2
$\begingroup$

This is related to my other question on tackling a gaussian integral for $f(w,u)=\frac{1}{w-u}$.

Q1 Suggestions on evaluating gaussian integrals with "nice" functions (not necessarily polynomials) $$\int_{\mathbb{R}^{d}} e^{-|x-c|^{2}}f(x)dx.$$

Are there any books on this class of integrals to help me get ideas? In the wiki article, there is an uncited statement:

$$ \int f({\vec {x}})e^{\left(-{\frac {1}{2}}\sum \limits _{i,j=1}^{n}A_{ij}x_{i}x_{j}\right)}d^{n}x={\sqrt {(2\pi )^{n} \over \det A}}\,\left.e^{\left({1 \over 2}\sum \limits _{i,j=1}^{n}(A^{-1})_{ij}{\partial \over \partial x_{i}}{\partial \over \partial x_{j}}\right)}f({\vec {x}})\right|_{{\vec {x}}=0}$$ for some analytic function f, provided it satisfies some appropriate bounds on its growth and some other technical criteria. (It works for some functions and fails for others. Polynomials are fine.) The exponential over a differential operator is understood as a power series.

Q2 Any ideas about which growth conditions they are referring to? Please include a reference if possible, so that I can include it at the wiki article.

Bookwise, I found:

1)Gaussian Integrals: Ultracold Quantum Fields,

2)Lectures on Gaussian Integral Operators by Neretin.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Isserlis%27_theorem $\endgroup$
    – Steve Huntsman
    Commented Jul 6, 2017 at 13:20
  • $\begingroup$ The Wikipedia article you mention seems suggestive that the above identity is derived from the "Wick" or "Isserlis" identity $$\int x^{k_1}\cdots x^{k_{2N}} \, e^{\left( -\frac{1}{2} \sum\limits_{i,j=1}^{n}A_{ij} x_i x_j \right)} \, d^nx =\sqrt{\frac{(2\pi)^n}{\det A}} \, \frac{1}{2^N N!} \, \sum_{\sigma \in S_{2N}}(A^{-1})_{k_{\sigma(1)}k_{\sigma(2)}} \cdots (A^{-1})_{k_{\sigma(2N-1)}k_{\sigma(2N)}}$$ Hence, you need to make sure that after expanding $f(\vec x)$ under the integral you can exchange summation with integrations and that the resulting series converge. $\endgroup$
    – alphanum
    Commented Feb 26, 2018 at 15:51
  • $\begingroup$ But I am not sure whether one can express this in a sharp and compact condition. $\endgroup$
    – alphanum
    Commented Feb 26, 2018 at 15:54

0

Reset to default

You must log in to answer this question.