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Feynman and gauss integral

  1. Feb 10, 2010 #1
    Hi all,

    This is just wild shot, since my memory is not what it used to be ...
    I remembered reading about Feynman pointing out an interesting
    fact, that the integration of the gaussian function

    \int_-\infty^\infty e^(-x x) dx = \sqrt[\pi]

    has to do with Pi. He then went on to show the connection. I
    couldn't find out if this is in one of his Lecture books or his
    autobiography. I could find out where he mentioned the
    connection between exp and trig functions, but that was as far
    as I could go.

    I am not even sure that it was Feynman.

    Any idea? Thanks in advance,

    Michuco

    Ps. google feynman and integral leads, no surpise, to many
    links that have to do with his path integral.
     
  2. jcsd
  3. Feb 11, 2010 #2
    This is true, but this was known for a long time before Feynman. It can be shown like this:

    Multiply two such gaussian integrals together. Combine them into one two-dimesional integral. Change to polar coordinates. Then both the angular and radial parts are easy to calculate. The result of this is pi. Since it was the square of the original integral, the answer is sqrt(pi).

    Torquil
     
  4. Feb 11, 2010 #3

    Matterwave

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    Gold Member

    I believe this is what you're talking about:
    http://en.wikipedia.org/wiki/Gaussian_integral

    The integral of a Gaussian is sqrt(pi). The computation is explained in that article also.
     
  5. Feb 11, 2010 #4
    Thanks for the replies. I know of the polar coord conversion proof which I asume that wiki took from Weinsstein's MathWorld. I was wondering about the Feynman connection, if there
    was one.

    Regards,

    Michuco
     
  6. Feb 18, 2010 #5
    See The Feynman Lectures on Physics, Vol I, section 40-4, "The distribution of molecular speeds," unnumbered equation between Eqs. (40.7) and (40.8).
     
  7. Feb 18, 2010 #6
    Thanks codelieb,

    This is exactly what I was looking for ...

    Michuco
     
  8. Feb 18, 2010 #7
    You're welcome.
     
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