Does this distribution make any sense? I met this distribution $u$ which acts like
$$
\langle u, \phi \rangle = \int _{\mathbb{R}} \frac{\phi (t)}{t^n}\,dt , \qquad \phi \in C_0^\infty (\mathbb{R})
$$
where $n\ge 2$ is an integer. Is this well-defined the way it stands? Or does it have to be written as $\lim_{\varepsilon \to 0} 1/(t\pm i\varepsilon )^n$?
Otherwise, could I say something about it in terms of $\phi (0)$?
 A: As written, the integral with $n=2$ does not converge for a smooth bump function $\phi$ with $\phi(0)\ne 0$. Replacing $t$ with $t+i\epsilon$ does not help much: it only tames the integral over $|t|\le \epsilon$, but the contributions of $t$ with $|t|\ge \epsilon$ are not changed much. In particular, the integral with $n=2$ will still have infinite limit as $\epsilon\to 0$, provided $\phi(0)\ne 0$. 
To get a well-defined distribution, let $P$ be the Taylor polynomial of $f$ at $0$ of degree $n-2$. Define
$$\langle u,\phi\rangle =p.v. \ \int_{\mathbb R} \frac{\phi(t)-P(t)}{t^n}\,dt$$ This integral converges absolutely at infinity. Near $0$,  the difference $\phi(t)-P(t)$ may have a term $c t^{n-1}$, which will not affect the principal value of the integral. The rest of $\phi(t)-P(t)$ is $O(t^n)$. 
The above distribution is also what you get by taking the distributional derivative of
$$\langle u,\phi\rangle =p.v. \ \int_{\mathbb R} \frac{\phi(t)}{t}\,dt$$ $n-1$ times, up to some constant factor.
