Why is the singularity not taken into account?

He replaces $$\mathbf{m}$$ with a volume element of magnetization $$\mathbf{M}\ dV$$ , integrates over $$V$$ , and lets the same integral define the magnetic potential inside as well as outside the magnetization to get in Art. $$385$$ equation $$8$$ (of Maxwell's treatise):

$$\displaystyle \psi_m (\mathbf {r})=\dfrac{1}{4 \pi} \int_{V'} \mathbf{M(r')}.\nabla' \left( \dfrac{1}{\left| \mathbf{r}-\mathbf{r'} \right|} \right) dV' \tag{22}$$

Why is the singularity in equation $$(22)$$ not taken into account?

Shouldn't we use the following improper integral instead of equation $$(22)$$:

$$\displaystyle \psi_m (\mathbf {r})=\dfrac{1}{4 \pi} \lim \limits_{\delta V' \to 0} \int_{V'-\delta V'} \mathbf{M(r')}.\nabla' \left( \dfrac{1}{\left| \mathbf{r}-\mathbf{r'} \right|} \right) dV'$$

How shall we show this integral converges (without using the spherical volume element) using Cartesian coordiante system?

• What do you mean by ignored? If it doesn't make the improper integral divergent, there's nothing to ignore. – Ruslan Mar 5 '19 at 10:48
• Have you checked if it really is a singularity if you take into account the suppressing factor in the volume element? – Qmechanic Mar 5 '19 at 10:52
• @Qmechanic: I think using spherical element, it would be removable singularity??? – N.G.Tyson Mar 5 '19 at 10:54
• @Ruslan: How can we show the improper integral $\displaystyle \psi_m (\mathbf {r})=\dfrac{1}{4 \pi} \lim \limits_{\delta V' \to 0} \int_{V'-\delta V'} \mathbf{M(r')}.\nabla' \left( \dfrac{1}{\left| \mathbf{r}-\mathbf{r'} \right|} \right) dV'$ converges? – N.G.Tyson Mar 5 '19 at 10:58
• Also posted (under a different user id?) at math.stackexchange.com/q/3135817 – user197851 Mar 5 '19 at 11:51