I have encountered the following representation for the Dirac delta function in a book, $$-2\pi i\delta(x-x') = \lim_{\epsilon \rightarrow 0}\left\{\frac{1}{x'-x + i\epsilon} + \frac{1}{x-x' +i\epsilon}\right\}.$$
I guess this make some intuitive sense when looking at the behaviour of the function on the right, but the fact that this gives a purely imaginary delta function confuses me. Also I wonder if the limit here is actually required. Can't one just substitute $\epsilon = 0$ and the representation still holds?
I can't seem to find any information on this representation outside of this book. Can anyone elaborate on where this representation comes from or maybe point me to a relevant text?
Thanks a lot!
Let $C(x)=\dfrac{1}{\pi}\dfrac{1}{1+x^2}$ (Cauchy function).
[Please note that $\int_{-\infty}^{+\infty} C(x)dx=1$].
Giving a common denominator to the right hand side of the equation expressing the limit, one gets the equivalent formulation:
$$2i \delta(x-x') \ = \ \lim_{\varepsilon \to 0}\ 2i \varepsilon C(\varepsilon (x-x'))$$
(the $2i$ cancel, of course) which is indeed true because the spike present in Cauchy's function at the origin will narrow while becoming higher and higher, the area being kept constant (equal to $1$, whatever $\varepsilon$), this area condition warranting that we have indeed a Dirac $\delta$ at the origin.
Edit: A different mode of derivation exists if you happen to know the existence and properties of distributions $\dfrac{1}{x+i0}$ and $\dfrac{1}{x-i0}$ different from "principal value" distribution $PV(\dfrac{1}{x})$ but connected to it by formula
$$\dfrac{1}{x+i0}=PV(\dfrac{1}{x})−i\pi\delta$$
and a similar one for $\dfrac{1}{x-i0}$ that I let you discover... (See page 3 of ueltschi.org/teaching/2012-MA433/distributions.pdf). An example of application of these distributions can be found here.