Re: Can Integrate[expr,{x,a,b}] give an incorrect result?
- To: mathgroup at smc.vnet.net
- Subject: [mg82516] Re: Can Integrate[expr,{x,a,b}] give an incorrect result?
- From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
- Date: Tue, 23 Oct 2007 05:30:58 -0400 (EDT)
- References: <20071020224211.967$wA@newsreader.com> <ffhrr4$5in$1@smc.vnet.net>
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.) > 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! 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. > 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