Classical limit of thermal capacity in Debye Solid

Question

In Debye Solid model thermal capacity with constant numer of particles $N$ (canonical ensemble) is $$C_{N}=\frac{9Nk_B}{(\beta \hbar \omega_D)^3 }\int_{0}^{T_D/T} dx \frac{x^4 e^x}{(e^x-1)^2}\tag{1}$$

Where $k_B$ is Boltzmann constant, $\omega_D$ is Debye frequency, $T_D=\hbar \omega_D/k_B$ is Debye temperature and $\beta=1/k_BT$.

For $T\to 0$ $(1)$ states that $C_N \sim T^3$ and I'm ok with that.

But how can one show that, if $T \to \infty$ (classical limit), then $C_N \to 3k_B N$, as stated in the classical energy equipartition theorem?

Answer 1

The key realization is that for $T \gg T_D$, the range of integration in the integral is only over values that are much less than 1. This means, in particular, that we can approximate $e^x \approx 1 + x$ inside the integral without affecting its value much: $$ \int_{0}^{T_D/T} dx \frac{x^4 e^x}{(e^x-1)^2} \approx \int_{0}^{T_D/T} dx \frac{x^4 (1)}{x^2} = \frac{1}{3} \left( \frac{T_D}{T} \right)^3. $$ This, combined with the missing factor of $T^3$ in your definition of $C_N$, implies that the heat capacity approaches $3 N k_B$ as expected.

Answer 2

Your formula does not look right, for example, a factor of $T^3$ is missing. See, e.g., https://en.wikipedia.org/wiki/Debye_model#Derivation