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Improper Integration

by Redoctober
Tags: improper, integration
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Redoctober
#1
Mar22-12, 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 !
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disregardthat
#2
Mar22-12, 01:26 PM
Sci Advisor
P: 1,741
Try to use that e^x >= x when x >= 1.
arildno
#3
Mar22-12, 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?

Redoctober
#4
Mar22-12, 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]
Char. Limit
#5
Mar22-12, 01:53 PM
PF Gold
Char. Limit's Avatar
P: 1,941
Compare it to g(x)=1, then.
Redoctober
#6
Mar22-12, 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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