Factorial as a Integration over a Gaussian I'm seeking a proof for a formula that I've affirmed through numerical checks. Let $n$ be a natural number. For
$$I_n = \frac{(-1)^n}{\sqrt{2\pi n}} \int\limits^{\infty}_{-\infty} dx\, x^n D^n (e^{-x^2/2n}),$$
where $D = d/dx$, how can we show that $I_n = n!$?
The formula can be checked for first few $n$'s with Wolfram alpha. n=6 case for identity
As an alternative question, what is $K_n$ for
$$K_n = \frac{(-1)^n}{\sqrt{2\pi n}} \int\limits^{\infty}_{-\infty} dx\, \left(\frac{d}{d\ln x}\right)^n e^{-x^2/2n},$$
where $(d/d\ln x)^nf(x) = x(d/dx)x(d/dx) \cdots x(d/dx) f(x)$. For example,
\begin{align} \left(\frac{d}{d\ln x}\right)^{3}x^{3} & = x\frac{d}{dx} \left[x\frac{d}{dx} \left[ x\frac{d}{dx}\left[ x^{3} \right] \right]\right]  \\
& = 27 x^3. 
\end{align}
 A: I interpret $\frac{\partial}{\partial \ln x}$ to mean $x \frac{\partial}{\partial x}$; I can't think of anything else it could mean, but with that interpretation I don't get your identity, and instead I get the following:
Consider the generating function
$$\sum_{n \ge 0} t^n \frac{(x \frac{\partial}{\partial x})^n}{n!} f(x) = e^{(x \frac{\partial}{\partial x})t} f(x) = f(e^t x)$$
(which follows formally from the observation that $xD$ acts by multiplication by $k$ on $x^k$ and hence $e^{(x \frac{\partial}{\partial x})t}$ acts by multiplication by $e^{kt}$). Setting $f(x) = \exp \left( - \frac{x^2}{2 \sigma^2} \right)$ gives
$$\sum_{n \ge 0} t^n \frac{(x \frac{\partial}{\partial x})^n}{n!} \exp \left( - \frac{x^2}{2 \sigma^2} \right) = \exp\left( - e^{2t} \frac{x^2}{2 \sigma^2} \right).$$
Integrating over $x$ gives
$$\sum_{n \ge 0} t^n \int_{-\infty}^{\infty} \frac{(x \frac{\partial}{\partial x})^n}{n!} e^{- \frac{x^2}{2 \sigma^2}} \, dx = \int_{-\infty}^{\infty} \exp\left( - e^{2t} \frac{x^2}{2 \sigma^2} \right) \, dx.$$
The $u$-substitution $u = e^t x, du = e^t \, dx$ gives that this is equal to
$$e^{-t} \int_{-\infty}^{\infty} e^{- \frac{u^2}{2 \sigma^2} } \, du = \sqrt{2\pi} \sigma e^{-t}$$
and extracting the coefficient of $\frac{t^n}{n!}$ gives
$$\int_{-\infty}^{\infty} \left( x \frac{\partial}{\partial x} \right)^n e^{- \frac{x^2}{2 \sigma^2}} \, dx = (-1)^n \sqrt{2 \pi} \sigma.$$
Setting $\sigma^2 = n$ then gives
$$\int_{-\infty}^{\infty} \left( x \frac{\partial}{\partial x} \right)^n e^{- \frac{x^2}{2n}} \, dx = (-1)^n \sqrt{2 \pi n}$$
which is off by a factor of $n!$ from your identity. I checked this explicitly for $n = 2$ though, so I don't think I just dropped the factorial somehow. Do you mean something else by $\frac{\partial}{\partial \ln x}$, and if so what?
A: Thanks @Professor Vector and @rrogers for the suggestions. I forgot that Hermite polynomials even exist! Integration by parts seems to be the best approach since you'd still need to use it for the Hermite approach, but I'll write out the two suggested solutions for posterity. Could probably be used as a fun calculus exercise for an integration by parts lesson!
Integration by parts
We rescale our $x$ variable by $\sqrt{n}$ so that $I_n$ is expressed as
$$ I_n = \frac{(-1)^n}{\sqrt{2\pi}} \int^{\infty}_{-\infty} dx\, x^n D^n(e^{-x^2/2}).$$
Applying integration by parts $n$ times on the RHS gives
$$ I_n = \frac{(-1)^{n+n}}{\sqrt{2\pi}} \int^{\infty}_{-\infty} dx\, D^n(x^n) e^{-x^2/2} = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} dx\, n! e^{-x^2/2} = n!\qquad \checkmark$$
Hermite Polynomials
For the Hermite polynomials approach, we note that for the Hermite polynomial definition
$$ \mathit{He}_n(x) = (-1)^n e^{{x^2}/{2}}D^n(e^{-{x^2}/{2}})$$
we have the ortgonolaity relation
$$ \int_{-\infty}^\infty \mathit{He}_m(x) \mathit{He}_n(x)\, e^{-{x^2}/{2}} \,dx = \sqrt{2 \pi}\, n!\, \delta_{nm}.$$
Thus we find
$$(-1)^n\int_{-\infty}^\infty \mathit{He}_n(x) D^n(e^{-{x^2}/{2}})\,dx = \sqrt{2 \pi}\, n!.$$
We note that the highest power term in $\mathit{He}_n(x)$ is $x^n$ (i.e., $He_n(x) = x^n + o(x^{n-1})$). All other terms are of lower powers and vanish when we apply $n$ integrations by parts on the integral. Thus we have the starting relation.
