Is $F(x)= \frac{1}{|x|^{r}}, (x\in \mathbb R)$ a distributiuon?

Question

Does it make sense to talk of $F(x)= \frac{1}{|x|^{r}}, (x\in \mathbb R)$ for some $r>0$ in the sense of distribution? (I am just confused about the origin) (I mean, Is $F$ a distribution? If yes, then I think we can take of its Fourier transform)

Answer 1

$F(x)=\frac{1}{|x|^r}\in L_{loc}^1(\mathbb R^n)$ iff $r<n$, where $x\in\mathbb R^n$ because $\frac{1}{|x|^\alpha}\in L_{loc}^1(\mathbb R)$ iff $\alpha<1$. So, indeed it defines a distribution by $\int\limits_{\mathbb R^n}{F(x)\varphi(x)dx},\,\forall \varphi\in C_0^\infty(\mathbb R^n)$