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Is it possible to express $$\exp\left[\mathrm{e}^{x^2/2}p\left(\frac{d}{dx}\right)\mathrm{e}^{-x^2/2}\right] f(x)$$ as an integral transform or something similar? $p(x)$ is a polynomial. $$\exp\left[\mathrm{e}^{x^2/2}p\left(\frac{d}{dx}\right)\mathrm{e}^{-x^2/2}\right]f(x) = \int k(x,t)f(t)\,dt.$$ By expanding the exponential, I get $$\exp\left[\mathrm{e}^{x^2/2}p\left(\frac{d}{dx}\right)e^{-x^2/2}\right] f(x) = f(x) + \mathrm{e}^{x^2/2}p\left(\frac{d}{dx}\right)\mathrm{e}^{-x^2/2}f(x) + \frac{1}{2} \mathrm{e}^{x^2/2}p\left(\frac{d}{dx}\right)^2\mathrm{e}^{-x^2/2} f(x) + \ldots,$$ which has operator coefficients similar to Hermite polynomials.

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You can of course write $$ k(x,t) = \exp \left[ e^{-t^2 /2} p\left( -\frac{d}{dt} \right) e^{t^2 /2} \right] \delta (x-t) $$ Not clear whether that affords you any sort of simplification you may be possibly hoping for.

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  • $\begingroup$ I would like to get ride of the operator. Not sure this solution helps. $\endgroup$
    – Mirar
    Commented Aug 16, 2023 at 6:37
  • $\begingroup$ Similar to en.m.wikipedia.org/wiki/Weierstrass_transform $\endgroup$
    – Mirar
    Commented Aug 16, 2023 at 6:39
  • $\begingroup$ I'm somewhat pessimistic. In the case of the Weierstrass transform, the operator is sufficiently simple such that one can identify it with a shift operator (after some manipulations). Once you start to go to more complicated operators, closed-form expressions quickly become unattainable. Maybe for selected polynomials $p$, it's still possible, but a formula for general $p$ ... I don't know. $\endgroup$
    – Michael Engelhardt
    Commented Aug 16, 2023 at 7:00
  • $\begingroup$ What about approximation? $\endgroup$
    – Mirar
    Commented Aug 16, 2023 at 13:10
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    $\begingroup$ @Mirar - depends on what your small parameters are. You've already written down something like a perturbative expansion for the case that $p$ has small coefficients. It's thinkable to generalize this to the case that only some of the coefficients can be regarded as small. If the other terms correspond to a sufficiently simple, tractable problem, you can resum those and are then left with the perturbative series in the small terms. $\endgroup$
    – Michael Engelhardt
    Commented Aug 18, 2023 at 1:44

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