Writing $1/(X+i\epsilon)$ in terms of principal value and imaginary part

I am looking to derive the relation $$\frac{1}{X + i\delta} = \text{P.V} \frac{1}{X} - i \pi \delta(X)$$ In particular, I don't see where the factor of $\pi$ comes from in the derivation. I proceed by computing the imaginary part of the l.h.s as a discontinuity: Take $X = 1-zt$, then, in complex $z$ plane, $$\text{Disc}_z \frac{1}{(1-zt)} = \text{lim}_{\epsilon \rightarrow 0 } \left( \frac{1}{(1-(z+i\epsilon)t)} - \frac{1}{(1-(z-i\epsilon)t)}\right)$$ $$= \text{lim}_{\epsilon \rightarrow 0 } \left( \frac{1}{(1-zt -i\epsilon)} - \frac{1}{(1-zt+i\epsilon)}\right) = \text{lim}_{\epsilon \rightarrow 0 } \frac{2i \epsilon}{(1-zt)^2 + \epsilon^2} = 2i \delta(1-zt)$$ The imaginary part is therefore $i \delta(1-zt)$ so I am off from the actual result by a minus and a factor of $\pi$. Can anyone see where I went wrong?

Thanks!

• ur last equality is not true, there should be a pi due to normailzation constraints – tired Mar 14 '16 at 9:20
• @tired: Many thanks! Could you elaborate on what these normalisation constraints are? – CAF Mar 14 '16 at 9:22

The best strategy is to rationalize $$\frac{1}{X + i\epsilon} = \frac{X-i \epsilon}{(X+i\epsilon)(X-i\epsilon)}=\frac{X}{X^2+\epsilon^2}+i\frac{-\epsilon}{X^2+\epsilon^2}\ .$$ The imaginary part is a nascent delta function, as the limit of a Lorentzian with an 'infinitely sharp' peak (see http://mathworld.wolfram.com/DeltaFunction.html) $$\lim_{\epsilon\to 0^+}\frac{\epsilon}{\pi (X^2+\epsilon^2)}=\delta(X)\ .$$ The factor of $$\pi$$ is due to normalization: $$\int_{-\infty}^\infty \delta(x-X)dX=1$$ by definition for all $$x$$, and the integral of the Lorentzian is normalized to $$1$$ only with that factor of $$\pi$$. See also https://en.wikipedia.org/wiki/Sokhotski%E2%80%93Plemelj_theorem