Proof without distributions 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? 
 A: I think the constant in the equality you wrote down depends on how exactly you normalize the Fourier transform.  Up to worrying about the constant, here's a proof that should work and may satisfy your criterion:
In dimension $d$, write 
$$ \int \frac{\widehat{f(\xi)}}{ |\xi|^{2}} d\xi = \int_0^\infty ds \int \hat{f}(\xi) e^{-s |\xi|^2} d\xi $$
Up to the normalization constant involving $(2 \pi)^d$ and justifying the convergence, the right hand side equals (by Plancharel)
$$ \int_0^\infty ds \int f(x) e^{- \frac{|x|^2}{4 s} } (2s)^{-d/2} dx  $$
Interchange the integrals and change the s variable such that $\tilde{s}^2 = (4s)^{-1}|x|^2$; the above equals $$ C \int \frac{f(x)}{|x|^{d - 2}} dx \int_0^\infty ~ e^{-\tilde{s}^2} \tilde{s}^{d - 3} d\tilde{s}   $$
for an explicit constant $C$.  Note that the power $d - 3 > -1$ since we are in dimension $d > 2$.  The value of the $d\tilde{s}$ integral is then known and that will give your formula.  (Equivalently, this answer gives a way to compute the Fourier transform of $|\xi|^{-2}$.)
A: @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)
$$
{1\over \Gamma(\alpha/2)}\int_0^\infty \int_{\mathbb R^n} \widehat{f}(x)\,e^{-t|x|^2}\,t^{{n-\alpha\over 2}}\;dx\;{dt\over t}
= {\Gamma({n-\alpha\over 2})\over \Gamma(\alpha/2)} \int_{\mathbb R^n} {\widehat{f}(x)\over |x|^{n-\alpha}}\;dx
$$
