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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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Why do you have $\left( y \frac{d}{y} \right)^2$ in the exponent? Wouldn't that just cancel out to $d^2$? –  Pedro Tamaroff Mar 4 '12 at 12:18
Sorry, it's a typo. I meant the derivative with respect to y. –  Slaviks Mar 4 '12 at 12:50
Considering $F$ as a function of both $y$ and $\beta$, we obtain the PDE $$\begin{cases} \left(\frac{d}{d\beta}-i\left(y\frac{d}{dy}\right)^2\right)F=0 \\ \\ \left. F\, \right|_{\beta=0}=ye^{-y^2/2}. \end{cases}$$ It's possible this could help. –  anon Mar 4 '12 at 13:20
@anon: With the substitution $y=e^{-x/2}$ it leads to the time-dependent Schrodinger equation of free 1d motion with a specific initial condition. Solving the latter using Green's function results in an integral which is the origin of my question. I keep a hope that progress can be made in explicit evaluation of $F$. –  Slaviks Mar 4 '12 at 14:13
I'm implenting the strategy, looks like these are Sloane A039755: oeis.org/A039755 , will post an answer if this leads to a solution. –  Slaviks Mar 4 '12 at 15:06
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1 Answer

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.


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.

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