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Re: RootSum

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74913] Re: RootSum
  • From: "Michael Weyrauch" <michael.weyrauch at gmx.de>
  • Date: Wed, 11 Apr 2007 01:55:17 -0400 (EDT)
  • References: <evfkhu$73g$1@smc.vnet.net>

Hello,

  I think, here is nothing very mysterious...

The antiderivative is
F = Integrate[(x^3 + 6*x + 7)/(x^6 + 4*x^4 + 3*x^3 + 6), x]

RootSum[6 + 3*#1^3 + 4*#1^4 + #1^6 & ,  (7*Log[x -    #1] + 6*Log[x - #1]*#1 + Log[x - #1]*#1^3)/(9*#1^2 + 16*#1^3 + 6*#1^5) & ]

The limit for x->Infinity is 0 as is easily obtained

RootSum[6 + 3*#1^3 + 4*#1^4 + #1^6 & , (7*Log[x] + 6*Log[x]*#1 + Log[x]*#1^3)/(9*#1^2 + 16*#1^3 + 6*#1^5) & ]

0

Therefore, the integral F is given by

-RootSum[6 + 3*#1^3 + 4*#1^4 + #1^6 & , (7*Log[-#1] + 6*Log[-#1]*#1 + Log[-#1]*#1^3)/(9*#1^2 + 16*#1^3 + 6*#1^5) & ]

according to Newton-Leibnitz.

That this is the correct result is easily checked numerically.

Directly calculating the definite integral yields

F = Integrate[(x^3 + 6*x + 7)/(x^6 + 4*x^4 + 3*x^3 + 6), {x, 0, Infinity}]

-RootSum[6 + 3*#1^3 + 4*#1^4 + #1^6 & , (Log[-#1]*#1)/(9 + 16*#1 + 6*#1^3) & ] -

6*RootSum[6 + 3*#1^3 + 4*#1^4 + #1^6 & , Log[-#1]/(9*#1 + 16*#1^2 + 6*#1^4) & ] -

7*RootSum[6 + 3*#1^3 + 4*#1^4 + #1^6 & , Log[-#1]/(9*#1^2 + 16*#1^3 + 6*#1^5) & ]

which is equivalent to the pedestrian Newton-Leibnitz result given above.

The only surprising thing may be the fact that the integrator in this example gets the limit x-> Infinity right, while in some other 
examples it does not do that. And also your way to calculate this limit using the build in command Limit[] does not seem to work 
properly... But this in not a problem of RootSum, but of Limit,  I think...

Regards    Michael





"dimitris" <dimmechan at yahoo.com> schrieb im Newsbeitrag news:evfkhu$73g$1 at smc.vnet.net...
> Working with integration of rational functions I came to the
> conclusion
> that when RootSum is generated in the course of the computation
> strange behavior of Mathematica (at least for me) takes place.
>
> In a recent thread I show one such example.
>
> As another example consider the integral of (x^3 + 6*x + 7)/(x^6 +
> 4*x^4 + 3*x^3 + 6)
> in [0,Infinity).
>
> The indefinite integral obtained by Mathematica is
>
> F=Integrate[(x^3 + 6*x + 7)/(x^6 + 4*x^4 + 3*x^3 + 6), x]
>
> RootSum[6 + 3*#1^3 + 4*#1^4 + #1^6 & , (7*Log[x - #1] + 6*Log[x -
> #1]*#1 + Log[x - #1]*#1^3)/(9*#1^2 + 16*#1^3 + 6*#1^5) & ]
>
> which is of course correct.
>
> Together[D[F, x]]
>
> (7 + 6*x + x^3)/(6 + 3*x^3 + 4*x^4 + x^6)
>
> Application of the Newton-Leibniz formula to get from the
> andiderivative the requested
> definite integral fails.
>
> Limit[RootSum[6 + 3*#1^3 + 4*#1^4 + #1^6 & , (7*Log[x - #1] + 6*Log[x
> - #1]*#1 + Log[x - #1]*#1^3)/
>      (9*#1^2 + 16*#1^3 + 6*#1^5) & ], x -> Infinity] -
>  Limit[RootSum[6 + 3*#1^3 + 4*#1^4 + #1^6 & , (7*Log[x - #1] +
> 6*Log[x - #1]*#1 + Log[x - #1]*#1^3)/
>      (9*#1^2 + 16*#1^3 + 6*#1^5) & ], x -> 0]
>
> Limit[RootSum[6 + 3*#1^3 + 4*#1^4 + #1^6 & , (7*Log[x - #1] + 6*Log[x
> - #1]*#1 + Log[x - #1]*#1^3)/
>      (9*#1^2 + 16*#1^3 + 6*#1^5) & ], x -> Infinity] - RootSum[6 +
> 3*#1^3 + 4*#1^4 + #1^6 & ,
>   (7*Log[-#1] + 6*Log[-#1]*#1 + Log[-#1]*#1^3)/(9*#1^2 + 16*#1^3 +
> 6*#1^5) & ]
>
> My question comes now...
>
> Since for integrals like this Mathematica uses (as it appears in a
> recent forum) the Newton-Leibniz formula
> how it can evaluate the definite integral?
>
> Integrate[(x^3 + 6*x + 7)/(x^6 + 4*x^4 + 3*x^3 + 6), {x, 0, Infinity}]
>
> -((Log[-Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 1]]*(7 + 6*Root[6 + 3*#1^3
> + 4*#1^4 + #1^6 & , 1] +
>      Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 1]^3))/(Root[6 + 3*#1^3 +
> 4*#1^4 + #1^6 & , 1]^2*
>     (9 + 16*Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 1] + 6*Root[6 +
> 3*#1^3 + 4*#1^4 + #1^6 & , 1]^3))) -
>  (Log[-Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 2]]*(7 + 6*Root[6 + 3*#1^3
> + 4*#1^4 + #1^6 & , 2] +
>     Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 2]^3))/(Root[6 + 3*#1^3 +
> 4*#1^4 + #1^6 & , 2]^2*
>    (9 + 16*Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 2] + 6*Root[6 + 3*#1^3
> + 4*#1^4 + #1^6 & , 2]^3)) -
>  (Log[-Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 3]]*(7 + 6*Root[6 + 3*#1^3
> + 4*#1^4 + #1^6 & , 3] +
>     Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 3]^3))/(Root[6 + 3*#1^3 +
> 4*#1^4 + #1^6 & , 3]^2*
>    (9 + 16*Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 3] + 6*Root[6 + 3*#1^3
> + 4*#1^4 + #1^6 & , 3]^3)) -
>  (Log[-Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 4]]*(7 + 6*Root[6 + 3*#1^3
> + 4*#1^4 + #1^6 & , 4] +
>     Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 4]^3))/(Root[6 + 3*#1^3 +
> 4*#1^4 + #1^6 & , 4]^2*
>    (9 + 16*Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 4] + 6*Root[6 + 3*#1^3
> + 4*#1^4 + #1^6 & , 4]^3)) -
>  (Log[-Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 5]]*(7 + 6*Root[6 + 3*#1^3
> + 4*#1^4 + #1^6 & , 5] +
>     Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 5]^3))/(Root[6 + 3*#1^3 +
> 4*#1^4 + #1^6 & , 5]^2*
>    (9 + 16*Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 5] + 6*Root[6 + 3*#1^3
> + 4*#1^4 + #1^6 & , 5]^3)) -
>  (Log[-Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 6]]*(7 + 6*Root[6 + 3*#1^3
> + 4*#1^4 + #1^6 & , 6] +
>     Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 6]^3))/(Root[6 + 3*#1^3 +
> 4*#1^4 + #1^6 & , 6]^2*
>    (9 + 16*Root[6 + 3*#1^3 + 4*#1^4 + #1^6 & , 6] + 6*Root[6 + 3*#1^3
> + 4*#1^4 + #1^6 & , 6]^3))
>
>
> What am I missing here?
>
> Dimitris
>
> 



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