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I am trying to solve: $$f_l(r)=\int_0^{\infty}e^{-k^2}k^4j_l(kr)dk,$$ where $j_l$ is the spherical Bessel function of the first kind, for any integer l >= 0.

Thanks in advance for any answers!

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    Start by looking at $\int_0^{\infty}e^{-s k^2}j_l(kr)dk$ from which you can derive your integral by differentiating twice w.r.t. $s$.2011-04-24

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After searching for a while, I realized there would be no simple expression for this integral. I found the following expressions in Abramowitz and Stegun, which you can download. They involve confluent hypergeometric functions $M(a,b,z)$ for the most general one.

$$\int_0^{\infty}e^{-a^2k^2} k^{\mu-1} J_{\nu}(kr)dk = \frac{\Gamma\left(\frac{\mu+\nu}{2}\right)\left(\frac{r}{2a}\right)^{\nu}}{2 a^{\mu} \Gamma(\nu + 1)} M\left(\frac{\mu+\nu}{2},\nu+1,-\frac{r^2}{4 a^2}\right) \; ,$$

with $J_{\nu}(z)$ the "ordinary" Bessel function of the first kind. It is related to the spherical one as follows:

$$j_{\nu}(z)= \sqrt{\frac{\pi}{2 z}}J_{\nu+1/2}(z) \; .$$

If $\mu=\nu+2$, then there is a simpler formula

$$\int_0^{\infty}e^{-a^2k^2} k^{\nu+1} J_{\nu}(kr)dk = \frac{r^{\nu}}{(2a^2)^{\nu+1}} e^{-\frac{r^2}{4 a^2}}\; .$$

There are some conditions on the range of values for the parameters for these formulae to hold, but I think that should not be a problem in your case. You can find them in the reference as well as further details on confluent hypergeometric functions.

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    Note that spherical Bessels are expressible in terms of trigonometric functions, so you might get away with much simple expressions.2011-04-25
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    Maybe, I'm not sure about that. The expressions are $$j_n(z)=(-z)^n(\frac{d}{zdz})^n\frac{\sin z}{z} \; ,$$ see [MathWorld](http://mathworld.wolfram.com/SphericalBesselFunctionoftheFirstKind.html).2011-04-25
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    @J.M.: Simplest I could manage to do with a bit of transforming will still involve confluent hypergeometric functions: http://www.wolframalpha.com/input/?i=integrate+0+to+infinity+e^%28-k^2%29+k^a+sin%28b+k%29+dk2011-04-25
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    I'm not on my *Mathematica* computer so I can't play around, but from the looks of the OP's integral, it seems to be expressible in terms of, for instance, error functions and exponentials (or parabolic cylinder functions alternatively). Of course, all are but special cases of the Kummer function.2011-04-25
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    ...and indulge me a tiny nitpick: it's "Abramowitz".2011-04-25
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    Fixed. I don't know why it always sounds as Abrahamowitz in my memories. It can't be too much bible thumping. :D2011-04-25
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    Thank you Raskolnikov! Using these expressions I find that: $\int_{0}^{\infty}e^{-a^2k^2}k^{\mu - 1/2}j_{\nu}(kr)dk=\sqrt{\frac{\pi}{2r}}\frac{\Gamma(\frac{\mu + \nu + 1}{2})(\frac{r}{2a})^{\nu + 1/2}}{2a^{\mu}\Gamma(\nu + 3/2)}M(\frac{\mu + \nu + 1}{2}, \nu + 3/2, -\frac{r^2}{4a^2})$, but how do I evaluate the confluent hypergeometric function M in practice? Andreas Werner Paulsen2011-04-25
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    @Andreas: For that particular case, it really depends on what $\mu$ and $\nu$ are. Since you did say that $\nu\in\mathbb N$, you might be able to use recursion relations (though you will have to check the numerical stability).2011-04-25
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    Also I did some experiments in Mathematica and it seems J.M. may be right: $\nu = 0: -\frac{1}{16}e^{-\frac{r^2}{4}}\sqrt{\pi}(-6 + r^2)$, $\nu = 1: \frac{r(r^2-2)+(4 + 4r^2 - r^4)DawsonF(r/2)}{8r^2}$ For larger $\nu$ the pattern is this: even $\nu$: $e^{-\frac{r^2}{4}}$ x a polynominal fraction of r + possibly a term with error function(r/2), for odd $\nu$: a polynominal in r x a Dawson integral(r/2). However I only tested up to $\nu=6$ so I do not know if this approach is generally applicable.2011-04-25
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    Thank you J.M! Using Mathematica I should be able to evaluate a few Ms and then use the recurrence relations in Abramowitz and Stegun to calculate all other Ms.2011-04-25