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Re: kelvin functions ker and kei

  • To: mathgroup at smc.vnet.net
  • Subject: [mg49141] Re: kelvin functions ker and kei
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Mon, 5 Jul 2004 04:54:20 -0400 (EDT)
  • Organization: The University of Western Australia
  • References: <cbrcik$ird$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <cbrcik$ird$1 at smc.vnet.net>, mss4 at duke.edu wrote:

> Does anyone know of a good way to way to integrate ker_2(r), kei_2(r),
> ker_0(r) and kei_0(r) from r=0 to r=Infinity?  

I think you will find it rather difficult to integrate kei_2(r) from r=0 
to r=Infinity since it has a 1/r^2 singularity:

  ke[n_][r_] = Exp[(-1/2) n Pi I] BesselK[n, Exp[(-1/4) Pi I] r]

  Series[ke[2][r], {r, 0, 3}]

> can it be done analytically?  

Yes. After removing the r^(-2) singularity in kei_2(r), all 4 integrals 
are the same and equal to Pi/Sqrt[8].

> I'm also a little unsure about how mathematica does this
> numerically.
> 
> ker_n(r) + I kei_n(r) = Exp[(-1/2)*n*Pi*I] BesselK[n, Exp[(-1/4)*Pi*I]*r]
>  
> any help would be greatly appreciated.

You can compute the indefinite integral for n=0 in closed form and then 
use the asymptotic formulae for BesselK and StruveH at

  http://functions.wolfram.com/BesselAiryStruveFunctions
  
Cheers,
Paul

-- 
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