TITLE: Does the derivative of log have a Dirac delta term?
QUESTION [15 upvotes]: Dirac writes down the following formula on page 61 of his "Principles of quantum mechanics":
$\frac{d}{dx}\log x = \frac{1}{x} -i\pi\delta(x)$, see http://adsabs.harvard.edu/abs/1947pqm..book.....D for the exact reference (but no text). What is the best way of formalizing this to a mathematician's satisfaction?

REPLY [5 votes]: Here is another context where something like this shows up, though I don't know how or if it's related to your question. Consider the simple left $\mathscr{D}$ module on $\mathbf{A}^1$ generated by $\text{log}x$, and similarly for $1/x$ and $\delta(x)$
$$\mathscr{D}\cdot\text{log}x\ =\  \mathscr{D}/\mathscr{D}\partial x\partial \ = \ k[x]\log x +k[x,x^{-1}],$$
$$\mathscr{D}\cdot 1/x\ =\  \mathscr{D}/\mathscr{D}\partial x \ = \ k[x,x^{-1}],$$
$$\mathscr{D}\cdot\delta(x)\ =\  \mathscr{D}/\mathscr{D}x \ = \ k[\partial]\delta(x)$$
where the global sections of $\mathscr{D}$ is the algebra $k[x,\partial]$ with $[x,\partial]=1$. There is an exact sequence of $\mathscr{D}$ modules
$$0 \ \longrightarrow \ \mathscr{D}\cdot 1/x \ \longrightarrow \ \mathscr{D}\cdot\log x\ \longrightarrow \ \mathscr{D}\cdot \delta(x)\ \longrightarrow \ 0$$
which in plain language just means that the subspace $k[x,x^{-1}]$ is preserved by $x$ and $\partial$.
Note that to get this $``\log=1/x+\delta"$ principle to work you really have to work with the whole $\mathscr{D}$ modules, not just consider elements of them. Indeed, the element $\log x\in\mathscr{D}\cdot\log x$ just satisfies $\partial \log x=1/x$.