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Re: Calculating a simple integral

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 at]
Inviato: gioved=A8=AC 13 giugno 2013 8.39
A: mathgroup at
Oggetto: Re: Calculating a simple integral

On 11 Jun., 08:23, Andrzej Kozlowski <akozlow... at> 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... at> 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... at
> >
> > From: dsmirno... at [mailto:dsmirno... at]
> > 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).


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