Question Number 83009 by mathmax by abdo last updated on 26/Feb/20
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({chx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx} \\ $$
Commented by mathmax by abdo last updated on 27/Feb/20
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({chx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\:\Rightarrow\mathrm{2}{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({chx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx} \\ $$$$={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{ichx}} }{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\right)\:{let}\:\varphi\left({z}\right)=\frac{{e}^{{ich}\left({z}\right)} }{{z}^{\mathrm{2}} \:+\mathrm{3}}\:\Rightarrow\varphi\left({z}\right)=\frac{{e}^{{ich}\left({z}\right)} }{\left({z}−{i}\sqrt{\mathrm{3}}\right)\left({z}+{i}\sqrt{\mathrm{3}}\right)} \\ $$$${residus}\:{theoem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\right)\:=\mathrm{2}{i}\pi×\frac{{e}^{{ich}\left({i}\sqrt{\mathrm{3}}\right)} }{\mathrm{2}{i}\sqrt{\mathrm{3}}} \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{3}}}\:{e}^{{i}\frac{{e}^{{i}\sqrt{\mathrm{3}}} +{e}^{−{i}\sqrt{\mathrm{3}}} }{\mathrm{2}}} \:=\frac{\pi}{\:\sqrt{\mathrm{3}}}{e}^{{i}\:{cos}\left(\sqrt{\mathrm{3}}\right)} \:=\frac{\pi}{\:\sqrt{\mathrm{3}}}\left\{\:{cos}\left({cos}\left(\sqrt{\mathrm{3}}\right)\right)+{isin}\left({cos}\left(\sqrt{\mathrm{3}}\right)\right)\right\} \\ $$$$\Rightarrow\mathrm{2}{A}\:=\frac{\pi}{\:\sqrt{\mathrm{3}}}{cos}\left({cos}\left(\sqrt{\mathrm{3}}\right)\right)\:\Rightarrow{A}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}{cos}\left({cos}\left(\sqrt{\mathrm{3}}\right)\right) \\ $$