a fraction containing a very small value is equal to Dirac's delta function

Question

When $\epsilon$ goes to infinite small value ($\epsilon\rightarrow 0$), how can I show $\sum_{k}\frac{\epsilon}{(E-E_k)^2+\epsilon^2}$ is equal to $\pi\sum_{k}\delta(E-E_k)$.

Answer 1

Distributions, like $\delta$, are formally defined by their behaviour on infinitely differentiable functions with compact support. Therefore we take $\phi \in C_c^\infty(\mathbb{R})$ and evaluate the following integral: $$ \int \frac{\epsilon}{E^2+\epsilon^2} \phi(E) \, dE = \{ E = \epsilon \hat{E} \} = \int \frac{\epsilon}{\epsilon^2 \hat{E}^2+\epsilon^2} \phi(\epsilon\hat{E}) \, \epsilon \, d\hat{E} \\ = \int \frac{1}{\hat{E}^2+1} \phi(\epsilon\hat{E}) \, d\hat{E} \to \int \frac{1}{\hat{E}^2+1} \phi(0) \, d\hat{E} = \int \frac{1}{\hat{E}^2+1} \, d\hat{E} \, \phi(0) \\ = \pi \phi(0) = \int \pi \, \delta(E) \, \phi(E) \, dE. $$ Thus, $$ \frac{\epsilon}{E^2+\epsilon^2} = \pi \, \delta(E), \\ \frac{\epsilon}{(E-E_k)^2+\epsilon^2} = \pi \, \delta(E-E_k), \\ \sum_k \frac{\epsilon}{(E-E_k)^2+\epsilon^2} = \sum_k \pi \, \delta(E-E_k). \\ $$