In an engineering setting, I reduced my problem to calculating the following sum:

$$\sum_{n=0}^\infty \frac{n!}{(k+n)!}\left[\int_0^a \left(\frac{x}{u}\right)^kL_n^{(k)}\left(\frac{x^2}{u^2}\right)\exp\left[-\frac{x^2}{2u^2}\right]J_n(vx)\mathrm{d}x\right]^2,$$

where $L_n^{(k)}(\cdot)$ is the generalized Laguerre polynomial, $J_n(\cdot)$ is $n$-th order Bessel function of the first kind, and $a>0$,$u>0$, $v>0$, and $k$ is a positive integer. I am not sure what to do with this beast. I will be happy with a tight upper bound, or a fast method to calculate this numerically (using numerical integration for each element in the sum is extremely slow, as I have to evaluate this sum for many values of $u$, $v$, and $k$).

Does anyone have any ideas?


This integral is in the literature (e.g. Bateman manuscript project Vol. 2, Equation 8.9(5)).

$ \int_0^\infty x^\mu \: L_n^{(\mu)}(\alpha x^2) \: J_\mu(xy) \: \mathrm{e}^{-\beta x^2} \: \mathrm{d}x = 2^{-\mu-1} \: \beta^{-\mu-n-1} \: (\beta - \alpha)^n \: y^\mu \: \mathrm{e}^{-y^2/(4\beta)} \: L_n^{(\mu)}\left(\frac{\alpha y^2}{4\beta(\alpha - \beta)}\right) $

So set $\alpha = 1/u^2$, $\mu = k$, $\beta = 1/(2u^2)$, $y = \nu$.

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  • $\begingroup$ Does $a=\infty$ in OP's question ? Also the order of the Bessel function doesn't seem to be the upper index of the Laguerre polynomial. $\endgroup$ – user111 May 29 '17 at 8:16
  • $\begingroup$ Hmm you're right. There probably isn't a nice expression for $a \neq \infty$. For the $a = \infty, k \neq n$ case, there are some other related integrals in that chapter of Bateman. I think the best bet would be to expand the Laguerre polynomial using its definition as a sum, and then apply eq. 8.6(13), $\int_0^\infty x^{2m+\mu+1} \: \mathrm{e}^{-x^2/4} \: J_\mu(xy)\mathrm{d}x = 2^{2m+\mu+1}n!y^\mu \: \mathrm{e}^{-y^2} \: L_n^{(\mu)}(y^2)$ $\endgroup$ – Edward Lilley May 29 '17 at 13:34

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