Convergence of numerical integration with parameter

Questions
1

Hi there, I would like to solve the integral:

NIntegrate[( Sqrt[12 ((x - y)^2/(x + y)^2 + (x - z)^2/(x + z)^2 + (y - z)^2/(y +  z)^2)] 
Boole[Sqrt[1/3 ((x - y)^2/(x + y)^2 + (x - z)^2/(x + z)^2 + (y - z)^2/(y + z)^2)] == a])/
(E^(x + y + z) Sqrt[(-((4 x (x - z))/(x + z)^3) - (4 y (y - z))/(y + z)^3)^2 + ((4 z (y - z))/(y + z)^3 - 
(4 x (x - y))/(x + y)^3)^2 + ((4 y (x - y))/(x + y)^3 + (4 z (x - z))/(x + z)^3)^2]), 
{x, 0, \[Infinity]}, {y, 0, \[Infinity]}, {z, 0, \[Infinity]}]

With parameter a\in [0,1]. But the expression above does not work.

I have also tried by creating an implicit region of the surface where I’m integrating (mainly the expression inside the Boole operation), but it fails to discretize the region.

Do you know what I might be doing wrong? Do you know any work-around? Any advice is welcome.

2

Hi,
Does the integral of this function even exist over infinite domains? You have an equation == in your integral?

3

Hi Key, thanks for your interest. Yes, it exists. It is actually a density function for random variable a.
The == is in the Boole command, representing the region of integration, a surface defined by \sqrt{\frac{1}{3} ((\frac{x-y}{x+y})^2+(\frac{y-z}{y+z})^2+(\frac{x-z}{x+z})^2)}=a.

Numerical integration on an infinite domain is possible with MonteCarlo techniques.

4

For problems of this sort, I find it easier to do a simpler case first. This is the case I tried

In[3]:= Integrate[
 Boole[x^2 + y^2 + z^2 == 1], {x, -2, 2}, {y, -2, 2}, {z, -2, 2}]

Out[3]= 0

It makes me think that you can’t do a surface integral by converting it to a volume integral in this manner.

5

Thank you Frank. I think you might be right.

I rephrase your simple case as follows:

In[84]:= 
reg = ImplicitRegion[x^2 + y^2 + z^2 == 1, {x, y, z}]

    Out[84]= ImplicitRegion[x^2 + y^2 + z^2 == 1, {x, y, z}]
    
    In[85]:= NIntegrate[1, {x, y, z} \[Element] reg]
    
    Out[85]= 12.5664

And it seems to work (4\pi is the answer). But if I do the same with my integral (for instance for a=0.5):

reg = ImplicitRegion[Sqrt[1/3 ((x - y)^2/(x + y)^2 + (x - z)^2/(x + z)^2 + (y - z)^2/(y + z)^2)] == .5 && x >= 0 && y >= 0 && z >= 0, {x, y, z}]

and then

NIntegrate[Sqrt[
 12 ((x - y)^2/(x + y)^2 + (x - z)^2/(x + z)^2 + (y - z)^2/(y + 
      z)^2)]/(E^(x + y + z)
   Sqrt[(-((4 x (x - z))/(x + z)^3) - (4 y (y - z))/(y + z)^3)^2 + ((
     4 z (y - z))/(y + z)^3 - (4 x (x - y))/(x + y)^3)^2 + ((
     4 y (x - y))/(x + y)^3 + (4 z (x - z))/(x + z)^3)^2]), {x, y, 
   z} \[Element] reg]

Then I get the error: "DiscretizeRegion was unable to discretize the region
ImplicitRegion[<<2>>]. "

Any further ideas?

6

I tried to do a ContourPlot3D of the Boole expression with a = 0.5 and got error messages about 1/0. I think that’s similar to the problem you are getting from NIntegrate and DiscretizeRegion. What is the surface you are trying to integrate over? Can you express it in a different way so you don’t get 0 in the denominator for certain subregions?

7

@Frank. Yes, I don’t know why it gives that error. The division by 0 occurs only at the origin (I consider only positive values of x, y and z), but then it is 0/0 and the limit is 1 for the whole region. The origin is part of the surface only when a==1.

If you use RegionPlot3D and change the == to \leq, it works. The region looks like this (for a=0.2 for instance):

Region1.jpg
Region1.jpg415×397 22.6 KB

I just assumed that the surface (==) is just the “skin” of this region (the orange surface). Isn’t it the case?

BTW: the region is a line (x=y=z) for a=0 and the whole first octant when a=1.

8

When you try to do the integral with the ImplicitRegion approach, does the error message show up right away? I’m trying that approach and not getting an error but the calculation has been evaluating for several minutes now with no result yet. I’m running version 10.3.

9

No, it takes about 10 seconds before giving the DiscretizeRegion error. I’m running version 10.2. Could that be the point?

If it takes too long, it might work better to use NIntegrate with the MonteCarlo (or any of its daughters) method of integration.

Tomorrow I’ll install 10.3 and try.

10

I think it is a problem of ImplicitRegion, because I was able to plot the surface using CountourPlot3D directly on the equation:

ContourPlot3D[
 Sqrt[1/3 ((x - y)^2/(x + y)^2 + (x - z)^2/(x + z)^2 + (y - z)^2/(y + 
        z)^2)] == .2, {x, 0, 10}, {y, 0, 10}, {z, 0, 10}]

surface.jpg
surface.jpg360×405 17.7 KB

And it confirms that the surface is the “skin” of the cone.

11

Perhaps you could transform the surface integral to a volume integral using something like Gauss’ law and then use NIntegrate, which works for an inequality constraint inside Boole.

12

Mmmhh… The problem is that my surface is not closed, which, according to my understanding, prevents it to apply Gauss’ law, is that right?

But maybe Stokes could be applied… The problem is how to interpret my integrand as (\nabla\times\underline{F})\cdot\underline{\rm{d}S}

Because already the normal to that surface is not something simple. Furthermore, I have no idea how to define the boundary curve, since this “cone” grows to \infty.