Question Number 68597 by Abdo msup. last updated on 14/Sep/19
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({e}^{{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}{dx} \\ $$
Commented by mathmax by abdo last updated on 14/Sep/19
$${residus}\:{method}\:{but}\:{need}\:{a}\:{proof} \\ $$$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({e}^{{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({e}^{{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}{dx} \\ $$$${let}\:{W}\left({z}\right)\:=\frac{{arctan}\left({e}^{{z}^{\mathrm{2}} } \right)}{{z}^{\mathrm{2}} \:+\mathrm{8}}\:\Rightarrow{W}\left({z}\right)\:=\frac{{arctan}\left({e}^{{z}^{\mathrm{2}} } \right)}{\left({z}−\mathrm{2}{i}\sqrt{\mathrm{2}}\right)\left({z}+\mathrm{2}{i}\sqrt{\mathrm{2}}\right)} \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}=\mathrm{2}{i}\pi\:{Res}\left({W},\mathrm{2}{i}\sqrt{\mathrm{2}}\right) \\ $$$$=\frac{{arctan}\left({e}^{\left(\mathrm{2}{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } \right)}{\mathrm{4}{i}\sqrt{\mathrm{2}}}\:=\frac{{arctan}\left({e}^{−\mathrm{8}} \right)}{\mathrm{4}{i}\sqrt{\mathrm{2}}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{{arctan}\left({e}^{−\mathrm{8}} \right)}{\mathrm{4}{i}\sqrt{\mathrm{2}}} \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:{arctan}\left({e}^{−\mathrm{8}} \right)\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:{arctan}\left({e}^{−\mathrm{8}} \right)\:. \\ $$