# Series Expansion of logarithm of integral by Landau and Lifshitz in Statistical Physics (First Part)

Let $$E(p,q) = E_0(p,q) + V(p,q)$$ where $$V$$ represents the small terms. To calculate the free energy of the body we put $$e^\frac{-F}{T} = \int' e^{-\frac{E_0(p,q) + V(p,q)}{T}} d\Gamma \approx \int' e^{-\frac{E_0(p,q)}{T}} \left(1 - \frac{V}{T} + \frac{V^2}{2T^2}\right) d\Gamma.$$ And then "taking logarithms and again expanding in series we have to the same accuracy" the following formula: $$F = F_0 + \int' \left(V - \frac{V^2}{2T}\right)e^{\frac{F_0 - E_0}{T}}d\Gamma + \frac{1}{2T}\left[\int' Ve^{\frac{F_0 - E_0}{T}}d\Gamma \right]^2.$$ I'm having trouble deriving the last formula from the previous. Seems like I would need to expand the logarithm of an integral but I have no idea on how to handle it. I know that $$F = -T \ln \int e^{-\frac{E(p,q)}{T}} d \Gamma$$ holds and I feel like I have to use it. Also note that $$d\Gamma = \frac{dqdp}{(2\pi \hbar)^s}.$$

• In a particular use case, you have to specify $E$ and $V$ and then you can calculate a truncated expansion of the free energy. Do you have a specific use case in mind? If not, I am afraid the formula that you show is as specific as you can get. Commented Aug 1 at 14:37
• I'm having trouble deriving the last formula which should be a consequence of 32.2 Commented Aug 1 at 14:41

Landau and Lifshitz used the usual formal expansion in a small parameter. If you do not know how to expand the logarithm of an integral, then let's introduce convenient notations and expand the logarithm of a series. Namely, we will use $$Z_0 = e^{-F_0/T} = \int' e^{-E_0/T} d\Gamma,$$ $$\langle V\rangle_0 = \frac1{Z_0}\int' V e^{-E_0/T} d\Gamma = \int' V e^{(F_0-E_0)/T} d\Gamma,$$ $$\quad \langle V^2\rangle_0 = \frac1{Z_0}\int' V^2 e^{-E_0/T} d\Gamma = \int' V^2 e^{(F_0-E_0)/T} d\Gamma,$$ where $$E_0\equiv E_0(p,q)$$. Now formula 32.2 can be written as $$e^{-F/T} = e^{-F_0/T}\left(1 - \frac{\langle V\rangle_0}T + \frac{\langle V^2\rangle_0}{2T^2}+\ldots \right)$$ Using the logarithm expansion $$\log(1+x) = x - x^2/2+\ldots$$ and preserving terms up to and including the second order in $$V$$, we obtain $$F = F_0 -T\log\left(1 - \frac{\langle V\rangle_0}T + \frac{\langle V^2\rangle_0}{2T^2}+\ldots \right) = F_0 + \langle V\rangle_0 - \frac1{2T}\left(\langle V^2\rangle_0 - \langle V\rangle_0^2 \right) + \ldots$$ Now the required formula is obtained after substituting the averages $$\langle V\rangle_0$$, $$\langle V^2\rangle_0$$ in the form of integrals.
It's just a use of the Newton-Mercator series, the fact that there are integrals plays no role. When $$x \ll 1$$ you have: $$\ln 1+x \simeq 1 + x - \frac{1}{2}x^2$$ The important thing you should note is that the second and third integrals in 32.2 are small compared to the first one (the whole section relies on the fact that $$V$$ is a small perturbation to the system without $$V$$). Starting from your equation for $$F$$, name the big integral $$Z_0$$ and use common factor, then you can expand in terms of the small parameter $$x$$ (the quotient between the second plus the third integral and $$Z_0$$).
$$F = -T\ln Z_0 \left( 1 + \frac{1}{Z_0}(\text{the other integrals}) \right) = -T\ln Z_0 - T\ln (1+x)$$ $$F \simeq -F_0 - T \left(x - \frac{x^2}{2} \right)$$