Question Number 36189 by prof Abdo imad last updated on 30/May/18
$${let}\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{e}^{−{x}^{\mathrm{2}} {t}} \sqrt{{t}}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$${calculate}\:{lim}_{{x}\rightarrow+\infty} \:{F}\left({x}\right)\:. \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 18/Aug/18
$${changement}\:\sqrt{{t}}={u}\:{give} \\ $$$${F}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{x}^{\mathrm{2}} {u}^{\mathrm{2}} } \:\:{u}}{\mathrm{1}+{u}^{\mathrm{4}} }\:\left(\mathrm{2}{u}\right){du} \\ $$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{u}^{\mathrm{2}} \:\:{e}^{−{x}^{\mathrm{2}} {u}^{\mathrm{2}} } }{\mathrm{1}+{u}^{\mathrm{4}} }\:{du} \\ $$$$=_{{xu}\:=\:\alpha} \:\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\alpha^{\mathrm{2}} }{{x}^{\mathrm{2}} }\:\:\frac{{e}^{−\alpha^{\mathrm{2}} } }{\mathrm{1}+\frac{\alpha^{\mathrm{4}} }{{x}^{\mathrm{4}} }}\:\:\frac{\mathrm{1}}{{x}}\:{d}\alpha \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\alpha^{\mathrm{2}} \:{e}^{−\alpha^{\mathrm{2}} } }{{x}^{\mathrm{3}} \:+\frac{\alpha^{\mathrm{4}} }{{x}}}{d}\alpha\:=\mathrm{2}{x}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\alpha^{\mathrm{2}} \:{e}^{−\alpha^{\mathrm{2}} } }{{x}^{\mathrm{4}} \:+\alpha^{\mathrm{4}} }\:{d}\alpha \\ $$$$\Rightarrow\:{F}\left({x}\right)\:\leqslant\frac{\mathrm{2}}{{x}^{\mathrm{3}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\alpha^{\mathrm{2}} \:{e}^{−\alpha^{\mathrm{2}} } {d}\alpha\:\:{but}\:\int_{\mathrm{0}} ^{\infty} \:\:\alpha^{\mathrm{2}} \:{e}^{−\alpha^{\mathrm{2}} } {d}\alpha\: \\ $$$${converges}\:\Rightarrow\:{lim}_{{x}\rightarrow+\infty} {F}\left({x}\right)=\mathrm{0}\:. \\ $$