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# Do all functions with singularities have divergent integrals?

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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!)

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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 $f(x) = \frac {1}{r}$(i.e. $\frac {1}{||x||}$) In dimension 3 0r above. Since the volume element in spherical coordinates is $r^{n-1}dr \times$ (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 $f(x)=\frac{1}{x^2}$ for $x \ge 1$ and $f(x)= \frac{1}{\sqrt x}$ for $0<x<1$

Edited by DrRocket
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Sure

Consider $f(x) = \frac {1}{r}$(i.e. $\frac {1}{||x||}$) In dimension 3 0r above. Since the volume element in spherical coordinates is $r^{n-1}dr \times$ (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 $f(x) = \frac {1}{r^2}$ in 3 dimensions?

SureSure. Take $f(x)=\frac{1}{x^2}$ for $x \ge 1$ and $f(x)= \frac{1}{\sqrt x}$ for $0<x<1$

Very interesting. I may need to crack out some math books and think about these things awhile before I understand all this.

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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 $f(x) = \frac {1}{r^2}$ in 3 dimensions?

You wind up with the divergent integrand $\frac {1}{r}$ -- 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

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