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Re: Infinity appears as a factor in Integrate result!

• To: mathgroup at smc.vnet.net
• Subject: [mg74885] Re: Infinity appears as a factor in Integrate result!
• From: "Michael Weyrauch" <michael.weyrauch at gmx.de>
• Date: Tue, 10 Apr 2007 05:13:47 -0400 (EDT)
• References: <evd3f6$59u$1@smc.vnet.net>

Hello,

  I think this example clearly supports the view I expressed in another recent thread on Integrate,
namely that Mathematica 5.2 has trouble calculating the limit of the antiderivative
for x -> Infinity.

First, I assume that this type of definte integral is evaluated by Mathematica by first calculating the
antiderivative (and not via a Mellin transform as was supposed by Dimitris recently). It was really very nice by Buvanesh of WRI 
that he confirmed this explicitely for a similar integral recently.

The antiderivative of the present integral reads

RootSum[1 + b*#1 + a*#1^2 + #1^3 & , Log[x - #1]/(b + 2*a*#1 + 3*#1^2) & ]

The limit for x->Infinity vanishes as can be shown by neglecting the constant under the Log

RootSum[1 + b*#1 + a*#1^2 + #1^3 & , Log[x]/(b + 2*a*#1 + 3*#1^2) & ]
which Mathematica simplifies to 0 immediately.

To my opinion, it is the Log[x+constant] which Mathematica puts explicitly to Infinity when evaluating this limit x-> Infinity, 
which is incorrect.

Evaluating the limit x->0 of the antiderivative given above e. g. for a->1 and b->2 reproduces the numerical values given by 
Dimitris.

Therefore, I am pretty sure that the problems of Integrate we see in such examples all have to do with an issue concerning limits 
for x-> Infinity. But I think we have good reason to hope that this issue will have disappeared in the forthcoming version of 
Mathematica.

Michael





"dimitris" <dimmechan at yahoo.com> schrieb im Newsbeitrag news:evd3f6$59u$1 at smc.vnet.net...
> Consider the integral
>
> In[13]:=
> f = HoldForm[Integrate[1/(x^3 + a*x^2 + b*x + 1), {x, 0, Infinity}]]
>
> Then
>
> In[17]:=
> Integrate[1/(x^3 + a*x^2 + b*x + 1), {x, 0, Infinity}]
>
> Out[17]=
> If[(Re[Root[1 + b*#1 + a*#1^2 + #1^3 & , 1]] <= 0 || Im[Root[1 + b*#1
> + a*#1^2 + #1^3 & , 1]] != 0) &&
>   (Re[Root[1 + b*#1 + a*#1^2 + #1^3 & , 2]] <= 0 || Im[Root[1 + b*#1
> + a*#1^2 + #1^3 & , 2]] != 0) &&
>   (Re[Root[1 + b*#1 + a*#1^2 + #1^3 & , 3]] <= 0 || Im[Root[1 + b*#1
> + a*#1^2 + #1^3 & , 3]] != 0),
>  -(Log[-Root[1 + b*#1 + a*#1^2 + #1^3 & , 1]]/(b + 2*a*Root[1 + b*#1
> + a*#1^2 + #1^3 & , 1] +
>      3*Root[1 + b*#1 + a*#1^2 + #1^3 & , 1]^2)) - Log[-Root[1 + b*#1
> + a*#1^2 + #1^3 & , 2]]/
>    (b + 2*a*Root[1 + b*#1 + a*#1^2 + #1^3 & , 2] + 3*Root[1 + b*#1 +
> a*#1^2 + #1^3 & , 2]^2) -
>   Log[-Root[1 + b*#1 + a*#1^2 + #1^3 & , 3]]/(b + 2*a*Root[1 + b*#1 +
> a*#1^2 + #1^3 & , 3] +
>     3*Root[1 + b*#1 + a*#1^2 + #1^3 & , 3]^2) +
>   Infinity*(1/(b + 2*a*Root[1 + b*#1 + a*#1^2 + #1^3 & , 1] +
> 3*Root[1 + b*#1 + a*#1^2 + #1^3 & , 1]^2) +
>     1/(b + 2*a*Root[1 + b*#1 + a*#1^2 + #1^3 & , 2] + 3*Root[1 + b*#1
> + a*#1^2 + #1^3 & , 2]^2) +
>     1/(b + 2*a*Root[1 + b*#1 + a*#1^2 + #1^3 & , 3] + 3*Root[1 + b*#1
> + a*#1^2 + #1^3 & , 3]^2)),
>  Integrate[1/(1 + b*x + a*x^2 + x^3), {x, 0, Infinity},
>   Assumptions ->  !((Re[Root[1 + b*#1 + a*#1^2 + #1^3 & , 1]] <= 0 ||
> Im[Root[1 + b*#1 + a*#1^2 + #1^3 & , 1]] != 0) &&
>      (Re[Root[1 + b*#1 + a*#1^2 + #1^3 & , 2]] <= 0 || Im[Root[1 +
> b*#1 + a*#1^2 + #1^3 & , 2]] != 0) &&
>      (Re[Root[1 + b*#1 + a*#1^2 + #1^3 & , 3]] <= 0 || Im[Root[1 +
> b*#1 + a*#1^2 + #1^3 & , 3]] != 0))]]
>
> In the results it appears Infinity!. For anyone who dosn't believe me
> try:
>
> In[18]:=
> %[[2]]
>
> Out[18]=
> -(Log[-Root[1 + b*#1 + a*#1^2 + #1^3 & , 1]]/(b + 2*a*Root[1 + b*#1 +
> a*#1^2 + #1^3 & , 1] +
>     3*Root[1 + b*#1 + a*#1^2 + #1^3 & , 1]^2)) - Log[-Root[1 + b*#1 +
> a*#1^2 + #1^3 & , 2]]/
>   (b + 2*a*Root[1 + b*#1 + a*#1^2 + #1^3 & , 2] + 3*Root[1 + b*#1 +
> a*#1^2 + #1^3 & , 2]^2) -
>  Log[-Root[1 + b*#1 + a*#1^2 + #1^3 & , 3]]/(b + 2*a*Root[1 + b*#1 +
> a*#1^2 + #1^3 & , 3] +
>    3*Root[1 + b*#1 + a*#1^2 + #1^3 & , 3]^2) +
>  Infinity*(1/(b + 2*a*Root[1 + b*#1 + a*#1^2 + #1^3 & , 1] + 3*Root[1
> + b*#1 + a*#1^2 + #1^3 & , 1]^2) +
>    1/(b + 2*a*Root[1 + b*#1 + a*#1^2 + #1^3 & , 2] + 3*Root[1 + b*#1
> + a*#1^2 + #1^3 & , 2]^2) +
>    1/(b + 2*a*Root[1 + b*#1 + a*#1^2 + #1^3 & , 3] + 3*Root[1 + b*#1
> + a*#1^2 + #1^3 & , 3]^2))
>
> The following term is multiplied with Infinity and the result of this
> product
> appeared in the results!
>
> In[19]:=
> Cases[%, (a_)*Infinity -> a]
>
> Out[19]=
> {1/(b + 2*a*Root[1 + b*#1 + a*#1^2 + #1^3 & , 1] + 3*Root[1 + b*#1 +
> a*#1^2 + #1^3 & , 1]^2) +
>   1/(b + 2*a*Root[1 + b*#1 + a*#1^2 + #1^3 & , 2] + 3*Root[1 + b*#1 +
> a*#1^2 + #1^3 & , 2]^2) +
>   1/(b + 2*a*Root[1 + b*#1 + a*#1^2 + #1^3 & , 3] + 3*Root[1 + b*#1 +
> a*#1^2 + #1^3 & , 3]^2)}
>
> Obviolusly Integarate algorithm failes for this integral.
>
> What is funnier is that
>
> In[20]:=
> Integrate[1/(x^3 + a*x^2 + b*x + 1), {x, 0, Infinity}]
>
> Out[20]=
> If[(Re[Root[1 + b*#1 + a*#1^2 + #1^3 & , 1]] <= 0 || Im[Root[1 + b*#1
> + a*#1^2 + #1^3 & , 1]] != 0) &&
>   (Re[Root[1 + b*#1 + a*#1^2 + #1^3 & , 2]] <= 0 || Im[Root[1 + b*#1
> + a*#1^2 + #1^3 & , 2]] != 0) &&
>   (Re[Root[1 + b*#1 + a*#1^2 + #1^3 & , 3]] <= 0 || Im[Root[1 + b*#1
> + a*#1^2 + #1^3 & , 3]] != 0), ComplexInfinity,
>  Integrate[1/(1 + x*(b + x*(a + x))), {x, 0, Infinity},
>   Assumptions -> (Im[Root[1 + b*#1 + a*#1^2 + #1^3 & , 1]] == 0 &&
> Re[Root[1 + b*#1 + a*#1^2 + #1^3 & , 1]] > 0) ||
>     (Im[Root[1 + b*#1 + a*#1^2 + #1^3 & , 2]] == 0 && Re[Root[1 +
> b*#1 + a*#1^2 + #1^3 & , 2]] > 0) ||
>     (Im[Root[1 + b*#1 + a*#1^2 + #1^3 & , 3]] == 0 && Re[Root[1 +
> b*#1 + a*#1^2 + #1^3 & , 3]] > 0)]]
>
> I.e. if you try again to obtain the integral Mathematica returns
> ComplexInfinity!
>
> Dimitris
>
> 

• Follow-Ups:
  • Re: Re: Infinity appears as a factor in Integrate result!
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>