I consider the Hermite functions $$\psi_n(x) = (-1)^n (2^n n! \sqrt{\pi})^{-1/2} e^{x^2/2} \dfrac{d^n}{dx^n} e^{-x^2}$$ and I would like to bound above, for $r \in (0, 2)$: $$I_r(n) = \int_{\mathbb{R}} |\psi_n(x)|^2 |x|^r dx \leq ? $$
I know that $I_0(n) = 1$ and $I_2(n) \lesssim n$ (using $x \psi_n(x) = \sqrt{\frac{n}{2}} \psi_{n-1} + \sqrt{\frac{n+1}{2}} \psi_{n+1}$), but I have no idea how to proceed for $r \in (0, 2)$.
I suspect the problem to be well known but I haven't found any answer online. I'd be happy to receive any indications or references.
EDIT : Thanks to Giuseppe Negro's comment I was able to obtain the bound $I_n(r) \lesssim n^{r/2}$ by using Holder's inequality : $$I_n(r) = \int_{\mathbb{R}} \big[|\psi_n(x)|^{r} |x|^{r}\big] |\psi_n(x)|^{2 - r} dx \leq \Big(\int_{\mathbb{R}} |\psi_n(x)|^{2} |x|^{2} dx\Big)^{r/2} \Big(\int_{\mathbb{R}} |\psi_n(x)|^{(2 - r)/(1 - r/2)} dx \Big)^{1 - r/2}$$ But, Holder's inequality seems not really tight in this context, so I wonder if this bound can be improved.
Let us start from
$$\psi_n(x) = \lambda_n \, e^{-\frac{x^2}{2}} H_n(x),$$
in which $\lambda_n=(2^n n! \sqrt{\pi})^{-\frac{1}{2}}$ and noting that
\begin{align}I_r(n) &= 2 \int_0^\infty |\psi_n(x)|^2 |x|^r dx \\ &= 2 \lambda_n^2 \int_0^\infty x^{r} e^{-x²}\, H_n(x)^2 dx \\ &= \lambda_n^2 \frac{n!^2}{ p!^2 } \Gamma(\frac{r+1}{2}+\delta) \times F_2\left(-p, -p,\frac{r+1}{2}+\delta; \frac{1}{2}+\delta, \frac{1}{2}+\delta; 1, 1\right), \end{align}
in which:
the latter formula resulting from Prudnikov et al.'s Integrals and Series [1992, Vol. 2 p. 503, eq. 2.20.16.12]. The Appell's function is actually a polynomial, therefore is defined everywhere, owing to its first two parameters being negative integers, so that there is no issue of convergence of $F_2$ at unity values (unlike in the general non-polynomial case).
There is a special case to consider if $r=0$. In this case, the Appell's function simplifies down to a Gauss hypergeometric function:
$$ F_2(-p, -p, \frac{1}{2}+\delta; \frac{1}{2}+\delta, \frac{1}{2}+\delta; 1, 1) = {_2}F{_1}(-p, -p; \frac{1}{2}+\delta; 1, 1) = \frac{(\frac{1}{2}+\delta+p)_n}{(\frac{1}{2}+\delta)_n}\;\text{ using the usual Pochhammer symbol, by the Chu-Vandermonde identity (NIST 2010 15.4.24) }$$
You can then either use the exact value in your work or apply whatever upper bound to the Gamma function you like best, there are numerous ones and I will not go further down this line.
Otherwise, if $r \neq 0$, there is no obvious simplication to $F_2$ but at least you can use the definition of the polynomial and apply the triangular inequality so that:
$$ \left|F_2\left(-p, -p, \frac{r+1}{2}+\delta; \frac{1}{2}+\delta, \frac{1}{2}+\delta; 1, 1\right)\right| \leq \sum_{(k,l) \in I} \frac{(-1)^l (-p)_{k+l}(-p)_k (\frac{r+1}{2}+\delta)_l}{(\frac{1}{2}+\delta)_k(\frac{1}{2}+\delta)_l \, k! \,l!},$$ in which $I$ is the set of integral index pairs $(k, l)$ such that $0\leq k < p \land 0\leq l < p-k$. This is a computable bound that depends only on parameters $n$ and $r$, lets us denote it as $M_{n, r}$.
Then the proposed upper bound is:
$$ I_r(n) \leq \frac{n!}{2^n \sqrt{\pi}\, p!^2 } \Gamma\left(\frac{r+1}{2}+\delta\right) M_{n,r}$$
When $n$ is even, then $p=\frac{n}{2}, \delta=0$ so that this inequality further simplifies by Legendre's duplication formula, down to:
$$I_r(n) \leq \frac{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(\frac{r+1}{2}\right)}{\pi\, \Gamma\left(\frac{n}{2}+1\right) } M_{n,r}$$
Noting further that $r \leq 2$, at least for $n$ even the coefficient before $M_{n, r}$ is uniformly lower than unity (but not that much), so that $M_{n,r}$ can after all be chosen as a reasonably simple bound. I leave it to you to work out what happens for $n$ odd.