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. $$m=(\theta-\text{mod}_{2\pi}(\theta))/2\pi$$
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
$$\frac{d\log(x)}{dx}\sim PV\left(\frac{1}{x}\right)-i\pi \delta(x)$$
Alternatively, let $\psi$ be the ditribution $\psi(x)=\left(\log(x)\right)$. Then, we can write for any $\phi\in C_C^\infty$
$$\langle \psi',\phi\rangle = - \langle \psi, \phi'\rangle$$
Hence, we have
$$\begin{align} -\int_{-\infty}^\infty \log(x)\phi'(x)\,dx&=-\int_{-\infty}^0 (\log(|x|)+i\pi)\phi'(x)\,dx-\int_0^\infty \log(x)\phi'(x)\,dx\\\\ &=-i\pi \phi(0)-\int_{-\infty}^\infty \log(|x|)\phi'(x)\,dx\\\\ &=-i\pi \phi(0)-\lim_{\delta\to 0^+}\int_{|x|\ge\delta} \log(|x|)\phi'(x)\,dx\\\\ &=-i\pi \phi(0)+\lim_{\delta\to 0^+}\int_{|x|\ge\delta}\frac{\phi(x)}{x}\,dx\\\\ &=-i\pi \phi(0)+\text{PV}\int_{-\infty}^\infty\frac{\phi(x)}{x}\,dx\\\\ \end{align}$$
Therefore, we find that in distribution
$$\frac{d}{dx}\log(x)=\text{PV}\left(\frac1x\right)-i\pi \delta(x)$$
as expected!