Question Number 25769 by abdo imad last updated on 14/Dec/17
$${a}−{nser}\:{to}\:{question}\:\mathrm{25765}…{we}\:{put}\:{I}=\int_{\mathrm{0}} ^{\infty} \left({cos}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}\:{and}\:{J}=\int_{\mathrm{0}} ^{\infty} {sin}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}…{we}\:{have}\:\mathrm{2}\left({I}+{iJ}\right)=\int_{{R}} {e}^{{ix}^{\mathrm{2}{n}} } \left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}…{let}\:{f}\left({z}\right)={e}^{{ix}^{\mathrm{2}{n}} } \left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {by}\:{residus}\:{therem}\:\int_{{R}} {f}\left({z}\right){dz}\:=\:\mathrm{2}{i}\pi{Res}\left({f}.{i}\right)\:{but}\:{Res}\left({f}.{i}\right)=\:{lim}_{{z}−{i}} \left({z}−{i}\right){f}\left({z}\right)={e}^{{i}\left({i}\right)^{\mathrm{2}{n}} } =\:{e}^{{i}\left(−\mathrm{1}\right)_{} ^{{n}} } \left(\mathrm{2}{i}\right)^{−\mathrm{1}} {then}\:\int_{{R}} {f}\left({z}\right){dz}\:=\:\pi{e}^{\left(−\mathrm{1}\right)^{{n}} } =\:\pi\left(\:{cos}\left(−\mathrm{1}\right)^{{n}} \:+{isin}\left(−\mathrm{1}\right)^{{n}} \right)\:{then}\:\int_{\mathrm{0}} ^{\infty} {cos}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}\:=\pi\mathrm{2}^{−\mathrm{1}} {cos}\left(−\mathrm{1}\right)^{{n}} {and}\:\int_{\mathrm{0}} ^{\infty} {sin}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}=\pi.\mathrm{2}^{−\mathrm{1}} {sin}\left(−\mathrm{1}\right)^{{n}_{} } <>….\right. \\ $$