$${find}\:{the}\:{values}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{\mathrm{2}} \right){dx}\:{and}\:\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{2}} \right){dx}\left({fresnel}\:{integrals}\right) \\$$$${by}\:{using}\:\Gamma\left({z}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{t}^{{z}−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:\: \\$$
$${we}\:{have}\:\int_{\mathrm{0}} ^{\infty} {cos}\left({x}^{\mathrm{2}} \right){dx}\:={Re}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ix}^{\mathrm{2}} } {dx}\right)\:\:{changement}\:{ix}^{\mathrm{2}} ={t} \\$$$${give}\:{x}^{\mathrm{2}} =−{it}\:\Rightarrow{x}\:=\left(−{it}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\left(−{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {t}^{\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(−{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \\$$$${so}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ix}^{\mathrm{2}} } {dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(−{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\int_{\mathrm{0}} ^{\infty} \:{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \:{e}^{−{t}} {dt} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:.\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right\}\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{\mathrm{2}} \right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\:\:{we}\:{have}\:\Gamma\left({x}\right).\Gamma\left(\mathrm{1}−{x}\right)=\frac{\pi}{{sin}\left(\pi{x}\right)} \\$$$$\Rightarrow\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\pi\:\Rightarrow\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{\mathrm{2}} \right){dx} \\$$$$=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\:=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{4}} \\$$$${also}\:\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{2}} \right){dx}\:=−{Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ix}^{\mathrm{2}} } {dx}\right)\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{2}} \right){dx}\:=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{4}} \\$$$$\\$$