I'm trying to solve the following exercise regarding limits of distributions:
Establish the following limit (on the distributional sense) $$\lim_{t\to 0\pm}\ln (\tau + it) = \ln |\tau| + i\pi H(-\tau),$$ where we are taking the principal branch of the log function, that is, $\arg(\tau+it)<\pi$ and $H(\tau)$ is the Heavside distribution. Relate this expression to the Plemelj-Sochozki's formula.
Now, since we want to compute the limit in the distributional sense, what we need to show is that if $\phi\in \mathcal{D}(\mathbb{R})$ then
$$\lim_{t\to 0}\int_{-\infty}^{\infty} \ln(\tau + it)\phi(\tau)d\tau = \int_{-\infty}^{\infty}\ln |\tau|\phi(\tau)d\tau + i\pi \int_{-\infty}^{\infty}H(-\tau)\phi(\tau)d\tau.$$
To do this, I did the following: first we write down $\ln$ explicitly:
$$\ln(\tau+it)=\ln(\tau^2+t^2)^{1/2}+i\arg(\tau+it),$$
in that case we have:
$$(\ln(\tau+it),\phi)=\int_{-\infty}^{\infty}\ln(\tau^2+t^2)^{1/2}\phi(\tau)d\tau+i\int_{-\infty}^{\infty}\arg(\tau+it)\phi(\tau)d\tau.$$
Let's call the first integral $I_1(t)$ and the second integral $I_2(t)$. On the second integral we see that we can rewrite it in the following way:
$$I_2(t)=\int_{-\infty}^0\left(\dfrac{\pi}{2}+\arctan\dfrac{t}{\tau}\right)\phi(\tau)d\tau+\int_0^\infty \arctan\dfrac{\tau}{t}\phi(\tau)d\tau.$$
We then see the following:
When taking $\lim_{t\to 0}I_1(t)$, if we move the limit inside the integral we get $\lim_{t\to 0}\ln(\tau^2+t^2)^{1/2}=\ln |\tau|$. So that if we move the limit inside the integral we have $\lim_{t\to 0}I_1(t)=\int_{-\infty}^{\infty}\ln|\tau| \phi(\tau)d\tau$.
When taking $\lim_{t\to 0}I_2(t)$, if we again move the limit inside the integrals, we get that $\lim_{t\to 0}I_2(t)=\pi\int_{-\infty}^0 \phi(\tau)d\tau$.
In that case we get the result we wanted:
$$\lim_{t\to 0}(\ln(\tau+it),\phi)=(\ln|\tau|,\phi)+(i\pi H(-\tau),\phi),$$
but it all relies on moving the limit inside the integrals. This can really be done in this case? Isn't anything wrong here?