# Evaluating the limit $y \to 0^+$

Given $t \in \mathbb{R}$ and $z = x + iy$ and $y>0$. $\lim_{y\to0^+} \frac{1}{t - z} = \frac{1}{t-x} + \pi i \delta(t-x)$

This limit is given in the book Integral Transforms and Their Applications - Debnath 2nd ed. (pg 379)

What precisely is $\delta$? Is $\delta(k) = 0$ for $k \neq 0$ and $\delta(0) = +\infty$? –  JavaMan Oct 24 '11 at 2:10
This has to mean that for reasonably well behaved functions $f$ (test functions or the like...) you'd have $\displaystyle\lim\limits_{y\to0+}\int_?^? \frac{f(t)}{t-z}\;dt = \int_?^? \frac{f(t)}{t-x}\;dt + \pi i f(x)$, where one hopes the bounds of integration are clear from the context. –  Michael Hardy Oct 24 '11 at 2:36
If I understand you correctly, the "original limit" $\displaystyle\lim\limits_{y\to0+}\frac{1}{t-z}$ is not to be understood the way limits of functions are usually understood, but rather its meaning concerns what happens when you multiply the exression by a test function and then integrate and then take the limit. –  Michael Hardy Oct 24 '11 at 3:48