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Homework Help: Improper Integral

  1. Jun 12, 2010 #1
    This is the problem:
    [tex]\int_{-\infty}^{\infty} {xe^{-x^{2}}dx}[/tex]

    I noticed the function was even, so I then did this:

    [tex]2\int_{0}^{\infty} {xe^{-x^{2}}dx}[/tex]

    I attempted to do integration by parts:

    [tex]u=e^{-x^{2}}, du=-2e^{-x^{2}}, dv=x, v=\frac{x^{2}}{2}[/tex]

    which still left me with this at the end:
    (I left out the bounds)
    [tex]\frac{x^{2}}{2}(e^{-x^{2}} - \int_{}^{} {e^{-x^{2}}dx})[/tex]

    and of course that integral can't be done. Doing parts the other way results with the same issue, and I don't see how a substitution could work. Thanks in advance for any help.
  2. jcsd
  3. Jun 12, 2010 #2


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    Science Advisor
    Homework Helper

    Why do you think x*e^(-x^2) is even?
  4. Jun 12, 2010 #3


    Staff: Mentor

    The integrand is not an even function - it is odd. This should make the problem a lot easier.
    That's not the best approach. An ordinary substitution is all that is required.
  5. Jun 12, 2010 #4
    I see. I'll post back once I'm done with the work.
    Last edited: Jun 12, 2010
  6. Jun 12, 2010 #5
    Bad logic. That's why. I thought about it more and it isn't.
  7. Jun 12, 2010 #6
    It worked.

    [tex]u=e^{-x^{2}}, du=-2e^{-x^{2}}dx \rightarrow \frac{du}{-2u}=xdx[/tex]

    Then got here:

    [tex]\lim_{t \rightarrow -\infty}\int_{t}^{0} {xe^{-x^{2}}dx} + \lim_{t \rightarrow \infty}\int_{0}^{t} {xe^{-x^{2}}dx}[/tex]

    [tex]\lim_{t \rightarrow -\infty}\int_{x=t}^{x=0} {-\frac{du}{2u}u} + \lim_{t \rightarrow \infty}\int_{x=0}^{x=t} {-\frac{du}{2u}u}[/tex]

    [tex]-\frac{1}{2}\lim_{t \rightarrow -\infty}\int_{x=t}^{x=0} {du} + -\frac{1}{2}\lim_{t \rightarrow \infty}\int_{x=0}^{x=t} {du}[/tex]

    Then I just got u as the integral, did the rest and found that it converged on zero. Thanks for the help.
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