Solving a problem in statistical mechanics, I obtained the following expression:


where $k_B$ is the Boltzmann constant, $\theta_D$ is the Debye temperature and $N$ the number of particles.

I wish to evaluate at high and low temperature limits, but I am running intro trouble. For $T\gg\theta_D$, the upper limit of the integral becomes 0, and thus the first term is 0. Then, the second term is divided by $T$, but $T\to \infty$, so it is 0.

How can this be? What am I doing wrong here?

In the second case, for $T\ll\theta_D$, the upper limit of the first integral goes to infinity, and thus the contribution of the first term is of the order $T^3$. However, the second therm is divided by $T$, which is almost 0 and an exponential that is infinite. What can I do?

I'm sure I am doing something very wrong, but I cannot figure out what

  • $\begingroup$ You might get more traction with this on Mathematics SE (and I'd suggest removing the parts not related to the integral). The integral is a Debye function. That WIkipedia link has some limiting information you shoudl eb able to apply to the high temperature region. $\endgroup$ May 26, 2019 at 22:54
  • $\begingroup$ For the high temperature limit you can do an expansion and for the low temperature limit you can use the Riemann zeta function. $\endgroup$
    – Silas
    Dec 18, 2022 at 16:14

2 Answers 2


If upper integration limit is small you can expand the integrand around x=0 and keep only the first order terms (x does not go too far from zero). You will end up with an easy to do integral and the result will be some power of $ (\theta_D/T) $ . For the second term, you can use the same approximation so the denominator will be just $ \theta_D $

For very high temperatures you could neglect the second term, I believe. It goes like $ x e^{-x} $ where $x =\theta_D/T $ when x goes to zero. Or you can expand it around x=0 and keep the linear term only.


Let us call the ratio $\theta_D/T$ to be $t$. Then your expression becomes $36Nk_b\frac{1}{t^3}\int_{0}^{t}dx\frac{x^3}{e^x-1}-9Nk_b\frac{t}{e^t-1}$. In the limit $t \to 0$, the integral becomes $\frac{t^4}{e^t-1}$, giving the expression $27Nk_b \lim_{t \to 0}\frac{t}{e^t-1}$, which can be evaluated by L'Hospital's rule to be 1.

In the low-temperature limit for the second term, you can see that $T(e^{\theta_d/T}-1)= T(\Sigma_{n=1}^\infty(\theta_D/T)^n)= \theta_D + \infty$ giving infinity, thus making the second term 0.


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