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, $J$ a Bessel function of the first kind, $\beta\geq0$ 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 but haven't found anything. The closest match seems to be a paper by A.D.Alhaidari (App. Math. Letters 2007, ScienceDirect link) however the exponent in the first term is different. Does anyone have any ideas?


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.

  • $\begingroup$ Thank you very much! $\endgroup$
    – Tom26
    Feb 14 '20 at 19:50

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