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Re: Can Integrate[expr,{x,a,b}] give an incorrect result?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg82517] Re: Can Integrate[expr,{x,a,b}] give an incorrect result?
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 23 Oct 2007 05:31:28 -0400 (EDT)
  • References: <20071023002439.352$aK@newsreader.com>

On 23 Oct 2007, at 13:24, David W. Cantrell wrote:

> [Message also posted to: comp.soft-sys.math.mathematica]
>
> Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
>> This is clearly a case of a misunderstanding, though I am not sure
>> that I am at fault here.
>
> Earlier in this thread, you said "In fact, with numerical limits
> it [Mathematica] correctly deals with the integral posted by the OP."
> That is clearly false. (And I wouldn't call it "a misunderstanding"
> either.)

O.K. I did not actually mean the integrand posted by the OP, because  
I never read his e-mail and only assumed what it contents were on the  
basis of Bobby's e-mail, which I read because of its comments on  
Simplify, D and Integrate (the latter actually he did not write  
about). Also, referring to symbolic limits was a mistake, caused by  
the fact that by the time I replied to your post I forgot the details  
of what I read in Bobby's post. Actually I meant "integration with  
symbolic parameters", whether limits or not. And what I meant was  
that when you use numerical values for everything, just as Bobby did,  
Mathematica deals with it correctly.

So indeed, I confess to:

1. Not having read the original post.
2. Not remembering exactly what the issues it raised were when I was  
replying to you (whence symbolic limits).

I was wrong to do that but the excuse is that, as I wrote, I am not  
interested in integration on the real line (or classical real  
analysis in all its aspects).

On the other hand, since my reply was entirely about Simplify, D,  
Integrate and the fact that the error was cause by Mathematica  
mistakenly applying the Newton-Leibniz rule to the result of  
Integrate, it was absurd for you to post your message as a reply to  
me, which made it look as if you disagreed with any of the points I  
made.


>
>> Note that I never try to reply to Craig's
>> original posting and have not even read it carefully. I am simply not
>> interested enough in the subject of definite integration and to tell
>> the truth I can see little point in all the great effort that is
>> being put into definite integration with symbolic limits (I do see a
>> point to definite integration with numerical limits).
>
> Unless I've overlooked something, you are the _only_ person in this  
> thread
> to have mentioned definite integration with symbolic limits!

I explained this above.
>
> Granted, from the title which Craig gave the thread, it is  
> reasonable to
> think that the thread _might_ have concerned definite integration with
> symbolic limits. But in fact, whenever something in the form
> Integrate[expr,{x,a,b}] has appeared here, a and b have always been
> specific numerical values.
>
>> Personally I
>> would be quite happy if Mathematica returned all definite integrals
>> with symbolic limits unevaluated.
>>
>> So, I only responded to Bobby's assertion that there was something
>> wrong with Simplify or D. I then changed D to Integrate in my
>> response, but even then I only asserted that Integrate is reliable in
>> producing anti-derivatives in the algebraic sense (That is, functions
>> g possibly with branch cuts etc in the complex plane such that D[g]=f
>> for a given f). What I found provocative in your response is that you
>> chose to respond to me rather than to Craig (or Daniel) although
>> nothing in your response is related to any of the assertions that I
>> made in mine.
>>
>> As for the algorithm that is implemented in "another CAS", you do not
>> make it clear what exactly it does.
>
> I could provide a reference. But you said in a subsequent post that
> "integration on the real line that has no attraction for me whatever"
> and so I know the reference would not interest you.

I am not interested enough to read it. But from your post get the  
idea that you are unable to distinguish between an algorithm (like  
the Risch algorithm) and heuristics, which is quite a different  
matter. I challenged you to provide  an algorithm that would be able  
to completely replace the the Risch algorithm for problems involving  
integration on the real line. If you can't do that than you haven't  
got an algorithm, you have only got some, possibly sophisticated and  
useful, heuristics.
If you can provide a "real" replacement for the Risch algorithm I  
will admit that I was quite wrong on this. But I am quite certain  
that you will not be able to do anything of the sort and I would be  
even to place a fairly high bet on this.

I think this also answers all your remaining points.

Andrzej Kozlowski

>
>> If you mean that it can deal with this particular case and with a few
>> similar ones I do not doubt that it can.
>
> Specifically, as far as this thread is concerned, it gets rid of the
> unnecessary discontinuities on the real line which arise as a  
> consequence
> of the Weierstrass substitution.
>
>> If, however, you are asserting that it can always find a
>> continuous integrand on an interval whenever it can be proved that it
>> exists than I simply don't believe you.
>
> Of course I'm not asserting that! That would be absurd.
>
> David W. Cantrell



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