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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?

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    $\begingroup$ You can't use the high-temperature approximation? $\endgroup$ – HiddenBabel May 25 at 1:59
  • $\begingroup$ Yes, but somehow I am still having trouble solving it. Do you know if what I posted is correct so far? $\endgroup$ – Nick Heumann May 25 at 2:01
  • $\begingroup$ 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 $\endgroup$ – HiddenBabel May 25 at 2:13
  • $\begingroup$ Thank you! I will try that approximation and see what I get $\endgroup$ – Nick Heumann May 25 at 2:24
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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.

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