I am confused about a derivation for the behaviour of electrons in a conductor with binary collisions, and scattering due to static impurities. The derivation begins as follows:
The distribution function $f(\mathbf r,\mathbf p)$ evolves according to:
$\frac{df}{dt}=\frac{\partial f}{\partial t}+\mathbf v\frac{\partial f}{\partial \mathbf r}+e\mathbf E\frac{\partial f}{\partial \mathbf p}=I_{imp}\{f\}+St\{f\}$,
where $St$ is the inelastic collision integral and $I_{imp}$ is the elastic collision integral. These take the following forms:
$St\{f_{\mathbf p}\}=-\sum_{\mathbf p_2,\mathbf p_3,\mathbf p_4}\{w(\mathbf p,\mathbf p_2;\mathbf p_3,\mathbf p_4)f_{\mathbf p}f_{\mathbf p_2}(1-f_{\mathbf p_3})(1-f_{\mathbf p_4})+w(\mathbf p_3,\mathbf p_4;\mathbf p,\mathbf p_2)f_{\mathbf p_3}f_{\mathbf p_4}(1-f_{\mathbf p})(1-f_{\mathbf p_2})\}\delta_{\mathbf p+\mathbf p_2-\mathbf p_3-\mathbf p_4}$ (collision of two electrons)
$I_{imp}\{f_{\mathbf p}\}=\sum_{\mathbf p_1}w(\mathbf p,\mathbf p_1)\{f_{\mathbf p_1}(1-f_{\mathbf p})-f_{\mathbf p}(1-f_{\mathbf p_1})\}$ (scattering with impurity)
The next step is to integrate over momentum $\mathbf p$, which gives:
$\frac{\partial n}{\partial t}+\frac{\partial \mathbf j}{\partial \mathbf r}+e\mathbf E\int\frac{\partial f}{\partial \mathbf p}d\mathbf p=\int I_{imp}\{f\}+St\{f\}\,d\mathbf p$,
from this you should get the continuity equation:
$e\frac{\partial n}{\partial t}+\nabla\cdot\mathbf j=0$
but I don't understand why the last three integrals disappear. Apparently $\int St\{f\}\,d\mathbf p=0$ due to particle conservation, but I don't understand the physical intuition for this. The impurities scattering also disappears, is this also due to particle conservation? Alongside this, I am unsure what happens to $e\mathbf E\int\frac{\partial f}{\partial \mathbf p}d\mathbf p=e\mathbf E f$ or where the factor of $e$ comes from in the continuity equation.
