$f(x) = 1 / \lvert x \rvert^2$, $x\in \mathbb{R}^3$ , for the Fourier transform F, prove by scaling: $ F(f) (y) = C \frac{1}{\lvert y\rvert}. $ Let $f(x) = 1 / \lvert x \rvert^2$, $x\in \mathbb{R}^3$, $\lvert x\rvert = \sqrt{x_1^2 + x_2^2 + x_3^2}$.
Let $F(f)$ denote the Fourier transform of $f$. Assume that $F(f)$ is an $L^1_{loc}$ function so it defines a distribution. By using scaling ( composing with $3x3$ matrices) prove that for some constant $C$:
$$ F(f) (y) = C  \frac{1}{\lvert y\rvert}. $$
My attempt: I've proven that $f$ indeed defines a tempered distribution so it makes sense to look at $F(f)$ but I'm really not sure how to proceed on this without computing it directly, which obviously isn't the inteded solution and I'm also not sure how to do it. I tried proving it by testing it on functinos from $S$ but it hasn't yielded anything useful. I proved what $F(f A)$ is for $u\in S$ and a matrix $A$, but I'm not sure how to use that information.
EDIT: I've also proven that $f$ is radial and $F(f)$ radial.
 A: So first $|x|^{-a}$ is in $L^1_{\mathrm{loc}}(\mathbb{R}^d)$ as soon as $a < d$ since by a radial change of variable
$$
\int_{|x|<1} \frac{\mathrm{d}x}{|x|^a} = \omega_d\int_0^1 r^{d-1-a} \mathrm{d}r = \frac{\omega_d}{d-a} < \infty
$$
where $\omega_d = \frac{2\pi^{d/2}}{\Gamma(d/2)}$ is the size of the unit sphere in $\mathbb{R}^d$. In particular, if $d=3$, $|x|^{-a}$ is a tempered distribution as soon as $a<3$.
Now, to get the form of the Fourier transform, just remark that since $f(x) = \frac{1}{|x|}$ is radial, its Fourier transform $\mathcal{F}(f)=\hat{f}$ is also radial. Moreover, for any $\lambda \in\mathbb{R}$ and $y\in\mathbb{R}^3$
$$
\hat{f}(\lambda\,y) = \frac{1}{|\lambda|^d}\mathcal{F}_x\left(f(x/\lambda)\right)(y) = \frac{1}{|\lambda|^d}\mathcal{F}_x\left(\frac{|\lambda|}{|x|}\right)(y) = \frac{1}{|\lambda|^{d-1}}\mathcal{F}_x\left(\frac{1}{|x|}\right)(y) = \frac{1}{|\lambda|^{d-1}}\hat{f}(y)
$$
In particular, taking $λ = |z|$ and $y = \frac{z}{|z|}$, one gets for $z≠0$
$$
\hat{f}(z)= \frac{1}{|z|^{d-1}}\hat{f}(\tfrac{z}{|z|}),
$$
and actually, the equality also holds as tempered distributions since this is the unique tempered distribution with this homogeneity. Since $\hat{f}$ is radial, $\hat{f}(\tfrac{z}{|z|}) = \hat{f}(e_1) = C$ is a constant. Therefore
$$
\hat{f}(z)= \frac{C}{|z|^{d-1}}
$$

Remark: one can get the constant by expressing $|x|^{-a}$ as an integral of Gaussian functions and using the known expression of the Fourier transform of a Gaussian. One can find this constant in the book Functional Analysis by Lieb and Loss for example.
With the convention $\mathcal{F}(f)(y) = \int_{\mathbb{R}^d} e^{-2iπx·y}\,f(x)\,\mathrm{d}x$, one gets for $a\in(0,d)$
$$
\mathcal{F}\left(\frac{1}{\omega_a|x|^a}\right) = \frac{1}{\omega_{d-a}|x|^{d-a}}
$$
In the case when $a=d$, one gets $\mathcal{F}\left(\frac{1}{\omega_d|x|^d}\right) = \frac{\psi(d/2)-\gamma}{2} - \ln(|πx|)$ as proved here The Fourier transform of $1/p^3$.
A: 
I started this before @ll-3-14 posted a much more deft answer, but I got far enough that I decided not to discard mine.


Tempered distribution boilerplate

Let $\varphi\in\mathcal{S}(\mathbb{R}^3)$, and let $\mathsf{T}$ be the tempered distribution defined by
\begin{equation}
\mathsf{T}[\varphi] = \int_{\mathbb{R}^3}\frac{\varphi(x)}{\left\Arrowvert x\right\Arrowvert^2}dV.
\end{equation}
As a tempered distribution, $\mathsf{T}$ has a Fourier transform $\widehat{\mathsf{T}}$ defined by\begin{equation}
\widehat{\mathsf{T}}[\varphi] = \mathsf{T}[\widehat{\varphi}] = \int_{\widehat{\mathbb{R}}^3}\frac{\widehat{\varphi}(k)}{\left\Arrowvert k\right\Arrowvert^2}dV
\end{equation}
for each $\varphi\in\mathcal{S}(\mathbb{R}^3)$. I write that $\widehat{\varphi}\in\mathcal{S}(\widehat{\mathbb{R}}^3)$ to indicate that the Fourier transform $\widehat{\varphi}$ is a function on the "conjugate space" $\widehat{\mathbb{R}}^3$.

Beginning the transformation

The goal is to express this as an integral (or limit of integrals) involving $\varphi$ instead of $\widehat{\varphi}$.
\begin{equation}
\begin{split}
\int_{\widehat{\mathbb{R}}^3}\frac{\widehat{\varphi}(k)}{\left\Arrowvert k\right\Arrowvert^2}dV(k)
=
\int_{\widehat{\mathbb{R}}^3}\frac{1}{\left\Arrowvert k\right\Arrowvert^2}\left[\int_{\mathbb{R}^3}\varphi(x)e^{-ik\cdot x}dV(x)\right]dV(k),
\end{split}
\end{equation}
where $dV(k)$ is the "differential volume" at $k\in\widehat{\mathbb{R}}^3$ and $dV(x)$ is the "differential volume" at $x\in\mathbb{R}^3$.
\begin{equation}
\begin{split}
&~\lim_{R\to\infty}\int_{\{k:\left\Arrowvert k\right\Arrowvert\leq R\}}\frac{1}{\left\Arrowvert k\right\Arrowvert^2}\left[\int_{\mathbb{R}^3}\varphi(x)e^{-ik\cdot x}dV(x)\right]dV(k)\\
&=~
\lim_{R\to\infty}\int_{\mathbb{R}^3}\varphi(x)\left[\int_{\{k:\left\Arrowvert k\right\Arrowvert\leq R\}}\frac{e^{-i x\cdot k}}{\left\Arrowvert k\right\Arrowvert^2}dV(k)\right]dV(x)
\end{split}
\end{equation}
Let $r = \left\Arrowvert k\right\Arrowvert$. In 3-dimensional spherical co-ordinates,
\begin{equation}
dV(k) = r^2\sin\theta d\theta~dr~d\phi,
\end{equation}
where $\theta$ is the polar angle (varying from 0 to $\pi$ radians) and $\phi$ is the azimuthal angle (varying from 0 to $2\pi$ radians).

The secret

The only angular dependence of the integrand is the dependence on the angle between $x$ and $k$. Define the co-ordinate system so that $x$ points along the $z$-axis, so that
\begin{equation}
x\cdot k = \left\Arrowvert x\right\Arrowvert\left\Arrowvert k\right\Arrowvert\cos\theta = \left\Arrowvert x\right\Arrowvert r\cos\theta,
\end{equation}
where $\theta$ is the polar angle. The integral of interest is now
\begin{equation}
\int_{0}^{2\pi}\left[\int_{0}^{R}\left[\int_{0}^{\pi}\frac{e^{-i\left\Arrowvert x\right\Arrowvert r\cos\theta}}{r^2}\sin\theta d\theta\right]r^2dr\right]d\phi
=
2\pi\int_{0}^{R}\left[\int_{0}^{\pi}e^{-i\left\Arrowvert x\right\Arrowvert r\cos\theta}\sin\theta d\theta\right]dr.
\end{equation}

Calculus!

Let $u = \left\Arrowvert x\right\Arrowvert r\cos\theta$, so that
\begin{equation}
du = -\left\Arrowvert x\right\Arrowvert r\sin\theta~d\theta,
\end{equation}
and
\begin{equation}
\begin{split}
\int_{0}^{\pi}e^{-i\left\Arrowvert x\right\Arrowvert r\cos\theta}\sin\theta d\theta
&=~
\int_{u=\left\Arrowvert x\right\Arrowvert r}^{u=-\left\Arrowvert x\right\Arrowvert r} e^{-iu}\frac{(-1)du}{\left\Arrowvert x\right\Arrowvert r}\\
&=~
\frac{1}{\left\Arrowvert x\right\Arrowvert r}\int_{u=-\left\Arrowvert x\right\Arrowvert r}^{u=\left\Arrowvert x\right\Arrowvert r}e^{-iu}du\\
&=~
\frac{1}{\left\Arrowvert x\right\Arrowvert r}\left(\left.\frac{e^{-iu}}{-i}\right|_{u=-\left\Arrowvert x\right\Arrowvert r}^{u=\left\Arrowvert x\right\Arrowvert r}\right)\\
&=~
\frac{1}{\left\Arrowvert x\right\Arrowvert r}\left(\frac{-2i\sin(\left\Arrowvert x\right\Arrowvert r)}{-i}\right)\\
&=~ \frac{2\sin(\left\Arrowvert x\right\Arrowvert r)}{\left\Arrowvert x\right\Arrowvert r}.
\end{split}
\end{equation}

A sincing feeling

The integral of interest is now
\begin{equation}
4\pi\int_{0}^{R}\textrm{sinc}(\left\Arrowvert x\right\Arrowvert r)dr,
\end{equation}
where we are using the un-normalized sinc function. We perform another change of variable:
\begin{equation}
v = \left\Arrowvert x\right\Arrowvert r,~~~\textrm{so that}~~~dr = \frac{dv}{\left\Arrowvert x\right\Arrowvert}.
\end{equation}
The integral is
\begin{equation}
\frac{4\pi}{\left\Arrowvert x\right\Arrowvert}\int_{0}^{R/\left\Arrowvert x\right\Arrowvert}\textrm{sinc}(v)dv.
\end{equation}
There was no other specific $R$-dependence, and we are taking limits anyway with  integrals "against" Schwartz functions, so we can replace $R/\left\Arrowvert x\right\Arrowvert$ with, say, $\rho$. We have
\begin{equation}
\widehat{\mathsf{T}}[\varphi]
=
\lim_{\rho\to\infty}\int_{\mathbb{R}^3}\varphi(x)\frac{4\pi}{\left\Arrowvert x\right\Arrowvert}\left[\int_{0}^{\rho}\textrm{sinc}(v)dv\right]dV(x).
\end{equation}
