# The following distributional limit is well-known: <munder> <mo movablelimits="true" form="pr

The following distributional limit is well-known:
$\underset{a\to {0}^{+}}{lim}\left(\frac{1}{x-ia}-\frac{1}{x+ia}\right)=2i\pi \delta \left(x\right).$
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barbesdestyle2k
Define ${f}_{\alpha }\left(y\right):=\frac{1}{\left(y-i{\right)}^{\alpha }}-\frac{1}{\left(y+i{\right)}^{\alpha }},\phantom{\rule{thinmathspace}{0ex}}{I}_{\alpha }:={\int }_{\mathbb{R}}{f}_{\alpha }\left(y\right)dy$ so distributionally
$\underset{a\to {0}^{+}}{lim}{a}^{\alpha -1}\left(\frac{1}{\left(x-ia{\right)}^{\alpha }}-\frac{1}{\left(x+ia{\right)}^{\alpha }}\right)=\underset{a\to {0}^{+}}{lim}\frac{1}{a}{f}_{\alpha }\left(\frac{x}{a}\right)={I}_{\alpha }\delta \left(x\right).$
In particular, ${I}_{\alpha }={\int }_{\mathbb{R}}\frac{\left(y+i{\right)}^{\alpha }-\left(y-i{\right)}^{\alpha }}{\left({y}^{2}+1{\right)}^{\alpha }}dy$ is finite for $\mathrm{\Re }\alpha \in \left(0,\phantom{\rule{thinmathspace}{0ex}}1\right)$, because its integrand is asymptotic to $2i\alpha {y}^{\alpha -1}$$|y|$. But you've asked about what we get by multiplying this by ${a}^{1-\alpha }$, which for $\mathrm{\Re }\alpha \in \left(0,\phantom{\rule{thinmathspace}{0ex}}1\right)$ gives 0.