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Integral: e^(-1/x)

  1. Jan 12, 2010 #1
    Can someone help me with this integral?

  2. jcsd
  3. Jan 12, 2010 #2
    Just follow the rules for exponential integration.

    [tex]\int[/tex] [tex]e^{u}[/tex] du = [tex]e^{u}[/tex] + C

    Last edited: Jan 12, 2010
  4. Jan 12, 2010 #3


    Staff: Mentor

    If this is the integral:
    [tex]\int e^{-1/x}dx[/tex]

    an ordinary substitution is not much help. Alexx1, can you show us the complete integral you're trying to do?
  5. Jan 12, 2010 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    If the problem were [itex]\int e^u du[/itex], but it isn't and there is no good way to change it to that form.

    It looks to me like [itex]\int e^{1/x} dx[/itex] does not have an anti-derivative in terms of elementary functions.
  6. Jan 13, 2010 #5
    in terms of an exponential integral function...

    \int \!{{\rm e}^{-{x}^{-1}}}{dx}=x{{\rm e}^{-{x}^{-1}}}-{\rm Ei}_1
    \left({x}^{-1} \right)

  7. Jan 13, 2010 #6

    Yes, now I see that a simple substitution is not the way to proceed. Thanks for correcting me.

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