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 $?$.

hint, what about $E<\mu$? – user619894 Jan 06 '20 at 11:02 — user619894, Jan 06 '20 at 11:02
Yes, it's a good hint. Thanks a lot. – Camillus Jan 06 '20 at 11:41 — Camillus, Jan 06 '20 at 11:41

Johnathan Howard Smith · Answer 1 · 2021-01-04T17:23:08.263

0

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 $I(0)=\mu^3$.

edited Jan 04 '21 at 17:23

answered Oct 27 '20 at 19:41

Johnathan Howard Smith

48

1

By definition, $I$ has energy dimension $3$, whereas your claimed result for $T\ne0$ ($T\to0^+$) has energy dimension $0$ ($2$). – J.G. Oct 27 '20 at 22:14
$\mu^3$ (just on the dimensional grounds @J.G. suggested) and also a factor of $\frac{1}{3}.$ As $T\to 0^+,$ the integrand becomes a parabola $E^2$ that cuts off to zero suddenly at $E=\mu.$ – spaceisdarkgreen Oct 28 '20 at 00:19

score 0 · Answer 2 · answered Oct 28 '20 at 07:36

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 $\int_0^\mu E^{2}dE=\tfrac13\mu^3$.

Fermi-Dirac integral

2 Answers2