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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. $s$. –  Raskolnikov Apr 24 '11 at 22:15
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1 Answer 1

up vote 3 down vote accepted

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. –  J. M. Apr 25 '11 at 9:56
    
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. –  Raskolnikov Apr 25 '11 at 10:03
    
@J.M.: Simplest I could manage to do with a bit of transforming will still involve confluent hypergeometric functions: wolframalpha.com/input/?i=integrate+0+to+infinity+e^%28-k^2%29+k^a+si‌​n%28b+k%29+dk –  Raskolnikov Apr 25 '11 at 10:36
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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. –  Andy Apr 25 '11 at 11:51
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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. –  Andy Apr 25 '11 at 12:11
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