3

Is there a general method for analytically approximating integrals of the form $$ \int_{\mathbb{R}^n} d^{n}\textbf{x} \, \exp(- F_n(\textbf{x})),$$ where $\textbf{x} = (x_1, x_2, \ldots, x_n)$, $F_n(\textbf{x})$ is a function of $\textbf{x}$ indexed by $n$, and $n \gg 1$?

I realize this question or one like it has been asked many times before (see below references), but here I have a concrete example which motivates a particular approach.

Context/Example

Laplace's method seems to be a naive possible approach but it results in large errors. For example, take the integral $$ I_n = \int_{\mathbb{R}^n} d^{n}\textbf{x} \, e^{-\textbf{x}^2/2}\Big(\sum_{j=1}^n x_j\Big)^n \prod_{j=1}^n x_j\equiv\int_{\mathbb{R}^n} d^{n}\textbf{x} \, \exp(- F_n(\textbf{x})) $$ If we use Laplace's method, we take $$F_n(\textbf{x}) = \textbf{x}^2/2 - \sum_{j=1}^n \ln x_j - n \ln \sum_{j=1}^n x_j,$$ and solving for $\{\bar{x}_j\}$ such that $\partial_{j} F_n(\bar{\textbf{x}}) = 0$, we find $\bar{x}_j = \sqrt{2}$ for all $j$. Which yields $$I_n \simeq c_n \exp(-F_n(\bar{\textbf{x}})) = c_n\,2^n \left(\frac{n}{e}\right)^n,$$ where $c_n$ is some $n$ dependent constant. This result makes sense because one can show that $I_n = n!$ and Stirling's approximation gives $n! \simeq \sqrt{2\pi n} (n/e)^n$. The problem comes when one tries to use the Hessian matrix to determine the $c_n$. The value one should find is $c_n = \sqrt{2\pi n}/2^n$ but one actually gets something quite different (I got $c_n = \sqrt{3/4}(2/3)^{n/2}$ ) which suggests that the $(n/e)^n$ factor derived from $F_n(\bar{\textbf{x}})$ is only coincidentally correct. Thus Laplace's method isn't the correct approach. Or maybe it is that Laplace's method should be applied in a special way? I'm looking for guidance either way.

Possibly relevant literature reference

Possibly relevant Stackexchange posts

  • 1
    Just how large is $n$ in the applications you are interested in? Or is this a more theoretical question and you are actually considering the case of $n \to \infty$? Just curious – Yuriy S Dec 29 '20 at 08:26
  • @YuriyS I'm considering $n$ on the order of $10^2$ and $10^4$. It is more of a theoretical question though, and it would be great if the approach worked for arbitrarily large $n$. – motherboard Dec 29 '20 at 09:48

0 Answers0