Passing from a discrete summation to a continous integal

I'm having trouble understanding the math behind a step in an explanation of BCS theory.

At one point the superconductor gap $\Delta$ is defined as

\begin{equation} 1 = V \sum_q \frac{1}{\xi_q^2+\Delta^2}, \end{equation} where $V$ is the potential and $\xi_q \equiv \epsilon_q - \mu$. The tricky part is when the summation is then recast as an integral:

\begin{equation} \boxed{\sum_q \frac{1}{(\epsilon_q - \mu)^2+\Delta^2} \mapsto \int \frac{1}{\sqrt{(\epsilon - \mu)^2+\Delta^2}}\color{red}{\rho(\epsilon)}d\epsilon.} \end{equation}

I didn't write the extrema for summation / integration for clarity, I don't think they matter in this instance. I am missing two points:

• How do we justify going from a summation to an integral in this case?
• Most importantly: why does the density of states $\rho(\epsilon)$ appear? I would (intuitively) say that the same integral without the density of states would look right.

In a way this looks more like a math problem than a physics problem, but I chose to post it here since there could also be some physical argument that comes into play for its solution.

• hint: $\mathrm dk=\frac{\mathrm dk}{\mathrm d\epsilon}\mathrm d\epsilon$. – AccidentalFourierTransform Dec 6 '16 at 18:24
• The index of the sum being $q$ (which is just an index for $k$ as in "wavevector $k$") I would get an integral in $dk$. So in order to get an integral over $d\epsilon$ I have to multiply it by $dk/d\epsilon$ which is just a density of states, is this it? – skyfish Dec 6 '16 at 18:46
• If the $\epsilon_q$ are getting dense in $\mathbb{R}$ (and hence necessarily not ordered), which is almost the only thing that seems to give sense to passing to the integral formulation, then some weight is missing in the sum, else it gets infinite when $q \to \infty$. At this point it would be helpful that you do write the extrema for summation / integration for clarity and further discussion. Nevertheless, if there are indeed some weight missing in the sum, those will naturally transform into the $\rho(\epsilon)$ density. – user130529 Dec 6 '16 at 21:05
• I think that energies $\epsilon$ would be indeed weighted, by the Fermi-Dirac distribution, such that the sum of all the probabilities of every energy is equal to 1 – skyfish Dec 6 '16 at 21:14
• OK, so then you have your missing term. – user130529 Dec 6 '16 at 21:18