h_f^4=h_u^4+h_d^4+h_s^4$, and we extend definition (\[I-i\]) for index $i=-1$ with $$\label{J_1}
     J_{-1}(M^2)=-\int_0^{M^2}\!\! J_0(M^2){\mathrm{d}}M^2
     =-\frac{1}{2}\left(M^2J_0(M^2)+\Lambda^4\ln
     \left(1+\frac{M^2}{\Lambda^2}\right)\right).$$ Here $J_0$ has the explicit form $$J_0(M^2)=\Lambda^2 -M^2\ln\left(1+\frac{\Lambda^2}{M^2}\right)$$ for the given choice of regulator.

The first integral in (\[effpot1\]) accounts for the leading order stationary phase contribution. The second integral describes the quark one-loop part. Since both integrands in (\[effpot1\]) are exact differentials, the line integrals depend only on the end points. The low limit of the integrals is adjusted so that $V(0)=0$ (to understand this, it is enough to notice that the power-series expansion of $V(M_f)$ at small $M_f$ starts from $V(0)$; this term does not depend on $M_f$ and, therefore, does not affect the physical content of the theory; so we simply subtract it, calculating the potential energy of the system with regard to the energy of the symmetric vacuum in the imaginary world of massless quarks).

Case of finite temperature and chemical potential
-------------------------------------------------

The extension to finite $T$ and $\mu$ of the bosonized Lagrangian (\[bos\]) is effected through the quark loop integrals $J_i$ (see eq.(\[Ji\])). Due to the recurrence relation (\[rec\]) it is sufficient to get it just for one of them, $J_0$, by introducing the Matsubara frequencies, $\omega_n$, and the 