Question Number 88929 by mathmax by abdo last updated on 13/Apr/20
$${cakculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({ch}\left({x}\right)\right)}{\mathrm{4}+{x}^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 14/Apr/20
$${residus}\:{method}\:\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({chx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({chx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:\:{let}\:{f}\left({z}\right)=\frac{{arctan}\left({chz}\right)}{{z}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow \\ $$$${f}\left({z}\right)\:=\frac{{arctan}\left({chz}\right)}{\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)}\:\Rightarrow\int_{−\infty} ^{+\infty} \:{f}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({f},\mathrm{2}{i}\right) \\ $$$$=\mathrm{2}{i}\pi×\:\frac{{arctan}\left({ch}\left(\mathrm{2}{i}\right)\right)}{\mathrm{4}{i}}\:=\frac{\pi}{\mathrm{2}}\:{arctan}\left({ch}\left(\mathrm{2}{i}\right)\right) \\ $$$${but}\:{ch}\left(\mathrm{2}{i}\right)\:=\frac{{e}^{\mathrm{2}{i}} \:+{e}^{−\mathrm{2}{i}} }{\mathrm{2}}\:={cos}\left(\mathrm{2}\right)\:\Rightarrow\mathrm{2}{I}\:=\frac{\pi}{\mathrm{2}}{arctan}\left({cos}\mathrm{2}\right)\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{4}}\:{arctan}\left({cos}\mathrm{2}\right) \\ $$$$ \\ $$