D Integration of Fermi Functions

    The Fermi function and its antiderivative is given by
$$\begin{align}f(E) &= \frac{1}{1+e^{\frac{E-E_F}{k_BT}}}\\ F(E) &= \int f(E)\,dE = -k_BT\ln\left(1+e^{-\frac{E-E_F}{k_BT}}\right) \end{align}$$
The product $f(E)(1-f(E+\Delta E))$ can be written as
$$\begin{gather}f(E)(1-f(E+\Delta E)) = \frac{f(E)-f(E+\Delta E)} {1-e^{-\frac{\Delta E}{k_BT}}}, \end{gather}$$
and the corresponding antiderivative is
$$\begin{gather}\int f(E)(1-f(E+\Delta E))\,dE = -\frac{k_BT}{1-e^{-\frac{\Delta E... ...-\frac{E-E_F}{k_BT}}} {1+e^{-\frac{E+\Delta E-E_F}{k_BT}}}\right). \end{gather}$$
A usual approximation for   metal tunnel junctions, where the Fermi energy is much bigger than the conduction band edge, $E_F \gg E_c$, is to set $E_c = -\infty$, thus
 $$\begin{gather} \int_{E_c}^{\infty} f(E)(1-f(E+\Delta E))\,dE \approx \int_{-\in... ...elta E))\,dE \approx\frac{\Delta E}{1-e^{-\frac{\Delta E}{k_BT}}}. \end{gather}$$