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Indefinate integral

  1. Aug 17, 2011 #1
    1. The problem statement, all variables and given/known data
    find the integral from (2 to infinity) of 1/(1+e^x)

    2. Relevant equations

    3. The attempt at a solution

    lim(t-->infinity) integral from (2 to t) of 1/(1+e^x)
    1/1+e^x = (1+e^x-e^x)/(1+e^x)
    = 1-e^x(1+e6x)

    when you solve the integral:
    x-ln(1+e^x) between 2 and t

    here we're confused, because when you plug in t=infinity, you end up with a limit that doesn't exist... help?
  2. jcsd
  3. Aug 17, 2011 #2


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    Well, first, of course, you don't find a limit by "pluggin in t= infinity". This of the form "infinity- infinity" so you should be able to use L'Hopital's rule.
  4. Aug 17, 2011 #3
    then i should hve 1-(e^x)/(1+e^x)?
    and use l'hospital's rule again and have
  5. Aug 17, 2011 #4


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    You are confronted at this point with the limit

    [tex]\lim_{t \rightarrow \infty} t - \ln | 1 + e^{t} | - 2 + \ln | 1 + e^{2} | [/tex]

    Plainly, that t - 2 is going to be a problem (and pulling the logarithms into a single term isn't much better). Try writing the other two terms as logarithms, e.g. , [tex]t \rightarrow \ln(e^{t}) ,[/tex] and write the entire limit expression as a single logarithm. With a little manipulation of the ratio in the argument of the logarithm, you'll find you won't even need l'Hopital...
    Last edited: Aug 17, 2011
  6. Aug 17, 2011 #5

    Ray Vickson

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    In order to get a finite result for int_{x=2..infinity} f(x) dx (f(x) > 0 for all x >= 2) we need f(x) to go to zero quickly enough as x --> infinity. Does this happen for your f(x)?

  7. Aug 17, 2011 #6


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    No. You need a quotient to apply the Hospital rule, so you'd need to rearrange your expression first.

    I think the most straightforward way to evaluate the limit is to use
    [tex]t-\log(1+e^t) = t-\log[e^t(1+e^{-t})][/tex]
    Use the properties of log to simplify your expression and then let t go to infinity.
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