# Derivative of the Logarithm - Dirac

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: $$\int_{-a}^{a}\frac{d}{dx} \log(x)\,dx=\int_{-a}^{a}\frac{1}{x}-i\pi\delta(x)dx$$ $$\log(-1)=-i\pi \bigg| \exp(.)$$ $$e^{\log(-1)}=e^{-i\pi}$$ $$-1=e^{-i\pi}$$

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: $$e^{i\theta}=\cos(\theta)+i\sin(\theta)+\hat{j}m$$ Where m indicates the mth Riemann surface and j is kind of a vector basis. $$m=(\theta-\text{mod}_{2\pi}(\theta))/2\pi$$

• What do you actually want to know? Jan 26 '16 at 22:45
• I would guess the question here is: the formulas Dirac writes are wrong in mathematics as conventionally understood. Is there a way to make sense of them in terms of conventional mathematics, perhaps one that involves Riemann surfaces? Jan 26 '16 at 23:33
• mathoverflow.net/questions/127601/… Aug 1 '16 at 1:55

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

$$\frac{d\log(x)}{dx}\sim PV\left(\frac{1}{x}\right)-i\pi \delta(x)$$