TITLE: Derivative of the Logarithm - Dirac
QUESTION [6 upvotes]: So I stumbled across P.Dirac's book Principles of Quantum Mechanics and I found something really peculiar on page 61 of the Fourth Edition.
He states that usually we accept that $$\frac{d}{dx}\log(x)=\frac
{1}{x}$$ for real positive x.
However the author states that it can be extended to $$\frac{d}{dx}\log(x)=\frac
{1}{x}-i\pi\delta(x)$$
He argues the following way:
\begin{equation}
\int_{-a}^{a}\frac{d}{dx} \log(x)\,dx=\int_{-a}^{a}\frac{1}{x}-i\pi\delta(x)dx
\end{equation}
\begin{equation}
\log(-1)=-i\pi \bigg| \exp(.)
\end{equation}
\begin{equation}
e^{\log(-1)}=e^{-i\pi}
\end{equation}
\begin{equation}
-1=e^{-i\pi}
\end{equation}
Intuitively I think it has to do with Riemann surfaces and I think one could somehow using this extend the definition of complex exponentials and write them the following way:
\begin{equation}
e^{i\theta}=\cos(\theta)+i\sin(\theta)+\hat{j}m
\end{equation}
Where m indicates the mth Riemann surface and j is kind of a vector basis.
$$m=(\theta-\text{mod}_{2\pi}(\theta))/2\pi$$

REPLY [5 votes]: We will define the distribution $\frac{d\log(x)}{dx}$ through the regularization given by
$$\frac{d\log(x)}{dx} \sim \lim_{\epsilon \to 0^+}\left(\frac{d\log(x+i\epsilon)}{dx}\right)$$
Then, for any $a<0<b$ and smooth test function $\phi(x)$, we have
$$\begin{align}
\int_a^b \frac{d\log(x)}{dx}\phi(x)\,dx&=\lim_{\epsilon \to 0^+}\int_a^b \left(\frac{d\log(x+i\epsilon)}{dx}\right)\phi(x)\,dx\\\\
&=\lim_{\epsilon \to 0^+}\int_a^b\left(\frac{1}{x+i\epsilon}\right)\phi(x)\,dx\\\\
&=\lim_{\epsilon \to 0^+}\left(\int_a^b\left(\frac{x}{x^2+\epsilon^2}\right)\phi(x)\,dx-i\int_a^b \left(\frac{\epsilon}{x^2+\epsilon^2}\right)\phi(x)\,dx\right)\\\\
&=\lim_{\epsilon \to 0^+}\left(\int_a^b\left(\frac{x}{x^2+\epsilon^2}\right)\phi(x)\,dx\right)-i\pi \phi(0)\\\\
&=PV\left(\int_a^b \frac{\phi(x)}{x}\,dx\right)-i\pi \phi(0)
\end{align}$$
where $PV$ denotes the Principal Value of the integral.
Therefore, we find that in distribution 
$$\frac{d\log(x)}{dx}\sim PV\left(\frac{1}{x}\right)-i\pi \delta(x)$$