Although this question probably should be moved to Math Stack Exchange, some remarks are easy:
For example, (the distribution) "$1/x^2$" is certainly not locally integrable near $0$, so ... it's not that we conclude that its Fourier transform blows up at $0$, although, indeed, the literal integral that would express this Fourier transform pointwise value does blow up... to make sense of this Fourier transform (rather than declaring that it has no sense) is to analytically continue in the family of tempered distributions $1/|x|^s$ (and, similarly, $\mathrm{sgn}\,x\cdot |x|^{-1}$). This is what Riesz showed that Hadamard did, in effect, in his "finie partie" device, c. 1930.
That is, to reiterate, many Fourier transforms (already even on $L^2$, as opposed to $L^1$) cannot possibly be the literal integral, but some sort of extension (by continuity) of that integral.
In particular, if your are wanting/needing tables/ideas of Fourier transforms that blow up in certain ways for distributions such as $1/x^2$, you will be frustrated to some degree by tables and explanations, because a big part of the underlying goal is to make sense of things even when the literal integral definition of Fourier transform gives a divergent integral.
So, yes, essentially all identities that do hold when the literal integral converges do also hold when Fourier transform is extended by continuity to larger classes of functions (or/and "generalized functions", so, here, tempered distributions).
And, I add, all these things are self-consistent, if one does not put too much stock in "pointwise values"... which, unsurprisingly, are not a good way to understand generalized functions (measure theory notwithstanding).