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Improper Integration

  1. Mar 22, 2012 #1
    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 !
     
    Last edited: Mar 22, 2012
  2. jcsd
  3. Mar 22, 2012 #2

    disregardthat

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    Try to use that e^x >= x when x >= 1.
     
  4. Mar 22, 2012 #3

    arildno

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    Dearly Missed

    Well, isn't e^x greater than 1 for ALL positive values for x above x=1?
     
  5. Mar 22, 2012 #4
    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. Mar 22, 2012 #5

    Char. Limit

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    Compare it to g(x)=1, then.
     
  7. Mar 22, 2012 #6
    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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