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Determinate the character of an improper integral

  1. Aug 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi. Well, the statement exhorts me to determinate the character of some improper integrals, and it begins with this one [tex]\displaystyle\int_{-\infty}^{\infty}xe^{-x^2}dx[/tex]

    I really don't know how to work this out. I think its not possible to integrate it, cause its not an elemental function. So I think I should use something, I know it converges to zero, cause I used mathematica to calculate it, but I don't know how to reason it, but maybe using something like the sandwich theorem for function, I mean squeezing it, knowing that integral is not bigger to some other one I could determinate its character. The other thing I thought watching at its graph was demonstrating that its symmetrical to the origin, so thats why it converges to zero. But I'm not sure.

    Bye there. And thanks.
    Last edited: Aug 6, 2010
  2. jcsd
  3. Aug 6, 2010 #2


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    That is an elementary integral and you can find an indefinite integral. Do the u substitution u=x^2.
  4. Aug 6, 2010 #3
    Ups. Thanks Dick.
  5. Aug 6, 2010 #4


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    Have you actually tried doing any of the things you suggested?

    Edit: whoops, I'm too slow.
    I'm pretty sure this would be faster. If nothing else, trying to reason this out is good practice.
  6. Aug 6, 2010 #5
    Maybe, but I don't know if I would realize of it if I wouldn't watched at the graph. I should buy a calculator that makes graphs :P
  7. Aug 6, 2010 #6


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    Symmetries are one of the kinds of things you should always be looking for -- when they exist, exploiting them often makes problems a lot simpler.

    Substituting u = -x is one of the more common things to try. And the more experience you get just trying the substitution and seeing what happens, the better you'll be able to quickly identify when it will be useful (and maybe even spotting other substitutions, like u = 1-x).
  8. Aug 7, 2010 #7


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    Let's be a little careful here. Obviously, by symmetry, [itex]\int_{-a}^a xe^{-x^2}dx= 0[/itex] but that, by itself, does not imply that [itex]\int_{-\infty}^\infty xe^{-x^2}dx[/itex] is equal to 0 or even converges.

    [tex]\lim_{\epsilon\to \infty}\int_{-\epsilon}^\epsilon xe^{-x^2}dx[/itex]
    is the "Cauchy principal value" and may exist even when the true integral:
    [tex]\lim_{\alpha\to\infty}\lim_{\beta\to -\inty}\int_\beta^\alpha xe^{-x^2} dx[/tex]
    does not.
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