Question Number 40160 by maxmathsup by imad last updated on 16/Jul/18
$${study}\:{the}\:{convergence}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}−{e}^{−{t}} }{{t}\sqrt{{t}}}\:{dt} \\ $$
Commented by math khazana by abdo last updated on 18/Jul/18
![let I = ∫_0 ^1 ((1−e^(−t) )/(t(√t)))dt changement (√t)=x give I = ∫_0 ^1 ((1−e^(−x^2 ) )/x^3 ) 2x dx = 2 ∫_0 ^1 ((1−e^(−x^2 ) )/x^2 )dx we have e^(−x^2 ) = Σ_(n=0) ^∞ (((−x^2 )^n )/(n!)) =1−x^2 +(x^4 /2) +o(x^4 ) ⇒1−e^(−x^2 ) = x^2 +(x^4 /2) +o(x^4 )⇒ ((1−e^(−x^2 ) )/x^2 ) =1+(x^2 /2) +o(x^2 ) ⇒lim_(x→0) ((1−e^(−x^2 ) )/x^2 ) =1 and the function is continue on]0,1] so the function is integrable on]0,1] so the integral is convergent .](https://www.tinkutara.com/question/Q40288.png)
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−{e}^{−{t}} }{{t}\sqrt{{t}}}{dt}\:\:{changement}\:\sqrt{{t}}={x}\:{give} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{3}} }\:\mathrm{2}{x}\:{dx}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx}\:\: \\ $$$${we}\:{have}\:\:{e}^{−{x}^{\mathrm{2}} } =\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−{x}^{\mathrm{2}} \right)^{{n}} }{{n}!} \\ $$$$=\mathrm{1}−{x}^{\mathrm{2}} \:\:+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{4}} \right)\:\Rightarrow\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } \:=\:{x}^{\mathrm{2}} \:+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{4}} \right)\Rightarrow \\ $$$$\frac{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }\:=\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{2}} \right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }\:=\mathrm{1}\: \\ $$$$\left.{a}\left.{nd}\:{the}\:{function}\:{is}\:{continue}\:{on}\right]\mathrm{0},\mathrm{1}\right]\:{so}\:\:{the}\: \\ $$$$\left.{f}\left.{unction}\:{is}\:{integrable}\:{on}\right]\mathrm{0},\mathrm{1}\right]\:{so}\:{the}\:{integral}\:{is} \\ $$$${convergent}\:. \\ $$