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

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 $$\pi\sum_{k}\delta(E-E_k)$$.

• Do you know the definition of Dirac's $\delta$? I mean the real one, which has to do with integrals. – Ivan Neretin Oct 30 at 8:02
• The minus sign should not be there. – md2perpe Oct 30 at 10:07

## 1 Answer

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, $$\frac{\epsilon}{E^2+\epsilon^2} = \pi \, \delta(E), \\ \frac{\epsilon}{(E-E_k)^2+\epsilon^2} = \pi \, \delta(E-E_k), \\ \sum_k \frac{\epsilon}{(E-E_k)^2+\epsilon^2} = \sum_k \pi \, \delta(E-E_k). \\$$

• Your solution is very interesting and you seems to be very intelligent!!! – Unbelievable Oct 30 at 14:45
• Actually I was a bit sloppy here. – md2perpe Oct 30 at 14:54