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I need to evaluate the following function and want to proceed analytically as far as possible:

$F(y) =e^{ i \beta \left ( y \frac{d}{d y} \right )^2} y \, e^{-y^2/2}$

My plan is to expand into power series in $\beta$ and indentify the polynomials

$\left (y \frac{d}{d y} \right )^k (y \, e^{-y^2/2})= p_k(y) \, y \, e^{-y^2/2}$

by the recursion relation I expect them to satisfy.

Is this a sound strategy? Is there a more direct way to identify polynomials $p_k(y)$ and compute their "generating function'' $F(y)$?

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    I'm implenting the strategy, looks like these are Sloane A039755: https://oeis.org/A039755 , will post an answer if this leads to a solution.2012-03-04

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Possible next step but not yet the final answer:

Working out first few terms in Mathematica and searching the OEIS, I found that my polynomials can be expressed in terms of "B-analogues of Stirling numbers of the second kind" $T(n,k)$, see sequence A039755 for definition and references.

Using the generating function from OEIS, I get

$p_k(y) = \sum_{m=0}^{k} T(k, m) (-1)^m y^{2 m} = \frac{\partial^k}{\partial z^k} g(z,y)$

where $g(z) = e^{z-(1/2)y^2 e^{2 z}} e^{y^2/2}$.

Thus $F(y) = \sum_{l=0}^{\infty} \frac{1}{l!} (i \beta)^l y e^{-y^2/2}\frac{\partial^{2 l}}{\partial z^{2 l}} g(z,y) $ .

Unfortunatelly, I have the square in the exponent and $l!$ does not match the dergee $2l$ in the expansion, and I don't see how to proceed.

UPDATE-1

Using "Dobinski-type identity" for the "row generating polynomials" (see OEIS entry), which in my notation is $p_k(y)=e^{y^2/2} \sum_{q=0}^{\infty} (q!)^{-1} (-2)^{-q} (2 q+1)^k y^{2 q}$, results in a series repsentation:

$F(y,\beta)=\sum_{q=0}^{\infty} (q!)^{-1} (-2)^{-q} e^{i \beta (2 q+1)^2} y^{2 q} $

This gives a useful asymptotic expansion of $F$ in terms of $\beta$ which, however does not capture correctly the asymptotics at large $y$.

Will try to manupulate this series, any advice is appreciated.