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Evaluating integrals with infinity as boundary

  1. Aug 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    Please take a look at this:

    [tex]\int_{ - \infty }^\infty {x \cdot \exp } \left( { - \left| x \right|} \right){\rm{d}}x = \left. {\left( { - \exp \left( { - \left| x \right|} \right) \cdot x} \right)} \right|_{ - \infty }^\infty + \int_{ - \infty }^\infty {\exp \left( { - \left| x \right|} \right){\rm{d}}x}[/tex]

    When evaluating with infinity, does the first time equal zero or infinity? I mean does the exponential function "win" (and then the terms goes to zero) or does x win (and the term the n goes to infinity)?

    The integral is supposed to equal zero.
     
    Last edited: Aug 20, 2008
  2. jcsd
  3. Aug 20, 2008 #2

    Dick

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    I would split it into two integrals -infinity to zero (where the integrand is x*exp(x)) and zero to +infinity (where the integrand is x*exp(-x)) just to get rid of the absolute value. Its very tricky to handle otherwise as your failed attempt shows. But you don't really even have to do that. f(x)=x*exp(-|x|) is an odd function (i.e. f(-x)=-f(x)). The integral of an odd function over a symmetric interval around the origin is always zero (if it exists). The plus part cancels the minus part.
     
  4. Aug 20, 2008 #3
    I actually laughed when reading that - you wrote it in a funny way :smile:

    The argument with the function being odd is good - but just so I have it in my "toolbox" for later, what would I do when I have to evaluate limits like that? What function "wins"?

    Thanks.
     
  5. Aug 20, 2008 #4

    Dick

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    The antiderivative for positive x is F(x)=(-x-1)e^(-x). So for the positive part do F(infinity)-F(0)=0-(-1)=1. (For the infinity part you'll need to take a limit as x -> infinity (-x-1)/e^x, use l'Hopital, you'll find the exponential 'wins' - is that what you mean?). For negative x the antiderivative is F(x)=(x-1)e^(x). F(0)-F(-infinity)=-1-0=-1. Plus and minus parts cancel. Does that help?
     
  6. Aug 20, 2008 #5
    Ahh, l'Hopital [The Hospital Rule, as my teacher used to say] - I had totally forgotten it.

    I solved it - thanks!
     
  7. Aug 20, 2008 #6

    statdad

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    You could also try this.

    [tex]
    \int_{-\infty}^\infty x e^{-|x|} \, dx = \int_{-\infty}^0 x e^{-|x|} \,dx
    +\int_0^\infty x e^{-|x|} \, dx
    [/tex]

    Since [tex] x [/tex] is always positive in the second integral, it is simply

    [tex]
    \int_0^\infty x e^{-x} \, dx
    [/tex]

    In the first integral, since [tex] x [/tex] is always negative,

    [tex]
    \int_{-\infty}^0 x e^{-|x|} \, dx = \int_{-\infty}^0 x e^x \,dx
    [/tex]

    Set u = -x; the second integral becomes

    [tex]
    \int_{\infty}^0 (-u) e^{-u} \, (-du) = \int_{-\infty}^0 u e^{-u} \, du = -\int_0^\infty u e^{-u} \, du
    [/tex]

    The original integral is the sum of these two integrals, and since they are negatives of each other, their sum is zero.
    This is a long use of the fact that the integrand is an odd function of [tex] x [/tex], but the idea of splitting a single integral at zero can be used in other problems.
     
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