Register to reply

Singularity in Integrand

Share this thread:
Teg Veece
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.
Phys.Org News Partner Science news on
'Office life' of bacteria may be their weak spot
Lunar explorers will walk at higher speeds than thought
Philips introduces BlueTouch, PulseRelief control for pain relief
Jun12-12, 04:44 AM
P: 4,579
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
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.

Register to reply

Related Discussions
Whats the difference between t=0 singularity and the singularity in a black hole ? Cosmology 5
Expand integrand Calculus 1
Integrand approaching sin(x)/x Calculus 2
Exploding Integrand Calculus 0
Integrand Simplification Calculus & Beyond Homework 2