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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!

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  • $\begingroup$ for $n=1$ Mathematica gives $C^{-1}B^l e^{\frac{B \rho^2}{2}} \rho^{l-1} \left(-B^2 \rho^2-2 B l+C l\right)$, which does not agree with your conjectured result $B^{l+1} e^{\frac{B \rho^2}{2}} \rho^{l-1} \left(l-\frac{\rho^2}{C}\right)$ $\endgroup$
    – Carlo Beenakker
    Aug 11, 2017 at 9:13
  • $\begingroup$ Hmmm, that's strange, because it's a well known form for the eigenvectors (which is what I'm trying to solve this problem for). It is certainly correct for specific values of B & C. Sorry, I thought it was more general than it obviously is. I'm trying to solve an eigenvector equation, and I'm 100% sure of the outcome - the form of the eigenvector has been well established, and numerically it works out perfectly (in Matlab). But I guess I need to put some constraints on the inputs for my assertion to hold true. I will have to check and get back to you. I'm sorry for wasting your time. $\endgroup$
    – Katie
    Aug 11, 2017 at 10:18
  • $\begingroup$ Perhaps a better question then is can you solve this integral, and then I can show under what conditions it reduces to my solution above? Thanks, katie. $\endgroup$
    – Katie
    Aug 13, 2017 at 0:11

1 Answer 1

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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.

$^\ast$ 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 $p>-1$, $p+q>-1$, $a>0$)

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  • $\begingroup$ Oh fantastic, thank you so much! Can we see the steps involved in getting the answer? I would so very much appreciate knowing how it's arrived at, having spent a lot of time on not being able to figure it out! $\endgroup$
    – Katie
    Aug 13, 2017 at 9:32
  • $\begingroup$ added the intermediate step that allows for evaluation of the integral for any given $n$ (the general form I wrote down for all $n$ is a "guess", but it agrees with all $n$ I tried) $\endgroup$
    – Carlo Beenakker
    Aug 13, 2017 at 12:47
  • $\begingroup$ That's wonderful, thank you so much. I will try to generalise this for an arbitrary Laguerre polynomial and post the answer here. Your help is much appreciated. $\endgroup$
    – Katie
    Aug 13, 2017 at 12:55
  • $\begingroup$ Also, slightly more general (and easily derived from what you gave above) is: $ \int x^{a-1} e^{-px^2} I_{\nu}(cx) dx = \frac{C^{\nu} \Gamma(\frac{a+\nu}{2})}{2^{\nu+1}p^{\frac{a+\nu}{2}}\Gamma(\nu+1)}{}_1F_1\left( \frac{a+\nu}{2}; \nu+1; \frac{C^2}{4p} \right)$ $\endgroup$
    – Katie
    Aug 15, 2017 at 1:03
  • $\begingroup$ You helped me hugely with this problem a while ago Carlo. I was wondering if you knew what the answer would be for the same integral but indefinite, between 0 & $\rho$, or between $\rho$ and $\infty$? $\endgroup$
    – Katie
    Mar 28, 2018 at 7:35

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