Let $x_1,\dots,x_p$ be $p$ points in $\mathbb{R}^n$ ($n\geq 2$) with $x_1=0$. Consider the symmetric matrix $M(x)=(m_{ij}(x))_{1\leq i,j\leq p}$ where $m_{ij}(x) = \exp(-\frac{1}{2}\Vert x_i - x_j\Vert^2)$. The norm is the standard Euclidean one. I would like to prove that $\det(M(x))^{-\frac{1}{2}}$ is locally integrable on $(\mathbb{R}^n)^{p-1}$.

I already know that $M(x)$ is positive definite if the $(x_i)_{1\leq i\leq p}$ are pairwise distinct. More precisely, if $x_1,\dots,x_p$ take exactly $q$ different values, then $M(x)$ has rank $q$.

I also managed to prove that, if $\lambda_1(x)$ is the smallest eigenvalue of $M(x)$ then there exists a constant $K_{n,p}>0$ depending only on $n$ and $p$ such that: $$ \frac{1}{\lambda_1(x)} \leq K_{n,p}\sum_{i=1}^p \frac{1}{\prod_{j\neq i}\Vert x_i-x_j\Vert^2} \exp(\Vert x_i\Vert^2) \prod_{j\neq i} (1+\Vert x_j\Vert^2).$$ This is enough to prove that $\frac{1}{\sqrt{\lambda_1}}$ is locally integrable but not enough to get the integrability of $\det(M(x))^{-\frac{1}{2}}$.

Does anyone know how to prove a similar inequality for $\frac{1}{\det(M(x))}$ instead of $\frac{1}{\lambda_1(x)}$? Alternatively, does anyone know of a suitable lower bound for $\det(M(x))$ in terms of the $(\Vert x_i-x_j\Vert)_{i\neq j}$? I know that this matrix appears in statistics, but I couldn't find anything concerning its determinant in the literature so far.