This is an exam question that I'm trying to solve. Here's the text:
A system is composed of N localised, but independent one-dimensional classical oscillators. Assume that the potential energy for an oscillator contains a small anharmonic term
$$ V(x) = \frac{k_0x^2}{2} + \alpha x^4 $$
where $\alpha < x4 << kT$. Write down an expression for the Canonical partition function for this system of oscillators. Then, to first order in the parameter $\alpha$ ($\alpha > 0$), derive an expression for the internal energy and the isochoric heat capacity for this system and show that the anharmonic correction tends to reduce the energy per oscillator compared to the equipartition result of a perfectly harmonic oscillator. Explain also why letting $T \rightarrow \infty$ in the expression for the heat capacity represents an unphysical result.
Attempted solution
The classical canonical partition function, $Q$, in this case will be
$$ Q = \iint \limits_{-\infty}^{\infty} e^{-\left( \frac{k_0}{2}x^2 + \alpha x^4 + \frac{p^2}{2m} \right)\frac{1}{kT}} \text{d}p\text{d}x $$
Now comes the difficult part. I have tried to find the Maclaurin series of $V(x)$ around $\alpha = 0$ to first order, but this just returns the original expression (weird, huh?). Then I tried to evaluate the integrals as they stand above, but this seems to not be possible:
\begin{align} Q &= \iint \limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} e^{-\frac{\alpha x^4}{kT}} e^{-\frac{p^2}{2mkT}} \text{d}p\text{d}x \\ &= \int \limits_{-\infty}^{\infty} e^{-\frac{p^2}{2mkT}} \text{d}p \int \limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} e^{-\frac{\alpha x^4}{kT}} \text{d}x \end{align}
The first integral I can do, but not the second. However, if I use that $a < x^4 << kT$, which logically leads to that $ax^4 << kT$, then I perhaps could simplify the second integral by using that $e^{-\frac{ax^4}{kT}} \approx e^{-0} = 1$. But this totally neglects the anharmonic term, which clearly I should draw some conclusions about later on.
Instead, the trick I think is to do the correct Maclaurin expansion. Should I expand the expression for $Q$? Using
$$ Q \approx Q \rvert_{\alpha = 0} + a\frac{\partial Q}{\partial a}\rvert_{\alpha = 0} $$
results in
\begin{align} Q &\approx \iint \limits_{-\infty}^{\infty} e^{-\left( \frac{k_0x^2}{2kT} + \frac{p^2}{2mkT} \right)} \text{d}p\text{d}x + a\left[ \frac{\partial}{\partial a} \left( \iint \limits_{-\infty}^{\infty} e^{-\left( \frac{k_0x^2}{2kT} + \frac{p^2}{2mkT} \right)} \text{d}p\text{d}x \right) \right] \\ &= \int \limits_{-\infty}^{\infty} e^{-\frac{p^2}{2mkT}} \text{d}p \int \limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} \text{d}x + 0 \end{align}
Since the integrals in the second term do not depend on $a$, the whole thing reduces to $0$, and the anharmonic contribution here is lost. I expected from the start that the anharmonic term would give a contribution to the final result, but it does not when I do it like this. Anyway, continuing, the integrals that are left are easily evaluated, and we get a first order approximation to the canonical partition function
\begin{align} Q &\approx 2kT \sqrt{\frac{m}{k_0 }} \end{align}
where the square root sort of resembles the angular frequency of the oscillator (assuming that $k_0$ is the force constant).
Now, finally, we can start to actually answer the question, and find an expression for the internal energy $U$
$$ U =kT^2 \left( \frac{\partial \ln Q}{\partial T} \right)_{N,V} $$
We need the logarithm of $Q$
$$ \ln Q = \ln T + \ln \left( 2k\sqrt{\frac{m}{k_0}}\right) $$
And then get the expression for $U$
\begin{align} U &= kT^2 \frac{1}{T} \\ &= kT \end{align}
Taking the derivative of $U$ with respect to $T$, should give the isochoric heat capacity:
$$ C_V = k $$
So in this case, the isochoric heat capacity does not depend on $T$ at all, so I can't really evaluate whether the high-temperature limit is meaningful or not. My approach may have been incorrect from the start.
