\simeq &
t + \frac12\int\limits_{\infty}^{t-\frac{{{\boldsymbol \Omega}}\cdot{\bf x}_{D0}}{{\rm v}}}h_{s}(v)\,dv \\
&&
- \left[ t - L - \frac{{{\boldsymbol \Omega}}\cdot{\bf n}_{ED}}{2 {\rm v}}
\int\limits_{t-L-\frac{{{\boldsymbol \Omega}}\cdot{\bf x}_{E0}}{{\rm v}}}^{t-\frac{{{\boldsymbol \Omega}}\cdot{\bf x}_{D0}}{{\rm v}}}h_{s}(v)\,dv
+ \frac12\int\limits_{\infty}^{t-L-\frac{{{\boldsymbol \Omega}}\cdot{\bf x}_{E0}}{{\rm v}}}h_{s}(v)\,dv\right] \nonumber \\
& = &
L + \frac12\left(1 + \frac{{{\boldsymbol \Omega}}\cdot{\bf n}_{ED}}{{\rm v}}\right)
\int\limits_{t-L-\frac{{{\boldsymbol \Omega}}\cdot{\bf x}_{E0}}{{\rm v}}}^{t - \frac{{{\boldsymbol \Omega}}\cdot{\bf x}_{D0}}{{\rm v}}}h_{s}(v)\,dv. \nonumber\end{aligned}$$ In Eq.(\[eq:sol1z\]) we have parametrized the constant phase surfaces $v$’s with the coordinate time $t'$ along the particle trajectory. Now we change the parametrization and will use a parameter $\lambda$ along the laser ray. To this end we notice that the trajectory $[{\rm t}_{0}(\lambda),{\bf x}_{0}(\lambda)]$ of the laser ray from ${\bf x}_{E0}$ to ${\bf x}_{D0}$ (end points are needed only in the $0$-th order in the argument of the integrand $h_{s}$ of Eq. (\[eq:tau\])) is given by $$\begin{aligned}
\label{eq:rays}
{\rm t}_{0}(\lambda) = \lambda, \qquad {\rm \bf x}_{0}(\lambda) = {\rm \bf x}_{D0} - {\bf n}_{ED}(t-\lambda).\end{aligned}$$ Thus, changing the variables, $v(\lambda) = {\rm t}_{0}(\lambda) - {{\boldsymbol \Omega}}\cdot{\rm \bf x}_{0}(\lambda)/{\rm v}$, one finds $$\begin{aligned}
\label{eq:DT3a}
\Delta \tau_{ED}(t)
& = &
L + \frac{1 - \left(\frac{{{\boldsymbol \Omega}}\cdot{\bf n}_{ED}}{{\rm v}}\right)^2}{2} \int\limits_{t-L}^{t}h_{s}\left[v(\lambda)\right]\,d\lambda \\
\label{eq:DT3b}
& = &
L + \frac{1 - \left({{\boldsymbol \Omega}}\cdot{\bf n}_{ED}\right)^2  + (1 - \frac{1}{{\rm v}^2})\left({{\boldsymbol \Omega}}\cdot{\bf 