Question Number 76353 by mathmax by abdo last updated on 26/Dec/19
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({sin}\left(\pi{x}^{\mathrm{2}} \right)\right)}{{x}^{\mathrm{2}} \:+\pi^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 29/Dec/19
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({sin}\left(\pi{x}^{\mathrm{2}} \right)\right)}{{x}^{\mathrm{2}} \:+\pi^{\mathrm{2}} }{dx}\:\Rightarrow\mathrm{2}{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({sin}\left(\pi{x}^{\mathrm{2}} \right)\right)}{{x}^{\mathrm{2}} \:+\pi^{\mathrm{2}} } \\ $$$${let}\:{W}\left({z}\right)\:=\frac{{arctan}\left({sin}\left(\pi{z}^{\mathrm{2}} \right)\right)}{{z}^{\mathrm{2}} \:+\pi^{\mathrm{2}} }\:\Rightarrow{W}\left({z}\right)\:=\frac{{arctan}\left({sin}\left(\pi{z}^{\mathrm{2}} \right)\right.}{\left({z}−{i}\pi\right)\left({z}+{i}\pi\right)} \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\pi\right) \\ $$$$=\mathrm{2}{i}\pi×\frac{\mid{arctan}\left({sin}\left(\pi\left(−\pi^{\mathrm{2}} \right)\right)\mid\right.}{\mathrm{2}{i}\pi}\:={arctan}\left({sin}\left(\pi^{\mathrm{3}} \right)\right) \\ $$$$\Rightarrow\:{A}\:=\frac{{arctan}\left({sin}\left(\pi^{\mathrm{3}} \right)\right)}{\mathrm{2}} \\ $$