In a quantum mechanics problem I encountered the following integral
\begin{equation\*}
 \int\_0^\infty t^{\nu+1}J\_\nu(\beta t)L\_{\mu-\nu}^{2\nu}(t)e^{-t/2}dt\,,
\end{equation\*}
where $L$ denotes the [associated Laguerre polynomial](https://en.wikipedia.org/wiki/Laguerre_polynomials), $J$ a [Bessel function of the first kind](https://en.wikipedia.org/wiki/Bessel_function#Bessel_functions_of_the_first_kind:_J.CE.B1), $\beta\geq0$ a real parameter, and $\mu\geq\nu\geq0$ positive integers. I've tried to find this integral in tables such as [Gradshteyn and Ryzhik
Table of Integrals, Series, and Products](https://www.mathtable.com/gr/) but haven't found anything. The closest match seems to be a paper by A.D.Alhaidari (App. Math. Letters 2007, [ScienceDirect link](https://www.sciencedirect.com/science/article/pii/S0893965906000954)) however the exponent in the first term is different. Does anyone have any ideas?



Integral involving associated Laguerre polynomial and Bessel function
Let $n=\mu - \nu$ be an integer, as specified in the problem. Then I'll indicate how to prove that


$$I:=\int\_0^\infty e^{-a\,t} t^{\nu+1} J\_\nu(b\,t) L\_n^{2v}(t) \,dt =\frac{(2b)^\nu \Gamma(\nu+1/2)}{\sqrt{\pi}}\sum\_{k=0}^n \frac{(-1)^k}{k!}\binom{n+2\nu}{n-k} $$
$$ 
 (a^2+b^2)^{-(\nu+k+3/2)}a^{k+1}{}\_2F\_1(-k/2, -k/2-1/2,1+\nu,-(b/a)^2)(2\nu+1)\_{k+1}$$


The ${}\_2F\_1$ is the Gauss hypergeometric and it is a polynomial, because either $-k/2$ or $-k/2-1/2$ is a negative integer or zero. The $(\cdot)\_n$ is a Pochhammer symbol.


The first step is to use the expansion for the Laguerre polynomials,
$$ L\_n^{2v}(t) = \sum\_{k=0}^n \frac{(-t)^k}{k!} \binom{n+2\nu}{n-k} $$
The parameter $a$ is not the specific $1/2$ of the problem, because we will differentiate with respect to it:
$$I=\sum\_{k=0}^n \frac{(-1)^k}{k!} \binom{n+2\nu}{n-k} (-1)^{k+1}\frac{d^{k+1}}{da^{k+1}} \int\_0^\infty e^{-a\,t} t^{\nu} J\_\nu(b\,t) \,dt
$$
This is done because Grashteyn 6.623.1 says
$$\int\_0^\infty e^{-a\,t} t^{\nu} J\_\nu(b\,t) \,dt = \frac{(2b)^\nu \Gamma(\nu+1/2)} {\sqrt{\pi}}(a^2+b^2)^{-(\nu+1/2)}$$
To complete the proof we need to show that
$$\frac{d^n}{da^n}(a^2+b^2)^{-(\nu+1/2)}=$$
$$= (a^2+b^2)^{-(\nu+n+1/2)}
(-a)^n{}\_2F\_1(-n/2, -n/2+1/2,1+\nu,-(b/a)^2)(2\nu+1)\_n$$


I did this by starting with the Rodrigues formula for the Gegenbauer polynomials
$$\frac{d^n}{da^n}(1-t^2)^{-(\nu+1/2)}=(1-t^2)^{-(\nu+n+1/2)}C\_n^{-\nu-n}(t)
\frac{{(1/2-\nu-n)}\_n\,2^n}{\binom{2(\nu+n)}{n}}$$ 
Now, scale $t \to ia/b,$ use a hypergeometric representation for the Gengenbaur polys, use several gamma function ID's, and simplify.


