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>