Orthogonal polynomials with respect to $e^{-|x|} \mathrm{d} x$ on the entire real line? The Laguerre polynomials
https://en.wikipedia.org/wiki/Laguerre_polynomials
form a system of orthogonal polynomials with respect to the measure $e^{ -x} \mathrm{d} x$ on $(0,\infty)$.
Is anything known about the orthogonal polynomials with respect to the measure $e^{ -|x|} \mathrm{d} x$ on $(-\infty,\infty)$?
Thank you!
 A: I'll quote from T. J. Sullivan's Introduction to Uncertainty Quantification (SpringerLink). Let $\mathfrak P$ be the set of polynomials (as a subset of $L^2(\mu)$, where $\mu$ is a measure). Then by Gram-Schmidt on monomials, we can get the existence of orthogonal polynomials for $\mu=\exp(-|x|)dx$-

Theorem 8.5. If the $L^{2}(\mu)$ inner product is positive definite on $\mathfrak{P},$ then there exists an infinite sequence of orthogonal polynomials for $\mu$.

The book also mentions that $\int \exp(a|x|)d\mu(x)<\infty$ for some $a>0$ implies the completeness of the above constructed set of orthogonal polynomials; thus this works for $\mu=\exp(-|x|)dx$.
Then, to find the polynomials, you can use the three-term recurrence-

Theorem
8.9. Let $\mathcal{Q}=\left\{q_{n} \mid n \in \mathcal{N}\right\}$ be the monic orthogonal polynomials for a measure $\mu .$ Then
$$
\begin{aligned}
q_{n+1}(x) &=\left(x-\alpha_{n}\right) q_{n}(x)-\beta_{n} q_{n-1}(x) \\
q_{0}(x) &=1 \\
q_{-1}(x) &=0
\end{aligned}
$$
where
$$
\begin{aligned}
\alpha_{n} &:=\frac{\left\langle x q_{n}, q_{n}\right\rangle_{L^{2}(\mu)}}{\left\langle q_{n}, q_{n}\right\rangle_{L^{2}(\mu)}}, & \text { for } n \geq 0 \\
\beta_{n} &:=\frac{\left\langle q_{n}, q_{n}\right\rangle_{L^{2}(\mu)}}{\left\langle q_{n-1}, q_{n-1}\right\rangle_{L^{2}(\mu)}}, & \text { for } n \geq 1, \\
\beta_{0} &:=\left\langle q_{0}, q_{0}\right\rangle_{L^{2}(\mu)} \equiv \int_{\mathbb{R}} \mathrm{d} \mu.
\end{aligned}
$$
Hence, the orthonormal polynomials $\left\{p_{n} \mid n \in \mathcal{N}\right\}$ for $\mu$ satisfy
$$
\sqrt{\beta_{n+1}} p_{n+1}(x)=\left(x-\alpha_{n}\right) p_{n}(x)-\sqrt{\beta_{n}} p_{n-1}(x)
$$
$$
\begin{aligned}
p_{0}(x) &=\beta_{0}^{-1 / 2} \\
p_{-1}(x) &=0.
\end{aligned}
$$

For instance, $q_{0,-1}$ are given, and to find $q_1$, we need to compute $\alpha_0$ and $\beta_0$, which are
$$\beta_0 = \int_{\mathbb R} d\mu = \int_{-\infty}^\infty \exp(-|x|)dx = 2,$$
and
$$ \alpha_{0} =\frac{\left\langle x q_{0}, q_{0}\right\rangle_{L^{2}(\mu)}}{\left\langle q_{0}, q_{0}\right\rangle_{L^{2}(\mu)}} = \frac12\int x\exp(-|x|)dx = 0 $$
so the first two in the list are
$$q_0=1,\quad q_1=x.$$
Then you can continue inductively. $\langle xq_1,q_1\rangle_{L^2(\mu)}$ is  zero (hence so is $\alpha_1$). In fact, a quick induction argument (once you have the three-term recurrence, and since $\mu$ is even; see comments) shows that $q_{2k}$ is an even polynomial, and $q_{2k+1}$ is an odd one. This also implies $\alpha_n\equiv 0$. A quick calculation shows that
$\beta_1 = \frac12\int x^2 d\mu = 2,$
and so
$$q_2=x q_1 -2 q_0 = x^2-2.$$
A less easy calculation gives
$ \quad \beta_2 = 10,$
which means that
$$ q_3 = xq_2 -20 q_1 = x^3 - 12x.$$
And it continues. Maybe there's a nice closed form for the recurrence, but I haven't tried.
Verifications: this link shows that $\langle q_0,q_2\rangle_{L^2(\mu)} = 0$. $\langle q_1,q_2\rangle_{L^2(\mu)} = 0 = \langle q_0,q_3\rangle_{L^2(\mu)}$  and also $\langle q_2 ,q_3\rangle_{L^2(\mu)} = 0$ by the parity of the polynomials. This link shows that $\langle q_1 ,q_3\rangle_{L^2(\mu)} = 0$.
