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Improper Integration 
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#1
Mar2212, 01:15 PM

P: 41

Help :@ !!
the question states that [tex]\int^{\infty}_{1}\frac{e^x}{x}.dx[/tex] Determine whether the integrand is convergent or divergent ?? It tried using the limit comparison test but i fail to select a g(x) to compare it with this i chose [tex]g(x) = \frac{e^x}{x +1}[/tex] but this hard to integrate too :S ! I can't find any other function to compare with cause i need to cancel out the e^x this seek for functions to compare the integral seem more like luck factor dependent lol ! 


#2
Mar2212, 01:26 PM

Sci Advisor
P: 1,741

Try to use that e^x >= x when x >= 1.



#3
Mar2212, 01:28 PM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Well, isn't e^x greater than 1 for ALL positive values for x above x=1?



#4
Mar2212, 01:32 PM

P: 41

Improper Integration
I do know that e^x > than x when x>1 thus i know that the integral is divergent . but the question is asking me to prove it via the comparison test
[tex]\lim_{x \to \infty } \frac{f(x)}{g(x)}[/tex] 


#6
Mar2212, 01:54 PM

P: 41

Its alright i found out the solution
It turns out, because f(x) is divergent , i shall choose a smaller function g(x) but diverge too i chose [tex]g(x) = \frac{e^x}{e^x + 1}[/tex] [tex]\int^{\infty}_{1} \frac{e^x}{e^x + 1}.dx [/tex] the integral is easy to calculate . It will give infinity . Because the g(x) < f(x) and g(x) integral is divergent , thus f(x) integral is divergant too :D ! 


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