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Integral of x^{3}e^{-x^2}dx

  1. Mar 10, 2007 #1

    I'm trying to show the following equation is correct:

    [tex]\int_{-\infty}^{\infty}x^{3}e^{-x^2}dx = 0[/tex]

    I obtained the result as 0 using Mathematica but couldn't figure out a way to evaluate the integral.

    I am just an unfortunate computer scientist who happens to follow a graduate course on statistical mechanics :)

    Also, I would appreciate if some one tells me how to query the forums to find an answer for a particular integration before posting it here.


    Last edited: Mar 10, 2007
  2. jcsd
  3. Mar 10, 2007 #2


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    Is the integrand an even or odd function?
  4. Mar 10, 2007 #3
    Unfortunately [tex]x[/tex] can take negative values if that's what you ask.
  5. Mar 10, 2007 #4
    I see, I am expected to show my own work when posting a question here. Actually that integral is just a small part of an answer to calculate the first four moments of a gaussian distribution with direct integration of the pdf.

    I tried to re-write the integral as
    [tex]\frac{1}{2}\int_{-\infty}^{\infty}t e^{-t}dt[/tex]
    by using the substitution
    [tex]t=u^2[/tex], [tex]dt=2u du[/tex]
    [tex]u du=\frac{dt}{2}[/tex].
    By this approach I can say that t is positive and evaluate the integral as a Gamma function by carrying it from 0 to infinity and multiplying by 2. The problem is the result is not 0 which I am expected to get.
  6. Mar 10, 2007 #5


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    No, my hint concerned the type of integrand you're having.
    With an EVEN function f, we mean that for any x, f(-x)=f(x)

    With an ODD function, we mean that for any x, we have f(-x)=-f(x)

    Suppose you are to integrate an odd function between the values -a and a.
    What should that integral be?
  7. Mar 10, 2007 #6
    I got it arildno, thanks very much.

    Could you also tell me how to search for the integrals before posting them here? I tried to use the latex code of the integral but it didn't help.
  8. Mar 10, 2007 #7

    Gib Z

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    its quite difficult to search for a specific integral, best bet is it search integral. no latex wont work with the searches.

    Basically when you have a function like arildno described, odd, then on the postive side of x, x>0, the values are the same as the values on the negative side, just with a different sign! so they cancel each other out.
  9. Mar 10, 2007 #8
    functions.wolfram.com has good functions search
  10. Mar 11, 2007 #9


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    Arildno's point was that you don't need to find an anti-derivative. The fact that that integrand is an odd function tells you that its graph is anti-symmetric. Any "area under the curve" for for x> 0 is cancelled by the "area above the curve" for x< 0. More specifically, the anti-derivative of any odd function is an even function. Evaluating it at any A and -A give the so subtracting results in 0.

    However there are other ways of integrating functions than "looking them up"! In this case you can write the integral as
    [tex]\int x^2 e^{-x^2} (xdx)[/tex]
    and make the substitution u= x2. That reduces the integral to
    [tex]2\int u e^{-u}du[/tex]
    which can be done by a simple integration by points.

    However, arildno's point, as I said, is that you don't need to actually do the integral to get the answer!
  11. Mar 11, 2007 #10

    Gib Z

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    That [tex]2\int u e^{-u} du[/tex] should be a [tex]\frac{1}{2}\int u e^{-u} du[/tex] I think halls.
  12. Mar 11, 2007 #11
    Yeah, as I said I got the idea of integrating odd functions between -a and a, thank you very much. I solved my problem.

    I tried to tell that I already did that substitution in my posting number 4. But I couldn't evaluate that integral either. I mistakenly argued that since [tex]u=x^2[/tex], I can say that u is always positive and the integral is equal to [tex]\int_{0}^{\infty}u e^{-u}du[/tex] (which happens to be the [tex]\Gamma[/tex] function). Then I realized that although [tex]u[/tex] is positive, [tex]du[/tex] is not. So it is still an odd function. Everything is fine now.
  13. Mar 12, 2007 #12


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    Yes, of course it is. I think I'll pretend the LaTex messed up!

    In fact, if you had converted the limits of integration to u when you changed variables, you would have found you were integrating from infinity to infinity! What is the integral of any function from a to a?
  14. Mar 12, 2007 #13
    Hmm, beautiful :)
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