Assume $x \in \mathbb{R}$. We already know that
$$\lim\_{\epsilon \to 0+} \frac{1}{x-i\epsilon} - \frac{1}{x+i\epsilon} = 2\pi i \delta\_x.$$
Here $\delta\_x$ denotes the Dirac distribution. If we consider a slightly more irregular limit
$$\lim\_{\epsilon \to 0+}\frac{\log^m(x-i\epsilon)}{x-i\epsilon} - \frac{\log^m(x+i\epsilon)}{x+i\epsilon},$$
what will be the result distribution? Here $m \ge 1$ is an integer.



What is the distribution of the following limit?
This question already has an impeccable answer but I would like to give a second one in the hope that it might be of interest to the OP. Before so doing, I will list the properties of distributions (on the line) that I require:


1. Any “reasonable” function can be regarded as a distribution (in our context, this means locally integrable);
2. Any “reasonable” notion of convergence for functions is stronger than distributional convergence (in our context, local $L^1$-convergence);
3. Any distribution is differentiable and we can always interchange the operations of taking limits and distributional convergence in a sequence of distributions.


I will start with the (familiar) case $m=0$ of the required equation since this displays the essence of the problem. We choose the branch of the logarithm function (on the plane minus the negative real axis) $\ln z= \ln |z| + i \arg z$ with the argument function ranging from $]-\pi,\pi[$.


Thus for positive $\epsilon$, $$\ln (x+ i \epsilon)= \ln |x+i\epsilon|+i\arg (x +i\epsilon).$$ If we substract this from the corresponding equation with $-\epsilon$, differentiate with respect to $x$ and take the limit we get the required expression on the LHS. On the right hand side, we first take the limit, which is $-2\pi i H (-x)$ (with $H$ the Heaviside function), and THEN differentiate. This gives the required equation.


The more general case can be proved similarly—starting from the equation
$$\ln^m(x+i\epsilon)=(\ln(|x+i\epsilon|)+i \arg(x+i\epsilon)^m,$$ subtracting this from the version for
$-\epsilon$, differentiating and taking the limit on the LHS, resp., taking the limit and then differentiating on the RHS.


Added as an edit: I should mention that I learned this proof during a lecture course on the elementary theory of distributions given by J. Sebastião e Silva in Lisbon over 50 years ago.


