The following distributional limit is well-known: $$\lim_{a\to0^+} \left(\frac{1}{x-ia} - \frac{1}{x+ia}\right) = 2i\pi \delta(x).$$ What if we consider the fractional power? In particular, for $\alpha \in (0,1)$, what is the distributional limit of the following?
$$\lim_{a\to0^+} \left(\frac{1}{(x-ia)^\alpha} - \frac{1}{(x+ia)^\alpha}\right) \; ?$$
Define $f_\alpha(y):=\frac{1}{(y-i)^\alpha}-\frac{1}{(y+i)^\alpha},\,I_\alpha:=\int_{\Bbb R}f_\alpha(y)dy$ so distributionally$$\lim_{a\to0^+}a^{\alpha-1}\left(\frac{1}{(x-ia)^\alpha}-\frac{1}{(x+ia)^\alpha}\right)=\lim_{a\to0^+}\frac1af_\alpha\left(\frac{x}{a}\right)=I_\alpha\delta(x).$$In particular, $I_\alpha=\int_{\Bbb R}\frac{(y+i)^\alpha-(y-i)^\alpha}{(y^2+1)^\alpha}dy$ is finite for $\Re\alpha\in(0,\,1)$, because its integrand is asymptotic to $2i\alpha y^{\alpha-1}$ ($2i\alpha y^{-\alpha-1}$) for small (large) $|y|$. But you've asked about what we get by multiplying this by $a^{1-\alpha}$, which for $\Re\alpha\in(0,\,1)$ gives $0$.