$F_{\epsilon} = e^{-\epsilon x}, x\geq 0$ , and it is easy to derive $\hat{F_{\epsilon}}(y) = \frac{1}{\epsilon + i y} $. I wonder whether this function has a limit function when $\epsilon$ tends to $0$.
I want to do the following in order to show its limit exists. Consider its action on a Schwarz function $g$, we have $\int \frac{1}{\epsilon+iy} g(y)dy = \int -i(ln(\epsilon +iy))'g = -i \int g dln(\epsilon +iy)$. This is not a valid argument because logarithm is not properly defined.
Let $1/y$ be the p.v. functional and let all the limits be for $\epsilon \to 0^+$. The distributional limit exists and can be found in any of these ways:
1) using the Sokhotski-Plemelj formula: $$\frac 1 {\epsilon + i y} = \frac i {-y + i \epsilon} \xrightarrow {\mathcal D} -\frac i y + \pi \delta(y);$$
2) using the fact that the limit and the Fourier transform are interchangeable: $$e^{-\epsilon x} H(x) \xrightarrow {\mathcal D} H(x), \\ (H(x), e^{-i y x}) = -\frac i y + \pi \delta(y);$$
3) writing the functionals as distributional derivatives of ordinary functions, as you suggest. The logarithm is an integrable function, therefore it just corresponds to a regular functional. It can be proved that the limit and the derivative of $\ln(\epsilon + i y)$ are the same in the ordinary sense and in the distributional sense, giving
$$\left( \frac 1 {\epsilon + i y}, \phi \right) = (i \ln(\epsilon + i y), \phi') \to \\ (i \ln(i y), \phi') = \left( i \ln |y| - \frac \pi 2 \operatorname{sgn} y, \phi' \right) = \\ \left( -\frac i y + \pi \delta(y), \phi \right).$$