Distributional expression for $\lim_{\epsilon\to 0^+}\frac{1}{(\epsilon-it)^n}$? I have computationally verified that the following equation holds
$$\lim_{\epsilon\to 0^+}\int_{-\infty}^{\infty}dt\frac{\phi(t)}{(\epsilon-it)^2}=i\pi\phi'(0)+PV\int_{-\infty}^\infty\frac{\phi(t)-\phi(0)}{t^2}dt$$
for various $\phi$, where $\phi$ is a test function. I thought this would follow as a consequence of the well known distributional relation
$$\lim_{\epsilon\to 0^+}\frac{1}{\epsilon-it}=\pi\delta(t)+iPV\frac{1}{t}$$
but since the two principal value prescriptions are different from each other, taking a derivative  of the last relation doesn't seem to yield the answer immediately. What is a good way to show this?
I have a feeling that complex analysis could work here, and even help with showing the more general conjecture (I verified with Mathematica up to $n=4$ as well)
$$i^{n-1}(n-1)!\lim_{\epsilon\to 0^+}\int_{-\infty}^{\infty}dt\frac{\phi(t)}{(\epsilon-it)^n}=(-1)^{n-1}\pi\phi^{(n-1)}(0)+iPV\int_{-\infty}^\infty\frac{\phi(t)-\sum_{k=0}^{n-2}\frac{\phi^{(k)}(0)}{k!}t^k}{t^n}dt$$
 A: Note that we have $\frac1{\varepsilon -it}=\frac{\varepsilon+it}{\varepsilon^2+t^2}$.  Hence, we can write
$$\begin{align}
\int_{-\infty}^\infty \frac{\phi(t)}{(\varepsilon-it)^2}\,dt&=-i\int_{-\infty}^\infty \phi(t) \frac{d}{dt}\left(\frac1{\varepsilon -it}\right)\,dt\\\\
&=i\int_{-\infty}^\infty \frac{\phi'(t)}{\varepsilon -it}\,dt\\\\
&=i\int_{-\infty}^\infty \frac{\varepsilon+it}{\varepsilon^2+t^2}\phi'(t)\,dt\\\\
&=-\int_{-\infty}^\infty \frac{t}{\varepsilon^2+t^2}\phi'(t)\,dt+i \int_{-\infty}^\infty \frac{\varepsilon}{\varepsilon^2+t^2}\phi'(t)\,dt\tag1
\end{align}$$
We will now evaluate the limit of the two integrals on the right-hand side of $(1)$.

Let $\delta>0$ be given and fixed.  Then, the limit as $\varepsilon\to 0^+$ of the first integral on the right-hand side of $(1)$ is
$$\begin{align}
\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{t}{\varepsilon^2+t^2}\phi'(t)\,dt&=\lim_{\varepsilon\to 0^+}\int_{|t|\le \delta} \frac{t}{\varepsilon^2+t^2}\phi'(t)\,dt+\lim_{\varepsilon\to 0^+}\int_{|t|\ge \delta} \frac{t}{\varepsilon^2+t^2}\phi'(t)\,dt\\\\
&=O(\delta)+\int_{|t|\ge \delta}\frac{\phi'(t)}{t}\,dt\\\\
&=O(\delta)-\int_{|t|\ge\delta}\frac{\phi(t)-\phi(0)}{t^2}\,dt
\end{align}$$
Letting $\delta \to 0^+$, we find that
$$\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{t}{\varepsilon^2+t^2}\phi'(t)\,dt=\text{PV}\int_{-\infty}^\infty \frac{\phi(t)-\phi(0)}{t^2}\,dt\tag2$$

The limit of the second integral on the right-hand side of $(1)$ is
$$\begin{align}
\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\varepsilon}{\varepsilon^2+t^2}\phi'(t)\,dt&=\pi \phi'(0)+\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\varepsilon}{\varepsilon^2+t^2}(\phi'(t)-\phi'(0))\,dt\\\\
&=\pi \phi'(0)+\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\phi'(\varepsilon t)-\phi'(0)}{t^2+1}\,dt\\\\
&=\pi \phi'(0)\tag3
\end{align}$$

Using $(2)$ and $(3)$ in $(1)$ yield the coveted result
$$\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\phi(t)}{(\varepsilon-it)^2}\,dt=-\text{PV}\int_{-\infty}^\infty \frac{\phi(t)-\phi(0)}{t^2}\,dt+i\pi \phi'(0)\tag4$$
from which we deduce that the distributional limit
$$\lim_{\varepsilon\to 0^+}\frac{1}{(\varepsilon-it)^2}=-\text{PV}\left(\frac1{t^2}\right)-i\pi \delta'(t)$$
where the $PV$ distribution is defined in $(4)$.


For the distributional limit of $\frac1{(\varepsilon-it)^n}$, we could follow an analogous procedure.   Instead, we proceed as follows.  Let $\psi_\varepsilon(t)=\frac1{\varepsilon-it}$.  Then, $\psi_\varepsilon^{(n-1)}(t)=\frac{(i)^{n-1}(n-1)!}{(\varepsilon -it)^n}$ and we  can write
$$\begin{align}
\lim_{\varepsilon\to 0^+}\langle (\psi_\varepsilon)^{n} ,\phi \rangle&=\lim_{\varepsilon\to 0^+} \frac{1}{i^{n-1}(n-1)!}\langle \psi_\varepsilon^{(n-1)},\phi \rangle \\\\
&=\lim_{\varepsilon\to 0^+}\frac{i^{n-1}}{ (n-1)!}\langle \psi_\varepsilon ,\phi^{(n-1)}\rangle\\\\
&=\frac{i^{n-1}}{(n-1)!}\left(\pi \phi^{(n-1)}(0)+i\text{PV}\int_{-\infty}^\infty \frac{\phi^{(n-1)}(t)}{t}\,dt\right)\tag5
\end{align}$$
Integrating by parts $n-1$ times the integral on the right-hand side of $(5)$ shows that in distribution we have
$$\lim_{\varepsilon\to 0^+}\frac1{(\varepsilon-it)^n} =\frac{(-i)^{n-1}\pi}{(n-1)!} \delta^{(n-1)}(t)+i^n \text{PV}\left(\frac1{t^n}\right)\tag6$$
The distribution $\text{PV}\left(\frac1{t^n}\right)$ in $(6)$ is defined as its action of a test function and can be written accordingly as
$$\int_{-\infty}^\infty \text{PV}\left(\frac1{t^n}\right)\phi(t)\,dt=\text{PV}\int_{-\infty}^\infty \frac{\phi(t)-\sum_{k=0}^{n-2}\frac{\phi^{(k)}(0)t^k}{k!}}{t^n}\,dt$$
