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Taylor expansion of an integral (for Thermodynamic Perturbations)

  1. Apr 25, 2012 #1
    In thermodynamic perturbation theory (chapter 32 in Landau's Statistical Physics) for the Gibbs (= canonical) distribution, we have [itex]E = E_0 + V[/itex], where [itex]V[/itex] is the perturbation of our energy.

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

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

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

    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 [itex]-T \log \int e^{E/T} = F[/itex], 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 [itex]f'(a) (x-a)[/itex] for first order, where exactly do I 'put' the [itex]x[/itex]? Inside the integral?
     
    Last edited: Apr 25, 2012
  2. jcsd
  3. Apr 25, 2012 #2
    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 [itex]V_i[/itex], albeit them lying infinitesimally close to each other. Therefore, one could view the expansion with respect to [itex]V[/itex] as expansions with respect to a lot of different [itex]V_i[/itex] in this 'sum'. After this expansion one could imagine making the sum into an integral once more, such that the orders of [itex]V[/itex] are inside the integrals.
     
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