Hermite polynomials $H\_k(x), x \in \mathbb{R}, k \in \mathbb{N}$ are defined by the formula
$$
H\_k(x)=(-1)^k e^{x^2} \frac{d^k}{d x^k}\left(e^{-x^2}\right) .
$$
Each $H\_k(x)$ is a polynomial of exact degree $k$. The Hermite polynomials are also given by the generating function
$$
e^{2 x w-w^2}=\sum\_{k=0}^{\infty} \frac{H\_k(x)}{k !} w^k
$$


Define the Hermite functions $\tilde{h}\_k(x)$ by
$$
\tilde{h}\_k(x)=H\_k(x) e^{-\frac{1}{2} x^2} .
$$


We have the Mehler's formula, for the Hermite functions $\tilde{h}\_k(x)$.


**Proposition** For $w \in \mathbb{C},|w|<1$ and $x, y \in \mathbb{R}$,
$$\sum\_{k=0}^{\infty} \frac{\tilde{h}\_k(x) \tilde{h}\_k(y)}{2^k k !} w^k=\left(1-w^2\right)^{-\frac{1}{2}} e^{-\frac{1}{2} \frac{1+w^2}{1-w^2}\left(x^2+y^2\right)+\frac{2 w}{1-w^2} x y}$$


One has $$\int\_\Bbb R\left(\tilde{h}\_k(x)\right)^2 d x=2^k k ! \sqrt{\pi} .$$


Thus we can normalise $\tilde{h}\_k(x)$ by defining
$$h\_k(x)=\left(2^k k ! \sqrt{\pi}\right)^{-\frac{1}{2}} \tilde{h}\_k(x)$$


This family $\{h\_k(x): k \in \mathbb{N}\}$ is an orthonormal system in $L^2(\mathbb{R})$. But we can say more.


**Theorem** The system $\{h\_k(x): k \in \mathbb{N}\}$ is an orthonormal basis for $L^2(\mathbb{R})$. Consequently, every $f \in L^2(\mathbb{R})$ has an expansion
$$f(x)=\sum\_{k=0}^{\infty}\left(f, h\_k\right) h\_k(x)$$
where the series converges to $f$ in the $L^2$ norm.


My question is there a close formula for this sum: $$\sum\_{k=0}^{\infty}\frac{1}{k+a}h\_k(x)h\_k(y)$$



Closed formula for Hermite polynomials
Up to some normalization, the harmonic oscillator $H$ is self-adjoint such that
$$
\langle Hu, u\rangle=\sum\_{k\ge 0}(\frac12+k) \vert u\_k\vert^2,
$$
and thus defining a self-adjoint $A$ by the equality
$$
\langle Au, u\rangle=\sum\_{k\ge 0}(a+k) \vert u\_k\vert^2,
\quad\text{implying}\ A=H+a-\frac12.
$$
As a result your sum is the kernel of the operator
$$
(H+a-\frac12)^{-1},
$$
which makes sense for $a>0$.


