problem with Integrate
- To: mathgroup at smc.vnet.net
- Subject: [mg23061] problem with Integrate
- From: Jack Michel CORNIL <jmcornil at club-internet.fr>
- Date: Sat, 15 Apr 2000 03:00:21 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Hello,
I am reading a paper about classical algorithmic methods
for integrating rational function and I am very
surprised that Mathematica is unable to give a "good"
expression of the following integral
p = 7*x^13 + 10*x^8 + 4*x^7 - 7*x^6 - 4*x^3 - 4*x^2 + 3*x + 3;
q = x^14 - 2*x^8 - 2*x^7 - 2*x^4 - 4*x^3 - x^2 + 2*x + 1;
Integrate[p/q, x]
1/2*RootSum[#1^14 - 2*#1^8 - 2*#1^7 - 2*#1^4 - 4*#1^3 -
#1^2 + 2*#1 + 1 & ,
(7*log*(x - #1)*#1^13 + 10*log*(x - #1)*#1^8 +
4*log*(x - #1)*#1^7 - 7*log*(x - #1)*#1^6 -
4*log*(x - #1)*#1^3 - 4*log*(x - #1)*#1^2 +
3*log*(x - #1)*#1 + 3*log*(x - #1))/
(7*#1^13 - 8*#1^7 - 7*#1^6 - 4*#1^3 - 6*#1^2 - #1 +
1) & ]
then the Rothstein/Trager method with a calculus
of resultant gives quickly a good expression with
two logarithmic functions.
Do I understand that for rational functions Mathematica
uses only the elementary methods teached to students
in an introductory course about integration ?
Or is there some option for the function Integrate
asking Mathematica using other algorithms ?
By the way, I found a bug (I suppose) when using twice
the function Integrate as follow
y = (-7*Sqrt[2]*x^6 + 7*x^6 + 2*Sqrt[2]*x - 4*x + 2*Sqrt[2] - 3)/
(2*(x^7 + Sqrt[2]*x^2 + Sqrt[2]*x - x - 1));
Integrate[y, x]
-(1/2)*RootSum[#1^7 + Sqrt[2]*#1^2 + Sqrt[2]*#1 - #1 -
1 & , (7*Sqrt[2]*log*(x - #1)*#1^6 -
7*log*(x - #1)*#1^6 - 2*Sqrt[2]*log*(x - #1)*#1 +
4*log*(x - #1)*#1 - 2*Sqrt[2]*log*(x - #1) +
3*log*(x - #1))/(7*#1^6 + 2*Sqrt[2]*#1 + Sqrt[2] -
1) & ]
Integrate[y, x]
-(1/2)*RootSum[#1^7 + Sqrt[2]*#1^2 + Sqrt[2]*#1 - #1 -
1 & , (-7*log*(x - #1)*#1^6 + 7*log*(x - #1)*
Integrate`A*1[6]*#1^6 + 4*log*(x - #1)*#1 -
2*log*(x - #1)*Integrate`A*1[6]*#1 +
3*log*(x - #1) - 2*log*(x - #1)*Integrate`A*1[6])/
(7*#1^6 + 2*Integrate`A*1[6]*#1 + Integrate`A*1[6] -
1) & ]
I don't understand the appearing of Integrate`A*1[6]
in the second expression .
Such a thing was not existing in Release 3
Jack-Michel CORNIL VERSAILLES - FRANCE