The umbrella legitimization of many such Fourier transforms is as tempered _distributions_ (where the sense of "distribution" is not the probability sense, but in the sense of Laurent Schwartz). The various "regularization" tricks amount to approaching the given distribution in the "weak *-topology" on distributions, by more tractable functions. Fourier transform on tempered distributions is (provably) continuous, so we conclude that all these trick must yield the same outcome.
[Edit in response to comment:] The "how to compute" (once we know that any device succeeds) is non-trivial, insofar as it is not clear a-priori how explicit an outcome could be expected. The first volume of Gelfand-Graev-et-alia's "Generalized Functions" does many illuminating examples, mostly computed via meromorphic continuation.
The simplest family of examples is probably $|x|^s$. Here, the homogeneity and rotational symmetry, and the fact that Fourier transform respects these (in suitable senses), promise that the Fourier transform of $|x|^s$ on $\mathbb R^n$ is a constant multiple of $|x|^{-n-s}$, for $-n<\Re(s)<0$ to assure local integrability (of both). The constant multiple is determined (for example) by integrating against Gaussians.
Then use the fact that the derivative of $|x|^s$ in $s$ multiplies it by $\log|x|$, and set $s=0$. This is the nice way logarithms can arise. The implicit claim that we can do complex analysis with distribution-valued functions was legitimized by Schwartz, and is pervasive in Gelfand-et-alia.
Products of $|x|^s$ by harmonic polynomials can be treated almost identically, using the repn theory of the orthogonal group on harmonic polynomials.
That is, very often, some sort of _unique_characterization_ of the tempered distribution, and of its image under Fourier Transform, reduce the computation to determination of the relevant constant!
Edit: oops, as Bazin notes, the exponent is not $n-s$ but $-n-s$, and adjust the local integrability assertion. (Adjusted above.)
The Fourier transform of a function with compact support is automatically holomorphic. This support can be deduced from growth conditions on the former, at least in the case when it is a euclidean ball and the function is square integrable. This is the celebrated Payley-Wiener theorem, upon which there is a comprehensive Wikipedia article for starters.
Best Answer
It is a difficut problem to compute the Fourier transform of the characteristic function of a union of open intervals (in the general case) and it is known that such Fourier transform can converge to 0 with a very slow growth rate. I am currently working on that problem. Probably the growth rate may depend on some arithmetical properties of the boundary of the set as it is the case in studying the Fourier dimension of sets.