Integral: magnetic field inside infinite cylindrical current

[DavideGenoa]

Let ##V\subset\mathbb{R}^3## be an infinitely high solid cylinder, or a cylindrical shell of radii ##R_1<R_2##, whose axis has the direction of the unit vector ##\mathbf{k}##.
For any point of coordinates ##\boldsymbol{r}\notin \bar{V}## external to ##V## the Lebesgue integral (which is proportional to the magnetic field generated by a uniform current flowing through ##V## in the ##\mathbf{k}## direction) $$\int_V\frac{\mathbf{k}\times(\boldsymbol{r}-\boldsymbol{x})}{\|\boldsymbol{r}-\boldsymbol{x}\|^3}d\mu_{\boldsymbol{x}},$$where ##\mu## is the 3-dimensional Lebesgue measure, converges, as it can be proved by using the fact that the asbsolute value of the components of ##\frac{\mathbf{k}\times(\boldsymbol{r}-\boldsymbol{x})}{\|\boldsymbol{r}-\boldsymbol{x}\|^3}## are less than or equal to
##\frac{\|\mathbf{k}\times(\boldsymbol{r}-\boldsymbol{x})\|}{\|\boldsymbol{r}-\boldsymbol{x}\|^3}\le\frac{1}{\|\boldsymbol{r}-\boldsymbol{x}\|^2}##, and the integral $$\int_V\frac{1}{\|\boldsymbol{r}-\boldsymbol{x}\|^2}d\mu_{\boldsymbol{x}}$$ exists finite, provided that ##\boldsymbol{r}\notin \bar{V}##. Nevertheless, if ##\boldsymbol{r}\in \bar{V}##, then ##\int_V\frac{1}{\|\boldsymbol{r}-\boldsymbol{x}\|^2}d\mu_{\boldsymbol{x}}## does not exist finite.

Can we prove that ##\int_V\frac{\mathbf{k}\times(\boldsymbol{r}-\boldsymbol{x})}{\|\boldsymbol{r}-\boldsymbol{x}\|^3}d\mu_{\boldsymbol{x}}## exists finite (by assigning to the integrand function an arbitrary value at ##\boldsymbol{x}=\boldsymbol{r}##) even if ##\boldsymbol{r}\in \bar{V}##?

I have not found a way to prove it and I am inclined to suspect, by an intuitive analogy with ##\int_V\frac{1}{\|\boldsymbol{r}-\boldsymbol{x}\|^2}##, that such an integral does not converge, but I would like to have a confirmation...
I heartily thank any answerer!

 

[Hawkeye18]

Both your integrals exist and finite for all ##x##. The function ##\|r-x\|^{-2}## is integrable in ##\mathbb R^3## near ##x=0## (in fact on any bounded measurable set). This function is not integrable at ##\infty## in whole ##\mathbb R^3##, but in your particular case the integral over ##V## is finite, because for large ##R## the measure of ##V\cap B_R## is bounded by ##CR##

 

[DavideGenoa]

Thank you so much, Hawkeye18!
The fact that the function ##x\mapsto \|r-x\|^{-2}## is integrable on a bounded domain even if it contains ##r## is shown by the use of spherical coordinates with ##dx_1dx_2dx_3=\rho^2\sin\varphi d\rho d\varphi d\theta##, if I am not wrong.
I am not sure how to see that the integral is finite on our unbounded cylinder ##V##: in particular what does your notation mean?

 

[DavideGenoa]

@Hawkeye18 Mmh... I suspect that you mean a ball of radius ##R##, by ##B_R##, probably centred in ##r\in\mathbb{R}^3##: OK?
Nevertheless, I cannot use this to see how the integral ##\int_{V\cap B_R}\|r-x\|^{-2}d\mu_x## remains bounded while ##R\to\infty##...

 

[Hawkeye18]

Yes, it can be shown by using polar coordinates, but I prefer to use distribution function. Namely, for a non-negative measurable function ##f## on a general measure space ##X, \mu## $$\int f d\mu = \int_0^\infty \mu(\{x\in X: f(x)>s\}) ds, $$ where the integral in the right hand side can be treated as Riemann integral. The function ##s\mapsto \mu(\{x\in X: f(x)>s\}) ## is usually called the distribution function of ##f##. Using the above formula one can also compute ##\int f^p d\mu##, ##p>0##, $$\int f^p d\mu = \int_0^\infty \mu(\{x\in X: f(x)^p>s^p\}) ds^p = p\int_0^\infty s^{p-1}\mu(\{x\in X: f(x)>s\}) ds .$$ In your case ##f## is ##x\mapsto \|r-x\|^{-1} 1 _V(x)##, where ##1_V## is the characteristic function of your cylinder ##V##, and you need to compute $$\int f^2 dm_3= 2\int_0^\infty s \cdot m_3(\{x\in X: f(x)>s\}) ds ,$$ where ##m_3## is the Lebesgue measure in ##\mathbb R^3##.

The set where ##f>s## is the intersection of ##V## and the ball ##B_{x,1/s}## of radius ##1/s## centered at ##x##. While it is possible to compute the exact volume of this set, we do not need to do that: we just need to estimate the measure of this set. First it can be estimated by the volume of the ball, i.e. by ##(4/3)\pi s^{-3}##. While this estimate is true for all radii, we will use it for small radii, for example for ##s\ge 1##. For ##s<1## we will use a different estimate: the set ##B_{x,1/s}\cap V## is contained in a piece of the cylinder ##V## of length ##2s^{-1}##, and the volume of this set is ##A 2s^{-1}##, where ##A## is the area of the base of the cylinder.

So you get 2 integrals $$2\int_0^1 s 2A s^{-1} ds $$ and $$ 2\int_1^\infty s (4/3)\pi s^{-3} ds$$ and both integrals converge.

 

[DavideGenoa]

I heartily thank you for your answer!

Yes, it can be shown by using polar coordinates

Mmh... I would find that very, very interesting, also because I know multivariable calculus tools much better than the measure theoretical facts that you have used (ex. why ##\int_X f^p d\mu ## ##= \int_0^\infty \mu(\{x\in X: f^p(x)>s^p\}) ds^p##, which I've never seen even for ##p=1##)...
Would it be more difficult or just less elegant to use polar coordinates?

 

[Hawkeye18]

Polar coordinates just less general, they work only when level sets are concentric spheres.
The formula for the integral via distribution function is unfortunately usually absent from the standard textbooks.

But the proof for the case ##p=1## is easy, you can just draw a picture. And for other ##p##, you just do the change of variables.

Integration via distribution functions is used a lot in probability, but in probability, the distribution function of a random variable ##f## is the probability of (measure) of the event ##f<s##. It is more convenient in some situations, but works only when the underlying measure is finite, which is the case in probability. But in analysis, one needs to treat the case of infinite measures, so the distribution function is defined as the measure of the set where ##f>s##.

 

[DavideGenoa]

@Hawkeye18 Thank you for your answer again! I think I've been able to understand why ##\int_X f^p d\mu ## ##= \int_0^\infty \mu(\{x\in X: f(x)>s\}) ds## by starting with simple functions and using Beppo Levi's theorem about monotone convergence.
When ##p>1## I'm not sure what you mean by change of variables, since I've only read Kolmogorov-Fomin's about Lebesgue integrals and measure theory and I know about the change of variables only for Riemann integrals of continuous functions...
To tell the truth I don't even know the notation ##\int_0^\infty ...ds^p## (I have only seen things like \int_0^\infty for triple Riemann integrals, or Lebesgue integrals where ##dx^3=dm_3##, but here we have only one variable of integration ##s## on the linear domain ##[0,+\infty)##)...

 

[Hawkeye18]

Notation ##d s^p## I used is pretty standard calculus notation used as a shorthand in change of variables, ##d s^p = d(s^p)=p s^{p-1} ds##, it is not a notation from the multivariable calc, just the standard notation for the differential in 1 variable calc.

As for the change of variables, you can also use simple functions and monotone convergence theorem. For a simple function the distribution function will be a piecewise constant function with bounded support, and for such functions the change of variables in the integral is trivial. So for simple functions you have the formula $$\int f^p d\mu = p \int_0^\infty s^{p-1} \mu(\{x\in X: f(x)>s\}) ds.$$ Using monotone convergence proves the formula in the general case.

 

[DavideGenoa]

Thank you so much! I have understood the proof (where I think you meant the ball to be ##B_{\boldsymbol{r},1/s}##) I have proved it by noticing that$$\int_{[0,+\infty)}ps^{p−1}μ(\{x\in X:f(x)>s\})d\mu_s=\int_{[0,+\infty)}\int_X ps^{p−1}\chi_{\{(x,s)\in X\times\mathbb{R}_{\ge 0}:f(x)>s\}}d\mu_x d\mu_s$$ $$=\int_X\int_{[0,+\infty)} ps^{p−1}\chi_{\{(x,s)\in X\times\mathbb{R}_{\ge 0}:f(x)>s\}}d\mu_s d\mu_x=\int_X\int_0^{f(x)}ps^{p-1} ds d\mu_x=\int_X f(x)^p d\mu_x$$