Let $f,g:\mathbb{R}\longrightarrow\mathbb{R_+}$ be Lebesgue integrable functions. We can show that if $g$ has compact support and $f$ has a Maclaurin series that converges absolutely in the support of $g$, then \begin{align*} \int_{\mathbb{R}} f(x) g(x) dx &= \sum_{k=0}^\infty \frac{\mu_k}{k!}f^{(k)}(0), \end{align*} where $\mu_k = \int_{\mathbb{R}} x^k g(x) dx$ is the $k$th moment of $g$.
I would like to know if it is possible to substitute the compact support hypothesis for a more general one that lets $\mathrm{supp}(g) = \mathbb{R}$ and includes the case $f(x) = g(x) = e^{-x^2}$.
Perhaps, it may help to know that the case $f(x) = e^{-x^2}$ and $g(x) = e^{-|x|}$ does not work.
My attempt: Since $f$ admits a Taylor expansion around $0$ that converges in $\mathrm{supp}(g)$, then \begin{align*} \int_{\mathbb{R}} f(x) g(x) dx &= \int_{\mathbb{R}} \sum_{k=0}^\infty \frac{x^k}{k!}f^{(k)}(0) g(x) dx. \end{align*} For the compact support case, we have that $\int_{\mathbb{R}} |x^k g(x)| dx \le M^k \|g\|_1$, for some positive constant $M$. Thus, \begin{align*} \sum_{k=0}^\infty \int_{\mathbb{R}} \left|\frac{x^k}{k!}f^{(k)}(0) g(x)\right| dx \le \|g\|_1 \sum_{k=0}^\infty \frac{M^k}{k!}|f^{(k)}(0)| . \end{align*} The right-hand side is finite because of the absolutely convergent hypothesis. Then, by Fubini-Tonelli theorem, we have that we can interchange the integral with the sum and the result holds.
On the other hand, if $g$ does not have compact support, I'm afraid we must ask that $|\mu_k| \le \|g\|_1 M^k$ for some positive constant $M$, otherwise there is no guarantee of convergence. The problem with this condition is that it does not even include the example I provided.
At some point, I did not think there would be an answer, because the series of the example is not absolutely convergent (thanks for bringing that up @StevenClark), and most of the results I'm familiar with require absolute convergence in some form.
However, I believe I have found a condition that includes, at least some, of the cases of conditional convergence. The condition that replaces compact support of $g$ is \begin{align} \lim_{n\to\infty} \sup_{x\in\mathbb{R}} \frac{|\mu_n|}{n!}|f^{(n)}(x)| = 0. \end{align}
To prove this, we use the mean-value form of the remainder to show that there exists a function $\xi:\mathbb{R} \longrightarrow \mathbb{R}$ such that \begin{align*} \int_\mathbb{R} f(x) g(x) dx = \sum_{k=0}^{n-1} \frac{\mu_k}{k!} f^{(k)}(0) + \int_\mathbb{R} \frac{f^{(n)}(\xi(x))}{n!} x^n g(x) dx. \end{align*} But, \begin{align*} \left|\int_\mathbb{R} \frac{f^{(n)}(\xi(x))}{n!} x^n g(x) dx \right| \le \frac{\mu_n}{n!} \sup_{x\in\mathbb{R}} |f^{(n)}(x)|. \end{align*} Therefore, if condition (1) holds, then the original identity must hold.
Mild verification: Let us apply the condition to the provided examples.
Let $f(x)=g(x)=e^{-x^2}$, $x\in\mathbb{R}$. From here, we have that condition (1) becomes \begin{align*} \lim_{n\to\infty} \sup_{x\in\mathbb{R}} \frac{|\mu_n|}{n!}|f^{(n)}(x)| \le \lim_{n\to\infty} \frac{\Gamma(\frac{1+n}{2})}{\Gamma(1+\frac{n}{2})} \to 0. \end{align*}
For the case $f(x)=e^{-x^2}$ and $g(x)=e^{-|x|}$, $x\in\mathbb{R}$, we have that $\mu_n = 2(n!)$ and the series coefficient of $f$ is $1/(n/2)!$ for even $n$. Then, condition (1) does not hold.
Checked!
Additional comment: Perhaps, there is a way to extend the result by removing the $\sup$ in condition (1), because the power series of $f$ is absolutely convergent and we can extend $f$ analytically to an entire function.