I am trying to prove that the integral 
\begin{align}
 \int\_{0}^{\infty } e^{-\frac{r^2}{2B}} r^{l-n} 
 L\_n^{l-n}\left(\frac{r^2}{C}\right) I\_{l-n}\left(\rho r \right) r dr
\end{align}


has the form 
\begin{align}
 B^{l+1} e^{\frac{B}{2}\rho^2} \rho^{l-n} L\_n^{l-n}\left(\frac{\rho^2}{C} \right),
\end{align}
where $L\_n^{l-n}$ is the generalised Laguerre function, and $I\_{l-n}$ is the modified Bessel function of the first kind. 


Expanding the modified Bessel function into an infinite sum, and then using Eq. 7, section 7.414 (pg. 809) from Tables of Integrals, Series & Products (Ed. 7) (by I.A. Gradshteyn & I.M.Ryzhik), which is
\begin{align}
 \int\_{0}^{\infty} e^{-st} t^{\beta} L\_n^{\alpha}(t) dt 
 = \frac{\Gamma(\beta+1) \Gamma(\alpha+n+1)}
 {n! \Gamma(\alpha+1) s^{\beta+1}}
 {}\_1F\_2\left(-n, \beta+1; \: \alpha+1; \: \frac{1}{s} \right),
\end{align}
I can get close, but not close enough! 


Does anyone know how to do this?


Please let me know if you need any further information. I haven't done this very often, but the few times I have I always seem to forget something pertinent!



Integral involving Laguerre, Gaussian and modified Bessel function
Mathematica can evaluate the integral$^\ast$
$$I\_{n,l}=\begin{align}
 \int\_{0}^{\infty } e^{-\frac{r^2}{2B}} r^{l-n} 
 L\_n^{l-n}\left(\frac{r^2}{C}\right) I\_{l-n}\left(\rho r \right) r dr
\end{align},\;\;B,C,\rho>0,$$
for any integer $n\geq 0$ as a function of $l>n-1$. The results are consistent with 
$$I\_{n,l}=\begin{align}
 B^{l+1} (1/B-2/C)^ne^{\frac{B}{2}\rho^2} \rho^{l-n} L\_{n}^{l-n}\left(\frac{B^2\rho^2}{C-2B}\right),\;\;l+1>n\geq 0,
\end{align}$$
Not quite what the OP suggested, but similar.




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$^\ast$ The evaluation of the integral for given $n$ follows from the general formula
$$\int\_0^\infty r^p e^{-a r^2} I\_q(r)\,dr=\frac{\Gamma \left(\frac{1}{2} (p+q+1)\right)}{2^{q+1}a^{\frac{1}{2} (p+q+1)}\,\Gamma (q+1)} \, \_1F\_1\left(\tfrac{1}{2} (p+q+1);q+1;\frac{1}{4 a}\right)$$
(for $p>-1$, $p+q>-1$, $a>0$)


