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Singularity in Integrand

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Teg Veece
#1
Jun12-12, 04:22 AM
P: 8
I have an equation that relates two variables:
[tex]k(\mathbf{x},\mathbf{x}') =exp(-(\mathbf{x}-\mathbf{x}')^2)[/tex]
If I want to determine the value of this equation where x' is kept constant and x is actually the set of every real number then I can express the function as the integral where the integrand relates x' to the integration variable u between the interval of minus infinity to infinity:
[tex]f(\mathbf{x}') = \int_{-\infty}^{\infty} exp(-(\mathbf{u}-\mathbf{x}')^2) d\mathbf{u}[/tex]
and the solution to this will be some sort of error function.

Now, a slight variation on this. I need to include an additional term that's like a weighting term which decays with distance from x. So I'm trying to find a solution for the following equation:
[tex]g(\mathbf{x},\mathbf{x}') = \int_{-\infty}^{\infty} \frac{exp(-(\mathbf{u}-\mathbf{x}')^2)}{|\mathbf{x}-\mathbf{u}|} d\mathbf{u}[/tex]
The problem I'm having is that when [tex]\mathbf{u} = \mathbf{x}, [/tex]
then the integrand goes to infinity. I think I can get around it by possibly converting to spherical coordinates (all of vectors here are 3-D vectors) but I also need to evaluate the function, h, when x' is also integrated from minus infinity to infinity and a second weighting term is introduced:
[tex]h(\mathbf{x},\mathbf{x}') = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{exp(-(\mathbf{u}-\mathbf{v})^2)}{|\mathbf{x}-\mathbf{u}||\mathbf{x}'-\mathbf{v}|} d\mathbf{u}d\mathbf{v},[/tex]
and here the spherical coordinate approach doesn't seem to help.

How do I deal with this singularity? Someone suggested complex analysis but I'm not very familiar with that area.
Any suggestions would be greatly appreciated. I can post how I evaluate g(.,.) using spherical coordinates if people think it'd help.
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chiro
#2
Jun12-12, 04:44 AM
P: 4,573
Hey Teg Veece and welcome to the forums.

One suggestion is to use a distance metric plus a constant. So instead of |x-u| you scale it by say |x-u|+c where c is a preferrably positive number (unless you want the behaviour of a negative). Something like c = 1 seems like a good initial one to try.
Teg Veece
#3
Jun12-12, 05:49 AM
P: 8
Thanks for the quick reply.

Wouldn't adding a constant to the denominator not have a significant effect on the final result depending on what I set c to be? Like c = 0.01 would be a very different solution from c=10.

I know that they have a similar problem with singularities when calculating gravitational potential but they get around that by using spherical coordinates:

[tex] \Phi(\mathbf{x}) = -G \int \frac{\rho(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}d \mathbf{x}'[/tex]

I think you can do a change of variables and, when you convert to spherical polar, the r^2 that appears above the line cancels with the denominator to leave you with just a r term multiplying the numerator.


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