# Calculating Total energy of 2D Debye monoatomic solid

I am trying to find the total energy of a mono-atomic 2D Debye solid. I started with the density of states: $$D(\omega)=\frac{A\omega}{\pi c^2}$$ where A is the area, $$\omega$$ the frequency and c the speed of sound.

Solving the following for the frequency of Debye: $$\int_0^{\omega_D} D(\omega)d\omega=2N$$ I obtained $$\omega_D=2c\sqrt{\frac{N\pi}{A}}$$ Now to obtain the total energy, I must integrate the following expression:

$$E= \int_0^{\omega_D}\frac{A\omega}{\pi c^2} \frac{\hbar\omega}{e^{\frac{\hbar\omega}{k_B T}} -1} d\omega$$

I use the change of variables $$x=\frac{\hbar\omega}{k_B T}$$ and $$dx=\frac{\hbar d\omega}{k_B T}$$

I finally arrive at $$E=\frac{A(k_BT)^2}{\pi c^2\hbar^2} \int_0^{\frac{\hbar \omega_D}{k_BT}} \frac{x^2}{e^x -1} dx$$

And I don't know hot to solve this. If the upper bound were infinity, it would be a know integral. How could I proceed?

• You can't use the high-temperature approximation? – HiddenBabel May 25 '19 at 1:59
• Yes, but somehow I am still having trouble solving it. Do you know if what I posted is correct so far? – Nick Heumann May 25 '19 at 2:01
• Well, I'm uncertain about the derivation in 2D, but $e^x - 1 \approx x$ in the high temp. limit. The actual value of that integral is called the Debye function en.wikipedia.org/wiki/Debye_function – HiddenBabel May 25 '19 at 2:13
• Thank you! I will try that approximation and see what I get – Nick Heumann May 25 '19 at 2:24

I think this might be a good idea to expand the $$1/e^x-1$$ as GP SERIES and then you can do individual $$x^2e^{nx}$$. I think it should work. Just like we do for packs formula for radiation.