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

$\begingroup$ WolframAlpha suggests it will be the same: wolframalpha.com/input/… $\endgroup$ – Matt F. May 9 at 0:34
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:
Any “reasonable” function can be regarded as a distribution (in our context, this means locally integrable);
Any “reasonable” notion of convergence for functions is stronger than distributional convergence (in our context, local $L^1$convergence);
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.
$\newcommand{\ep}{\epsilon}\newcommand{\de}{\delta}\newcommand{\R}{\mathbb{R}}$Let \begin{equation*} K_\ep(x):=\frac{\ln^m(xi\ep)}{xi\ep}\frac{\ln^m(x+i\ep)}{x+i\ep}. \end{equation*} Let us show that, in an appropriate weak sense, \begin{equation*} \frac{K_\ep}{\ln^m\ep}\to2\pi i\de_0 \tag{1} \end{equation*} (as $\ep\to0$).
Take any bounded continuous function $g\colon\R\to\R$. Then \begin{align*} &\int_\R dx\,K_\ep(x)g(x)=\int_\R du\,[L(ui)L(u+i)]g(\ep u), \end{align*} where \begin{equation*} L(z):=\frac{(\ln\ep+\ln z)^m}z=\sum_{k=0}^m\binom mk \ln^k\ep\,\frac{\ln^{mk}}z. \end{equation*} So, \begin{align*} &\int_\R dx\,K_\ep(x)g(x)=\sum_{k=0}^m\binom mk \ln^k\ep\; I_{mk}(\ep), \tag{2} \end{align*} where \begin{equation*} I_j(\ep):=\int_\R du\,\Big(\frac{\ln^j(ui)}{ui}\frac{\ln^j(u+i)}{u+i}\Big)g(\ep u). \end{equation*} For each $j=0,1,\dots,m$ and all real $u$ with $u\ge2$, \begin{equation*} \ln^j(u\pm i)\ln^j u=\int_0^1 i\,dt\,j\frac{\ln^{j1}(u\pm ti)}{u\pm ti}\ll \frac{\ln^{j1}u}{u}, \end{equation*} where $a\ll b$ means $a=O(b)$; here and in what follows, the constant in $O(\cdot)$ may depend only on $g$ and $m$. So, \begin{align*} &\int_{u\ge2} du\,\Big(\frac{\ln^j(ui)}{ui}\frac{\ln^j(u+i)}{u+i}\Big)g(\ep u) \\ &=\int_{u\ge2} du\,\ln^j u\,\Big(\frac{1}{ui}\frac{1}{u+i}\Big)g(\ep u)+O(1) \\ &=\int_{u\ge2} du\,\frac{2i\ln^j u}{u^2+1}\,g(\ep u)+O(1) =O(1). \end{align*} Also, clearly, \begin{equation*} \int_{u<2} du\,\Big(\frac{\ln^j(ui)}{ui}\frac{\ln^j(u+i)}{u+i}\Big)g(\ep u)=O(1). \end{equation*} So, \begin{equation*} I_j(\ep)=O(1) \tag{3} \end{equation*} for each $j=0,1,\dots,m$.
Moreover, \begin{equation*} I_0(\ep)=2i\int_\R \frac{du}{u^2+1}g(\ep u), \end{equation*} \begin{equation*} \int_{u\le1/\sqrt\ep} \frac{du}{u^2+1}g(\ep u) =\int_{u\le1/\sqrt\ep} \frac{du}{u^2+1}(g(0)+o(1))\to\pi g(0), \end{equation*} \begin{equation*} \int_{u>1/\sqrt\ep} \frac{du}{u^2+1}g(\ep u) \ll\int_{u>1/\sqrt\ep} \frac{du}{u^2+1}\to0. \end{equation*} So, \begin{equation*} I_0(\ep)\to 2i\pi g(0). \tag{4} \end{equation*}
Collecting (2), (3), and (4), we get \begin{align*} &\frac1{\ln^m\ep}\,\int_\R dx\,K_\ep(x)g(x)\to2i\pi g(0). \end{align*} Thus, (1) is proved.