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MathGroup Archive 2007

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Re: Integrate modified in version 6?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg78127] Re: Integrate modified in version 6?
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sat, 23 Jun 2007 07:15:22 -0400 (EDT)
  • References: <20070622094436.388$kC_-_@newsreader.com>

On 22 Jun 2007, at 22:44, David W. Cantrell wrote:

> [Message also posted to: comp.soft-sys.math.mathematica]
>
> Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
>> My version of Mathematica 6.0 gives:
>>
>> Integrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}]
>>
>> (1/2)*(MeijerG[{{-(1/2), 0}, {1}}, {{0, 0, 1/2}, {}}, 1] -
>>        MeijerG[{{-(1/2), 0}, {1, 1}}, {{0, 0, 1/2, 1/2}, {}}, 1]/Sqrt
>> [Pi])
>>
>> I don't know whether this answer is right or wrong (Mathematica takes
>> for ever to evaluate this numerically) but one thing is clear: it
>> certainly is not -Infinity.
>
> How is that clear?


To borrow a quote from a former president:   It depends on what the  
meaning of "is" is.

It was ( and still is) "clear" to me that my version does not return - 
Infinity, in other words: that the answer it returns "is not" - 
Infinity in the ordinary, common-sensical meaning of "is" -   
completely different from the mathematical "equal to -Infinity" or  
whatever "is" is to you.
I still think the same:  the answer  return by my Mathematica "is  
not" -Infinity. Is it clear now?   If still not I suggest consulting  
a higher authority, namely Humpty Dumpty in Alice in Wonderland, who  
is also concerned with the meaning of "is" (in connection with the  
name of a song).

I also assumed that since my answer was different (not the same, not  
identical, SameQ returns False, etc., etc, ) from the one posted by  
Dimitris and since the person from whom Dimitris got this information  
had several times in the past posted, on this very forum,  bugs found  
in beta versions of Mathematica, claiming that they were bugs "in  
Mathematica", he must have done the same thing again. Apparently this  
was not the case, so I apologize to him for this suggestion, which at  
that time seemed to me almost a certainty. I did not imagine that  
this sort of thing could be platform dependent just as I  never  
imagine that there are still out there people who are going to make  
an issue over the meaning of "is".

Andrzej Kozlowski



>
> Also note that Bob H. and Curtis O. both report getting -Infinity  
> as the
> result in version 6 (which I don't have).
>
> FWIW, using version 5.1, FullSimplify applied to those two MeijerG  
> terms
> yields Indeterminate and ComplexInfinity, resp.
>
> My guess is that the output from version 6 involving MeijerG might  
> be pure
> nonsense. If so, maybe that's why it takes "for ever" to evaluate
> numerically.
>
> In any event, version 5.1 gets a nice answer easily:
>
> In[1]:= Integrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}]
> Out[1]= -1 - 2*Catalan + Pi*(1/4 + Log[2])
>
> In[2]:= N[%]
> Out[2]= 0.131053
>
> In[3]:= NIntegrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}]
> Out[3]= 0.131053
>
> and so the behavior of version 6 on this particular integral is  
> clearly not
> as good.
>
> David W. Cantrell



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