Is the function $\int_0^{1} e^{-t H}e^{-2x^2}dt$ in $L^2$ with respect to the gaussian measure?

Question

Consider the function $g(x)=\int_0^{1} e^{-t H}e^{-2x^2}dt$, where $H$ is the differential operator $x\partial_x-\partial_x^2$.

1. Is $g(x)$ in the $L^2$ space with respect to the gaussian measure; i.e. $$\int_{-\infty}^\infty g(x)^2\frac{e^{-x^2/2}}{\sqrt{2\pi}}dx<\infty?$$
2. What if $H=x^2\partial_x^2-2x\partial_x^3-2\partial_x^2+\partial_x^4+x\partial_x$? Is $g(x)$ still in $L^2$ space with respect to the gaussian measure?

Things I have tried:

1. I have tried computing $e^{-tH}e^{-2x^2}=\sum\limits_{i=0}^\infty \frac{(-tH)^i}{i!}e^{-2x^2}$ using mathematical softwares. However, computing beyond the fifth order becomes almost impossible.

2. I also tried fourier transformations and inversions. However, that didn't make the problem any easier.