I have an integral of the form
$$ \int\_R^{\infty} e^{-x} x^n \vert L\_m^{\alpha}(x) \vert^2 \ dx,$$
where $L\_m^{\alpha}$ is the generalized Laguerre polynomial and $n \ge 0.$


I would to get a nice explicit exponential bound on this integral in terms of $R$ (I intend to take $R$ large). However, most upper bounds on these polynomials that I could find involve exponentials themselves, see for instance here [on DLMF.](https://dlmf.nist.gov/18.14)


Does anybody see a way to get a nice estimate on this integral for $R$ large? The exponential decay is clear, as we are integrating an exponential against a polynomial. However, I would like to have a bound that is a good tradeoff between compact and efficient.



Exponential decay bound on integral
Let $a:=\alpha$. The integral in question is
\begin{equation\*}
 I(R):=\int\_R^\infty e^{-x} x^n L\_m^a(x)^2\, dx, 
\end{equation\*}
[where](https://en.wikipedia.org/wiki/Laguerre_polynomials#Explicit_examples_and_properties_of_the_generalized_Laguerre_polynomials)  

\begin{equation\*}
 L\_m^a(x)=\sum\_{i=0}^m b\_i,\quad b\_i:=(-1)^i \binom{m+a}{m-i}\frac{x^i}{i!},
\end{equation\*}
so that for $i=0,\dots,m-1$ and $x\ge R>0$
\begin{equation\*}
 \frac{|b\_i|}{|b\_{i+1}|}=\frac{i+1}{(m-i)x}\,|i+1+a|\le\frac{c\_{m,a}}R,\quad 
 c\_{m,a}:=m\max(|m+a|,|a+1|)
\end{equation\*}
and hence for $R>c\_{m,a}$
\begin{equation\*}
 L\_m^a(x)^2\le\frac{b\_m^2}{(1-c\_{m,a}/R)^2}
 =\frac1{(1-c\_{m,a}/R)^2}\frac{x^{2m}}{(m!)^2}. 
\end{equation\*}
Thus, for $R>\max(c\_{m,a},n+2m)$, by [Proposition 2.7](https://www.semanticscholar.org/paper/Exact-lower-and-upper-bounds-on-the-incomplete-Pinelis/df79633882e070bc48bda463ccbbc9919a256030),
\begin{equation\*}
\begin{aligned}
 I(R)&\le J(R):=\frac1{(1-c\_{m,a}/R)^2}\frac1{(m!)^2}\,\frac{e^{-R}R^{n+2m}}{1-(n+2m)/R} \\ 
 &\sim\frac{e^{-R}R^{n+2m}}{(m!)^2} 
\end{aligned}
\tag{1}\label{1}
\end{equation\*}
as $R\to\infty$. So, we do get an exponential decrease in $R$.


The upper bound $J(R)$ on $I(R)$ in \eqref{1} is asymptotically exact (as $R\to\infty$), since $L\_m^a(x)^2\sim\dfrac{x^{2m}}{(m!)^2}$ as $x\to\infty$.


