# Simplification of this complicated Fermi-Dirac integral

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.

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.

• I understand your solution but I first want to evaluate the integral in terms of $f_0$, let's say by partial integration. We know that Fermi-Dirac integrals are PolyLog functions, so maybe get that in that form and then take the limit of high $\beta$. Commented Aug 5, 2020 at 2:36
• I am not sure what you are getting at. Are you looking for a solution that is not an expansion in $\beta$? Commented Aug 5, 2020 at 13:21
• Yes, I am looking for an expansion in $B$ and not $\beta$. Commented Aug 6, 2020 at 16:26
• Can you solve the large B or small B case at all? Commented Aug 6, 2020 at 16:31
• How exactly should I do that? Commented Aug 6, 2020 at 17:05