Question Number 76352 by mathmax by abdo last updated on 26/Dec/19
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({arctan}\left({x}^{\mathrm{2}} +\mathrm{2}\right)\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$
Commented by mathmax by abdo last updated on 29/Dec/19
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({arctan}\left({x}^{\mathrm{2}} +\mathrm{2}\right)\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:\Rightarrow \\ $$$$\mathrm{2}{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{sin}\left({arctan}\left({x}^{\mathrm{2}} +\mathrm{2}\right)\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:={Im}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{{iarctan}\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)} }{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\right) \\ $$$${let}\:{W}\left({z}\right)=\frac{{e}^{{iarctan}\left({z}^{\mathrm{2}} +\mathrm{2}\right)} }{{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow{W}\left({z}\right)\:=\frac{{e}^{{iarctan}\left({z}^{\mathrm{2}} +\mathrm{2}\right)} }{\left({z}−{i}\right)\left({z}+{i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\right)\:=\mathrm{2}{i}\pi×\frac{{e}^{{iarctan}\left(\mathrm{1}\right)} }{\mathrm{2}{i}}\:=\pi\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$$=\pi\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right\}\:\Rightarrow\:\mathrm{2}{A}\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\Rightarrow{A}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$