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Re: Volume integration pb

  • To: mathgroup at smc.vnet.net
  • Subject: [mg104092] Re: Volume integration pb
  • From: Leonid Shifrin <lshifr at gmail.com>
  • Date: Mon, 19 Oct 2009 07:10:27 -0400 (EDT)
  • References: <200910180924.FAA17110@smc.vnet.net>

Fred,

your integral is ill-defined.

In the first of the methods you list, this shows up when you look at the
particular point z = 0 - you get a logarithmic divergence. In the second
method, this shows up for a particular point r = 0 -
you get a quadratic divergence in the term -2 z^2/(3 (r^2 + z^2)^(5/2).

Most explicitly, you see the problem  when you go from cylindrical  to
polar coordinates where you expression becomes

dtheta dphi d rho*rho^2*rho^2*Sin[theta]*(Sin[theta]^2 - 2 Cos[theta]^2)/(3
rho^5),

with rho = Sqrt[z^2+r^2], and theta being the usual polar angle
(such that r = rho*Sin[theta], h = rho*Cos[theta]). Integrating over theta
in the small neibourhood of the origin from 0 to Pi gives zero, while
integration over rho is logarithmically divergent (at the origin) integral
d rho/rho. Thus, we have a 0*Infinity situation.
What this means is that the value of the integral will depend on the way it
is computed - that is, the procedure which we use to compute the integral
becomes a part of the definition. One may say that the integral must be
somehow "regularized" at the origin, then computed, then regularization
removed. Since it diverges only logarithmically, removing the regularization
gives a finite answer, but different in each case.

Your two methods of computing this integral can be considered two different
regularizations, each giving a different meaning to the integral.

The bottom line: neither result is "correct" or "wrong", because the
integral  is  ill - defined. Similar situation happens in computing various
anomalies in Quantum Field Theory, for example. In that case, the way you
regularize the integral is a part of the definition of  the model
(determines the physics of the problem). Which regularization is correct in
your case must be decided by you.

Regards,
Leonid


On Sun, Oct 18, 2009 at 2:24 AM, Fred Bartoli <@news.free.fr> wrote:

> Integrating (r^2 - 2 z^2)/(3 (r^2 + z^2)^(5/2)) over a cylinder of R
> radius and -h/2 to +h/2 height.
>
> (* Integrating first over radius then over height : *)
> Integrate[2\[Pi] r (r^2-2 z^2)/(3
> (r^2+z^2)^(5/2)),{r,0,R},Assumptions->R>0&&z>0];
> int1=Integrate[%,{z,-h/2,h/2},Assumptions->R>0&&h>0]
>
>
> (* Then integrating first over height then over radius : *)
> Integrate[(r^2 - 2 z^2)/(3 (r^2 + z^2)^(5/2)), {z, -h/2, h/2},
>  Assumptions -> h > 0 && r > 0]
> int2 = Integrate[2 \[Pi] r %, {r, 0, R},
>   Assumptions -> h > 0 && R > 0]
>
> int2-int1//Simplify
>
> -> 4 Pi/3
>
> Am I missing something?
>
>
> --
> Thanks,
> Fred.
>
>


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