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Has anybody seen a version of the saddle point method for a real-value multivariate function with multiple local minima ?

For a real-value function with a single variable which has a unique minimum the approximation result is usually presented as follow.

We want to approximate a integral of the form $I_\alpha = \int_\mathbb{R}e^{-f_\alpha (x)dx}$ where $f_\alpha$ is twice differentiable. For every $f_\alpha$, $f_\alpha$ has a unique minimum reached at $x_\alpha$. As $\alpha$ goes to $\infty$, $f_\alpha$ is more localised around $x_\alpha$: $$ \frac{d^2 f}{dx^2}(x_\alpha) \xrightarrow[\alpha \rightarrow \infty]{} + \infty $$ And we use the approximation: $$ I_\alpha \approx e^{-m_\alpha} \sqrt{\frac{2\pi}{\frac{d^2 f}{dx^2}(x_\alpha)}} $$

Thanks

stackoverflower
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  • If you mean Laplace's method, see for instance de Bruijn, Asymptotic Methods in Analysis. Under certain assumptions, the integral of $e^{\alpha f(\boldsymbol x)}$ is asymptotically equivalent to $\alpha^{−d/2} e^{\alpha f(0)}$ times the integral of $\exp(\sum a_{i, j} x_i x_j)$, where the exponent is the sum of the second-order terms in the Taylor expansion of $f$ around the maximum point $\boldsymbol x = 0$. – Maxim Jul 30 '19 at 21:14
  • It's not clear what happens when there are multiple minimums and the function "overlaps" around the minimums: eg for the function $exp(- \frac{x^2}{2})+exp(- \frac{(x-\epsilon)^2}{2})$ where $\epsilon$ is close to $0$. – stackoverflower Aug 01 '19 at 07:38
  • Consider the one-dimensional case first. The conditions that you listed are not sufficient. If you take $f_\alpha(x) = \ln(1 + \alpha x^2)$, then the second derivative at $x_\alpha = 0$ is $2 \alpha$ but the integral is not asymptotically equivalent to $\sqrt {\pi/\alpha}$. There are various generalizations of Laplace's method that cover functions that are not of the form $\alpha f(x)$, including functions with coalescing stationary points, but there isn't one general formula ("it's a recipe, not a blueprint"). – Maxim Aug 01 '19 at 16:14
  • thanks, the Bruijn seems to be focused on complex analysis. – stackoverflower Aug 06 '19 at 14:57
  • For those interested, you can one recipe p71 of the Bruijn. – stackoverflower Sep 09 '19 at 08:26

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