Specific heat capacity of a 2D free electron gas I have got so far the 2D density of states as $g(\epsilon)=\frac{Am}{\pi\hbar^2}$ where $A$ is the area of the "square" and $m$ is the the electron mass. Then I have found an expression for the the chemical potential for the gas using:
$N={\int_0^\infty}g(\epsilon)n_Fd\epsilon$,
 where $n_F$ is the fermi distribution function. This spits out $N=\frac{Am}{\pi\hbar^2}k_BT\ln(1+e^{\frac{\mu}{k_BT}})$,
 and taking the limit of $T$ goes to zero we get $\epsilon_F=\frac{N\pi\hbar^2}{Am}$, so $\mu=k_BT\ln(e^{\frac{\epsilon_F}{k_BT}}-1)$. This is where I start to get a little stuck. I am thinking I should next work out $<E>=\int_0^\infty{n_F}{\epsilon}g(\epsilon){d}\epsilon$ However I think this is going down the wrong route (or I can't get the algebra right) as I have a hint to use the result $\int_{-\infty}^{\infty}dx\frac{x^2e^x}{(e^x+1)^2}=\frac{\pi^2}{3}$. If anyone has a hint on where I should be going then it'd be much appreciated!
 A: The procedure to obtain the specific heat of a free electron gas in arbitrary dimension is related to the so-called Sommerfeld expansion that is applied to integrals of the form
$$I(\mu,T)=\int_{-\infty}^{+\infty} d\epsilon \, f(\epsilon) n_F(\epsilon) $$
where $n_F(\epsilon)$ is the Fermi-Dirac distribution and $f(\epsilon)$ is an arbitrary function which vanishes at $\epsilon \rightarrow -\infty$ and diverges as a power law when $\epsilon \rightarrow +\infty$. 
This technique uses the fact that the derivative of the Fermi function is peaked around the chemical potential $\mu$ and broadened by $k_BT$ [and indeed collapses to $-\delta(\epsilon-\mu)$ at $T=0$]. Therefore, if the function behaves as above, which is true for $f(\epsilon)=\epsilon g(\epsilon$), we can integrate by parts to get
$$\int_{-\infty}^{+\infty} d\epsilon \, f(\epsilon) n_F(\epsilon)=-\int_{-\infty}^{+\infty} d\epsilon \, F(\epsilon) \dfrac{\partial n_F}{\partial \epsilon}$$
where
$$ F(\epsilon)=\int_{-\infty}^{\epsilon}d\epsilon' f(\epsilon')$$
and expand $F(\epsilon)$ around $\epsilon=\mu$ in Taylor series
$$F(\epsilon)=F(\mu)+\sum_{k=1}^{+\infty} \dfrac{F^{(k)}(\mu)}{k!}(\epsilon-\mu)^k$$
with $F^{(k)}(\mu)$ being the $k$-th derivarive of the function $F$. Observing that $(-\partial n_F/\partial \epsilon)$ is normalized to unity and that is an even function (so that only terms even in $k$ contribute) we can go back to the original function $f(\epsilon)$ and using $x=\beta(\epsilon-\mu)$
$$I(\mu,T)=\int_{-\infty}^{\mu}d\epsilon f(\epsilon)+\sum_{k=1}^{+\infty} \dfrac{f_k}{\beta^{2k}}\dfrac{d^{2k-1}}{d\epsilon^{2k-1}}f(\epsilon)\Bigg|_{\epsilon=\mu}$$
with coefficients
$$f_k=\int_{-\infty}^{+\infty} dx \dfrac{x^{2k}}{(2k)!}\left(-\dfrac{d}{dx} \dfrac{1}{e^x+1}\right)dx$$
Using this you should not have any problem to find the mean kinetic energy for the free 2DEG and from there the specific heat.
