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Gaussian integral as an average

  1. Jun 20, 2007 #1
    OK so we have:

    [tex]\int f(z) e^{a g(z)} dz^3[/tex]

    integerated over all space.

    Now there is a identity for this integral as an average, or something like that, right? What is it? Or perhaps you have suggestions where I could read up on that kind of thing?

    (I'm not looking for the integral in terms of a determinant by the way, or am I? :/ ).

    Any help would be much appreciated. Thanks.
  2. jcsd
  3. Jun 24, 2007 #2
    Maybe if I say expectation value rather than average?
  4. Jun 24, 2007 #3


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    How is that a Gaussian integral? What is with the z cubed? And it is not clear at all what kind of identity you are asking for. Are you saying that by definition, this integral is the expectation value of something, and you are asking us what that something is?

    Also, how could an integral be a determinant?
  5. Jun 25, 2007 #4
    Well the g(z) has a term of order z^2 as the highest order. The z cubed is just d^3z (I wrote it the wrong way round. It just means its a volume integral... where z = z_i + z_j + z_k ...

    Yeah that is what I'm asking. I'm pretty sure there is a standard identity for such a 3D gaussian integral as an average or an expectation value, (I'm thinking statistical physics here.. ? )I had it once, but lost the paper I wrote it on, and can't remember where it came from.

    Basically though I just want to be able to integrate this without having to exapnd in terms of z_i, z_j, z_k
  6. Jun 25, 2007 #5


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    The integral as stated is much to general to say anything about. If g(z) is quadratic, you had better spell that out. You'll also need to say something about f(z).
  7. Jun 25, 2007 #6
    Yeah I can't edit the post.

    g(z) is definately quadratic.

    f(z) is a mononomial in z of order >= 0.
  8. Jun 25, 2007 #7


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    Complete the square for g(z). Ie write g(z)=-w^2+A where w is a linear function of z. Write f(z) using the same variable. Now you just have to integrate things like w^n*exp(-w^2). To do these think about the function f(A)=integral(exp(-A*x^2)) which you can do, and differentiate wrt A. The odd powers are easy.
  9. Jun 25, 2007 #8
    Thanks dick!
  10. Jun 26, 2007 #9
    Hmm, still struggling to be honest.

    Can we work through an example please?

    [tex]\int^{\infty}_{-\infty}d^3 \mathbf{z}e^{-az^2+b\mathbf{z}\cdot \mathbf{x} -c}[/tex]

    I tried breaking it up into 3 integrals in terms of polar coords, but that resulted in a singular integral... However if we look at the integral as I've written it there, we can see that it isn't singular... I'm sure there's a way to tackle this without breaking it up into polars. But how?
  11. Jun 26, 2007 #10


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    I am not sure if this is what you are looking for. But it's fairly easy to rpove the following formula

    [itex] \int dq_1 dq_2 \ldots dq_N ~ e^{-1/2 \sum_{nm} q_n M_{nm} q_m + \sum_n J_n q_n} = \frac{(2 \pi)^{N/2}}\sqrt{det M}} e^{1/2 \sum_{nm} J_n M_{nm}^{-1} J_m} [/itex]

    In your case, you would have an integral over [itex] dz_1 dz_2 dz_3 [/itex] and the matrix M is diagonal.

    Is that what you are looking for?
  12. Jun 26, 2007 #11


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    I gave a very general formula but your case is much simpler. To see how to do it, consider breaking the integral as a product of three integrals over [itex] z_1, z_2 [/itex] and [itex] z_3 [/itex]. Let's just look at the integral over [itex] z_1 [/itex], ok? It is

    [itex] \int dz_1 e^{-az_1^2 + b z_1 x_1 } [/itex]
    (the constant piece [itex] e^{-c} [/itex] can trivially be factored out of the intgeral of course so I won't worry about it).

    Now complete the square by writing
    [itex] -az_1^2 + b z_1 x_1 = -a(z_1 - \frac{b x_1}{2a})^2 + \frac{x_1^2 b^2}{4 a} [/itex]

    Then you shift your variable of integration following [itex] z_1 \rightarrow z_1 + \frac{b x_1}{2a} [/itex] And you then end up with a trivial integral to do.

    Hope this helps.

  13. Jun 26, 2007 #12
    Yes thank you very much nrqed. that does help a lot. :)
  14. Jun 26, 2007 #13


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    You are very welcome.
    Going back to your original question, now you could easily do the integral of an arbitrary power of z times your exponential. For example, if you have [itex] z_1^3 z_3 [/itex] as part of the integrand, you may use that
    [itex] z_1^3 z_3 \times e^{\ldots} = \frac{1}{b^4} \partial_{x_1}^3 \partial_{x_3} e^{\ldots} [/itex]

    (I put dots in the exponential because I di dnot feel like rewriting the whole thing!)

    Now, you pull out the partial derivatives from the integral, integrate as before, and then apply the derivatives on the final (integrated) result.

    By the way, this whole trick is fundamental in quantum field theory where it is used to calculate Feynman diagrams (in the path integral approach).

    Best luck!
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