Is it true that the Fermi-Dirac distribution is a probability? Is it true that the Fermi-Dirac distribution is a probability ?
I checked by doing the integral, and this does not give 1.

Also, for $T=0 K$, since it is a step function, it is easy to see that the integral will be $E_F$, different to 1.
 A: What @Manuel Gallego is saying that for each state of energy $E$ we have
$$
P({\rm empty})= \frac{1}{(1+e^{-\beta (E-\mu)})}, \quad P({\rm occupied})= \frac{e^{-\beta (E-\mu)}}{(1+e^{-\beta (E=\mu)})},
$$
so $P({\rm empty})+P({\rm occupied})=1$, and for $n=0$ or $n=1$ we have
$$
\langle n\rangle = \frac{0\times 1}{(1+e^{-\beta (E-\mu)})}+ \frac{1 \times e^{-\beta (E-\mu)}}{(1+e^{-\beta (E-\mu)})}
\\
= \frac{1}{1+e^{\beta (E-\mu)}}
$$
which is the F-D distribution.
A: The Fermi occupation function
$$ n(\epsilon)= \frac{1}{e^{-\beta(\epsilon-\mu)}+1},$$
is not a probability density function. The normalisation condition is that:
$$N = \sum_i \bar{n}_i, $$
where $\bar{n}_i=n(\epsilon_i)$ is the mean occupation number of a 1-particle state with energy $\epsilon_i$, and where the sum is over all one particle states. Thus the numbers
$$ p_i = \frac{\bar{n}_i}{N},$$
are indeed probabilities, which is why the Fermi occupation factor might sometimes be referred to as giving the occupation probabilities. Note however that it is strictly speaking a probability mass function not a density function. To get a density you have to multiply by the density of states. That is, it is not correct to write:
$$\sum_i \bar{n}_i \underbrace{\approx}_{\rm \color{red}{wrong}} \int_0^\infty \frac{{\rm d}\epsilon}{e^{-\beta(\epsilon-\mu)}+1}. $$
Instead, what is correct is that but with an additional factor, known as the density of states $g(\epsilon)$ included in the integral on the right hand side. In the case of a 3-D non-relativistic ideal gas for instance there are more states at per energy interval at larger energies, scaling like $g(\epsilon)\sim \sqrt{\epsilon}$.
A: The Fermi-Dirac distribution (also the Bose-Einstein's) does not give the probability of finding the fermion (boson) in a given energy region. What this distribution provides is the probability of finding a state with energy E being occupied by a fermion (boson), so It doesn't have to give 1 when integrated in the whole system. Note how the FD function is limited to one, remember that two fermions cannot occupy the same quantum state (actually, you should multiply the distribution f(E_i) by g_i, being g_i the degeneracy of the state with energy E_i, i.e this would take into consideration the spin), in the case of BE statistics this does not happen, since bosons can actually share a quantum state!
Ahother point I wanted to make, you should use the quimical potential, \mu, instead of E_F. The Fermi energy is fixed for the system and it's defined as E_F=\mu(T=0).
I hope this helps!
