# Taylor expansion of an integral (for Thermodynamic Perturbations)

1. Apr 25, 2012

### mSSM

In thermodynamic perturbation theory (chapter 32 in Landau's Statistical Physics) for the Gibbs (= canonical) distribution, we have $E = E_0 + V$, where $V$ is the perturbation of our energy.

When we want to calculate the free energy, we have:
$$e^{-F/T} = \int e^{-(E+V)/T} \mathrm{d}\Gamma$$

We can expand the expression inside the integral for small values of $V$ to second order we obtain:
$$e^{-F/T} = \int e^{E/T} (1 - V/T + (V/T)^2) \mathrm{d}\Gamma$$

Now, the next step is to logarithmize this expression and then expand it in terms of $V$ again, to second order:
$$F = -T \log \bigl( \int e^{E/T} (1 - V/T + (V/T)^2) \mathrm{d}\Gamma \bigr)$$

But now I am stuck. I don't understand how to expand this in a series. The zeroth order still works great, and I obtain $-T \log \int e^{E/T} = F$, which is just what I want. But from the first order onwards I am stuck:
What is the rule for Taylor expanding an integral? Especially, e.g., if I have $f'(a) (x-a)$ for first order, where exactly do I 'put' the $x$? Inside the integral?

Last edited: Apr 25, 2012
2. Apr 25, 2012

### mSSM

So, I thought about this a little with a friend. I believe that I could see the integral as a sum, with a lot of different $V_i$, albeit them lying infinitesimally close to each other. Therefore, one could view the expansion with respect to $V$ as expansions with respect to a lot of different $V_i$ in this 'sum'. After this expansion one could imagine making the sum into an integral once more, such that the orders of $V$ are inside the integrals.