Let $S$ be in $\mathcal{T}$, the set of tempered distributions, and $\mathcal{F}S$ be its Fourier Transform. Is there some relationship for such distributions, analogous to the Plancherel Theorem for $f \in L^2$?
$$\int_{-\infty}^{\infty} \lvert f(x) \rvert^2 \ dx = \int_{-\infty}^{\infty} \lvert \mathcal{F}f(k) \rvert^2 dk $$
I suspect that perhaps this is not possible to define in general due to the difficulties in defining the convolution of two distributions with non-compact support and/or associative multiplication with distributions. However, are there some important restricted cases in which it works out?
What follows is not at all rigorous, but just a kind of attempt at exploring how one would compute such things. I have noticed some coincidences, for instance that if we let (using the appropriate convention such that the Plancherel theorem works out as above) $$ S = P.V. \frac{\cos{x}}{x}, \mathcal{F}S = \frac{i}{2} \sqrt{\frac{\pi}{2}} \Big(\text{Sign}(k+1) \ + \text{Sign}(k-1) \Big) $$ forging ahead foolishly and treating these as functions with point-wise multiplication then we have to evaluate $$ \int_{- \infty}^{\infty} \frac{\cos^2{x}}{x^2} dx$$ and $$\frac{\pi}{2}\bigg( \int_{-\infty}^{-1} 1 \ dk + \int_{1}^{\infty} 1 \ dk\bigg)$$
The most 'natural' thing I could think to do is take the product of two principal-value distributions to lead to the Hadamard regularization $$\mathcal{H} \int_{- \infty}^{\infty} \frac{\cos^2{x}}{x^2} = -\pi $$ and cut off the second integral by hand $$\lim_{a \to \infty} \frac{\pi}{2}\bigg( \int_{-a}^{-1} 1 \ dk + \int_{1}^{a} 1 \ dk\bigg) \sim \pi \ a - \pi $$ and somehow the finite parts of both 'expressions' (what you get when you just throw away the divergent parts) are the same. Is there some good reason that these finite parts should be the same, or is it merely coincidental?