in the non-equilibrium, TRANSIESTA's approach

From eqs.(90) and (95),

$$\rho_{ij} = \int_{-\infty}^{\infty} d\omega \; \rho^L_{ij}(\omega ) f(\omega -\mu_L) + \int_{-\infty}^{\infty} d\omega \; \rho^R_{ij}(\omega ) f(\omega -\mu_R)$$ (97)

$$= \int_{-\infty}^{\infty} d\omega \; \left( \rho^L_{ij}(\omega ) + \rho^R_{ij}(\omega ) \right) f(\omega -\mu_R)$$ (98)

$$+ \int_{-\infty}^{\infty} d\omega \; \rho^L_{ij}(\omega ) \left( f(\omega -\mu_L) - f(\omega -\mu_R) \right)$$ (99)

Assuming $\mu_L > \mu_R$, we can replace $\rho^L_{ij}(\omega ) + \rho^R_{ij}(\omega )$ with $A(\omega )$ for $\omega \le \mu_R$.

$$\rho_{ij} = \int_{-\infty}^{\infty} d\omega \; A(\omega ) f(\omega -\mu_R)$$ (100)

$$+ \int_{-\infty}^{\infty} d\omega \; \rho^L_{ij}(\omega ) \left( f(\omega -\mu_L) - f(\omega -\mu_R) \right)$$ (101)

We evaluate the first term in eq.(81). $\rho^L(\omega )$ in the second term is not analytical.