How does one define the complex distribution $1/z$? 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.
 A: Let $\phi(x)$ be a suitable test function ($\phi(x)\in C^\infty$ and has compact support on $(-\infty,\infty)$).  
Then, write
$$\begin{align}
\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{1}{x+iy}\,dx&=\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{x-iy}{x^2+y^2}\,dx\\\\
&=\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{x}{x^2+y^2}\,dx-i\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{y}{x^2+y^2} \,dx\\\\
\end{align}$$

To evaluate the first limit, $\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{x}{x^2+y^2}\,dx$, we first write$$\begin{align}\int_{-\infty}^\infty \phi(x)\frac{x}{x^2+y^2}\,dx&=\int_{-\infty}^{-\epsilon}\phi(x)\frac{x}{x^2+y^2}\,dx\\\\&+\int_{-\epsilon}^\epsilon \phi(x)\frac{x}{x^2+y^2}\,dx\\\\&+\int_{\epsilon}^\infty \phi(x)\frac{x}{x^2+y^2}\,dx\tag 1\end{align}$$Inasmuch as $\phi$ is of compact support and $\phi(x)\frac{x}{x^2+y^2}$ is continuous, we have the limits$$\lim_{y\to 0^+}\int_{-\infty}^{-\epsilon} \phi(x)\frac{x}{x^2+y^2}\,dx=\int_{-\infty}^{-\epsilon} \frac{\phi(x)}{x}\,dx \tag2$$and$$\lim_{y\to 0^+}\int_{\epsilon}^\infty \phi(x)\frac{x}{x^2+y^2}\,dx=\int_{\epsilon}^\infty \frac{\phi(x)}{x}\,dx\tag3$$Next, we use Taylor's theorem to write $\phi(x)=\phi(0)+\phi'(0)x+o(x)$ for $x\in [-\epsilon,\epsilon]$.Then,$$\begin{align}\int_{-\epsilon}^\epsilon \phi(x)\frac{x}{x^2+y^2}\,dx&=\color{red}{\int_{-\epsilon}^\epsilon \phi(0)\frac{x}{x^2+y^2}\,dx}+\color{blue}{\int_{-\epsilon}^\epsilon \phi'(0)\frac{x^2}{x^2+y^2}\,dx}+\color{orange}{\int_{-\epsilon}^\epsilon \frac{o(x^2)}{x^2+y^2}\,dx}\\\\
&=\color{red}{0}+\color{blue}{\phi'(0)\left(2\epsilon +2y\arctan\left(\frac{\epsilon}{y}\right)\right)}+\color{orange}{o(\epsilon)}\tag 4
\end{align}$$Letting $y\to 0^+$ in $(4)$ we obtain$$\lim_{y\to 0^+}\int_{-\epsilon}^\epsilon \phi(x)\frac{x}{x^2+y^2}\,dx=2\phi'(0)\epsilon+o(\epsilon) \tag5 $$Finally using $(2)-(5)$, and letting $\epsilon\to 0^+$ yields$$\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{x}{x^2+y^2}\,dx=\lim_{\epsilon\to 0^+}\left(\int_{-\infty}^{-\epsilon} \frac{\phi(x)}{x}\,dx +\int_{\epsilon}^\infty \frac{\phi(x)}{x}\,dx  \right)\equiv \text{PV}\left(\int_{-\infty}^\infty \frac{\phi(x)}{x}\,dx\right)$$
as was to be shown!

To evaluate the second limit, $\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{y}{x^2+y^2} \,dx$, we write
$$\begin{align}
\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{y}{x^2+y^2} \,dx&=\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(yx)\frac{1}{x^2+1} \,dx\tag6
\end{align}$$
whence applying the Dominated Convergence Theorem to $(6)$ yields the coveted result
$$\begin{align}
\lim_{y\to 0^+}\int_{-\infty}^\infty \phi(x)\frac{y}{x^2+y^2} \,dx&=\int_{-\infty}^\infty \lim_{y\to 0^+}\left(\phi(yx)\right)\frac{1}{x^2+1} \,dx\\\\
&=\phi(0)\int_{-\infty}^\infty \frac{1}{x^2+1}\,dx\\\\
&=\pi \phi(0)
\end{align}$$
as expected!

And we are done!
A: $\text{pv}.(\frac{1}{x})$ 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
$$\boxed{\text{pv}.(\frac{1}{x}) = \frac{d}{dx}\log |x| = \lim_{y \to 0^+} \frac{d}{dx}(\log (x+iy)- i \pi 1_{x < 0})= i \pi \delta(x)+\lim_{y \to 0^+}\frac{1}{x+iy}}$$
