Fermi-Dirac integral

Question

I'm trying to do some integral calculation for Fermi-Dirac distribution, specifically for:

$$ \int_{0}^{\infty}{E^{2} \over 1 + \exp\left(E - \mu \over k_{B}\,T\right)} \, dE $$

I know that it can be only solved numerically, but I got information that when we set $T = 0$ it can be solved analytically.

And here is my question: When we will do that directly, all is left is $0$. So, how can I approach such sort of problem $?$.

Answer 1

You have

$$I(T)=\int\limits_0^\infty {E^2\over 1+\exp{E-\mu\over k_BT}}\mathrm{d}E = -\operatorname{Li}_3\left(-e^{\mu\over k_BT}\right)$$

in general. However, for your case of interest where $T\rightarrow 0$, you will have a Dirac impulse for $E=\mu$ and the solution is just $I(0)=\mu^3$.

Answer 2

Applying $E=\mu+k_BTx$ for $E\ge\mu$,$$\int_0^\infty\tfrac{E^2dE}{1+\exp\frac{E-\mu}{k_BT}}=\int_0^\mu\tfrac{E^2dE}{1+\exp\frac{E-\mu}{k_BT}}+\ln2\cdot\mu^2k_BT+\frac{\pi^2}{6}\mu k_B^2T^2+\tfrac32\zeta\left(3\right)k_B^3T^3.$$The $T\to0^+$ right-hand limit is $\int_0^\mu E^{2}dE=\tfrac13\mu^3$.