Question Number 91621 by abdomathmax last updated on 02/May/20
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{6}} \right){dx} \\ $$
Commented by mathmax by abdo last updated on 02/May/20
$$\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{6}} \right){dx}\:={Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{{ix}^{\mathrm{6}} } {dx}\right)\:\:{we}\:{have}\:{by}\:{changement}\:{ix}^{\mathrm{6}} \:=−{t} \\ $$$$\Rightarrow{x}^{\mathrm{6}} \:={it}\:\Rightarrow\:{x}\:=\left({it}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} \:={e}^{\frac{{i}\pi}{\mathrm{12}}} \:{t}^{\frac{\mathrm{1}}{\mathrm{6}}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{{ix}^{\mathrm{6}} } {dx}\:={e}^{\frac{{i}\pi}{\mathrm{12}}} \int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{t}} \:×\frac{\mathrm{1}}{\mathrm{6}}\:{t}^{\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}{e}^{\frac{{i}\pi}{\mathrm{12}}} \:\:\int_{\mathrm{0}} ^{\infty} \:{t}^{\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:\:\:\:\:{we}\:{have}\:\Gamma\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)\left({cos}\left(\frac{\pi}{\mathrm{12}}\right)+{isin}\left(\frac{\pi}{\mathrm{12}}\right)\right)\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{6}} \right){dx}=\frac{\mathrm{1}}{\mathrm{6}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right){sin}\left(\frac{\pi}{\mathrm{12}}\right) \\ $$