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Taylor expansion of gaussian integral with respect to variance

  1. May 7, 2010 #1
    Hi everyone. The problem I have to face is to perform a taylor series expansion of the integral

    [tex] \int_{-\infty}^{\infty}\frac{e^{-\sum_{i}\frac{x_{i}^{2}}{2\epsilon}}}{\sqrt{2\pi\epsilon}^{N}}\cdot e^{f(\{x\})}dx_{i}\ldots dx_{N} [/tex]

    with respect to variance [tex]\epsilon[/tex]. I find some difficulties because the function in the integral has a singularity in [tex]\epsilon=0[/tex] and this makes calculation a bit strange, but it must be possible to calculate the taylor series coefficients (I need them only at second order). In fact, for zero variance the entire integral reduces to f(0) besause the gaussian function becomes a Dirac's delta distribution centered in 0.

    The function f is a polynomial of degree 4 in all x variables and contains only even powers and terms like x_i x_j, so it is not possible to factor it into product of single terms dependending only from i index.

    I have tried to expand f in series and use the moment generating function of the gaussian, but neither this trick seems to work. Nor I have had much luck expanding the gaussian. Anyone has some ideas? Thank you.
     
  2. jcsd
  3. May 7, 2010 #2

    Mute

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    Homework Helper

    You're right that as [itex]\epsilon \rightarrow 0[/itex] you get a delta function that picks out f(0). So, what I might do is use the fact that as [itex]\epsilon[/itex] gets very small most of the contribution to the integral is going to come from the region just around the origin, so you can Taylor expand your function f close to the origin (easy to do since it's a polynomial), and just have to keep the lowest order terms. Since you only have even powers, you essentially get a Gaussian. That is, for [itex]\epsilon[/itex] small,

    [tex]\int_{-\infty}^{\infty}\frac{e^{-\sum_{i}\frac{x_{i}^{2}}{2\epsilon}}}{\sqrt{2\pi \epsilon}^{N}}\cdot e^{f(\{x\})}dx_{i}\ldots dx_{N} \approx e^{f(0)}\int_{-\infty}^{\infty}e^{-\sum_{i}\frac{x_{i}^{2}}{2\epsilon}}\cdot e^{\frac{1}{2}x_iA_{ij}x_j}dx_{i}\ldots dx_{N} [/tex].
    where
    [tex]A_{ij} = \frac{\partial^2 f}{\partial x_i \partial x_j}[/tex]
    and the denominator has been neglected for a reason I will explain below. Note also that I'm using the summation convention: repeated dummy indices are summed over.

    One potential danger: I'm assuming that the integrand is most sharply peaked at the origin. To be really careful what I ought to do is look for the minima of the exponent,

    [tex] -\frac{1}{\epsilon}\left(\frac{x_i x_i}{2} - \epsilon f(x) + \frac{N \epsilon}{2}\ln(2\pi \epsilon)\right)[/tex]

    However, unless f(x) contains a 1/epsilon, then as epsilon gets small the minimum should be close to the origin, and as long as N doesn't tend to infinity the constant logarithmic term won't matter and can be neglected compared to the other terms as epsilon tends to zero, hence why I neglected it above. If you wanted to do higher order corrections, though, you would need to consider the minima of the above expression.

    If N does get large in any point of the calculation, you'll need to constrain it such that [itex]N\epsilon[/itex] is fixed as N goes to infinity and epsilon goes to zero in order to have a well defined expression.

    Finally, note that the result is an asymptotic expression, not from an exact series expansion.

    This method is called the Method of Steepest Descent (look it up, but note that http://en.wikipedia.org/wiki/Method_of_steepest_descent isn't really useful). The complex-form is also useful, called the Method of stationary phase.
     
    Last edited: May 7, 2010
  4. May 7, 2010 #3
    Thank you very much. I have fixed this problem by simply rescale [tex]x_i[/tex] with [tex]\sqrt \epsilon[/tex]. this makes [tex]\epsilon[/tex] disappear from any fraction and the structure of even powers of the polynomial makes everything analytical and taylor expandeable.

    The idea of the steepest descent method is a smart one. I'll think about it for some future applications.
     
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