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Consider the Fermi-Dirac expansion for an arbitrary function $f(\epsilon)$:

$I(f)=\int_0^\infty d\epsilon\frac{f(\epsilon)}{e^{\beta(\epsilon-\mu)}+1}$

The large $\beta$ expansion of this quantity is called the Sommerfeld expansion, and is well known to have the following all orders in $1/\beta$ expansion:

$I(f)=\int_0^\mu f(\epsilon)d\epsilon+\sum_{n=1}^\infty\frac{1}{\beta^{2n}}(2-2^{2-2n})\zeta(2n)f^{(2n-1)}(\mu)+O(e^{-\beta \mu})$

Let us now define the analogous 2-dimensional function

$J(f)=\int_0^\infty d\epsilon_1\int_0^\infty d\epsilon_2 \frac{f(\epsilon_1,\epsilon_2)}{e^{\beta(\epsilon_1-\mu)}+1}\frac{1}{e^{\beta(\epsilon_2-\mu)}+1}$

Is there a similar asymptotic expansion for large $\beta$ for this when $f(\epsilon_1,\epsilon_2)\neq g(\epsilon_1)h(\epsilon_2)$ for some functions $g,h$?

Of course in the factorizable case, one can just do the Sommerfeld expansion to each function $g,h$, seperately. But when the function does not factorize, how is this expansion done? For concreteness, we can consider the simple case

$J(|{\epsilon_1-\epsilon_2}|)=\frac{\mu^3}{3}+\frac{\pi^2\mu}{3\beta^2}-\frac{4\zeta(3)}{\beta^3}+O(e^{-\beta\mu})$

where I computed this by doing the integral exactly and then expanding in large $\beta$. How can one recover these perturbative terms from a Sommerfeld-like expansion?

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