Yes, there are transformations that transform finite integrals to doubly infinite integrals, like the $\tanh$-substitution
$$\int_a^b f(x) \mathrm dx=\frac{b-a}{2}\int_{-\infty}^\infty f\left(\frac{a+b}{2}+\frac{b-a}{2}\tanh\,u\right)\operatorname{sech}^2 u \,\mathrm du$$
and yes, you can apply Gauss-Hermite quadrature to any suitably transformed doubly infinite integral:
$$\int_{-\infty}^\infty f(v)\mathrm dv\approx\sum_{k=1}^n w_k \exp(x_k^2)f(x_k)$$
BUT
Gauss-Hermite doesn't work very well if the functions you are integrating are not of the form $\exp(-x^2)f(x)$, where $f(x)$ is a function that is well approximated by a polynomial. Remember that the idea of Gaussian quadrature in general is to "factor out" unruly behavior in your integrands, and keep that behavior to the nodes and weights of that quadrature rule. That supposedly leaves you with a remainder that is much more well-behaved for quadrature than the original integrand. In effect, trying to use Gauss-Hermite to integrate things that don't have the factor $\exp(-x^2)$ is like trying to use a toothbrush to scrub a toilet: you'll manage to finish, but it'll take you a while.
To drive the point home, consider the two integrals
$$\begin{align*}
\mathcal I_1&=\int_{-\infty}^\infty \operatorname{sech}\,u;\mathrm du=\pi\\
\mathcal I_2&=\int_{-\infty}^\infty \exp(-u^2)\operatorname{sech}\,u\;\mathrm du\approx 1.4790611714495759
\end{align*}$$
Here are a few short Mathematica routines for generating the nodes and weights for Gauss-Hermite quadrature:
(* Golub-Welsch algorithm *)
golubWelsch[d_?VectorQ, e_?VectorQ] :=
Transpose[
MapAt[(First[e] Map[First, #]^2) &,
Eigensystem[
SparseArray[{Band[{1, 1}] -> d, Band[{1, 2}] -> Sqrt[Rest[e]],
Band[{2, 1}] -> Sqrt[Rest[e]]}, {Length[d], Length[d]}]], {2}]]
(* generate nodes and weights for Gauss-Hermite quadrature *)
ghq[n_Integer, prec_: MachinePrecision] :=
Transpose[
Sort[golubWelsch[ConstantArray[0, n],
N[Prepend[Range[n - 1]/2, Sqrt[Pi]], prec]]]]
Here's a short table of the relative error in using $n$-point Gauss-Hermite quadrature for numerically evaluating these two integrals (done with Mathematica):
$$\begin{array}{c|c|c}
n&-\log_{10}\left|\frac{\text{estimate}}{\mathcal I_1}-1\right| &-\log_{10}\left|\frac{\text{estimate}}{\mathcal I_2}-1\right|\\\hline
2&0.58153&1.3066\\
5&1.0804&2.6953\\
10&1.6518&4.3750\\
15&2.0926&5.6971\\
20&2.4643&6.8235\\
25&2.7917&7.8217\\
\end{array}$$
Clearly, the convergence of Gauss-Hermite quadrature for $\mathcal I_1$ is rather shabby compared to the relatively quicker convergence for $\mathcal I_2$. Sometimes, you'll get lucky and find a function where Gauss-Hermite performs well even if it does not have an explicit $\exp(-x^2)$ factor, but those things aren't that common.
From the given representation, we can express
\begin{equation}
L_n(\frac{2r}{n+1/2})= \frac{2}{n!} e^{\frac{2r}{n+1/2}} \int_0^\infty e^{-t^2} t^{2n+1} J_0\left(2t \sqrt{\frac{2r}{n+1/2}}\right) \,dt
\end{equation}
changing $ t=\sqrt{u\left( n+1/2 \right)}$,
\begin{equation}
L_n(\frac{2r}{n+1/2})= \frac{(n+1/2)^{n+1}}{n!} e^{\frac{2r}{n+1/2}}\int_0^\infty e^{-u\left( n+1/2 \right)}u^{n+1/2}J_0\left( 2\sqrt{2ru} \right)\frac{du}{\sqrt{u}}
\end{equation}
or
\begin{equation}
L_n(\frac{2r}{n+1/2})= \frac{(n+1/2)^{n+1}}{n!} e^{\frac{2r}{n+1/2}}\int_0^\infty e^{-\left( n+1/2 \right)\left( u-\ln u \right)}J_0\left( 2\sqrt{2ru} \right)\frac{du}{\sqrt{u}}
\end{equation}
The argument of the exponential is minimum at $u=1$. We can use the Laplace method to derive an asymptotic approximation for the integral. Near $u=1$,
$u-\ln(u)\sim 1+(u-1)^2/2$ and $u^{-1/2}J_0\left( 2\sqrt{2ru}\right)\sim J_0\left( 2\sqrt{2r}\right)$, then
\begin{equation}
L_n(\frac{2r}{n+1/2})\sim \sqrt{2\pi}\frac{(n+1/2)^{n+1/2}}{n!} e^{\frac{2r}{n+1/2}}
e^{-(n+1/2)}J_0\left( 2\sqrt{2r}\right)
\end{equation}
Now, plugging the Stirling approximation for $n!$ and an expansion for the exponential term, we obtain the expected result $ L_n(\frac{2r}{n+1/2})\sim J_0\left( 2\sqrt{2r}\right)$. To improve the approximation, the above formula and/or higher orders in the Laplace expansion can be used.
Best Answer
Note that the integrand is an even function--$4x^2-2$--times an odd function--$8x^3-12x$--so is an odd function. This function is a polynomial, and so is continuous. Given any $L>0$, we have that the integral over $[-L,L]$ of an odd function (that's continuous there) is $0$.