Simplification of this complicated Fermi-Dirac integral

Question

I want to simplify the following integral:

$$ \int_0^{\infty} \frac{E^{a}}{1+E^b (\tau(E))^2B^2}\cdot (\tau(E))^c\cdot \frac{\partial f_0}{\partial E} \ \mathrm{d}E $$

where $f_0 = \frac{1}{1+e^{\beta(E-\mu)}}$. Here $a,b,c, \mu, \beta$ are fixed constants and $\tau(E)$ is a general function of $E$. I basically want to solve for as general form as possible before making any approximations. If not, I would like to solve it in the region $B\ll1$ (possibly expand it in terms of $B$) or $B\gg1$. Finally we can make the approximation $\tau(E)=E^r$ where $r$ is again a constant, when we cannot generalize even further to get a more closed for answer. I tried partial integration but I am not able to move forward much. I looked up Somerfeld expansion to help me with this integral but it didn't help much. Any help or direction is appreciated.

Answer 1

The Sommerfeld expansion treats f as a Heavyside function to first approximation ( for large $\beta$). Since you have a derivative, you should treat it as a Dirac delta to first approximation. So basically you are expanding the integrand near $E=\mu$, at least for large $\beta$.

As an example, consider the large $\beta$ limit, where we may replace $f_0(E)\rightarrow \Theta(\mu-E)$, thus ${df_0(E)\over dE}\sim -\delta(\mu-E)$ and the integral becomes

$\int_0^{\infty} \frac{E^{a}}{1+E^b (\tau(E))^2B^2}\cdot (\tau(E))^c\cdot \frac{\partial f_0}{\partial E} \ \mathrm{d}E \rightarrow -\frac{\mu^{a}}{1+\mu^b (\tau(\mu))^2B^2}\cdot (\tau(\mu))^c$

Higher order terms in $\beta$ are more complicated, but the expansion is still straightforward.