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How can we prove that the Fourier transform of: $(1/|x|^\alpha)$ is $C(1/|\xi|)^{n-\alpha}$ as a tempered distribution (in $\mathcal{S}'(\mathbb{R}^n)$) , where $C$ is a constant.

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Find the Fourier transform of $f(\omega) = \frac{1}{|\mathbf{\omega}|^{\alpha}}$. This distribution is radial and homogeneous of degree $-\alpha$. You can use the scaling properties of the Fourier transform to show that this means that its Fourier transform is radial and homogeneous of degree $-n + \alpha$. Then you can show that this only happens when this Fourier transform is of the form $c_\alpha \frac{1}{|\mathbf{\xi}|^{n - \alpha}}$.

So it remains to determine $c_\alpha$. For this you can use Plancherel's theorem in conjunction with the fact that $e^{-\pi |\omega|^2}$ is its own Fourier transform, so that you have $$\int_{R^n} \frac{1}{|\mathbf{\omega}|^{2\alpha}}e^{-\pi|\omega|^2}\,d\omega = c_{\alpha} \int_{R^n} \frac{1}{|\mathbf{\xi}|^{n - 2\alpha}}e^{-\pi|\xi|^2}\,d\xi$$ You can turn both integrals into one dimensional integrals using polar coordinates and then solve for $c_{\alpha}$. The result will be a ratio of gamma functions.

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