total charge using modifed matsubara summaiton

$$e^Z \simeq (1+Z/n)^n$$ (84)

then

$$f(Z) = \frac{1}{1+e^{\beta (Z-\mu) }} \simeq \bar{f}(Z) =\frac{1}{1+\left( 1+\frac{\beta (Z-\mu)}{2M} \right)^{2M} }$$ (85)

when

$\bar{f}(Z)$ has poles at

$$E_p = \mu + \frac{2M}{\beta } (z_p -1)$$ (86)

$$z_p = \exp\left( \frac{i\pi (2p+1)}{2M} \right)$$ (87)

total charge is

$$\rho \simeq \int_{-\infty}^{\infty} d\omega A(\omega ) \bar{f}(\omega )$$ (88)

From the counter integration of the upper-half plane,

$$\int_{-\infty}^{\infty} d\omega A(\omega ) \bar{f}(\omega ) = 2 \pi i \sum_{p=0}^{M-1} A(E_p) R_p$$ (89)

where $R_p$ is residue and equals .