Assume the following relationship between the Hilbert and Fourier transforms: $$ \mathcal{H}(f) = {\mathcal{F}^{-1}}(-i ~ \text{sgn}(\cdot) \cdot \mathcal{F}(f)), $$
where $ \displaystyle [\mathcal{H}(f)](x) \stackrel{\text{def}}{=} \text{p.v.} \frac{1}{\pi} \int_{- \infty}^{\infty} \frac{f(t)}{x - t} ~ d{x} $.
What happens when $ f(x)$ is a distribution? We know that the Fourier transform exists for distributions, but what about the Hilbert transform?
For example, take $ f(x) = x^{n} $. Its Fourier transform exists as the $ n $-th derivative of the delta function $ \delta(x) $. However, the integral $$ \text{p.v.} \frac{1}{\pi} \int_{- \infty}^{\infty} \frac{x^{n}}{x - t} ~ d{x} $$ is divergent. :(
The Hilbert transform is anti-self-adjoint. Therefore, it is natural to define it on distribution by passing $\mathcal H$ to the test functions, similar to "pass the hat" definition of the Fourier transform. In fact, the Wikipedia article already says this.
Since the stated relation between $\mathcal{F}$ and $\mathcal H$ holds for test functions, the duality-based definition implies that it holds for distributions as well.