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

  1. May 2, 2008 #1
    I ran into an integral while working on response of a signal processing filter, it looks like:

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

    While trying integration by parts u = [tex]x^{2}[/tex] we get du = 2xdx but can't proceed with dv = [tex]e^{-x^{2}}[/tex] because then
    v = [tex]\int e^{-x^{2}}[/tex]
    can't be integrated unless we use the limits.

    Can anyone suggest an approach for this?

  2. jcsd
  3. May 2, 2008 #2


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    What exactly are you trying to do? As you point out, your v is not any elementary function, and that tells you that neither is
    [tex]\int x^2e^{-x^2} dx[/tex]
    You might be able to do that in terms of the "error function", Erf(x), which is defined to be
    [tex]\int e^{-x^2} dx[/tex]
  4. May 2, 2008 #3
    If you just want to calculate the definite integral, I don't see why you wouldn't want to include the limits when integrating by parts?
  5. May 2, 2008 #4


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    His point, about the limits of integration, was that it is well known that
    [tex]\int_{-\infty}^{\infty} e^{x^2}dx= 2\sqrt{\pi}[/itex]
    while the anti-derivative is not an elementary function. in other words, he could, theoretically, do it as a definite integral with the right limits but not as an indefinite integral.
  6. May 2, 2008 #5
    Was that adressed to me?

    Anyway, Maple tells me that:
    \int_{-\infty}^{\infty} x^{n} e^{-x^{2}}\rm{d} x=\frac{1}{2}\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)\left(1+(-1)^n\right)
    [/tex] ,
    which should be possible to prove by induction.

    PS. HallsofIvy: You have forgotten the minus-sign in your integrand.
  7. May 2, 2008 #6


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    A cute way to solve this is to recall that


    Then use Feynman's favorite trick and differentiate both sides with respect to a, and evaluate at a = 1.
  8. May 2, 2008 #7
    For one, it's [itex]e^{-x^2}[/itex] and for two, it's [itex] \sqrt{\pi}[/itex].
  9. May 2, 2008 #8
    I think I should use this as a guideline for future problems. Both u and v should be elementary functions otherwise integration by parts becomes too messy(perhaps impossible).

    Yes, that was what I had in mind. That's why I got stuck there.

    Thanks!! That will help me move forward.

    Thanks to everyone who replied, I learnt a lot from this thread.

  10. May 2, 2008 #9
    Using lots of substitutions and integration by parts I get this:

    [tex]\int x^{2}e^{-{x^2}}dx=xe^{x^{2}}\left[1-\sum_{n=1}^{\infty}\frac{\prod_{k=2}^{n}\left(2k-3\right)}{2^{n}x^{2n}}\right][/tex]

    I would go over the derivation but LaTex is killing me.
  11. May 3, 2008 #10
    Quite frankly, I think using Leibniz' rule, as suggested by Nicksauce, would by far be the simplest method in this case.
  12. May 3, 2008 #11
    Quite true.

    But, if doing by parts, then the proper selection of u an dv is


    [tex]dv= x*e^{-x^2}dx[/tex]

    and then things won't be so messy - however, it will involve the definite integral [itex]\int^{\infty}_{-\infty}{ e^{-{x^2}}dx}[/itex] which we know equals [itex]\sqrt{\pi}[/itex].
  13. May 3, 2008 #12
    umm mathematica gives me [tex]\frac{\sqrt{\pi}}{2}[/tex]

    and for the indefinite :

    \frac{1}{4} \sqrt{\pi } \text{erf}(x)-\frac{1}{2} e^{-x^2} x
  14. May 3, 2008 #13
    Well, perhaps the very simplest approach is to recognize that the integral is [itex]\sqrt{\pi}[/itex] times the variance of a Gaussian random variable with mean 0 and standard deviation [itex]\frac{1}{\sqrt{2}}[/itex]. That's certainly all I'd bother doing in the signal processing context the OP mentioned.
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