Proving $|x|^a$ and $ |x|^a ln|x| \in S'(\mathbb{R}^n)$ for $a \in (-n, \infty)$

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My attempt so far:

Case $a \in [0,\infty)$: choose $\phi \in S(\mathbb{R}^n)$:

$\mid{<|x|^a, \phi>}\mid$ $\leq \int_{\mathbb{R}^n} \mid \frac{1}{1+x^2}\ (x^a + x^{2a}) \ \phi(x)\mid dx$ $\leq (\rho_{a,o} + \rho_{2a,o})\ \pi$

Linearity is clear, continuity follows from a theorem we proved in class.

Case $a \in (-n,0):$

Set $k = -a$, $\phi$ as in previous case:

$\mid{<|x|^{-k}, \phi>}\mid$ $\leq \int_{\mathbb{R}^n} \mid \frac{\phi(x)}{x^k} \mid dx$

At this point, I'm not sure how to proceed. I know I need to use $k < n$ at some point, but I can't figure out how.

Edit: Having the same problems trying to show $|x|^a ln|x|$ and $ln|x|$ in $S'(\mathbb{R}^n)$.

Edit 2: I now can show that $|x|^a ln|x|$ and $ln|x|$ are locally integrable (i.e. distributions). Just missing the Schwartz space part.

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For $a<0$ split the function in two parts: $|x|^a$ restricted to $|x|\ge 1$ is bounded and hence belongs to $S'$ (recall that $L^\infty\subset S'$).

On the other hand, if you restrict on $|x|\le 1$ you have $|x|^a\in L^1 $ (a simple change of variables to spherical ones proves that), and hence you have again a distribution in $S'$ (recall that $L^1\subset S'$)

Finally, a sum of two tempered distributions is tempered.