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