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

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  • Subject: [mg82639] Re: Can Integrate[expr,{x,a,b}] give an incorrect result?
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Fri, 26 Oct 2007 05:34:14 -0400 (EDT)
  • References: <20071025164458.769$l2@newsreader.com> <9005304.1193373452141.JavaMail.root@m35> <op.t0sjlpotqu6oor@monster.gateway.2wire.net>

Why "defending" and why "so much"? I just pointed out where the real  
problem lied (at that stage you had not noticed it yet) and left it  
at that. I only added that these two functions are very unlikely to  
fail. This can be useful knowledge when looking for the cause of a  
problem.
That was all I did and thought it was the end of it. Then I got the  
weird reply from David which seemed to me tobe saying that I somehow  
wrote something wrong and that the problem was that Mathematicas  
Integrate (indefinite Integral) did not return the anti-derivative  
which he provided. Later he tried ot withdraw from this claim, but he  
it was not the first time he worte this. So, I finally decied to  
explain (roughly) what the Risch algorithm was, which is something  
American educated analysts rarely known about (In the years when I  
was an assistant professor at a large state university I found that  
most analysts had not the faintest idea about algebra and most  
algebraists new little analyis. The only people who knew both were  
algebraic geometers and lagebraic topologists. I doubt that there has  
been any change in this respect. The point of it all is that the  
Risch algorithm is pure algebra).
So I am curious where you found all this "defending" of Simplify and D?

Andrzej Kolzowski


On 26 Oct 2007, at 15:14, DrMajorBob wrote:

>>> So, from that, D and/or Simplify must be wrong.
>
> Not to beat a dead horse, but I was going through a thought  
> process, investigating the problem, not giving my final conclusion.  
> I didn't stop there, after all, and I got to the actual problem in  
> the end.
>
> Besides, I'm not sure why D and Simplify need so much defending.
>
> Bobby
>
> On Thu, 25 Oct 2007 19:26:26 -0500, Andrzej Kozlowski  
> <akoz at mimuw.edu.pl> wrote:
>
>> *This message was transferred with a trial version of CommuniGate 
>> (tm) Pro*
>> 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
>>
>>
>
>
>
> -- 
> DrMajorBob at bigfoot.com



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