we use the normalization $\int \rho_{\rm nuc} ({\bf r})d^3r=Z$. Taking the nuclear density to be spherically symmetric, and after angular integrations, the well-known finite-nucleus expression for the Uehling potential is obtained

$$\label{eq:double}
V_{\rm Ueh}^{\rm fin}(r)=\frac{2}{3}\frac{\alpha^2}{r}\int_0^{\infty}dr'\int_1^{\infty}dt\sqrt{t^2-1}\Big(
\frac{1}{t^3}+\frac{1}{2t^5}\Big) \rho_{\rm nuc} (r') r' \Big(
e^{-2t|r-r'|/\alpha}-e^{-2t(r+r')/\alpha}
\Big) \ .$$

There are publicly available procedures for performing this integral. For example, the code of Hnizdo [@hnizdo] or the expansion of Fullerton and Rinker [@FR]. We will instead consider the case of the step-function density (homogeneous charge distribution) and reduce the double integration, Eq. (\[eq:double\]), to a single integration. The Uehling potential then reduces to the following form, divided into parts valid inside and outside the nucleus,

$$\label{eq:step}
V_{\rm Ueh}^{\rm step}(r) =
\Bigg\{
\begin{array}{ll}
\frac{\alpha}{\pi} \frac{Z}{r}
\int_1^{\infty}dt \sqrt{t^2-1} \Big( \frac{1}{t^2} +\frac{1}{2t^4}
\Big) \Big(\frac{2}{\varkappa^3}\Big)\Big[ \frac{r}{r_n}\varkappa -
e^{-\varkappa}(1+\varkappa) \sinh(2tr/\alpha)\Big] \ , & \qquad r\leq 
r_n \\
\frac{\alpha}{\pi}\frac{Z}{r}\int_1^{\infty}dt
\sqrt{t^2-1}\Big(\frac{1}{t^2}+\frac{1}{2t^4}\Big) \Big(\frac{2}{\varkappa^3}\Big) e^{-2tr/\alpha}
\Big[ \varkappa \cosh (\varkappa)-\sinh(\varkappa)\Big] \ , & \qquad
r> r_n \ \ ,
\end{array}$$

where $r_n$ is the nuclear radius and $\varkappa=2 t r_n/\alpha$. The first term inside the square brackets of the upper line of Eq. (\[eq:step\]) may be simplified by integrating over $t$ analytically, yielding $\frac{\alpha}{\pi}\frac{Z}{r}\big(\frac{\alpha}{2r_n}\big)^3\Big(\frac{8r}{5\alpha}\Big)$, where all factors have been included. Eq. (\[eq:step\]) agrees with 