 Solution

A general-purpose high-dimensional choice is Monte Carlo integration. Draw independent vectors $W_r$ uniformly from the unit cube and evaluate $H(W_r)={\sum }_{i=1}^{100}{e}^{\Vert W_r-a_i{\Vert }^2}$. The cube has volume $1$, so its integral is the expected value of $H(W_1)$. The iterative estimator isThe integrand is continuous on a compact set, hence bounded with finite variance. The strong law of large numbers gives almost sure convergence, and the square root of the mean squared error is exactly $\sigma /\sqrt{N}$, where ${\sigma }^2=\mathrm{Var}(H(W_1))$. Thus the standard stochastic accuracy rate is $O(N^{-1/2})$; the dimension affects the variance and evaluation cost, not this exponent.

There is also an important structure-specific improvement here: $e^{\Vert x-a_i{\Vert }^2}={\prod }_{j=1}^{100}{e}^{(x_j-a_{ij}{)}^2}$ is a separable integrand. The Fubini theorem reduces the requested value exactly toOne can therefore replace high-dimensional sampling by $10000$ one-dimensional composite Simpson's rule calculations. For each factor use even $n$, $h=1/n$, andthen iterate $n\leftarrow 2n$ and form the sum of the products. These smooth factors give $O(n^{-4})$ deterministic discretization error for fixed centres. Thus Monte Carlo integration is a valid generic answer, while exploiting separability is the more efficient specialized alternative; a large dimension alone does not force a tensor-product grid.