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Evaluate the following integral

  1. Sep 25, 2005 #1

    quasar987

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    Any idea on how to evaluate

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

    where n is even???
     
    Last edited: Sep 25, 2005
  2. jcsd
  3. Sep 25, 2005 #2

    Tom Mattson

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    What's that mean?
     
  4. Sep 25, 2005 #3

    quasar987

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    oh sorry, pair = french for even.

    Edited.
     
  5. Sep 25, 2005 #4

    Tom Mattson

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    I would integrate by parts.

    Hint:

    [tex]
    u=x^{n-1}
    [/tex]

    [tex]
    dv=xe^{-ax^2}dx
    [/tex]

    Try that and see what you come up with.
     
  6. Sep 25, 2005 #5

    quasar987

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    How could have I missed that?! :grumpy:

    Thx Matt!

    Btw - are you still learning diff forms?
     
  7. Sep 25, 2005 #6

    Tom Mattson

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    Can happen to anyone.

    The name's Tom, not Matt. How would you like it if I called you "nebula"? :biggrin:

    Yeah, my next set of notes is in preparation. I stopped posting to the thread because I was asked to teach a couple of summer courses. They're condensed into 6 weeks, so I didn't have time for anything else.

    Keep watching...
     
  8. Sep 25, 2005 #7

    quasar987

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    Sorry Tom. :biggrin:

    I think I'm still missing an important point in your idea though.. in tegrating by parts like you advised I get

    [tex]\int_{0}^{\infty} x^n e^{-ax^2} dx = \left[-\frac{x^{n-1}}{2a} e^{-ax^2} \right]_0^{\infty} + \frac{n-1}{2a} \int_0^{\infty} x^{n-2} e^{-ax^2} dx[/tex]

    So we can perform the same technique of integration on the last integral, and so on, such that after n/2 times of performing this trick, the integral we have to evaluate is

    [tex]\int_0^{\infty} e^{-ax^2} dx [/tex]

    But what is that?! the integrator says its a constant times [itex]\mbox{Erf}[\sqrt{5}x][/itex] The heck is Erf ?
     
  9. Sep 25, 2005 #8

    Tom Mattson

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    That is perhaps the most famous integral ever taught to students of multivariable calculus, at least in America. It is a standard way to introduce polar coordinates in double integration.

    [tex]I=\int_0^{\infty} e^{-ax^2} dx [/tex]

    [tex]I=\int_0^{\infty} e^{-ay^2} dy [/tex]

    [tex]I^2=\int_0^{\infty}\int_0^{\infty} e^{-a(x^2+y^2)} dx [/tex]

    Now convert to polar coordinates and evaluate.

    Erf(x) is the error function.
     
  10. Sep 25, 2005 #9

    quasar987

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    RIIIIGHT! I remember browsing my calculus book last year, and for some random reason, I began reading this random problem. And I found it so cute I had to do it, and of course it was proving that this integral is worth [itex]\sqrt{\pi}/2[/itex] by the method you outlined!
     
  11. Sep 25, 2005 #10
    If you want to skip the integration by parts, you can choose to use the differentiation under the integral sign:

    [tex] \int_{-\infty}^{\infty} x^{n} dx e^{-ax^{2}} = \int_{-\infty}^{\infty} dx \frac{d^{\frac{n}{2}}}{da^{\frac{n}{2}}}(-1)^{\frac{n}{2}}e^{-ax^{2}} [/tex]

    So now you can take the derivative with respect to a out in front:

    [tex] (-1)^{\frac{n}{2}} \frac{d^{\frac{n}{2}}}{da^{\frac{n}{2}}}\int_{-\infty}^{\infty} dx e^{-ax^{2}} [/tex]

    So, once you've figured out what [tex] \int_{-\infty}^{\infty} dx e^{-ax^{2}} [/tex] is, you could also do it in this way.
    Notice of course, that this is only useful if n is even, which I think you stipulated, and if the integral is a definite integral.
     
  12. Sep 25, 2005 #11

    quasar987

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    I'm not familiar with the use of this method TOKAMAK (I'm not Feynman).

    btw, if anyone carried out the problem til the end and want to compare, I arrive at

    [tex]\sqrt{\frac{\pi}{2^n a^{n+1}}}\prod_{j=1,3,5,...}^{n-1} (n-j)[/tex]
     
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