It is known that the Gauss-Hermite quadrature $Q_n(f) := \sum_{i=1}^n w_i f(x_i)$ is exact for all polynomials up to degree 2n-1, i.e. $$ \int_\mathbb{R} f(x) e^{-x^2} dx = Q_n(f) \quad \text{ for all } f \in span\{1,x,x^2,\ldots,x^{2n-1}\} $$
Now assume $$ f(x) := \exp(-a x^2) , \quad a>0. $$
What can be said about the error $$ R_n(f) := \left| \int_\mathbb{R} f(x) e^{-x^2} dx - Q_n(f) \right| $$
Numerical experiments suggest that $$ R_n(f) \leq \exp(-c n) \quad \text{ for some } c >0 , $$ for all $a > 0$, i.e. we have exponential convergence.
However, all theoretical results I am aware of only yield arbitrary large algebraic convergence, i.e. $R_n(f) \leq n^{-s}$ for all $s > 0$. See e.g. Paper by Xiang or Paper by Mastroianni .
Are any further results to deal with entire functions of order $2$ and yield sharp error bounds for Gauss-Hermite quadrature rules? Thanks a lot!
I believe Lemma 3.1 in the preprint "A General Expression for Hermite Polynomials with Applications" can help.
Lemma 3.1: The error term for Gauss-Hermite quadrature $$ E\left[f\right] = \frac{N!}{(2N)!}f^{(2N)}(\xi') $$ is bounded by $$ l\frac{N!}{(2N)!}\leq E[f] \leq u\frac{N!}{(2N)!} $$ where $$ l=\inf\{f^{(2N)}(x)|x\in[-x_{N-1},x_{N-1}]\}, \\ u=\sup\{f^{(2N)}(x)|x\in[-x_{N-1},x_{N-1}]\} $$ and $x_{N-1}$ is the largest zero of $H_N(x)$.
Proof: Theorem 2.1.5.9 in Stoer and Bulirsch states that for every $\overline{x}\in[-\infty,\infty]$ there exists a point $\xi$ within an interval of the support abscissae such that the error of the interpolating polynomial is $$ f(\overline{x}) - h(\overline{x}) = \frac{H_N^2(\overline{x})f^{(2N)}(\xi)}{(2N)!}. $$ This $\xi$ is therefore bounded by the maximum and minimum support abscissae, which in the case of Gauss-Hermite quadrature, are the zeros of the $N^\textrm{th}$ Hermite polynomial. The error is integrated against the Gaussian weighting function results in the error of the integral \begin{align} E[f] = \frac{1}{(2N)!}\int_{-\infty}^\infty f^{(2N)}(\xi(x))H_N(x)^2\omega(x)dx. \end{align} The next step involves the mean value theorem of integral calculus, which results in the derivative term being evaluated at an unknown point $\xi'(x)$ where $x\in[-\infty,\infty]$. However, since $\xi'$ is bounded by the maximal root of $H_N(x)$, a maximum and minimum error can be determined by finding the infimum and supremum of the function inside this range, resulting in the bounds presented above.
Applying this in your case, where $f(x)=\exp(-ax^2)$ gives the $f^{(2n)}(x) \propto H_{2n}(x)\exp(-ax^2)$, which you can find the maximum and minimum in the range spanned by the lowest/highest root (negatives of each other by symmetry) to find your error bounds.
The dependence on $N$ (your $n$) is given by the factorial in front of the $2N$ derivative $N!/(2N)!$ which, by Stirling's approximation, can be written $$ R\approx \frac{e^{-N}N^{-N}}{2^{2N+\frac{1}{2}}}H_{2N}(\xi)\exp(-a\xi^2) $$ for some $\xi\in[-x_N,x_N]$.