TITLE: How does one define the complex distribution $1/z$?
QUESTION [6 upvotes]: I have read the following formula in a quantum mechanics book, supposedly attributed to Dirac
$$ \lim_{y\,\searrow\, 0} \frac 1 {x+iy} = \operatorname{p.v.} \left(\frac 1 x\right) - i \pi\delta (x)$$
Could anyone elucidate how this formula should be read (or derived)? Is it analogous to the real case, where we integrate against a test function? How should this integral be taken, as an area integral over the complex plane or as some kind of contour integral?
I would be grateful for any good references on complex distribution theory from a mathematical standpoint. I learnt real distribution theory from Friedlanders book.

REPLY [4 votes]: $\text{pv}.(\frac{1}{x})$ is easily understood as the distributional derivative of $\log |x|$, which means for any $\phi \in C^\infty$ with $\phi,\phi'$ decreasing faster than $1/x^2$ :
$$\int_{-\infty}^\infty \phi(x) \text{pv}.(\frac{1}{x}) dx = -\int_{-\infty}^\infty \phi'(x) \log |x| dx$$
Now
$$\log |x| = \log x- i \pi 1_{x < 0}= \lim_{y \to 0^+}\log (x+iy)- i \pi 1_{x < 0}$$
where the convergence is in $L^1_{loc}$. Therefore, as distributions
$$\boxed{\text{pv}.(\frac{1}{x}) = \frac{d}{dx}\log |x| = \lim_{y \to 0^+} \frac{d}{dx}(\log (x+iy)- i \pi 1_{x < 0})= i \pi \delta(x)+\lim_{y \to 0^+}\frac{1}{x+iy}}$$