- #1

- 154

- 5

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!