A_j (\mbox{\footnotesize{\mathversion{bold}${\xi}$}}) +
{1\over 2}\, F_{jk}\left(\xi_k-x^\prime_k\right)] \; d\xi_j\}}$}\; ,$$ where the line integral is assumed to be along curve $\Gamma$, to be specified in a moment. As we shall see, this line integral does not depend on the curve $\Gamma$ joining ${{\bf r}}^{\;\prime}$ and ${\bf r}$, as expected, since the l.h.s. of (\[dlnC\]) is an exact differencial.

In order to determine the differential equation for $C({{\bf
r}}^{\;\prime},{{\bf r}}^{\;\prime})$ we must substitue expression (\[integracaoCrr’1\]) into equation (\[Equacao2ParaC\]). Doing that and using carefully the fundamental theorem of differential calculus, it is straightforward to show that $$\label{integracaoCrr'2}
{\partial C\over\partial x^\prime_j}({{\bf r}}^{\;\prime},{{\bf
r}}^{\;\prime})=0\; ,$$ which means that $C({\bf r}^{\,\prime},{{\bf r}}^{\;\prime})$ is a constant, $C_0$, independent of ${{\bf r}}^{\;\prime}$. Noting that $$[{\bf B}\times\left(\mbox{{\mathversion{bold}${\xi}$}}- {{\bf
r}}^{\;\prime}\right)]_j =
 -F_{jk}\left(\xi_k-x^\prime_k\right)\, ,$$ equation (\[integracaoCrr’1\]) can be written as $$\label{Cintegrado1}
C({\bf r},{{\bf r}}^{\;\prime}) \!= C_0\; \exp\left\{ {ie\over
\hbar}_{\;\;\Gamma}\!\! \int_{\;\;\atop{{{\bf
r}}^{\;\prime}}}^{\;{\bf r}}\!\! \bigl[{\bf A}
(\mbox{{\mathversion{bold}${\xi}$}})-{1\over 2}\, {\bf B}\times\left
(\mbox{{\mathversion{bold}${\xi}$}}-{{\bf
r}}^{\;\prime}\right)\bigr]\! \cdot\!
d\mbox{{\mathversion{bold}${\xi}$}}\right\} .$$ Observe, now, that the integrand in the previous equation has a vanishing curl, $$\mbox{{\mathversion{bold}${\nabla}$}}_{\mbox{{\mathversion{bold}${\xi}$}}}
\times\biggl[{\bf A} (\mbox{{\mathversion{bold}${\xi}$}})-{1\over
2}\, {\bf B} \times\left(\mbox{{\mathversion{bold}${\xi}$}}-{{\bf
r}}^{\;\prime}\right)\biggr]= {\bf B}-{\bf B}={\bf 0}\; ,$$ which means that the line integral in (\[Cintegrado1\]) is path independent. Choosing, for convenience, the straightline from ${{\bf r}}^{\;\prime}$ to ${\bf 