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Re: Re: Infinity appears as a factor in Integrate result!
*To*: mathgroup at smc.vnet.net
*Subject*: [mg74945] Re: Re: Infinity appears as a factor in Integrate result!
*From*: "Michael Weyrauch" <michael.weyrauch at gmx.de>
*Date*: Thu, 12 Apr 2007 04:51:28 -0400 (EDT)
*References*: <evd3f6$59u$1@smc.vnet.net> <200704100913.FAA07254@smc.vnet.net> <evhtl6$2s2$1@smc.vnet.net>
Hello,
I agree with you that the trick "ignoring the constant inside the Log"
is probably not very useful for programming an integration routine in a CAS.
But this trick is also not relevant to my point as can be seen in the following.
Let us take again the integral suggested by Dimitris.
Integrate[1/(x^3 + a*x^2 + b*x + 1), {x, 0, Infinity}]
As was pointed out by Dimitris, Mathematica returns a nonsensical result
which explicitly contains Infinity.
Now let me calculate this using a pedestrian Newton-Leibnitz method without using
the trick but using Adrezej Kozlowski's suggestion, namely a power series
expansion. This should be a fairly general method to calculate limits.
The antiderivative reads
int=RootSum[1 + b*#1 + a*#1^2 + #1^3 & , Log[x - #1]/(b + 2*a*#1 + 3*#1^2) & ]
so it has the following structure
fff = Log[x + b1]/a1 + Log[x + b2]/a2 + Log[x + b3]/a3
Now let me expand this using the built in Series command
Series[fff, {x, Infinity, 2}]
SeriesData[x, Infinity, {-(Log[x^(-1)]/a1) - Log[x^(-1)]/a2 - Log[x^(-1)]/a3,
b1/a1 + b2/a2 + b3/a3, -b1^2/(2*a1) - b2^2/(2*a2) - b3^2/(2*a3)}, 0, 3, 1]
So obviously, the Limit x->Infinity will be finite, and in particular 0, if
1/a1+1/a2+1/a3=0
(To see this more easily, it is probably good, to rewrite the above result in StandardForm again!)
Now this is the case for the coefficients in our antiderivative...
RootSum[1 + b*#1 + a*#1^2 + #1^3 & , 1/(b + 2*a*#1 + 3*#1^2) & ]
0
(If you like you can make such arguments more general.)
So, as far as I am concerned, it should be EASILY possible for Mathematica
to obtain 0 for the Limit x-> Infinity with a standard procedure.
So the question is: Does it handle series expansions of such "exotic" structures like RootSums properly?
My answer is: Yes and No...
If I expand the RootSum e.g. around 0 it works nicely...
Series[int, {x, 0, 2}] (*int is the RootSum given above*)
SeriesData[x, 0, {RootSum[1 + b*#1 + a*#1^2 + #1^3 & ,
Log[-#1]/(b + 2*a*#1 + 3*#1^2) & ], 1, -b/2}, 0,
3, 1]
But if I expand around Infinity I get the error message
Series::nmer: RootSum[...] is not a meromorphic function of x at Infinity.
But in the light of the above analysis of the sum of three Logs, this error message is irrelevant to the point. I would expect that
Mathematica produces a RootSum in terms of Log[1/x] as it has done before without complaining.
So, I come to the conclusion that the (probably ??) standard method to calculate such limits, which normally works and should work
here as well, is somehow broken, eventually because of the RootSum structure. (??)
Regards Michael
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