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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 $\delta(0) = +\infty$?2011-10-24
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    @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.2011-10-24
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    @DJC Yes it is the Delta function.2011-10-24
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    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.2011-10-24
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    @MichaelHardy I can see you that result works out based on the limit. But I'm confused about how the original limit evaluates. Any ideas?2011-10-24
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    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.2011-10-24

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