Is it valid to apply a local harmonic approximation to integrals over all space in thermal expansion?

Question

I´m currently studying solid state physics. I have been using the book Kittel's Introduction to solid state physics for so long, but came up with a question when studying the Chapter 5: Phonons II: Thermal Properties more precisely, the section Thermal Expansion on p.120.

We calculate the average displacement by using the Boltzmann distribution function, which weights the possible values of $x$ according to their thermodynamic probability $$ \langle x \rangle = \frac{ \int_{-\infty}^{\infty} dx \, x \, e^{-\beta U(x)} } { \int_{-\infty}^{\infty} dx \, e^{-\beta U(x)} }, $$ where $\beta \equiv 1 / k_B T$. For displacements such that the anharmonic terms in the energy are small in comparison with $k_B T$, we may expand the integrands as: $$ \int dx \, x \, e^{-\beta U} \approx \int dx \, [e^{-\beta c x^2}] \left(x + \beta g x^4 + \beta f x^5 \right) = \left( \frac{3\pi^{1/2}}{4} \right) \left( \frac{g}{c^{5/2}} \right) \beta^{-3/2} ; $$ $$ \int dx \, e^{-\beta U} \approx \int dx \, e^{-\beta c x^2} = \left( \frac{\pi}{\beta c} \right)^{1/2}. \tag{39} $$

However, I’m a bit confused about the last two approxiamtions. In our problem sessions, our instructor pointed out that this approximation is only valid near $x = 0$, and therefore it might not make sense to apply it to integrals over the full range $ (-\infty, \infty) $.

So my question is:
If the approximation is only valid around $ x = 0 $, how can we correctly compute the average displacement $ \langle x \rangle $ without relying on this expansion over the whole real line?

Edit:

With my question I also refer to the integral $$\int dx \, x \, e^{-\beta U} \approx \int dx \, [e^{-\beta c x^2}] \left(x + \beta g x^4 + \beta f x^5 \right).\tag{1}$$

Now I understand that Equation (39) is valid, but it's still unclear to me what happens with Equation (1).

Our teacher tried to explain this using a simple example: we know that for small $ x $,$ \sin(x) \approx x $. However, if we evaluate the integral $\int_0^{\pi} \sin(x) dx$, we can't use this approximation, because it only holds near $ x = 0 $. I'm not sure if my point is coming across clearly, but my concern is about applying approximations outside of their valid range.

Answer 1

1. OP has a point. Kittel considers the potential $$\beta U(x) ~=~ \beta (cx^2 - gx^3 - fx^4 )~\stackrel{x=\beta^{-1/2}y}{=}~cy^2- \beta^{-1/2}gy^3 - \beta^{-1}fy^4,\tag{38}$$ where $c,f,g>0$. This is an unstable potential. The $x$-integral $$\int_{\mathbb{R}}\! dx~\exp(-\beta U(x))$$ is divergent in it current form.

2. Presumably the potential $$\beta U(x)~\stackrel{x=x_0+\beta^{-1/2}y}{=}~\beta U(x_0)+U^{\prime\prime}(x_0)y^2+{\cal O}(\beta^{-1/2})$$ contains stable higher-order terms that Kittel doesn't write explicitly.

3. In the low temperature limit $\beta\to \infty$, the Gaussian term dominates, cf. the saddle-point approximation.

   [The parameter $\beta$ plays the same role as $1/\hbar$ in the semiclassical approximation, cf. e.g. this related Phys.SE post.]

4. Finally, let us consider finite $\beta$ and address OP's question. Kittel's argument seems to be that the higher-order terms only become dominant when the harmonic approximation/quadratic term $U^{\prime\prime}(x_0)y^2$ is already much bigger than 1, so that these contributions to the Boltzmann factor $\exp(-\beta U(x))$ [and its $x$-integral] is negligible anyway.

Answer 2

The Gaussian factor goes to zero very rapdly as $|x|$ gets large, so you only need the values of the term with $f$ and $g$ for small $|x|$. Therefore it is safe to take the integral over the full range.