I would like to compute (or estimate) this integral of 2D Hermite function, where $n, m \in \mathbb{N}^2$ and $r \in [0, 2]$ :
$$I_{n, m}(r) := \int_{\mathbb{R}^2} u_m(x) u_n(x) |x|^r dx$$
where, for $n = (n_1,n_2) \in \mathbb{N}^2$, and $x = (x_1, x_2) \in \mathbb{R}^2$ :
$$u_n(x) = (\pi n_1! n_2! 2^{n_1} 2^{n_2})^{-1/2} H_{n_1}(x_1) H_{n_2}(x_2) e^{-\frac12 (x_1^2 + x_2^2)}$$
and for $k \in \mathbb{R}$ and $y \in \mathbb{R}$, $H_k(y)$ is the Hermite polynomial defined by :
$$H_{k}(y) = (-1)^{k} e^{y^2} \dfrac{d^k}{dy^k} e^{-y^2}$$
Finally, for $x = (x_1, x_2) \in \mathbb{R}^2$ and $r \in \mathbb{R}$, $|x|^r = (|x_1|^2+|x_2|^2)^{r/2}$.
By parity argument, $I_{n, m}(r) = 0$ when :
- $n_1$ is even and $n_2$ is odd / $n_2$ is even and $n_1$ is odd
- $m_1$ is even and $m_2$ is odd / $m_2$ is even and $m_1$ is odd
But, I would like to understand what happend for the other case.
According to the equation (52) of this page https://mathworld.wolfram.com/HermitePolynomial.html, there is a closed form for a related integral for the 1D case and integer $r$.
I suspect that there is no closed form for the non integer case in the 1D and 2D settings, but an upper bound of $I_{n, m}(r)$ is ok for me.
QUESTION : How to upper bound $|I_{n, m}(r)|$ for $r \in (0, 2]$ ?
I am not used to Hermite function, and I don't know wich properties of theses function could be useful, so I don't know how to start. Do you have any hint or reference ?
CONTEXT : I came across with integral in a signal processing problem. I the best world, I hope that $I_{n,m}(r)$ is "small".
EDIT :
Using the fact that $(u_n)_{n \in \mathbb{N}^2}$ is an hermitian basis of $L^2(\mathbb{R}^2)$, I was able to obtain that, $\forall m \in \mathbb{N}^2, I_{n, m} \underset{|n| \to \infty}{\to} 0$. Indeed the function $u_m(x)|x|^r \in L^2(\mathbb{R}^2)$ and thus $\sum_{n \in \mathbb{N}^2} \Big|\int_{\mathbb{R}^2} u_n(x) u_m(x)|x|^r dx\Big|^2 = \int_{\mathbb{R}^2} |u_m(x)|^2 |x|^{2r} dx < \infty$.
But, I am interested in the speed of decreasing of $I_{n, m}(r)$ as $|n|, |m| \to \infty$.
For $r=0$ there is no problem because this is just an integral over decoupled Cartesian Gaussians, so the integral is a product of integrals. Otherwise, omitting the unrelated constants, $$ \int_{-\infty}^\infty dx \int_{-\infty}^\infty dy H_{n_1}(x)H_{n_2}(y)e^{-(x^2+y^2)/2} H_{m_1}(x)H_{m_2}(y)e^{-(x^2+y^2)/2} (x^2+y^2)^{r/2} $$ $$ = \int_0^{2\pi} d\phi \int_0^{\infty}\rho d\rho H_{n_1}(\rho\cos\phi)H_{n_2}(\rho\sin\phi)e^{-\rho^2} H_{m_1}(\rho\cos\phi)H_{m_2}(\rho\sin\phi) \rho^r $$ $$ = \sum_{l_1=0}^{\lfloor n_1/2\rfloor} \sum_{l_2=0}^{\lfloor n_2/2\rfloor} \sum_{s_1=0}^{\lfloor m_1/2\rfloor} \sum_{s_2=0}^{\lfloor m_2/2\rfloor} \int_0^{2\pi} d\phi \int_0^{\infty}\rho d\rho n_1!\frac{(-)^{l_1}}{l_1!(n_1-2l_1)!}(2\rho\cos\phi)^{n_1-2l_1} $$ $$ \times n_2!\frac{(-)^{l_2}}{l_2!(n_2-2l_2)!}(2\rho\sin\phi)^{n_2-2l_2} $$ $$ \times m_1!\frac{(-)^{s_1}}{s_1!(m_1-2s_1)!}(2\rho\cos\phi)^{m_1-2s_1} $$ $$ \times m_2!\frac{(-)^{s_2}}{s_2!(m_2-2s_2)!}(2\rho\sin\phi)^{m_2-2s_2} e^{-\rho^2} \rho^r $$ Define $R_n\equiv \int_0^\infty x^n e^{-x^2}=\frac12\Gamma(\frac{n+1}{2})$ $$ \ldots = \sum_{l_1=0}^{\lfloor n_1/2\rfloor} \sum_{l_2=0}^{\lfloor n_2/2\rfloor} \sum_{s_1=0}^{\lfloor m_1/2\rfloor} \sum_{s_2=0}^{\lfloor m_2/2\rfloor} R_{1+r+n_1-2l_1+n_2-2l_2+m_1-2s_1+m_2-2s_2} $$ $$ \times \int_0^{2\pi} d\phi n_1!\frac{(-)^{l_1}}{l_1!(n_1-2l_1)!}(2\cos\phi)^{n_1-2l_1} $$ $$ \times n_2!\frac{(-)^{l_2}}{l_2!(n_2-2l_2)!}(2\sin\phi)^{n_2-2l_2} $$ $$ \times m_1!\frac{(-)^{s_1}}{s_1!(m_1-2s_1)!}(2\cos\phi)^{m_1-2s_1} $$ $$ \times m_2!\frac{(-)^{s_2}}{s_2!(m_2-2s_2)!}(2\sin\phi)^{m_2-2s_2} $$ $$ \ldots = \sum_{l_1=0}^{\lfloor n_1/2\rfloor} \sum_{l_2=0}^{\lfloor n_2/2\rfloor} \sum_{s_1=0}^{\lfloor m_1/2\rfloor} \sum_{s_2=0}^{\lfloor m_2/2\rfloor} R_{1+r+n_1-2l_1+n_2-2l_2+m_1-2s_1+m_2-2s_2} $$ $$ \times P(n_2-2l_2+m_2-2s_2,n_1-2l_1+m_1-2l_2) n_1!\frac{(-)^{l_1}}{l_1!(n_1-2l_1)!}(2)^{n_1-2l_1} \times n_2!\frac{(-)^{l_2}}{l_2!(n_2-2l_2)!}(2)^{n_2-2l_2} $$ $$ \times m_1!\frac{(-)^{s_1}}{s_1!(m_1-2s_1)!}(2)^{m_1-2s_1} \times m_2!\frac{(-)^{s_2}}{s_2!(m_2-2s_2)!}(2)^{m_2-2s_2}. $$ This uses the shortcut $$ P(p,q)\equiv \int_0^{2\pi} \cos^q\phi \sin^p\phi d\phi = \int_0^{\pi} \cos^q\phi \sin^p\phi d\phi +\int_{\pi}^{2\pi} \cos^q\phi \sin^p\phi d\phi $$ $$ = \int_0^{\pi} \cos^q\phi \sin^p\phi d\phi -(-)^p \int_{\pi}^{0} \cos^q\phi \sin^p\phi d\phi = \int_0^{\pi} \cos^q\phi \sin^p\phi d\phi +(-)^p \int_{0}^{\pi} \cos^q\phi \sin^p\phi d\phi $$ so $P(p,q)=0$ if $p$ is odd. For $p$ even $$ P(p,q) = 2\int_0^{\pi} \cos^q\phi \sin^p\phi d\phi = 2\int_0^{\pi/2} \cos^q\phi \sin^p\phi d\phi +2\int_{\pi/2}^{\pi} \cos^q\phi \sin^p\phi d\phi $$ $$ = 2\int_0^{\pi/2} \cos^q\phi \sin^p\phi d\phi -2(-)^q \int_{\pi/2}^{0} \cos^q\phi \sin^p\phi d\phi = 2\int_0^{\pi/2} \cos^q\phi \sin^p\phi d\phi +2(-)^q \int_{0}^{\pi/2} \cos^q\phi \sin^p\phi d\phi $$ so $P(p,q)=0$ if $q$ is odd. So if $p$ and $q$ are even $$ P(p,q)= 4\int_0^{\pi/2} \cos^q\phi \sin^p\phi d\phi = \frac{(p-1)(q-1)}{(p+q)(p+q-2}P(p-2,q-2) $$ recursively starting at $P(0,0)=2\pi$, $P(0,2k)=P(2k,0)=2\pi \frac{(2k-1)!!}{(2k)!!}$ where $!!$ is the double factorial. For $r=1$ one can use the Maple program
which generates the following table of $n_1$, $n_2$, $m_1$, $m_2$, $I_{n,m}$ (analytic and floating point)
where only cases for $I_{n,m}\neq 0$, $n_1+n_2+m_1+m_2\le 12$, $n_1\le n_2\le m_1\le m_2$ have been tabulated.