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Question Number 66169 by mathmax by abdo last updated on 10/Aug/19

find the values of ∫_0 ^∞ cos(x^2 )dx and ∫_0 ^∞ sin(x^2 )dx(fresnel integrals) by using Γ(z) =∫_0 ^∞ t^(z−1) e^(−t) dt

$${find}\:{the}\:{values}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{\mathrm{2}} \right){dx}\:{and}\:\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{2}} \right){dx}\left({fresnel}\:{integrals}\right) \\ $$$${by}\:{using}\:\Gamma\left({z}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{t}^{{z}−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:\: \\ $$

Commented by mathmax by abdo last updated on 10/Aug/19

we have ∫_0 ^∞ cos(x^2 )dx =Re(∫_0 ^∞ e^(−ix^2 ) dx) changement ix^2 =t give x^2 =−it ⇒x =(−it)^(1/2) =(−i)^(1/2) t^(1/2) ⇒dx =(1/2)(−i)^(1/2) t^((1/2)−1) so ∫_0 ^∞ e^(−ix^2 ) dx =(1/2)(−i)^(1/2) ∫_0 ^∞ t^((1/2)−1) e^(−t) dt =(1/2)e^(−((iπ)/4)) .Γ((1/2)) =(1/2)Γ((1/2)){(1/(√2))−(i/(√2))} ⇒ ∫_0 ^∞ cos(x^2 )dx =(1/(2(√2)))Γ((1/2)) we have Γ(x).Γ(1−x)=(π/(sin(πx))) ⇒Γ^2 ((1/2)) =π ⇒Γ((1/2))=(√π) ⇒∫_0 ^∞ cos(x^2 )dx =((√π)/(2(√2))) =((√(2π))/4) also ∫_0 ^∞ sin(x^2 )dx =−Im(∫_0 ^∞ e^(−ix^2 ) dx) ⇒ ∫_0 ^∞ sin(x^2 )dx =((√(2π))/4)

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

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