 Solution

Put $\lambda =(r/a{)}^2$, so the inverted observation point is $\overline{\mathbf{r}}=\mathbf{r}/\lambda$ and $\overline{r}=r/\lambda$. Let $\mathbf{R}=\overline{\mathbf{r}}-{\mathbf{r}}_0$. The Kelvin transform correction in the first bracket is transformed usingThus its dipole contribution is exactlyFor the remaining correction use the dipole line-integral identity. Writing ${\mathbf{P}}_c=\mathbf{r}-c{\mathbf{r}}_0$ and $P_c=|{\mathbf{P}}_c|$, it isIt can be proved by differentiating the right side with respect to $c$ and checking its zero value at $c=0$. Alternatively resolve into directions parallel and perpendicular to ${\mathbf{r}}_0=b\mathbf{e}$: if $\mathbf{r}=\rho {\mathbf{e}}_{\rho }+z\mathbf{e}$, the two integrals are $(z/r-(z-bc)/P_c)/(b\rho )$ and $(1/P_c-1/r)/b$, respectively, which combine to the displayed expression. The limiting cases follow by continuity.

Set $c=\lambda$. Then ${\mathbf{P}}_c=\lambda \mathbf{R}$, $P_c=\lambda R$, $\mathbf{r}=\lambda \overline{\mathbf{r}}$ and $r=\lambda \overline{r}$. Substitution givesDotting with $\mathbf{Q}/(4\pi \sigma )$ reproduces the second correction. Including the primary dipole therefore proves the Spherical Neumann dipole images representation,The upper limit is $(r/a{)}^2$ in the PDF; extracting the small stacked fraction as text can incorrectly invert it. The extra terms are one scaled dipole and a continuous dipole line in this parametrized method of images. Their positions and strengths depend on the observation radius, so they should not be treated as a fixed set of independent physical sources. For an interior point $r<a$, the entire image segment satisfies $t|{\mathbf{r}}_0|\le (r/a{)}^2|{\mathbf{r}}_0|<r$, so it introduces no new interior singularity. The only physical interior dipole singularity remains at ${\mathbf{r}}_0$.