Question about analytic continuation of a function defined by an integral

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Suppose I have a function $f$ define \begin{align} f(x)= \int_{-\infty}^\infty e^{-x^2t} g(t) dt. \end{align} where $g(t)$ is some positive, integrable fucntion

We also know that $f(x)$ has a power series representation \begin{align} f(x)=\sum_{i=1}^\infty a_n x^n \end{align} which is valid for all $x\in \mathbb{R}$.

The existance of a power series also implies that there exists and entire function $h(z)$ such that \begin{align} h(z)=\sum_{i=1}^\infty a_n z^n \end{align} for all $z \in \mathbb{C}$ and $f(x)=h(x)$ for $x\in \mathbb{R}$.

My question now concerns the integral part of the defintion of $f(x)$.

Can I related $\int_{-\infty}^\infty e^{-x^2t} g(t) dt$ to $h(z)$.

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Yes, the analytic continuation is $\int_{-\infty}^{\infty}e^{-z^2 t}\ g(t)\ dt$ wherever the integral converges.

(Note that this is indeed an analytic function of $z$, since $e^{-z^2t}$ is analytic for each $t$, and if one makes suitable assumptions on $g(t)$ such that the integral converges then those same assumptions will allow us to interchange the derivative and integral to verify the Cauchy-Riemann equations for analyticity.)