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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg82634] Re: Can Integrate[expr,{x,a,b}] give an incorrect result?
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Fri, 26 Oct 2007 05:31:40 -0400 (EDT)
  • References: <20071025164458.769$l2@newsreader.com>

I think you are just making excuses, which are not worth replying to.  
If I blamed Bobby for anything it was for suggesting that Simplify or  
D or indefinite integration was at faullt. To prove this I will  
simply quote from my original mail:


First is a quote from Bobby:

> So, from that, D and/or Simplify must be wrong.
>

Must they? Even without any investigation I would say that it is much
more likely that this integral should not be evaluated by means of
the Leibniz rule (substituting the limits into the anti-derivative
and subtracting).

Actually, as you know very well since he sent this message also to  
you,Bobby has since apologized for claiming that, (although I don'ts  
see why one would actually have to apologize for anything of this  
kind). I see I did not actually say this happens often but he words  
"even without nay investigation" obviously sugges that this sort of  
behaviour is fairly common in Mathemaitca.

Further more, I wrote:

Now let's compute the indefinte integral:

indef = Integrate[int, t];

Using the Liebniz rule gives clearly the wrong answer:

indef = Integrate[int, t];

but why should it give the right answer? The function indef is
clearly discontinuous in the interval 0 to 2Pi

Plot[indef, {t, 0, 2 Pi}]

so the Leibniz rule does not apply. There is no reason at all to
suspect the very reliable function Simplify and almost as realiable
Interate.

Where is there here any mention of definite integration except in  
saying that the Newton-Leibniz rule should not be applied here?

What is worse, you wrote that Mathematica as well as every CAS known  
to you uses the t=Tan[th/2] substituiton (which you call the  
Weierstrass substitution) even though you have no gounds for  
beleiving that and even after I wrote to you that the Risch algorithm  
(wich you clearly do not know) does not use it because it does not  
need it - it can deal with such integrals by converting then to  
Exponentials and Logs (TrigToExp). Now Daniel Lichblau who knows the  
source code has confirmed that Mathematica indeed does not use the  
"Weierstrass substitution". So much for your "certianty" about this  
matter.

Andrzej Kozlowski







On 26 Oct 2007, at 05:44, David W. Cantrell wrote:

> [Message also posted to: comp.soft-sys.math.mathematica]
>
> Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
>> As I don't see anything concrete to reply to in your post I will just
>> restate my version of this whole argument and leave it at that.
>>
>> I originally replied to Bobby's post in which he seemed to question
>> the reliability of Simplify or D (or, perhaps, of indefinite
>> integration). I replied that Mathematica's Simplify and D were
>> practically 100% reliable and Integrate (by which I  have always
>> meant the function that computes indefinite integrals) was nearly as
>> reliable except when its heuristics failed or perhaps when it entered
>> an unimplemented branch of the Risch algorithm. I attributed the
>> problem the OP was having to the way Mathematica evaluated the
>> definite integral by applying the Newton-Leibniz rule without
>> detecting that it should't do so in this case.
>
> You may well have made that attribution in your head. But what you  
> _wrote_
> made it seem to me as though you thought Mathematica was faultless  
> in this
> matter, at least in relation to the specific problems at hand, and  
> that
> Bobby or the OP was instead at fault. I have now re-read your original
> reply, and it still seems that way to me. That, in large measure,  
> is why I
> replied to you, rather than to Bobby or the OP.
>
> BTW, if I had seen Daniel Lichtblau's reply, I most likely would  
> never have
> replied to you. Unlike you, he made it clear that Mathematica was  
> at fault.
> (I overlooked Daniel's post only because its severely curtailed title,
> "give an", made it seem as though it were not part of this thread.)
>
>> I did not comment in any way on why the latter problem occured,
>> except for remarking that it happened rather often.
>
> You did not make that remark in your reply to Bobby; look at what you
> wrote. OTOH, I made such a remark when replying to you: "This happens
> fairly often, the subject of this thread being a case in point."
>
>> At this point you entered, with a reply to me, whose purpose I still
>> do not understand.
>
> I explained that above.
>
>> When I read it it seemed to me that you must be
>> disagreeing with something I had written. As there was nothing at all
>> about Simplify or D in your message, I naturally concluded that you
>> were criticizing the way Mathematica performed indefinite integration
>> - what I call Integrate. You seem to be suggesting that Integrate
>> should somehow find a different anti-derivative.
>
> I did not say that _per se_, but perhaps it's reasonable that you  
> thought I
> seemed to suggest such.
>
> But your "should" is too strong a word since there are at least two  
> options
> which would keep Mathematica from making the error in integration  
> from 0 to
> 2Pi, mentioned by the OP:
> 1.  Leave the result of indefinite integration as
> it is, but then, since that result is not an antiderivative on the  
> desired
> interval, improve Mathematica's detection of discontinuities so  
> that it
> will not apply Newton-Leibniz incorrectly.
> or
> 2.  Change the result of indefinite integration in problems in  
> which the
> classic Weierstrass substitution, u = tan(x/2), had been used so that
> discontinuities introduced as an artifact of that substitution are not
> present in the result.
>
> My preference is option 2. In problems such as that mentioned by  
> the OP,
> since option 2 gives an antiderivative on R, it allows us to use
> Newton-Leibniz naively, there being then no discontinuities with  
> which to
> be concerned. But again, I cannot say that the people at Wolfram  
> Research
> should choose option 2, instead of option 1. For all I know, they  
> may have
> an excellent reason for rejecting the option I prefer.
>
>> That's why I asked
>> you for another algorithm - (for indefinite integration of course)
>> that would return a different anti-derivative than the Risch
>> algorithm. But now it turns out that you had nothing at all to add to
>> what I had written, except perhaps the claim, which you did not make
>> in your first post, that there is a reliable method of detecting
>> discontinuities in the antiderivative.
>
> I didn't say that in my first post because one need not detect
> discontinuities if you know that there are none to detect! The  
> algorithm
> performs as I described previously. Thus, for the OP's problem, it  
> returns
> an antiderivative on R.
>
>> But actually, if you re-read
>> your own post you will see that in it you are (or at least seem to
>> be) suggesting that a different anti-derivative should have been
>> found rather than the discontinuities in the one actually used by
>> Mathematica should have been detected. These are two rather different
>> claims.
>
> Indeed, they are different. The former corresponds to my "option 2"  
> above,
> the latter to my "option 1".
>
>> In any case, even assuming that what you meant was "that Mathematica
>> should be better at detecting discontinuities in the anti-derivative"
>
> Option 1.
>
>> I don't see in what way that is different from what I originally
>> wrote?
>
> It differs significantly. You didn't say, or even hint, "that  
> Mathematica
> should be better at" anything. It seemed -- and still seems -- to  
> me that
> you were blaming Bobby or perhaps the OP, rather than Mathematica.  
> As I
> said previously, that is why I responded to you, rather than to  
> someone
> else.
>
> -------------------------------------------
>
> The following examples and comments will likely be of interest to  
> anyone
> following this thread. (It should also answer a few questions  
> raised in
> some private emails.)
>
> The classic Weierstrass substitution is u = tan(x/2). Its use  
> typically
> introduces spurious discontinuities in the result of an indefinite
> integration.
>
> Consider
>
> In[4]:= Integrate[1/(2 - Cos[alpha - x]), x]
>
> Out[4]= -((2*ArcTan[Sqrt[3]*Tan[(alpha - x)/2]])/Sqrt[3])
>
> Although the function to be integrated is continuous on R, the  
> result of
> Integrate has discontinuities on R, due to the presence of
> Tan[(alpha - x)/2], surely caused by a Weierstrass substitution.
>
> BTW, someone had questioned my assertion that Mathematica and other  
> CASs
> use Weierstrass substitution. That's certainly a reasonable question
> because I have no way of proving that they use it. I based my  
> assertion on
> (1) what I had read in the literature and (2) the form of answers,  
> such as
> Out[4] above, which clearly seem to be the results of Weierstrass
> substitutions. In a court of law, (1) might be called mere  
> "hearsay" and
> (2) might be called "circumstantial evidence". In any event, until  
> someone
> from Wolfram Research assures me that Weierstrass substitution is  
> not used,
> I shall assume that it is used.
>
> Now consider the definite integration
>
> In[5]:= defint = Integrate[1/(2 - Cos[alpha - x]), {x, 0, 2*Pi}];
> defint /. alpha -> 20
>
> Out[5]= 0
>
> That is incorrect, presumably due to an improper use of Newton- 
> Leibniz. Not
> surprisingly, if we set alpha to 20 in the integration, we do get the
> correct answer:
>
> In[6]:= Integrate[1/(2 - Cos[20 - x]), {x, 0, 2*Pi}]
>
> Out[6]= 2*Pi/Sqrt[3]
>
> The algorithm which I had mentioned takes a result such as Out[4],  
> having
> spurious jump discontinuities on R caused by the Weierstrass  
> substitution,
> and produces an antiderivative without those jumps.
> For 1/(2 - Cos[alpha - x]), such an antiderivative on R is
>
> #    (x - 2 ArcTan[Sin[alpha - x]/(2 + Sqrt[3] - Cos[alpha - x])])/ 
> Sqrt[3]
>
> Note that # is a little messier than Out[4]; perhaps that's why  
> Wolfram
> Research prefers Out[4], despite its discontinuities on R. OTOH, for a
> definite integration on the real line, it is impossible to misuse
> Newton-Leibniz with #, and so no error such as Out[5] could occur.
>
> David W. Cantrell



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