sources $\rho$ and $\textbf{J}$ evaluated at the retarded time. We then enclose the terms of the left of (1) in the retardation symbol $[\;\;]$ which indicates that the enclosed quantity is to be evaluated at the source point $\textbf{x}'$ at the retarded time $t'=t-R/c$, $$[\nabla'\cdot\!\textbf{ J}]+\bigg[\frac{\partial\rho}{\partial t'}\bigg]=0.$$ We now multiply the first term of (2) by the factor $\mu_0/(4\pi R)$ and the second term of (2) by the equivalent factor $1/(4\pi\epsilon_0 R c^2)$, where $\epsilon_0$ and $\mu_0$ are constants satisfying the relation $\epsilon_0\mu_0\!=\!1/c^2$, and integrate over all space, obtaining the equation $$\frac{\mu_0}{4\pi}\!\int\!\frac{[\nabla'\cdot\!\textbf{ J}]}{R}d^3x' + \frac{1}{c^2}\frac{1}{4\pi\epsilon_0}\!\int\! \frac{[\partial\rho/\partial t']}{R}d^3x'=0.$$ With the idea of taking out the derivative operators from the integrals in (3), we perform an integration by parts in the first term of (3), in which we use the result [@20]: $[\nabla'\cdot\!\textbf{ J}]/R=\nabla\cdot([\textbf{ J}]/R)+ \nabla'\cdot([\textbf{ J}]/R)$ and the fact that the surface integral arising from the term $\nabla'\cdot([\textbf{ J}]/R)$ vanishes at spatial infinity because $\textbf{J}$ is localised. Next we use the result [@6]: $[\partial\rho/\partial t']\!=\!\partial[\rho]/\partial t$ in the second term of (3). After performing the specified operations, the final result reads $$\nabla\cdot\Bigg\{\frac{\mu_0}{4\pi}\!\int\!\frac{[\textbf{ J}]}{R}d^3x'\Bigg\} + \frac{1}{c^2}\frac{\partial}{\partial t}\Bigg\{\frac{1}{4\pi\epsilon_0} \!\int\!\frac{[\rho]}{R}d^3x'\Bigg\} =0.$$ The terms within the curly braces $\{...\}$ are determined by the retarded values 