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Integral of x²e^-x²

  1. Feb 17, 2008 #1
    I am working on an integral I am finding tricky, and I think I'm missing something.
    I need to integrate on the interval 0 to infinity, x²e^-x².
    We have proved that on the interval of -∞ to ∞, e^-x²=√∏ so from o to ∞, it equals √∏/2. I can use this in my proof, but I don't see how. When I try integrating by parts I have trouble getting a finite answer. I would love some help,
  2. jcsd
  3. Feb 17, 2008 #2
    The functions symmetric so I don't see your issue. Out of curiosity how did you prove the first integral as if I recall correctly the improper integral is done using residues. The proper integral can also be solved if you don't mind solutions in terms of error functions.
    Last edited: Feb 17, 2008
  4. Feb 17, 2008 #3


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    Well, then show what you did!
  5. Feb 17, 2008 #4
    its the euler poisson gaus integral and evaluate it as a double integral, to be honest the proof is a bit over my head.
  6. Feb 17, 2008 #5
    I've heard of that method, I believe it uses polar coordinates. That said the better way to solve it is via residues:
  7. Feb 17, 2008 #6
    ok here goes:
    from 0-∞∫ x²e^-x²dx
    u= x² dv = e^-x²dx
    du=2xdx v=∫e^-x²dx

    (x²∫e^-x²dx) – (2∫xdx)(∫e^-x²dx)
    since I have that from 0-∞∫ e^-x²dx = √∏/2
    I have x²(√∏/2) -√∏ which when evaluated on the interval 0-∞ goes to ∞. . .
    This makes me think I’ve done it incorrectly
  8. Feb 17, 2008 #7
    cant one just go like this

    integ(x^2e^-x^2)dx=integ(x*x*e^-x^2)dx, and then take u=x, du=dx, and v=integ(x*e^-x^2)dx, this last one can be done using the sub -x^2=t=> -2xdx=dt, xdx=-dt/2.

    THis is easier i guess!
  9. Feb 17, 2008 #8
    I think that may help.
  10. Feb 17, 2008 #9
    that is my all time favorite xkcd comic.
  11. Feb 17, 2008 #10
    I'm not sure what you have done is valid as this:
    seems illegal.
  12. Feb 17, 2008 #11
    i think you're right, and I appreciate the feedback. . .so I'll try to hit it from another angle. thanks
  13. Feb 17, 2008 #12

    George Jones

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    Did you try to work through post #7?
  14. Feb 17, 2008 #13
    The first substitution you did does not change the integral. You made a mistake.
  15. Feb 17, 2008 #14
    Why not write out how you solved it for the limits from -oo to oo and we can see if we can extend the method. If you're interested I can show you how to solve the equation for the proper integral but I don't think that is what the teacher is looking for as you haven't learned about the error function yet.
  16. Feb 17, 2008 #15
    Yes you are right that, indeed, it does not change the integral at all, it ends up with
    integ e^(-x^2)dx. Besides this what mistake are you talking about?
  17. Feb 17, 2008 #16
    no the integrand remains unchanged. That is you now have


    and I wasn't talking about any other mistake.
  18. Feb 17, 2008 #17
    integ(x^2e^-x^2)dx=integ(x*x*e^-x^2)dx, and then take u=x, du=dx, and v=integ(x*e^-x^2)dx, this last one can be done using the sub -x^2=t=> -2xdx=dt, xdx=-dt/2.
    now v=-1/2e^(-x^2), now going back to integ by parts we have

    -1/2 xe^-x^2 +1/2 integ (e^-x^2) dx

    i do not see how you will end up with the same integrand.!!
  19. Feb 17, 2008 #18
    Oh sorry, my mistake.
  20. Feb 17, 2008 #19
    now i guess he can express the last part in terms of the error function,
  21. Feb 18, 2008 #20


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    If you've proved

    [tex]\int_0^\infty dx~e^{-x^2} = \sqrt{\pi}[/tex]

    you should be able to extend that result to the more arbitrary case

    [tex]\int_{-\infty}^\infty dx~e^{-\alpha x^2} = \sqrt{\frac{\pi}{\alpha}}[/tex]
    by substitution.

    Now, the Liebniz rule for integrals says that (in the case of constant limits)

    [tex]\frac{d}{d\alpha}\int_a^b dx f(x,\alpha) = \int_a^b dx\frac{\partial f(x,\alpha)}{\partial \alpha}[/tex]

    With this in mind, can you figure out how to evaluate your integral?

    (I don't know if you're allowed to do it this way if this is homework, but it's a clever trick...)
    Last edited: Feb 18, 2008
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