# Why does the following define a distribution and of which order?

I want to show that $$\phi\mapsto\underset{\varepsilon\searrow 0}{lim}\int_{-\infty}^{\infty}\frac{\phi(x)}{x+i\varepsilon}dx$$ defines a distribution on $\mathcal{D}(\mathbb{R})$ but I just don't come up with a good idea to show that. Second part would be to determine the order of it. Thanks for any suggestions.

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$x + i\epsilon \sim x$. Split the problem into parts: one part where the approximation is useful, and one part where it is not. – Hurkyl Jun 27 '12 at 18:14
I wanted to split it into $x\in [-\delta,\delta]$ for $\delta$ small and the part away from zero. The part away from zero should be nice and the limit can be evaluated under the integral, but what to do with the part containing $0$? – user34623 Jun 27 '12 at 18:20
The usual story these things follow is that you make that part small or otherwise easy to analyze. The key idea is probably going to be that $\phi(x) \sim \phi(0)$ on that part, though you might need a Taylor approximation with a few more terms. – Hurkyl Jun 27 '12 at 22:53
Do you understand the arguments in this math.stackexchange.com/questions/162931/… recent question? – Vobo Jun 28 '12 at 17:13