# Singularity in Integrand

by Teg Veece
Tags: cauchy integral, complex analysis, integration, singularity
 P: 8 I have an equation that relates two variables: $$k(\mathbf{x},\mathbf{x}') =exp(-(\mathbf{x}-\mathbf{x}')^2)$$ 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: $$f(\mathbf{x}') = \int_{-\infty}^{\infty} exp(-(\mathbf{u}-\mathbf{x}')^2) d\mathbf{u}$$ 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: $$g(\mathbf{x},\mathbf{x}') = \int_{-\infty}^{\infty} \frac{exp(-(\mathbf{u}-\mathbf{x}')^2)}{|\mathbf{x}-\mathbf{u}|} d\mathbf{u}$$ The problem I'm having is that when $$\mathbf{u} = \mathbf{x},$$ 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: $$h(\mathbf{x},\mathbf{x}') = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{exp(-(\mathbf{u}-\mathbf{v})^2)}{|\mathbf{x}-\mathbf{u}||\mathbf{x}'-\mathbf{v}|} d\mathbf{u}d\mathbf{v},$$ 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.
 P: 4,570 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.
 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: $$\Phi(\mathbf{x}) = -G \int \frac{\rho(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}d \mathbf{x}'$$ 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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