A laplace duality for integration
Jean B Lasserre
TL;DR
Problem: compute the univariate integral $v(y)=\int_{K_y} f(\mathbf{x})\,d\mathbf{x}$ with $K_y=\{\mathbf{x}: g(\mathbf{x})\le y\}$, where $g$ is nonnegative and $K_y$ is compact. Approach: develop a Laplace-transform based duality that yields a 'Laplace dual variable' $\lambda_y>0$ satisfying $v(y)=\int_{\mathbb{R}^d} f(\mathbf{x}) e^{-\lambda_y g(\mathbf{x})}\,d\mathbf{x}$, effectively tilting the measure from $K_y$ to $\mathbb{R}^d$. Key results: (i) existence of $\lambda_y$ under mild conditions; (ii) when $f,g$ are positively homogeneous of degrees $d_f,d_g$, $y\lambda_y=(\Gamma(1+(d+d_f)/d_g))^{d_g/(d+d_f)}$ and a decomposition $f=\sum f_k$ yields $v(y)=\sum_k \int f_k e^{-\lambda_{y,k} g}$ with explicit $\lambda_{y,k}$. Significance: provides a duality parallel to Legendre-Fenchel duality in optimization, enabling practical computation via Gaussian-like cubatures and offering explicit forms in the homogeneous and quadratic cases.
Abstract
We consider the integral v(y) = Ky f (x)dx on a domain Ky = {x $\in$ R d\,: g(x) $\le$ y}, where g is nonnegative and Ky is compact for all y $\in$ [0, +$\infty$). Under some assumptions, we show that for every y $\in$ (0, $\infty$) there exists a distinguished scalar $λ$y $\in$ (0, +$\infty$) such that which is the counterpart analogue for integration of Lagrangian duality for optimization. A crucial ingredient is the Laplace transform, the analogue for integration of Legendre-Fenchel transform in optimization. In particular, if both f and g are positively homogeneous then $λ$y is a simple explicitly rational function of y. In addition if g is quadratic form then computing v(y) reduces to computing the integral of f with respect to a specific Gaussian measure for which exact and approximate numerical methods (e.g. cubatures) are available.