$\phi \mapsto \lim_{\epsilon \to 0} \int_{\mathbb{R}^n} \frac{\phi (x)}{\left \Vert x \right \Vert-1 + i\epsilon}$ is a distribution

Question

I want to show that $$\phi \mapsto \lim_{\epsilon \to 0} \int_{\mathbb{R}^n} \frac{\phi (x)}{\left \Vert x \right \Vert-1 + i\epsilon}$$ is a distribution for $x \in \mathbb{R}^n$, $\phi \in \mathcal{S}(\mathbb{R}^n)$ in the Schwartz space and $\left \Vert \cdot \right \Vert$ the euclidean norm. I think that for the one dimensional case this should follow from the theory of Cauchy principle values but I'm not sure.

Answer 1

First write the integral in polar coordinates, then one can argue that we only have to worry about the one dimensional case: $$ \phi \mapsto \lim_{\epsilon\to 0} \int_0^\infty \frac{\phi(r)}{r-1+i\epsilon} dx ,$$ where $\phi:(0,\infty) \to \mathbb C$ is zero outside a small neirghborhood of $0$.

Let $\psi$ be a function which is $1$ in a small neighborhood of $1$, and zero outside a slightly larger neighborhood of $1$. Write $$ \phi(r) = \phi_1(r) + \phi_2(r) ,$$ where $$ \phi_2(r) = \psi(r) \left[ \sum_{k=0}^m \frac{\phi^{(k)}(1)}{k!} (r-1)^k \right] ,$$ $$ \phi_1(r) = \phi(r) - \phi_2(r) ,$$ and compute the integral on each of the two parts.

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