Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » Software » comp.soft-sys.math.mathematica

Topic: Calculating a simple integral
Replies: 10   Last Post: Jun 14, 2013 4:50 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Dr. Wolfgang Hintze

Posts: 195
Registered: 12/8/04
Re: Calculating a simple integral
Posted: Jun 13, 2013 2:33 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On 11 Jun., 08:23, Andrzej Kozlowski <akozlow...@gmail.com> wrote:
> No, it's similar to:
>
> Integrate[(1 -
> Cos[x])/(x^2*(x^2 - 4*Pi^2)^2), {x, -Infinity, Infinity}]
>
> 3/(32*Pi^3)
>
> On 10 Jun 2013, at 10:11, djmpark <djmp...@comcast.net> wrote:
>
>
>

> > Doesn't this have a singularity at 2 Pi that produces non-convergence? It's
> > similar to:

>
> > Integrate[1/x^2, {x, \[Epsilon], \[Infinity]},
> > Assumptions -> \[Epsilon] > 0]

>
> > 1/\[Epsilon]
>
> > That diverges as epsilon -> 0.
>
> > Are you sure you copied the integral correctly?
>
> > David Park
> > djmp...@comcast.net
> >http://home.comcast.net/~djmpark/index.html

>
> > From: dsmirno...@gmail.com [mailto:dsmirno...@gmail.com]
>
> > If there is a way to calculate with Mathematica the following integral:
>
> > in = -((-1 + Cos[kz])/(kz^2 (kr^2 + kz^2)^2 (kz^2 - 4 \[Pi]^2)^2))
> > Integrate[in, {kz, -Infinity, Infinity}, Assumptions -> kr > 0]

>
> > Another system calculates the same integral instantly. :)
>
> > Thanks for any suggestions.

Sorry, but I made indeed a calculation error!
Correcting it the partial fraction decomposition leads to Dmitry's
result.
Furthermore, calculating first the indefinite integral and then taking
limits leads to a false result.
Direct calculation of the integral leads to MeierG functions which are
useless because we cannot enter any numerical value.
So, rather than provding the correct result Mathematica comes up with
different false result depending on the method used, and we cannot tel
which one is correct without "research" work.
Summarizing, I need to restate my criticism of Mathematica with
respect to integration (I'm using version 8).

Regards,
Wolfgang




Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.