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

*To*: mathgroup at smc.vnet.net*Subject*: [mg82638] Re: Can Integrate[expr,{x,a,b}] give an incorrect result?*From*: DrMajorBob <drmajorbob at bigfoot.com>*Date*: Fri, 26 Oct 2007 05:33:43 -0400 (EDT)*References*: <20071025164458.769$l2@newsreader.com> <9005304.1193373452141.JavaMail.root@m35>*Reply-to*: drmajorbob at bigfoot.com

>> 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: > 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 >> 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