D^{-1}(\overline{p}^*)\right].$$

Here $D^{-1}(\overline{p}^*)$ is the neutral boson propagator given by (\[propagator\]), $\beta = 1/T$ denotes the inverse temperature, $\mu$ the boson chemical potential and ${\overline{p}}^*=(ip_{4}-\mu,0,p_{\perp},p_{3})$. After doing the Matsubara sum, Eq.(\[Grand-Potential-4\]) becomes

$$\Omega(B,\mu,T)= \Omega_{st}+\Omega_{vac},$$

where $\Omega_{st}$ is the statistical contribution of bosons/antibosons that depends on $B, T$ and $\mu$

$$\Omega_{st}(B,\mu,T)=  \sum_{s=-1,0,1} \frac{1}{\beta}\left(\int\limits_{0}^{\infty}\frac{p_{\perp}dp_{\perp}dp_3}{(2\pi)^2} \ln \left((1-e^{-(\varepsilon(p_3,p_{\perp}, B,s)- \mu)\beta})(1-e^{-(\varepsilon(p_3,p_{\perp}, B,s)+ \mu)\beta})\right)  \right ),$$

and $\Omega_{vac}$ is the vacuum term which is only B-dependent [^6]

$$\Omega_{vac}=\sum_{s=-1,0,1}\int\limits_{0}^{\infty}\frac{p_{\perp}dp_{\perp}dp_3}{(2\pi)^2}\varepsilon(p_3,p_{\perp}, B,s).$$

We can rewrite $\Omega_{st}$ as

$$\Omega_{st}(B,\mu,T)=  \sum_{s=-1,0,1} \Omega_{st}(s),$$

with

$$\label{Grand-Potential-sst}
\Omega_{st}(s)=\frac{1}{\beta}\left(\int\limits_{0}^{\infty}\frac{p_{\perp}dp_{\perp}dp_3}{(2\pi)^2} \ln \left((1-e^{-(\varepsilon(p_3,p_{\perp}, B,s)- \mu)\beta})(1-e^{-(\varepsilon(p_3,p_{\perp}, B,s)+ \mu)\beta})\right)  \right )$$

being the contribution of each spin state to the statistical part of the potential.

Using the Taylor expansion of the logarithm, Eq. (\[Grand-Potential-sst\]) can be written as

$$\label{Grand-Potential-sst1}
\Omega_{st}(s)= - \frac{1}{4 \pi^2 \beta}  \sum_{n=1}^{\infty} \frac{e^{n \mu \beta}+e^{- n \mu \beta }}{n} \int\limits_{0}^{\infty} p_{\perp}  dp_{\perp} \int\limits_{-\infty}^{\infty} dp_3 e^{-n \beta \varepsilon(p_3,p_{\perp}, B,s)}.$$

The term $e^{n \mu \beta}$ stands for the particles and the term $e^{- n \mu \beta}$ for the antiparticles.

In Eq. (\[Grand-Potential-sst1\]) the integration in $p_3$ can be completely carried out, while the integration in $p_{\perp}$ can be only partially done and we obtain for the thermodynamical potential the expression

$$\begin{aligned}
\label{Grand-Potential-sst2}
\Omega_{st}(s)= - \frac{z_0^2}{2 \pi^2 \beta^2}  \sum_{n=1}^{\infty} 