TITLE: Proof without distributions
QUESTION [5 upvotes]: I was wondering whether there is a way to show this identity 
$$\pi \int_{\mathbb{R}^3} \frac{f(x)}{|x|} dx = \int_{\mathbb{R}^3} \frac{\widehat{f(x)}}{|x|^2} dx $$ without using distributions for $f \in \mathcal{S}(\mathbb{R}^n)$.
The reason I ask is the following: Obviously this identity makes perfect sense in the classical way, but the most obvious proof would use the fourier transform of $\frac{1}{|x|^2}$ which is clearly not defined in $L^1$ or $L^2$.
Thus, I would be interested, whether one can proof this identity with classical methods, i.e. no distribution theory,  too?

REPLY [2 votes]: @Phil Isett's device also does apply to arbitrary $\int_{\mathbb R^n} f(x)/|x|^\alpha\;dx$ with $0< \Re(\alpha)<n$, as follows, disregarding constants and normalization of Fourier transforms:
$$
\int_{\mathbb R^n} {f(x)\over |x|^\alpha}\;dx
\;=\;
{1\over \Gamma(\alpha/2)}\int_0^\infty \int_{\mathbb R^n} f(x)\,e^{-t|x|^2}\,t^{\alpha/2}\;dx\;{dt\over t}
$$
$$
= 
{1\over \Gamma(\alpha/2)}\int_0^\infty \int_{\mathbb R^n} \widehat{f}(x)\,e^{-{1\over t}|x|^2}\,t^{{\alpha-n\over 2}}\;dx\;{dt\over t}
$$
by Plancherel and by knowing how to take Fourier transform of Gaussians. Replace $t$ by $1/t$ to obtain (up to constants)
$$
{1\over \Gamma(\alpha/2)}\int_0^\infty \int_{\mathbb R^n} \widehat{f}(x)\,e^{-t|x|^2}\,t^{{n-\alpha\over 2}}\;dx\;{dt\over t}
= {\Gamma({n-\alpha\over 2})\over \Gamma(\alpha/2)} \int_{\mathbb R^n} {\widehat{f}(x)\over |x|^{n-\alpha}}\;dx
$$