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   sci.math.symbolic      Symbolic algebra discussion      10,432 messages   

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   Message 9,265 of 10,432   
   antispam@math.uni.wroc.pl to clicliclic@freenet.de   
   Re: alarum: Risch integrator fails to di   
   07 Feb 17 04:09:45   
   
   clicliclic@freenet.de wrote:   
   >   
   > clicliclic@freenet.de schrieb:   
   > >   
   > > antispam@math.uni.wroc.pl schrieb:   
   > > >   
   > > > My calculations indicate that points of order 3 correspond to   
   > > > u^2 = 1 +- 2*sqrt(3)/3.  FriCAS says that integal is nonelementary,   
   > > > so something is wrong.  Theory behind example is simple, so   
   > > > I believe it.  I am not sure about points, but this is probably   
   > > > correct.  So it remains to check FriCAS :(   
   > > >   
   > >   
   > > A simple theory should allow to read off simple antiderivatives for   
   > > the order-3 integrands.   
   > >   
   > > I suppose you have tried the integral on Axiom as well - the problem   
   > > would have to be quite old then.   
   > >   
   > > The points of order 3, 6 and 8 determined by me correspond to the   
   > > following three pairs of integrals:   
   > >   
   > > INT(x/((3*x^2 + 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?   
   > >   
   > > INT(x/((3*x^2 - 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?   
   > >   
   > > INT(x/((x^2 + 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?   
   > >   
   > > INT(x/((x^2 - 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?   
   > >   
   > > INT(x/((x^2 + 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),   
   > > x) = ?   
   > >   
   > > INT(x/((x^2 - 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),   
   > > x) = ?   
   > >   
   > > So your result for order 3 is confirmed. I derived these x^2 - u^2 from   
   > > a recursive definition of the division polynomials and then checked the   
   > > points [u,v] by iterating the group operation.   
   > >   
   > > I should mention that my statements about Goursat integrability   
   > > referred to Moebius mappings realizing the dihedral group D_2   
   > > (isomorphic to the Klein four-group K_4) only; the special nature of   
   > > the radicand x^3 - x may open other possibilities (thus, x <-> 1/x   
   > > swaps two roots and preserves the remaining two). Compare Goursat pp.   
   > > 113-118 for x^3 - 1.   
   > >   
   >   
   > Actually, Goursat considered the special radicand x^3 - x as well (on   
   > pp. 118-119) and concludes that no special pseudo-elliptic cases arise.   
   > I tend to believe him.   
      
   Hmm, you checked that for order 3 one gets special case, Goursat   
   says that there are none and you believe Goursat?  I do not   
   understand.   
      
   > Meanwhile I can confirm that the order-3 integrals are elementary. The   
   > antiderivatives take less than 1000 Bytes to write down.   
   >   
   > Martin.   
      
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