r}$, it can be readly shown that $${\;}_{\;\;\;\atop{\mbox{$\Gamma_{sl}$}}}\!\int_{\;\;\atop{{{\bf
r}}^{\;\prime}}}^{\;{\bf r}} [{\bf B}\times\left(
\mbox{{\mathversion{bold}${\xi}$}}-{{\bf r}}^{\;\prime} \right)]
\cdot d\mbox{{\mathversion{bold}${\xi}$}}=0\; ,$$ where $\Gamma_{sl}$ means a straightline from ${{\bf r}}^{\;\prime}$ to ${\bf r}$. With this simplification, the $C({{\bf r}}^{\;\prime},{\bf r})$ takes the form $$\label{Cintegrado1}
C({\bf r},{{\bf r}}^{\;\prime}) \!= C_0\; \exp\left\{ {ie\over
\hbar}_{\;\;\Gamma_{sl}}\!\! \int_{\;\;\atop{{{\bf
r}}^{\;\prime}}}^{\;{\bf r}}\!\! {\bf A}
(\mbox{{\mathversion{bold}${\xi}$}}) \cdot
d\mbox{{\mathversion{bold}${\xi}$}}\right\} .$$ Substituting last equation into (\[ProtoPropagador\]) and using the initial condition (\[CondicaoInicial\]), we readly obtain $C_{0}=\frac{m\omega}{2\pi i\hbar}$. Therefore the complete Feynman propagator for a charged particle under the influence of a constant and uniform magnetic field takes the form $$\begin{aligned}
\label{PropagadorFinalBUniforme}
K({\bf x},{{\bf x}}^{\prime};\tau)\hspace{6.0cm}\nonumber\\
={m\, \omega\over 2\pi i\hbar\,\sin{(\omega\tau)}}  \sqrt{{m\over 2\pi i\hbar\tau}}
\exp\left\{ {ie\over \hbar}
 \int_{{\bf r}^{\prime}}^{{\bf r}}\!\!\!\!\!{\bf A}
  (\mbox {{\mathversion {bold}${\xi}$}}) \cdot d\mbox{{\mathversion{bold}${\xi}$}}\right\}\nonumber\\
\exp\biggl\{{im\omega\over 2\hbar}\cot(\omega\tau)({\bf r}-{{\bf r}}^{\;\prime})^2\biggr\}
 \exp\biggl\{{im\over 2\hbar}{\left( x_3 - x_3^\prime\right)^2\over
 \tau}\biggr\}\, ,\end{aligned}$$ where in the above equation we omitted the symbol $\Gamma_{sl}$ but, of course, it is implicit that the line integral must be done along a straightline, and we brought back the free propagation along the ${\cal OX}_3$ direction. A few comments about the above result are in order.

1.  Firstly, we should emphasize that the line integral which appears in the first exponencial on the r.h.s. of (\[PropagadorFinalBUniforme\]) must be 