Algebra Questions

Question Number 151449 by mathdanisur last updated on 21/Aug/21

$$\Omega\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \right)\:\mathrm{dx}\:=\:? \\$$

Answered by puissant last updated on 21/Aug/21

$${f}\left({u}\right)={e}^{−{u}^{\mathrm{2}} } \\$$$$\Rightarrow\:\oint{f}\left({u}\right){du}=\int_{\mathrm{0}} ^{{a}} {e}^{−{t}^{\mathrm{2}} } {dt}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {iae}^{{it}} {e}^{−{a}^{\mathrm{2}} {exp}\left(\mathrm{2}{it}\right)} {dt}−\int_{\mathrm{0}} ^{{a}} {e}^{{i}\frac{\pi}{\mathrm{4}}} {e}^{−{it}^{\mathrm{2}} } {dt} \\$$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{{a}} {e}^{−{t}^{\mathrm{2}} } {dt} \\$$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {iae}^{{it}} {e}^{−{a}^{\mathrm{2}} {exp}\left(\mathrm{2}{it}\right)} {dt} \\$$$${I}_{\mathrm{3}} =\int_{\mathrm{0}} ^{{a}} {e}^{{i}\frac{\pi}{\mathrm{4}}} {e}^{−{it}^{\mathrm{2}} } {dt} \\$$$$\mid{I}_{\mathrm{2}} \left({a}\right)\mid\leqslant\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ae}^{−{a}^{\mathrm{2}} {cos}\left(\mathrm{2}{t}\right)} {dt}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{a}}{\mathrm{2}}{e}^{−{a}^{\mathrm{2}} {cosz}} {dz} \\$$$$\forall{u}\in\left[\mathrm{0};\frac{\pi}{\mathrm{2}}\right],\:\mathrm{1}−\frac{\mathrm{2}{z}}{\pi}\leqslant{cosz}\leqslant\mathrm{1} \\$$$$\Rightarrow\:\forall{u}\in\left[\mathrm{0};\frac{\pi}{\mathrm{2}}\right],\:{e}^{−{a}^{\mathrm{2}} {cosz}} \leqslant{e}^{{a}^{\mathrm{2}} \left(\frac{\mathrm{2}{z}}{\pi}−\mathrm{1}\right)} \\$$$$\Rightarrow\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{a}}{\mathrm{2}}{e}^{−{a}^{\mathrm{2}} {cosz}} {dz}\leqslant\frac{\pi}{\mathrm{4}{a}}\left(\mathrm{1}−{e}^{−{a}^{\mathrm{2}} } \right) \\$$$${lim}_{{a}\rightarrow\infty} {I}_{\mathrm{2}} \left({a}\right)=\mathrm{0}\:\left({gendarme}\:{theoreme}\right) \\$$$${I}_{\mathrm{1}} \left({a}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\:\left({Gauss}\:{integrale}\right) \\$$$${lim}_{{a}\rightarrow\infty} {I}_{\mathrm{3}} \left({a}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{{i}\frac{\pi}{\mathrm{4}}} {e}^{−{it}^{\mathrm{2}} } {dt}\:=\:\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} {e}^{−{it}^{\mathrm{2}} } {dt} \\$$$${f}\:{parcout}\:{une}\:{distance}\:{ferme}, \\$$$$\Rightarrow\:\oint{f}\left({u}\right){du}=\mathrm{0} \\$$$$\Rightarrow\:\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} {e}^{−{it}^{\mathrm{2}} } {dt}=\frac{\sqrt{\pi}}{\mathrm{2}} \\$$$$\Rightarrow\:\int_{\mathrm{0}} ^{\infty} {e}^{−{it}^{\mathrm{2}} } {dt}=\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)\sqrt{\frac{\pi}{\mathrm{2}}} \\$$$${Q}=\int_{\mathrm{0}} ^{\infty} {sin}\left({x}^{\mathrm{2}} \right){dx}={Im}\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{it}^{\mathrm{2}} } {dt}\right)=\sqrt{\frac{\pi}{\mathrm{8}}}..\blacksquare \\$$$${Remarque}:\:\int_{\mathrm{0}} ^{\infty} {sin}\left({x}^{\mathrm{2}} \right){dx}=\int_{\mathrm{0}} ^{\infty} {cos}\left({x}^{\mathrm{2}} \right){dx}=\sqrt{\frac{\pi}{\mathrm{8}}}..\bigstar \\$$$${appele}\:{Integrale}\:{de}\:{FRESNEL}.. \\$$$$\\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:..............\mathscr{L}{e}\:{puissant}.............. \\$$

Commented by mathdanisur last updated on 21/Aug/21

$$\mathrm{Thank}\:\mathrm{yoy}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{ans}\:\sqrt{\pi/\mathrm{8}}\:\mathrm{or}\:\sqrt{\pi/\mathrm{8}}\:{i}.? \\$$

Commented by puissant last updated on 21/Aug/21

$$\sqrt{\frac{\pi}{\mathrm{8}}}\:\:{because}\:{e}^{{ix}} ={cosx}+{isinx} \\$$$${Re}\left({e}^{{ix}} \right)={cosx}\:\:{and}\:{im}\left({e}^{{ix}} \right)={sinx} \\$$

Answered by Ar Brandon last updated on 26/Aug/21

$$\Omega=\int_{\mathrm{0}} ^{\infty} \mathrm{sin}\left({x}^{\mathrm{2}} \right){dx},\:{u}={x}^{\mathrm{2}} \\$$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {u}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{sin}\left({u}\right){du} \\$$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\pi}{\mathrm{2}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}\right)} \\$$$$\:\:\:\:=\frac{\pi}{\:\mathrm{2}\sqrt{\mathrm{2}\pi}}=\sqrt{\frac{\pi}{\mathrm{8}}} \\$$