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Question Number 68597 by Abdo msup. last updated on 14/Sep/19

calculate ∫_0 ^∞    ((arctan(e^x^2  ))/(x^2  +8))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({e}^{{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}{dx} \\ $$

Commented by mathmax by abdo last updated on 14/Sep/19

residus method but need a proof  let I =∫_0 ^∞   ((arctan(e^x^2  ))/(x^2  +8)) ⇒2I =∫_(−∞) ^(+∞)  ((arctan(e^x^2  ))/(x^2  +8))dx  let W(z) =((arctan(e^z^2  ))/(z^2  +8)) ⇒W(z) =((arctan(e^z^2  ))/((z−2i(√2))(z+2i(√2))))  residus theorem give ∫_(−∞) ^(+∞)  W(z)dz=2iπ Res(W,2i(√2))  =((arctan(e^((2i(√2))^2 ) ))/(4i(√2))) =((arctan(e^(−8) ))/(4i(√2))) ⇒∫_(−∞) ^(+∞)  W(z)dz =2iπ×((arctan(e^(−8) ))/(4i(√2)))  =(π/(√2)) arctan(e^(−8) ) ⇒ I =(π/(2(√2))) arctan(e^(−8) ) .

$${residus}\:{method}\:{but}\:{need}\:{a}\:{proof} \\ $$$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({e}^{{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({e}^{{x}^{\mathrm{2}} } \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}{dx} \\ $$$${let}\:{W}\left({z}\right)\:=\frac{{arctan}\left({e}^{{z}^{\mathrm{2}} } \right)}{{z}^{\mathrm{2}} \:+\mathrm{8}}\:\Rightarrow{W}\left({z}\right)\:=\frac{{arctan}\left({e}^{{z}^{\mathrm{2}} } \right)}{\left({z}−\mathrm{2}{i}\sqrt{\mathrm{2}}\right)\left({z}+\mathrm{2}{i}\sqrt{\mathrm{2}}\right)} \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}=\mathrm{2}{i}\pi\:{Res}\left({W},\mathrm{2}{i}\sqrt{\mathrm{2}}\right) \\ $$$$=\frac{{arctan}\left({e}^{\left(\mathrm{2}{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } \right)}{\mathrm{4}{i}\sqrt{\mathrm{2}}}\:=\frac{{arctan}\left({e}^{−\mathrm{8}} \right)}{\mathrm{4}{i}\sqrt{\mathrm{2}}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{{arctan}\left({e}^{−\mathrm{8}} \right)}{\mathrm{4}{i}\sqrt{\mathrm{2}}} \\ $$$$=\frac{\pi}{\sqrt{\mathrm{2}}}\:{arctan}\left({e}^{−\mathrm{8}} \right)\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:{arctan}\left({e}^{−\mathrm{8}} \right)\:. \\ $$

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