Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2004
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Integratecrashes kernel

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50768] Re: Integratecrashes kernel
  • From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
  • Date: Sun, 19 Sep 2004 21:39:24 -0400 (EDT)
  • References: <cije7o$hjs$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Bill Rowe <readnewsciv at earthlink.net> wrote:
> On 9/18/04 at 5:48 AM, math5bug at lmm.jussieu.fr (B. Oudoli) wrote:
>
> >However, Integrate[x Cos[1. x], x]
> >crashes the kernel, and it's impossible to abort or interrupt the
> >calculation, which lasts forever.

> In this specific case
>
> Integrate[x Cos[a x],x]
>
> seems to work fine. So, you could do
>
> Integrate[x Cos[a x],x]/.a->1

Yes, it works fine in all but the a=0 case. Since 1. is clearly not close
to 0, I doubt that the comment below is related to the bug mentioned above,
but I hope it might be of some interest nonetheless.

The antiderivative given is Cos[a x]/a^2 + x Sin[a x]/a. Not only does this
misbehave _at_ a=0, it also misbehaves as a _approaches_ 0. But in fact an
antiderivative can be given which is correct for all nonzero a and which
also behaves as we would like in the limit as a->0:

  (Cos[a x] - 1)/a^2 + x Sin[a x]/a

Of course, the difference between this antiderivative and the one given by
Mathematica is only a constant (with respect to x). But as a->0, its limit
is x^2/2, as it should be.

David Cantrell


  • Prev by Date: Re: Forcing a Derivative
  • Next by Date: Re: Bilinear Transforms
  • Previous by thread: Re: Integratecrashes kernel
  • Next by thread: ScatterPlot3D Plot Symbol