md65536 Posted March 27, 2011 Share Posted March 27, 2011 I haven't done integrals for over a decade and I'm having trouble with them and my math skills are inadequate The function f(x) = 1/x^2 has a singularity at x=0. The definite integral of 1/x^2 is divergent, if it includes x=0. However, the integral from 1 to infinity, of 1/x^2, is 1. Are there examples of functions that have a singularity (where the function approaches infinity), with a convergent integral? For example of what I'm trying to get is... 1/x^2 remains non-zero for all finite values of x. Along the x axis, I imagine there's basically an infinitesimally tall rectangle that is infinitely wide, and yet it has 0 volume. Yet along the y axis at x=0, 1/x^2 is undefined and a similar infinitesimally wide rectangle has infinite volume. Is there any function, or any way, to basically "take what we have on the x axis and get it on the y axis as well", so that we have a function that stretches to infinity along both axises but has a convergent integral everywhere? (If you know of related Sage expressions that would also be appreciated thanks!) Link to comment Share on other sites More sharing options...

DrRocket Posted March 27, 2011 Share Posted March 27, 2011 (edited) I haven't done integrals for over a decade and I'm having trouble with them and my math skills are inadequate The function f(x) = 1/x^2 has a singularity at x=0. The definite integral of 1/x^2 is divergent, if it includes x=0. However, the integral from 1 to infinity, of 1/x^2, is 1. Are there examples of functions that have a singularity (where the function approaches infinity), with a convergent integral? Sure Consider [math]f(x) = \frac {1}{r}[/math](i.e. [math]\frac {1}{||x||}[/math]) In dimension 3 0r above. Since the volume element in spherical coordinates is [math] r^{n-1}dr \times [/math] (other stuff involving angles cosines and sines) the integral converges on finite balls centered at 0. This is why random walks in dimensions 1 and 2 are recurrent but random walks in dimension 3 and above are non-recurrent. (With reasonable constraints on the probability measure). Or if you want to stick to dimension 1 see below. Is there any function, or any way, to basically "take what we have on the x axis and get it on the y axis as well", so that we have a function that stretches to infinity along both axises but has a convergent integral everywhere? (If you know of related Sage expressions that would also be appreciated thanks!) Sure. Take [math] f(x)=\frac{1}{x^2}[/math] for [math] x \ge 1[/math] and [math]f(x)= \frac{1}{\sqrt x}[/math] for [math]0<x<1[/math] Edited March 27, 2011 by DrRocket 2 Link to comment Share on other sites More sharing options...

md65536 Posted March 27, 2011 Author Share Posted March 27, 2011 Sure Consider [math]f(x) = \frac {1}{r}[/math](i.e. [math]\frac {1}{||x||}[/math]) In dimension 3 0r above. Since the volume element in spherical coordinates is [math] r^{n-1}dr \times [/math] (other stuff involving angles cosines and sines) the integral converges on finite balls centered at 0. This is why random walks in dimensions 1 and 2 are recurrent but random walks in dimension 3 and above are non-recurrent. (With reasonable constraints on the probability measure). Or if you want to stick to dimension 1 see below. Fascinating! Thanks! No, actually I'm only interested in dimension 3 but assumed it made more sense to figure it out in one dimension first. Oops. I don't get why the integral converges on finite balls centered at 0. I don't get the connection with random walks. Is the integral related to the probability of eventually returning to a finite segment in 1D, or area in 2D, or volume in 3D? Does that mean that for any arbitrarily small value of epsilon, a random walk starting at location x,y will return to within a distance of epsilon away from x,y, with infinite probability (given infinite time) -- but as soon as you add in a third dimension the probability becomes finite? What happens with [math]f(x) = \frac {1}{r^2}[/math] in 3 dimensions? SureSure. Take [math] f(x)=\frac{1}{x^2}[/math] for [math] x \ge 1[/math] and [math]f(x)= \frac{1}{\sqrt x}[/math] for [math]0<x<1[/math] Very interesting. I may need to crack out some math books and think about these things awhile before I understand all this. Link to comment Share on other sites More sharing options...

DrRocket Posted March 27, 2011 Share Posted March 27, 2011 Fascinating! Thanks! No, actually I'm only interested in dimension 3 but assumed it made more sense to figure it out in one dimension first. Oops. I don't get why the integral converges on finite balls centered at 0. Look at the volume element. The"r" in the volume element cancels the "1/r" in the integrand. I don't get the connection with random walks. It is not obvious. It takes some work to show the connection. I don't have a ready reference. Is the integral related to the probability of eventually returning to a finite segment in 1D, or area in 2D, or volume in 3D? Does that mean that for any arbitrarily small value of epsilon, a random walk starting at location x,y will return to within a distance of epsilon away from x,y, with infinite probability (given infinite time) ?????????? Nothing has infinite probability. What is infinite, in dimensions 1 and 2 is the expected number of returns to a neighborhood of a point in infinitely many steps. -- but as soon as you add in a third dimension the probability becomes finite? Yes. A drunk on the street with continually walk into the lamp pole, but a drunk astronaut will wander off to infinity. What happens with [math]f(x) = \frac {1}{r^2}[/math] in 3 dimensions? You wind up with the divergent integrand [math]\frac {1}{r}[/math] -- look at the volume element. Very interesting. I may need to crack out some math books and think about these things awhile before I understand all this. good idea Link to comment Share on other sites More sharing options...

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