Eq. (\[Particle-Density-2Dcondensate\]) has a divergence when $\mu'\rightarrow 0$ but the particle density must remain finite, thus its right interpretation is as a manner to compute $\mu^{\prime 1D}$

$$\label{miuprima-1D}
\mu^{\prime 1D} = -\varepsilon(0,B)\frac{T^2}{ N^2}= -m \sqrt{1-b}\frac{T^2}{ N^2}.$$

From Eq. (\[miuprima-1D\]) follows that $\mu^\prime$ is a decreasing function of T, as occurs in the usual Bose-Einstein condensation, but also that $\mu^{\prime 1D}$ is a decreasing function of the field that goes to zero when $B \rightarrow B_c$. Since $\mu^{\prime 1D} = 0$ is the condition for the BEC to occur, $\mu^{\prime 1D} \sim 0$ means that $p_3 \sim 0$. In this regard, by the use of Eq. (\[miuprima-1D\]) we can approximate the Bose-Einsten distribution $n^{+}(p_3)$ in a vicinity of $p_3 =0$ as

$$\begin{aligned}
n^+(p_3) = \frac{1}{e^{\beta(\varepsilon(p_3,B)-\mu)-1}} \simeq \frac{2 \varepsilon(0,B) T}{p^2 +\varepsilon(0,B)^2 - \mu^2} \simeq \frac{2 \varepsilon(0,B) T}{p^2 -2 \varepsilon(0,B) \mu^{\prime 1D}}, \\\nonumber\end{aligned}$$

$$\begin{aligned}
\label{nmas}
n^+(p_3) \simeq 2 N \frac{\gamma}{p_{3}^2 + \gamma^2},\end{aligned}$$

where $\gamma = \sqrt{- 2 \varepsilon(0,B) \mu^{\prime 1D}}$. Eq.(\[nmas\]) means that for $p_3 \approx 0$, $n^+(p_3)$ tends to a Cauchy distribution centered in $p_3=0$ (as happens for the charged vector bosons distribution [@ROJAS1996148]). Now the equivalent to the limits $T \rightarrow 0$ or $B\rightarrow B_c$ is the limit $\gamma \rightarrow 0$ 