Singularity in Integrandby Teg Veece Tags: cauchy integral, complex analysis, integration, singularity 

#1
Jun1212, 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 3D 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. 



#2
Jun1212, 04:44 AM

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 xu you scale it by say xu+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. 



#3
Jun1212, 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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