of mass $\Lambda$ contribute to $\nu_\Lambda(T)$ $$\label{intconst-2} 
     \nu_\Lambda (T)=\frac{90 N_c}{\pi^4 T^4}\int_0^\infty |\vec{p}_E|^4
     {\mathrm{d}}|\vec{p}_E|\left(1-\Lambda^2\frac{\partial}{\partial\vec{p}_E^{\,2}}   
     \right)\frac{n_{q\Lambda}+n_{\bar q\Lambda}}{E_p(\Lambda )}.$$ This contribution is given by an integral with a positive integrand. Here we have two dimensionful parameters, $T$ and $\Lambda$. Therefore, at large $T$ the series for $\nu_{\Lambda}(T)$ can be organized in powers of the dimensionless ratio $\Lambda/T$, i.e. $$\label{sb-1}
     \nu_\Lambda (T)=\frac{180N_c}{\pi^4}\int_{\frac{\Lambda}{T}}^\infty\!{\mathrm{d}}x
     \frac{(x^2-\frac{\Lambda^2}{T^2})^{\frac{3}{2}}}{1+e^x}\left[1+
     \frac{\Lambda^2}{2T^2x^2}\left(1+\frac{xe^x}{1+e^x}\right)\right]=   
     \frac{21N_c}{2}\left[1+
     {\cal O}\left(\frac{\Lambda}{T}\right)\right].$$ A useful observation is that although the heavy mass states determine the value of the integral, it still has the correct asymptotics at large $T$, which does not depend on $\Lambda$. In other words, the integral $\sim (n_{q0}+n_{\bar q0})$ in eq. (\[J-1med\]) must vanish at $T\to\infty$, in order that the function $\nu (T)$ has the right asymptotic behavior. This really happens, due to the Pauli-Villars regulator $\hat\rho_{\Lambda \vec{p}_E}$. In fig. \[fig-1a\] one can see that up to around $T\sim 100\,\mbox{MeV}$ the contribution to the effective number of degrees of freedom is practically constant and nearly zero, then it increases and reaches asymptotically the value $31.5$. The constant dashed curve shows the same quantity if the integral is not regularized. Such a big difference in the behavior of $\nu_\Lambda (T)$ translates in $\nu (T)$ (see fig. \[fig-1\]) to 