Question Number 35237 by abdo.msup.com last updated on 17/May/18
$${study}\:{the}\:{convergence}\:{of} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}} \:−{e}^{−{x}^{\mathrm{2}} } }{{x}}{dx}\:. \\ $$
Commented by prof Abdo imad last updated on 20/May/18
$${we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} −{e}^{−{x}^{\mathrm{2}} } }{{x}}{dx}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(...\right){dx}+\int_{\mathrm{1}} ^{+\infty} \left(...\right){dx}\:{and}\:{e}^{−{x}} \:=\mathrm{1}−{x}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{3}} \right) \\ $$$${e}^{−{x}^{\mathrm{2}} } \:=\mathrm{1}−{x}^{\mathrm{2}} \:\:+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\:+\:{o}\:\left({x}^{\mathrm{6}} \right)\:\Rightarrow \\ $$$${e}^{−{x}} \:−{e}^{−{x}^{\mathrm{2}} } \:\:\:\sim\:\:{x}^{\mathrm{2}} −{x}\:\:+\frac{{x}^{\mathrm{2}} \:−{x}^{\mathrm{4}} }{\mathrm{2}}\:\left({x}\rightarrow\mathrm{0}\right)\Rightarrow \\ $$$$\frac{{e}^{−{x}} \:\:−{e}^{−{x}^{\mathrm{2}} } }{{x}}\:\sim\:{x}−\mathrm{1}\:\:+\frac{{x}\:−{x}^{\mathrm{3}} }{\mathrm{2}}\:\left({x}\rightarrow\mathrm{0}\right) \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{e}^{−{x}} \:−{e}^{−{x}^{\mathrm{2}} } }{{x}}\:\:=−\mathrm{1}\:{so}\:{the}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{e}^{−{x}} −{e}^{−{x}^{\mathrm{2}} } }{{x}}{dx} \\ $$$${converges}\:\:{also}\:{we}\:{have}\:{lim}_{{x}\rightarrow+\infty} {x}^{\mathrm{2}} \frac{{e}^{−{x}} \:−{e}^{−{x}^{\mathrm{2}} } }{{x}} \\ $$$$=\mathrm{0}\:{so}\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{e}^{−{x}} \:−{e}^{−{x}^{\mathrm{2}} } }{{x}}\:{converges}. \\ $$