# 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)

I don't understand how this limit was evaluated. Please help out.

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What precisely is $\delta$? Is $\delta(k) = 0$ for $k \neq 0$ and $\delta(0) = +\infty$? – JavaMan Oct 24 '11 at 2:10
@DJC, I'm assuming it's the Dirac delta function, but the solution only seems to make sense (if at all) if one is integrating over the expression and the limit is really sloppy notation for a limit of the integrals. – Henning Makholm Oct 24 '11 at 2:15
@DJC Yes it is the Delta function. – sauravrt Oct 24 '11 at 2:15
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