm{3}}} +{e}^{−\sqrt{\mathrm{3}}} }{\mathrm{2}}\right). \\ $$

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