(This may sound too simple a question to be asked in MO. But it is a research problem but not an exercise, so I post it here. If it is no appropriate, I will move it to Stackexchange.)
Let $0 < a < b$ be real numbers and $p$ a polynomial. The Fourier transforms of $\exp(-ax^2/2)$ and $\exp(-bx^2/2)$ are (up to constant factors) $\exp(-\xi^2/(2a))$ and $\exp(-\xi^2/(2b))$ respectively. The Fourier transform of $p(x) \exp(-ax^2/2)$ is $q(\xi) \exp(-\xi^2/(2a))$ where $q$ is a polynomial of the same degree as $p$'s.
Suppose we know that a function $f$ has its Fourier transform $$ \hat{f}(\xi) = \left( q(\xi) e^{-\frac{\xi^2}{2a}} \right) e^{-\frac{\xi^2}{2b}}, $$ then $f$ has a simple expression as the convolution (up to a constant) $$ f(x) = \int^{\infty}_{-\infty} \left( p(y) e^{-\frac{ay^2}{2}} \right) e^{-\frac{b(x - y)^2}{2}} dy. $$
My question is: if a function $g$ has its Fourier transform $$ \hat{g}(\xi) =\left( q(\xi) e^{-\frac{\xi^2}{2a}} \right) e^{\frac{\xi^2}{2b}}, $$ is there a simple formula of $g$ in $p$, analogous to the convolution formula above? A naive extrapolation $$ \int^{\infty}_{-\infty} \left( p(y) e^{-\frac{ay^2}{2}} \right) e^{\frac{b(x - y)^2}{2}} dy. $$ does not converge at all.