Question Number 74348 by mathmax by abdo last updated on 22/Nov/19
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({sin}\left({x}^{\mathrm{2}} \right)\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$
Commented by mathmax by abdo last updated on 30/Nov/19
$${let}\:{I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({sin}\left({x}^{\mathrm{2}} \right)\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({sinx}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${use}\:{of}\:{dedmos}\:{give}\:\frac{{arctan}\left({sin}\left({x}^{\mathrm{2}} \right)\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:\geqslant\mathrm{0}\:\:\:\:{I}\:>\mathrm{0} \\ $$$${let}\:\varphi\left({z}\right)=\frac{{arctan}\left({sinz}^{\mathrm{2}} \right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow\varphi\left({z}\right)=\frac{{arctan}\left({sinx}^{\mathrm{2}} \right)}{\left({z}−{i}\right)\left({z}+{i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:=\mathrm{2}{i}\pi×\frac{\mid{arctan}\left({sin}\left(−\mathrm{1}\right)\right)\mid}{\mathrm{2}{i}} \\ $$$$=\pi\:{arctan}\left({sin}\mathrm{1}\right)\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{2}}\:{arctan}\left({sin}\left(\mathrm{1}\right)\right) \\ $$