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|    sci.math.symbolic    |    Symbolic algebra discussion    |    10,432 messages    |
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|    Message 9,726 of 10,432    |
|    Nasser M. Abbasi to Richard Fateman    |
|    Re: question on Rubi and special functio    |
|    02 Dec 17 20:32:09    |
   
   From: nma@12000.org   
      
   On 12/1/2017 10:20 AM, Richard Fateman wrote:   
      
   >   
   > I think the smart way to do a lot of these, and what is done   
   > at least in part by Maxima's specint, and I think inside Mathematica,   
   > is to change the integrand to a different representation as   
   > a Meyer G-function or hypergeometric function, and then   
   > mess around with it (probably algorithmically, but maybe   
   > amenable to Rubi approach). Then convert back to more   
   > common functions as possible (again using Rubi?)   
   >   
      
   If Rubi could integrate MeijerG, then this will convert   
   many special functions including Bessel*   
      
   restart;   
   convert( BesselY(1,x),'MeijerG');   
    MeijerG([[], [-1]], [[1/2, -1/2], [-1]], (1/4)*x^2)   
      
   convert( BesselI(1,x),'MeijerG');   
    -I*MeijerG([[], []], [[1/2], [-1/2]], -(1/4)*x^2)   
      
   convert( BesselJ(1,x),'MeijerG');   
    MeijerG([[], []], [[1/2], [-1/2]], (1/4)*x^2)   
      
   convert( BesselK(1,x),'MeijerG');   
    (1/2)*MeijerG([[], []], [[1/2, -1/2], []], (1/4)*x^2)   
      
   convert( BesselY(1/3,x),'MeijerG');   
    MeijerG([[], [-2/3]], [[1/6, -1/6], [-2/3]], (1/4)*x^2)   
      
   convert( BesselI(1/3,x),'MeijerG');   
    -(-1)^(5/6)*MeijerG([[], []], [[1/6], [-1/6]], -(1/4)*x^2)   
      
   convert( BesselJ(1/3,x),'MeijerG');   
    MeijerG([[], []], [[1/6], [-1/6]], (1/4)*x^2)   
      
   convert( BesselK(1/3,x),'MeijerG');   
    (1/2)*MeijerG([[], []], [[1/6, -1/6], []], (1/4)*x^2)   
      
      
   Therefore   
    int(BesselY(1,x),x)   
   would be the same as   
    int(MeijerG([[], [-1]], [[1/2, -1/2], [-1]], (1/4)*x^2),x)   
      
   Verified in Mathematica:   
      
   r1 = Integrate[MeijerG[{{}, {-1}}, {{1/2, -2^(-1)}, {-1}}, (1/4)*x^2], x]   
    -((Sqrt[x^2]*BesselY[0, Sqrt[x^2]])/x)   
      
   r2=Integrate[BesselY[1,x],x]   
    -BesselY[0,x]   
      
   And r1-r2 is zero verified by plotting.   
      
   So I think if Rubi has rules to integrating MeijerG   
   special function then this will cover many other special function   
   that it currently do not have rules for by converting them   
   to MeijerG first. But may be having rules to integrate MeijerG   
   might be very complicated.   
      
   --Nasser   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   
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