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