I am faced with a problem while computing the Fourier transform of $||x||^{-\alpha}$, $x\in \mathbb R^n$ and $\alpha\in(0,n)$. I consider the measurable function $f(x)=||x||^{-\alpha}$ and show it induces a tempered distribution as follows.
Let $\phi\in \mathcal S(\mathbb R^n)$, the space of Schwartz functions and we have the following estimate. $$\left |T_f(\phi) \right |=\left | \int _{\mathbb R^n} f(x)\phi(x)dx \right | \leq \int_{\mathbb R^n} \left |phi(x) \right | ||x||^{-\alpha}dx$$ $$=\int_{||x||>1}\left |phi(x) \right | ||x||^{-\alpha}dx \ + \ \int_{||x||\leq1}\left |phi(x) \right | ||x||^{-\alpha}dx$$ $$\leq \int_{||x||>1}\left |phi(x) \right | dx \ + \ ||\phi||_{\infty}\int_{||x||\leq1} ||x||^{-\alpha}dx$$ $$\leq ||\phi||_1+ C||\phi||_{\infty}\int_0^1r^{-\alpha}r^{n-1}dr=||\phi||_1+ \frac{C||\phi||_{\infty}}{n-\alpha}<\infty $$
This further shows $T_f\in \mathcal S'(\mathbb R^n)$, the space of tempered distributions so we can calculate the Fourier transform as follows: \ $$\hat{T_f}(\phi)=T_f(\hat \phi)=\int _{\mathbb R^n} f(x)\hat \phi(x)dx=\int _{\mathbb R^n} f(x)\left [ \int_{\mathbb R^n} \phi(y)e^{-2\pi i x\cdot y} \ dy \right]dx$$ $$=\int_{\mathbb R^n}\int_{\mathbb R^n} f(x)\phi(y)e^{-2\pi i x\cdot y} \ dy \ dx $$ I cannot apply Fubini at this stage and I am lost as how to proceed after this. I know the Fourier transform of radial, homogeneous functions in $L^1(\mathbb R^n)$ but $f$ is not in $L^1$ so if I can show the Fourier transform is induced by some element in $L^1_{loc} (\mathbb R^n)$, I would be able to complete the proof. Some hints would be helpful.
