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 ]
Brambilla Roberto Luigi (RSE)

Posts: 25
Registered: 2/21/12
Re: Calculating a simple integral
Posted: Jun 14, 2013 4:48 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

Using Mathematica 5.1 or 8 I obtained:

in = -((-1 + Cos[z])/(z^2 (r^2 + z^2)^2 (z^2 - 4 =A6=D0^2)^2));

r = 0.1;
NIntegrate[in, {z, -Infinity, Infinity}]

f1[x_] := MeijerG[{{1, 1,3/2},{}},{{1, 1, 3/2}, {0, 1/2}}, x, 1/2]
f2[x_] := MeijerG[{{1, 1,3/2},{}},{{1, 3/2, 2}, {0, 1/2}}, x, 1/2]

A = (3r^7+28r^5Pi^2) + 16Pi^(5/2) (5r^2+4Pi^2) f1[ I r/2] + 32Pi^(5/2) (r^2+4Pi^2) f2[ I r/2];

B = 32r^5Pi^3 (r^2+4Pi^2);



Why to hate Meijer functions with parameter 1/2?

Roberto Brambilla

-----Messaggio originale-----
Da: Dr. Wolfgang Hintze [mailto:weh@snafu.de]
Inviato: gioved=A8=AC 13 giugno 2013 8.39
A: mathgroup@smc.vnet.net
Oggetto: Re: Calculating a simple integral

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).


RSE SpA ha adottato il Modello Organizzativo ai sensi del D.Lgs.231/2001, in=
forza del quale l'assunzione di obbligazioni da parte della Societ=A8=A4 av=
viene con firma di un procuratore, munito di idonei poteri.
RSE adopts a Compliance Programme under the Italian Law (D.Lgs.231/2001). Ac=
cording to this RSE Compliance Programme, any commitment of RSE is taken by=
the signature of one Representative granted by a proper Power of Attorney.=
Le informazioni contenute in questo messaggio di posta elettronica sono ris=
ervate e confidenziali e ne e' vietata la diffusione in qualsiasi modo o for=
ma. Qualora Lei non fosse la persona destinataria del presente messaggio, La=
invitiamo a non diffonderlo e ad eliminarlo, dandone gentilmente comunicazi=
one al mittente. The information included in this e-mail and any attachments=
are confidential and may also be privileged. If you are not the correct rec=
ipient, you are kindly requested to notify the sender immediately, to cancel=
it and not to disclose the contents to any other person.

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

[Privacy Policy] [Terms of Use]

© The Math Forum 1994-2015. All Rights Reserved.