# 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.

• Also, this result has a name and origins: Sokhotski-Plemelj theorem, from Sokhotski 1871 and Plemelji 1908. Google-able. Also, see "Hilbert transform". Commented Aug 7, 2017 at 13:42

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!

• And to answer one of the OP's main questions, the integral should be over the real line.
– user14972
Commented Aug 6, 2017 at 15:47
• @Hurkyl Indeed; great comment. (+1) Commented Aug 6, 2017 at 15:53
• Great answer! I was expecting just a hint not so much detail, thanks! :) Commented Aug 6, 2017 at 20:46
• @AerinmundFagelson I originally did leave a hint only. But I thought there was quite a bit of detail to fill in ... so I filled it in. ;-) Commented Aug 7, 2017 at 13:25

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

• This is solid (+1) Commented Aug 7, 2017 at 13:24
• Why $-i\pi$ and not $+i\pi$ or $-3i\pi$? Commented Aug 11, 2017 at 20:00
• @md2perpe Because it is the branch of $\log(z)$ analytic on $\Im(z) > 0$ whose derivative is $\frac{1}{x+iy}$ and $\lim_{y \to 0^+} \log(x+iy) = \ln|x|+i\pi 1_{x < 0}$ Commented Aug 12, 2017 at 6:00