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Question Number 88929 by mathmax by abdo last updated on 13/Apr/20

cakculate ∫_0 ^∞   ((arctan(ch(x)))/(4+x^2 ))dx

$${cakculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({ch}\left({x}\right)\right)}{\mathrm{4}+{x}^{\mathrm{2}} }{dx} \\ $$

Commented by mathmax by abdo last updated on 14/Apr/20

residus method  I =∫_0 ^∞  ((arctan(chx))/(x^2  +4))dx ⇒  2I =∫_(−∞) ^(+∞)  ((arctan(chx))/(x^2  +4))dx  let f(z)=((arctan(chz))/(z^2  +4)) ⇒  f(z) =((arctan(chz))/((z−2i)(z+2i))) ⇒∫_(−∞) ^(+∞)  f(z)dz =2iπ Res(f,2i)  =2iπ× ((arctan(ch(2i)))/(4i)) =(π/2) arctan(ch(2i))  but ch(2i) =((e^(2i)  +e^(−2i) )/2) =cos(2) ⇒2I =(π/2)arctan(cos2) ⇒  I =(π/4) arctan(cos2)

$${residus}\:{method}\:\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({chx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({chx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:\:{let}\:{f}\left({z}\right)=\frac{{arctan}\left({chz}\right)}{{z}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow \\ $$$${f}\left({z}\right)\:=\frac{{arctan}\left({chz}\right)}{\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)}\:\Rightarrow\int_{−\infty} ^{+\infty} \:{f}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({f},\mathrm{2}{i}\right) \\ $$$$=\mathrm{2}{i}\pi×\:\frac{{arctan}\left({ch}\left(\mathrm{2}{i}\right)\right)}{\mathrm{4}{i}}\:=\frac{\pi}{\mathrm{2}}\:{arctan}\left({ch}\left(\mathrm{2}{i}\right)\right) \\ $$$${but}\:{ch}\left(\mathrm{2}{i}\right)\:=\frac{{e}^{\mathrm{2}{i}} \:+{e}^{−\mathrm{2}{i}} }{\mathrm{2}}\:={cos}\left(\mathrm{2}\right)\:\Rightarrow\mathrm{2}{I}\:=\frac{\pi}{\mathrm{2}}{arctan}\left({cos}\mathrm{2}\right)\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{4}}\:{arctan}\left({cos}\mathrm{2}\right) \\ $$$$ \\ $$

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