Completeness of an orthogonal sequence of functions is a bit tricky on unbounded intervals, while it is relatively straightforward on bounded intervals. In the case of Laguerre and Hermite polynomials, there is a nice trick due to von Neumann that allows the reduction to bounded intervals.
There seems to be a bit of confusion about the interval in the statement of the question. Here's a correct statement:
For any real number $\alpha \gt -1$ the functions $\langle e^{-x/2} x^{\alpha/2} L_{n}^{(\alpha)}(x)\rangle_{n=0}^\infty$ obtained from the Laguerre polynomials $L_{n}^{(\alpha)}(x)$ are a complete orthogonal system in $L^2(0,\infty)$. The Hermite polynomials $H_n(x)$ yield the complete orthogonal system $\langle e^{-x^2/2} H_n(x)\rangle_{n=0}^\infty$ in $L^2(\mathbb{R})$.
This is proved in detail in the classic book Gábor Szegő, Orthogonal polynomials, Chapter 5. The entire chapter discusses the main properties oft the Laguerre polynomials $L^{(\alpha)}_n(x)$ for an arbitrary real number $\alpha \gt -1$ and proves their completeness in Section 5.7.
More precisely, Szegő shows in Theorem 5.7.1 on pages 108f that for fixed $\alpha \gt -1$ the functions $f_n(x) = e^{-x/2}x^{\alpha/2} x^n$ span a dense subspace of $L^2(0,\infty)$.
The first idea is to use a change of variables $y = e^{-x}$ in order to use the case of $L^2(0,1)$ where density of the span of $(\log1/y)^{\alpha/2} y^n$ is not too hard to prove (see Theorem 3.1.5).
Write a function in $L^2(0,\infty)$ as $e^{-x/2} x^{\alpha/2} f(x)$. Then we have that $(\log1/y)^{\alpha/2} f(\log(1/y)) \in L^2(0,1)$ can be approximated by functions of the form $(\log1/y)^{\alpha/2} p(y)$ where $p$ is a polynomial. Transforming back to $(0,\infty)$ this shows that
$$
\int_{0}^\infty e^{-x} x^\alpha (f(x) - p(e^{-x}))^2 \,dx \lt \varepsilon
$$
for a suitable polynomial $p$. This reduces the task to proving that for all natural $k$ there exists a polynomial $q$ such that
$$\tag{$\ast$}
\int_{0}^\infty e^{-x} x^\alpha (e^{-kx} - q(x))^2\,dx
$$
is as small as we wish.
To do this, von Neumann's trick is to use the generating function
of the Laguerre polynomials $L_{n}^{(\alpha)}(x)$
$$
(1-w)^{-\alpha-1} \exp\left(-\frac{xw}{1-w}\right) = \sum_{n=0}^\infty L_n^{(\alpha)}(x) w^n.
$$
Choosing $w = \frac{k}{k+1}$ we have $\exp\left(-\frac{xw}{1-w}\right) = \exp{(-kx)}$.
Thus, a natural choice for $q$ is $q_N(x) = (1-w)^{\alpha+1} \sum_{n=0}^N L_n^{(\alpha)}(x) w^n$ with large enough $N$. Plugging this into $(\ast)$ we obtain using the orthogonality relations
$$
\begin{align*}
\int_{0}^\infty e^{-x} x^\alpha (e^{-kx} - q_N(x))^2\,dx & = (1-w)^{2\alpha+2} \int_{0}^\infty e^{-x} x^\alpha \left(\sum_{n=N+1}^\infty L_{n}^{(\alpha)}(x) w^{n}\right)^2\,dx \\
&= (1-w)^{2\alpha+2} \Gamma(\alpha+1) \sum_{n=N+1}^\infty \binom{n+\alpha}{n} w^{2n}
\end{align*}$$
where term-wise integration is justified using an application of Cauchy-Schwarz. It remains to observe that the last expression tends to $0$ as $N \to \infty$.
Another reference discussing the case of $\alpha = 0$ nicely is Courant and Hilbert, Methods of mathematical physics, I, §9, sections 5 and 6. They discuss ordinary Laguerre and Hermite polynomials and their completeness.
Very roughly speaking: Laguerre polynomials look like the family of trigonometric functions ($\sin nx$ or $\cos nx$) in the region where the weight is concentrated. That is, they have moderate size and slowly increasing oscillation with $n$. Outside of this region, they look pretty much like any random collection of polynomials.
When $\alpha=0$, the weight $e^{-x}$ is concentrated near $0$. On the graph below the weight is shown in black, and the first five Laguerre polynomials $L_n$ are in various colors. You can see that they look like $1-\sin nx$ near $0$, and then wildly diverge when the weight becomes small.
For comparison, I did the same with $\alpha=2$. The polynomials have civilized appearance where the weight is not too small, roughly in the interval $[1,4]$. They wildly diverge both to the left and to the right.
do these polynomials still converge using the Least Square approximant
When $\alpha=0$, the Laguerre polynomial form on orthonormal basis of $L^2([0,\infty),e^{-x})$ (reference here). Therefore, for any function $f\in L^2([0,\infty),e^{-x})$ we have $\sum c_n L_n\to f$ in the norm, where $c_n=\langle f,L_n\rangle$.
When $\alpha>0$, the Laguerre polynomial form on orthogonal (but not normalized) basis* of $L^2([0,\infty),x^\alpha e^{-x})$. Therefore, for any function $f\in L^2([0,\infty),x^\alpha e^{-x})$ we have $\sum c_n L_n^{(\alpha)}\to f$ in the norm, where $c_n=\langle f,L_n^{(\alpha)}\rangle / \|L_n^{(\alpha)}\|$. (The value of $\|L_n^{(\alpha)}\|$ is given on Wikipedia).
(*) Disclaimer: I don't have a reference for the the completeness of $\{L_n^{(\alpha)}\}$, but I just put a bounty on unanswered question On the completeness of the generalized Laguerre polynomials in the hope someone does. Or you can consult the books G. Polya Orthogonal polynomials and G. Sansone Orthogonal functions, in case they treat $L_n^{(\alpha)}$.
Best Answer
Using the substitution $u=az\to dz=\frac1adu$ gives the integral $$\int_0^\infty (\frac{u}a)^\frac12e^{-u}L_m^\frac12(\frac{u}a)(\frac1a)du=\frac{1}{a\sqrt{a}}\int_0^\infty e^{-u}u^\frac12L_m^\frac12(\frac{u}a)du$$
Then by using the generating function $$\sum_{k=0}^\infty t^ke^{-u}u^\frac12L_k^\frac12(\frac{u}{a})=\frac{u^\frac12}{(1-t)^\frac32}e^{-\frac{u(t-at+a)}{a(1-t)}}$$ we can integrate every polynomial simultaneously giving $$\int_0^\infty\sum_{k=0}^\infty t^ke^{-u}u^\frac12L_k^\frac12(\frac{u}{a})du=\int_0^\infty\frac{u^\frac12}{(1-t)^\frac32}e^{-\frac{u(t-at+a)}{a(1-t)}}du$$ $$=\frac1{(1-t)^\frac32}\int_0^\infty u^\frac12e^{-\frac{u(t-at+a)}{a(1-t)}}du$$ Then we can use the substitution $v=\frac{u(t-at+a)}{a(1-t)}\to du=\frac{a(1-t)}{t-at+a}dv$ $$=\frac1{(1-t)^\frac32}\int_0^\infty \Big(\frac{a(1-t)}{t-at+a}v\Big)^\frac12e^{-v}\Big(\frac{a(1-t)}{t-at+a}\Big)dv$$ $$=\frac1{(1-t)^\frac32}\Big(\frac{a(1-t)}{t-at+a}\Big)^\frac32\int_0^\infty v^\frac12e^{-v}dv$$ $$=\frac1{(1-t)^\frac32}\Big(\frac{a(1-t)}{t-at+a}\Big)^\frac32\Gamma(\frac32)$$ $$=\Big(\frac{a}{t-at+a}\Big)^\frac32\frac{\sqrt{\pi}}{2}$$ $$=(t(1-a)+a)^{-\frac32}\frac{a\sqrt{a\pi}}{2}$$ $$=(1+\frac{t(1-a)}{a})^{-\frac32}\frac{\sqrt{\pi}}{2}$$ $$=\frac{\sqrt{\pi}}{2}\sum_{k=0}^\infty\binom{-\frac32}{k}\Big(\frac{t(1-a)}{a}\Big)^k$$ $$=\sum_{k=0}^\infty\frac{\sqrt{\pi}}{2}\binom{-\frac32}{k}\Big(\frac{1-a}{a}\Big)^kt^k$$
So the value of the integral for a given value of $k$ (or $m$ in your case) is $$\int_0^\infty z^{1/2}e^{-a\space z}L_k^{1/2}(z)\space dz = \frac{1}{a\sqrt{a}}\Big(\frac{\sqrt{\pi}}{2}\binom{-\frac32}{k}\Big(\frac{1-a}{a}\Big)^k\Big)=\boxed{\frac1{2a}\sqrt{\frac{\pi}{a}}\binom{-\frac32}{k}\Big(\frac{1-a}{a}\Big)^k}$$