\pm 1)^2} \ dx~,$$ where the signs $+$ and $-$ correspond to the Fermi-Dirac and Bose-Einstein cases, respectively, and $x_0 = -\mu/(k_\B T)$ is some reference parameter. Here, $x_0
\in \mathbb{R}$ in MB and FD, but $x_0
> 0$ in BE.

[Although $C(n)$ depends on $x_0$, we will see that the transport properties of the system are essentially independent of $x_0$. This explains why we do not write $C(n,x_0)$.]{.nodecor}

In terms of $C(n)$, one can write $$\begin{aligned}
L_{ij}^{(0)} &=& \frac{e^2}{h} C(0)  \Gamma_{ij}~,\label{L0}\\
L_{ij}^{(1)} &=&  \frac{e}{h} k_\B T C(1) \Gamma_{ij}~, \label{L1}\\
L_{ij}^{(2)} &=& \frac{(k_\B T)^2}{h} C(2) \Gamma_{ij}~. \label{L2}\end{aligned}$$ In Appendix A, we show that $C(0)$, $C(1)$ and $C(2)$ are positive for all $$x_0 = -\frac{\mu}{k_\B T} \in \left\{\begin{array}{l}(-1,\infty)
\mbox{ in MB}\\ 
(-\infty,\infty) \mbox{ in FD}\\ (0,\infty) \mbox{ in BE}\end{array}\right.~.$$ In the Maxwell-Boltzmann case, when $x_0 \in (-\infty,-1]$, we find that $C^{\rm MB}(0)$ and $C^{\rm MB}(2)$ are positive, but $C^{\rm
  MB}(1)$ is non-positive. As one may easily check, all the subsequent results also hold in this situation. Nevertheless, in order to be able to treat the three considered cases on the same footing we make the following restricting assumption.

In the Maxwell-Boltzmann case: $x_0 = -\mu/(k_\B T)\in (-1,\infty)$.\[Assumption xo\]

[Note that the 