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Question Number 76350 by mathmax by abdo last updated on 26/Dec/19
calculate ∫_0 ^∞   ((arctan(e^x^2  ))/(x^2  +4))dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({e}^{{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$
Commented by mathmax by abdo last updated on 29/Dec/19
let I =∫_0 ^∞   ((arctan(e^x^2  ))/(x^2  +4))dx ⇒2I =∫_(−∞) ^(+∞)  ((arctan(e^x^2  ))/(x^2  +4))  let ϕ(z) =((arctan(e^z^2  ))/(z^2  +4)) ⇒ϕ(z)=((arctan(e^z^2  ))/((z−2i)(z+2i)))  ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,2i) =2iπ×((arctan(e^(−4) ))/(4i)) =(π/2) arctan(e^(−4) )  =(π/2) arctan((1/e^4 )) =(π/2)((π/2) −arctan(e^4 ))=(π^2 /4) −(π/2) arctan(e^4 )
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({e}^{{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({e}^{{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{{arctan}\left({e}^{{z}^{\mathrm{2}} } \right)}{{z}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow\varphi\left({z}\right)=\frac{{arctan}\left({e}^{{z}^{\mathrm{2}} } \right)}{\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\mathrm{2}{i}\right)\:=\mathrm{2}{i}\pi×\frac{{arctan}\left({e}^{−\mathrm{4}} \right)}{\mathrm{4}{i}}\:=\frac{\pi}{\mathrm{2}}\:{arctan}\left({e}^{−\mathrm{4}} \right) \\ $$$$=\frac{\pi}{\mathrm{2}}\:{arctan}\left(\frac{\mathrm{1}}{{e}^{\mathrm{4}} }\right)\:=\frac{\pi}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}\:−{arctan}\left({e}^{\mathrm{4}} \right)\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:−\frac{\pi}{\mathrm{2}}\:{arctan}\left({e}^{\mathrm{4}} \right) \\ $$

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