Evaluating a definite integral with hypergeometric function

Question

I want to evaluate $$ \int_0^\infty e^{-x} x^{m+1} |{}_{1}F_1 (-n,m,x)|^2dx $$ where $ {}_{1}F_1 $ is the hypergeometric confluent function, that is defined by the series $$ {}_{1}F_1 (a,b,z)=\sum_{s=0}^\infty \frac{(a)_s z^s}{(b)_s s!} = 1 + \frac{a}{b}z + \frac{a(a+1)}{b(b+1)} \frac{z^2}{2}+\dots $$ Thank you.

Answer 1

Using the representation of the Laguerre polynomials as hypergeometric functions DLMF, \begin{equation} L^{(\alpha)}_{n}\left(x\right)=\frac{{\left(\alpha+1\right)_{% n}}}{n!}{{}_{1}F_{1}}\left(-n,\alpha+1,x\right) \end{equation} the integral can be written as \begin{equation} I=\left[\frac{n!}{{\left(m\right)_{% n}}}\right]^2\int_0^\infty e^{-x}\left[L^{(m-1)}_{n}\left(x\right)\right]^2 x^{m+1}\,dx \end{equation} Now, one may use twice the recurence relation \begin{equation} L^{(\alpha)}_{n}\left(x\right)=L^{(\alpha+1)}_{n}\left(x\right)-L^{(\alpha+1)}% _{n-1}\left(x\right) \end{equation} to obtain, for $n\geq2$, \begin{equation} L^{(m-1)}_{n}\left(x\right)=L^{(m+1)}_{n}\left(x\right)-2L^{(m+1)}_{n+1}\left(x\right)+L^{(m+1)}_{n-2}\left(x\right) \end{equation} In the integral, the factor $ e^{-x} x^{m+1}$ is just the weight for the $^{(m+1)}_{k}$. Developping the square, the cross-products give a vanihing contribution, for orthogonality reasons. With the normalisation factor \begin{equation} I=\left[\frac{n!}{{\left(m\right)_{n}}}\right]^2\left[\frac{\Gamma(n+m+2)}{n!}+\frac{4\Gamma(n+m+1)}{(n-1)!}+\frac{\Gamma(n+m)}{(n-2)!}\right] \end{equation} which can be simplified as \begin{equation} I=\frac{\Gamma(n+1)\Gamma^2(m)}{\Gamma(n+m)}\left( m(m+1)+6n(n+m)\right) \end{equation} The extension for $n=0,1$ is simple, with $L^{(m-1)}_{0}\left(x\right)=1$ and $L^{(m-1)}_{1}\left(x\right)=-x+m-1$.