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Re: Re: Folding Deltas, and evaluation scheme of Mathematica
*To*: mathgroup at smc.vnet.net
*Subject*: [mg56976] Re: [mg56906] Re: [mg56876] Folding Deltas, and evaluation scheme of Mathematica
*From*: zhengji.li at t3gt.com
*Date*: Wed, 11 May 2005 05:25:34 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
I can not agree with you.
Integrate[1/x, {x, 1, Infinity}] is Infinity,
and Integrate[1/x, {x, Infinity, 1}] = - Integrate[1/x, {x, 1, Infinity}] =
- Infinity.
These are true.
But before evaluate Integrate[1/x, {x, 1, Infinity}] - Integrate[1/x, {x,
Infinity, 1}]
= Infinity - Infinity, the following rules can be applied:
(* this comes from the definition of definite integral*)
Integrate[exp_, {x_, a_, b_}] := - Integrate[exp, {x, b, a}] /; a > b;
Subtract[exp_, exp_] = 0;
Sadly, Mathematica will not apply the second rule at first and it will do
the definite integral at first.
In my opinion, if Mathematica apply the rule "Subtract[exp_, exp_] = 0"
before evaluate
exp will free you from applying limit wrappers to improper integrals.
More generally, for some functions, such as Subtract, Plus, Times, Power,
before evaluate their parameters
some rule should be applied if possible. for example:
Plus[exp_, exp_] := 2 a; Times[0, exp_] := 0;
This is just what "human beings" do when simplify expression. :)
gauerkk at uregina.c
a To: Zhengji Li <zhengji.li at gmail.com>
cc:
05, 11, 2005 Subject: [mg56976] Re: [mg56906] Re: [mg56876] Folding Deltas
04:29 AM
Recently reading the mathgroup postings, I saw you post below a
generalization which seems to be false.
Quoting Zhengji Li <zhengji.li at gmail.com>:
> ..............
> But, Integrate[Anything, {t, a, b}] + Integrate[Anything, {t, b, a}]
> =
> 0, so you will get the result.
set Anything -> 1/x, a -> 1, b -> Infinity. My understanding is that
Infinity-Infinity does not evaluate without a warning message, and
instead does not vanish, as you indicate (compare to the other
standard improper integral used for calculating the volume of the Horn
of Gabriel, Anything -> 1/x^2, where this time, the differences are
equal and finite (and also zero), although the intgrals are improper,
as below - in the original question).
Of course, instead choosing b -> k (a fixed finite real), others as
above, and running Limit[Integrate[Anything, {t, a, b}] + Integrate
[Anything, {t, b, a}], k -> Infinity] instead should return zero (for
most evaluatable functions, excluding Cantor and DiracDelta functions
which leap around on a semi-infinite box).
I personally always try to apply a limit wrapper to any improper
integral which I may have to evaluate in version 3, and first operate
symbolically on my limit in question, followed by, only if necessary,
the numerical evaluation. I wonder if they'd fixed in v 5 yet the
capability of calculating limits from multiple directions. My guess is
probably not, based on a reply to one of the archives messages, circa
last decade.
> On 5/9/05, baermic at yahoo.com <baermic at yahoo.com> wrote:
> > Can anyone help to verify in Mathematica the expression given by
> Rota
> > (http://xoomer.virgilio.it/maurocer/Text07.htm):
> >
> > Convolution ( Sum of DiracDeltaFct ** Sum of DiracDeltaFct) == Sum
> > (DiracDeltaFct + Values).
> >
> > I tied
> >
> > Integrate[DiracDelta[t] DiracDelta[t - 2] , {t, -3, 3} ]
> > which does not evaluate;
> > but
> >
> > Integrate[DiracDelta[t] DiracDelta[t - 2] , {t, -3, 1} ] +
> > Integrate[DiracDelta[t] DiracDelta[t - 2] , {t, 1, 3} ] == 0
> > True
> >
> > ( I use ver 5.1 with W2k)
> >
>
>
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