Consider the function defined over $\mathbb{R}$ by $$f(t) = \frac{1}{1 + \mathrm{i} \ a \ \mathrm{sign} (t)}$$ with $a \neq 0$ and $\mathrm{sign}(t) = 1$ if $t \geq 0$, $(-1)$ otherwise. This function can be seen as a tempered distribution, so that its Fourier transform $F = \mathcal{F} \{f\} \in \mathcal{S}'$ is well-defined.
Question: Is there a known closed form for $F$?
This can be derived from the Fourier transform of the step function. We obtain $$\mathcal{F}\{f\}(\omega) =- \frac{2 a}{1+a^2} \mathcal{P} \left(\frac1{\omega}\right) + \frac{2\pi}{1+a^2} \delta(\omega); $$ here, $\mathcal{P}$ denotes the principal value and I have defined the Fourier transform as $\mathcal{F}\{f\}(\omega) = \int\!dt f(t) e^{-i\omega t}$.