Is there an analytical way to get average energy of a Fermi-Dirac gas?

[seflyer]

The average particle energy of a Fermi-Dirac gas, with zero chemical potential, is about 3.15T, where T is the temperture of this gas. To get the average energy, one needs to do an integration. The integrand is something like
\frac{x^3}{e^{x/k_BT}+1}.

I could get the result numerically. But is there a way to do it analytically? Thanks.

[updated] Sorry. The expression is corrected now. So basically I would like to know whether the following can be integrated analytically:

\int_0^\infty \frac{x^3}{e^{x/k_BT}+1} dx

 

[CompuChip]

Erm, it's been a while since I have done thermal physics, so correct me if I'm wrong, but I think
$$\langle E \rangle = \int \varepsilon f(\varepsilon) g(\varepsilon) \, \mathrm d\varepsilon$$
where
$$f(\varepsilon) = \frac{1}{e^{E/k_B T} + 1}$$

I'm not sure what g is anymore, I just Googled and found $g(\varepsilon) \propto \varepsilon^{1/2}$. This would give
$$\langle E \rangle = \int \frac{\varepsilon^{3/2}}{\exp(\varepsilon / k_B T) + 1} \, \mathrm d\varepsilon$$
which is like your expression but with $x = E = \varepsilon$.

In short, I'm not sure you have posted the correct integral so maybe you can post the exact integral you are trying to do.

 

[seflyer]

Erm, it's been a while since I have done thermal physics, so correct me if I'm wrong, but I think
$$\langle E \rangle = \int \varepsilon f(\varepsilon) g(\varepsilon) \, \mathrm d\varepsilon$$
where
$$f(\varepsilon) = \frac{1}{e^{E/k_B T} + 1}$$

I'm not sure what g is anymore, I just Googled and found $g(\varepsilon) \propto \varepsilon^{1/2}$. This would give
$$\langle E \rangle = \int \frac{\varepsilon^{3/2}}{\exp(\varepsilon / k_B T) + 1} \, \mathrm d\varepsilon$$
which is like your expression but with $x = E = \varepsilon$.

In short, I'm not sure you have posted the correct integral so maybe you can post the exact integral you are trying to do.

To CompuChip: Thanks. Now I've corrected the expression ( Not exactly like yours. ) The question is more clear now.

 

[CompuChip]

Hmm, a cubic? Interesting.
I asked Wolfram Alpha and it says that the integral can be done, but it involves some ugly polylog functions.

I think you would have better luck trying contour integration. You can get rid of the constant:
$$I = \int_0^\infty \frac{x^3}{\exp(x) + 1} \, dx$$
which has a pole at $x = i\pi$. I haven't done contour integrals in a while so I would have to work this out, but usually you would integrate over a quarter circle with radius R and then down the imaginary axis, going around the pole. Putting in the integration boundaries in Wolfram Alpha gives a nice answer involving pi, so this is probably going to work :)

Would that help you?

PS I also found this page which does not help you in this case, but I thought the trick was cool so I'd share it.

 

[CompuChip]

Hmm, actually the "nice" result of $\pi^2/12$ was for x/(exp(x) + 1), so the "linear" variant. I have to leave now, but maybe you can work that out by hand and see if you can generalize to higher powers, either by doing the same computation, or by some differentiation trick.

 

[Avodyne]

$$\int_0^\infty \frac{x^n}{e^x+ 1} dx = (1-2^{-n})\,\Gamma(n{+}1)\,\zeta(n{+}1)$$
for n>-1, where Γ is the Euler gamma function and ζ is the Riemann zeta function.