What "goes wrong" in a Sommerfeld expansion?

Question

Let $f$ be the Fermi function and $H$ be a function which which vanishes as $\epsilon \to -\infty$ and which diverges at $\infty$ no worse than some power of $\epsilon$. In the Sommerfeld expansion of solid state physics (see e.g. Ashcroft and Mermin Appendix C), one writes an integral of the form $$\int_{-\infty}^\infty H(\epsilon)f(\epsilon)\, d\epsilon$$ instead as, for some $\mu \in \mathbb{R}$, $$\int_{-\infty}^\infty H(\epsilon)f(\epsilon)\, d\epsilon = \int_{-\infty}^\mu H(\epsilon)f(\epsilon)\, d\epsilon + \sum_{n=1}^{\infty}\int_{-\infty}^\infty \frac{(\epsilon - \mu)^{2n}}{2n!}(-\partial f/\partial \epsilon)\, d\epsilon \left(\frac{d^{2n-1}H}{d\epsilon^{2n-1}}\right)(\mu).$$ The key step is doing a Taylor expansion of $K(\epsilon) := \int_{-\infty}^\epsilon H({\epsilon'})\,d\epsilon'$ about $\epsilon = \mu$ and then exchanging the infinite sum and integral.

My question is, when is this exchange justified? In particular, $H$ usually has some sort of lack of smoothness at $\epsilon = 0$ (e.g. in 2D, $H$ can be a positive constant for $\epsilon>0$ and $0$ for negative numbers). I would have thought that the expansion would be rigorously valid as long as we restrict ourselves to $\mu >0$ (since usually $H = 0$ for $\epsilon<0$ so that $\int_{-\infty}^\infty H(\epsilon)f(\epsilon)\, d\epsilon = \int_{0}^\infty H(\epsilon)f(\epsilon)\, d\epsilon$ anyway), and yet I have just worked a problem which shows that this is not so; that somehow the exchange "knows about" the singularity at $0$ in spite of the fact that $\epsilon < 0$ "doesn't matter". I'm hoping someone can help me understand what's going on.

Further, some of my confusion lies in that, in 3D (where $H$ is not differentiable) the expansion works whereas in 2D (where $H$ is not continuous) it does not work. Does the fact that the expansion is valid to one extra term in the 3D case really change things that materially?

Answer 1

For a while I kept thinking that you were asking something about Sommerfeld expansion. Then I realised that you are actually asking something about mathematics, really.

https://math.stackexchange.com/questions/83721/when-can-a-sum-and-integral-be-interchanged

I think it is important to point out an important fact about the Sommerfeld expansion. It is known that the Sommerfeld expansion is an asymptotic expansion. It has a radius of convergence of zero. It behaves a little like the renormalisation problem as in QFT. But you can easily sidestep this problem because you can do a variant of this that would not have the problem---manipulate the integrals analytically exactly until you have the same initial terms and the final small remainder term is an integral, and just evaluate this last integral numerically.

Note, however, that just because it is looks mathematically nonsense to do such asymptotic expansions, it is actually perfectly fine. These approximations are highly motivated by physics, and so the maths would work hard to give us something sensible and incredibly accurate, even though the expressions may be formally silly.