Re: Integrate[ x^2 * Erfc[a x] *(BesselJ[1, b x])^2 ,
- To: mathgroup at smc.vnet.net
- Subject: [mg100914] Re: [mg100842] Integrate[ x^2 * Erfc[a x] *(BesselJ[1, b x])^2 ,
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Thu, 18 Jun 2009 04:52:22 -0400 (EDT)
- References: <200906170149.VAA25659@smc.vnet.net>
ralf.schaa wrote:
> Hi group,
>
> The integral
>
> Integrate[ x^2 * Erfc[a x] *(BesselJ[1, b x])^2 , {x, 0, Inf}]
>
> has a solution in Mathematica (in terms of HypergeometricPFQ)
>
> My question: how did Mathematica know?
>
> I tried the usual suspects:
> i) for example Gradshteyn and Ryzhik (1965), p.769, 6.784 is very
> close, but not close enough
> ii) Abramowitz & Stegun hasn't got it on the menu either
>
> Then how about expressing Erfc[a*x] in terms of...
> Erfc(z)=-1/(sqrt(pi)) * IncompleteGamma[1/2,x^2]+1 (Grads.+Ryzh. p.
> 942)
> ...and hoping to get lucky? ... no alas!
>
> sorry, this is a cross-post (also in sci.math.symbolic)...
>
> Help!
> -Ralf
As was suggested in the s.m.s thread, MeijerG functions are used in a
convolution approach. Here isn your example with certain assumptions on
parameter ranges.
Integrate[x^2 * Erfc[a*x] * BesselJ[1,b*x]^2, {x,0,Infinity},
Assumptions->{b>0,a>0}]
In this case, up to multiplicative constants the Erfc becomes
MeijerG[{{}, {1}}, {{0, 1/2}, {}}, a^2*x^2]
and the BesselJ square becomes
MeijerG[{{1/2}, {}}, {{1}, {-1, 0}}, b^2*x^2]
The code subsequently uses these MeijerG representations in convolving
to get an integral from 0 to infinity.
Daniel Lichtblau
Wolfram Research
- References:
- Integrate[ x^2 * Erfc[a x] *(BesselJ[1, b x])^2 , {x, 0, Inf}]
- From: "ralf.schaa" <ralf.schaa@gmail.com>
- Integrate[ x^2 * Erfc[a x] *(BesselJ[1, b x])^2 , {x, 0, Inf}]