Question Number 26107 by abdo imad last updated on 19/Dec/17
$${answer}\:{to}\:\mathrm{26024}\:\:\:\:{let}\:\:{put}\:{c}=\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left({ax}^{\mathrm{2}} \right){dx}\:\:\:{and}\:\:\:{c}\:=\:\:\int_{\mathrm{0}} ^{\infty} \:{sin}\left({ax}^{\mathrm{2}} \right){dx} \\ $$$${ew}\:{have}\:\:{c}−{is}\:\:=\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{iax}^{\mathrm{2}} } {dx}\:\:\:=\mathrm{2}^{−\mathrm{1}} \:\:\int_{{R}} \:{e}^{−{iax}^{\mathrm{2}} } {dx}\:\:\:{and}\:\:{i}\:{put} \\ $$$${x}^{\mathrm{1}/\mathrm{2}} \:={r}\left({x}\right)\left({notation}\right)\:\:{so}\:\:\mathrm{2}\left({c}−{is}\right)\:\:=\:\:\:\int_{{R}} \:\:{e}^{−\left({r}\left({ia}\right){x}\right)^{\mathrm{2}} } {dx} \\ $$$${and}\:{by}\:{the}\:{changement}\:\:\:{t}=\:{r}\left({ia}\right)\:{x}\:\:\:{we}\:{find} \\ $$$$\mathrm{2}\left({c}+{is}\right)\:\:=\:\:\:\left({r}\left({ia}\right)\right)^{−\mathrm{1}} \:\int_{{R}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:\:\:=\:\:{r}\left(\pi\right)/{r}\left({ia}\right)\:\:{but} \\ $$$${r}\left({ia}\right)\:\:={r}\left({i}\right)\:{r}\left({a}\right)\:\:=\:\:{r}\left({a}\right)\:{e}^{} \:\:\:−−>\mathrm{2}\left({c}+{is}\right)\:\:\:\:=\:\:{r}\left(\pi\right)\:{r}\left({a}\right)^{−\mathrm{1}} \:\:{e}^{−{i}\pi/\mathrm{4}} \:\:^{\left.\right)} \\ $$$$−−>\:\:{c}\:=\:\:\:{r}\left(\mathrm{2}\pi\right)/_{\mathrm{4}{r}\left({a}\right)} \:\:{and}\:\:\:{s}\:\:=\:\:{r}\left(\mathrm{2}\pi\right)/_{\mathrm{4}{r}\left({a}\right)} \\ $$