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Consider fermion DM with g internal degrees of freedom and the statistical distribution results in:

$$f(E) = \frac{g}{ \exp[(E − µ)/T] + 1}$$

with $g = 1$ when $E < µ$ and $g = 0$ when $E > µ$

When I integrate $f(E)$ the phase-space, in the relativistic limit, distribution I get this:

$n = \frac{gT^3}{2\pi^2} \int_0^{\infty} \frac{x^2}{(e^x +1)} dx =\frac{3 \zeta(3)gT^3}{4\pi^2}$ (1)

But the answer I'm looking for is

$n = \frac{g m^3 v^3}{6\pi^2}$ (2)

where $v$ is the fermi velocity $v = \sqrt{2E_f/m}$

I don't see how to get from (1) to (2), what am I missing? I'm sure it must be something dumb..

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  • $\begingroup$ Yes, I mean zeta, thanks $\endgroup$ – Silvia_Arer Jun 30 at 5:37
  • $\begingroup$ Have you computed $E_f$ and thus $v$ as a function of $T$? $\endgroup$ – G. Smith Jun 30 at 5:58
  • $\begingroup$ I get $n = \frac{m^3 v^3}{3 \pi^2}$ when computing $\E_f $ and then computing $\v^3$ and then clear n. $\endgroup$ – Silvia_Arer Jun 30 at 6:16
  • $\begingroup$ I think your condition for $g$ is only true at $T=0$. $\endgroup$ – ProfM Jun 30 at 7:00
  • $\begingroup$ You seem to have forgotton $\mu$ in your integral. It will shift the limits of integration Also why is $g$ determined by $\mu$? This is not the usual case. $\endgroup$ – mike stone Jun 30 at 12:12

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