This comes from physics, but since it's a question on computing Fourier transforms, it probably is more accurately asked here.
To slowly get back into math shape after the summer break, I am currently redoing some of the math in an old paper by Nelson and Fisher, https://journals.aps.org/prb/abstract/10.1103/PhysRevB.16.4945 . Mostly smooth sailing, but there is a step of Fourier transforming $$S(x,t) = \exp\left[-\frac{\kappa}{2} (|x-ct| + |x + ct|)\right]\tag{3.1}$$ into $$S(q,\omega) = \frac{8\kappa^2/c}{\left(\kappa^2 + \left(\frac{\omega}{c} - q\right)^2\right) \left(\kappa^2 + \left(\frac{\omega}{c} + q\right)^2\right)}.\tag{3.2}$$ They claim it is "very easy", so I am embarassed for not getting there. Is there a simple trick I'm missing?
Botched attempt
To give a measure of where I'm stuck, here's what I have. I guess, however, that there is an elegant solution and all this math is pretty unnecessary.
However, going from $(x,t)$ to $(q,t)$ with the convention for the Fourier transform given in (2.6a), $$ \begin{aligned} S(q,t) &= \int_{-\infty}^\infty dx \, e^{iqx}e^{-\frac{\kappa}{2} (|x-ct| + |x + ct|)}\\ &= \int_{-\infty}^{-ct} dx \, e^{iqx}e^{-\frac{\kappa}{2} ((ct-x) + (ct-x))} + \int_{-ct}^{ct} dx \, e^{iqx}e^{-\frac{\kappa}{2} ((ct-x) + (x+ct))} + \int_{ct}^{\infty} dx \, e^{iqx}e^{-\frac{\kappa}{2} ((x-ct) + (x+ct))}\\ &= e^{-\kappa ct} \frac{1}{iq+\kappa}e^{-iqct - \kappa ct} + e^{-\kappa ct}\frac{2}{q}\sin(cqt) - e^{-\kappa ct}\frac{1}{iq-\kappa}e^{iqct-\kappa ct}\\ &= e^{-\kappa ct} \frac{1}{q^2+\kappa^2}\left[(\kappa - iq)e^{-iqct} + (\kappa + iq)e^{iqct}\right] + e^{-\kappa ct}\frac{2}{q}\sin(cqt)\\ &= 2 e^{-\kappa ct} \left[ \frac{\kappa}{q^2+\kappa^2} \cos(cqt) + \left(\frac{1}{q} - \frac{q}{q^2+\kappa^2}\right)\sin(cqt)\right] \end{aligned} $$ I don't find reasonable simplifications beyond this point.
Proceeding forward, one could transform $$ \begin{aligned} &\int_0^\infty d t\, e^{-i\omega t} e^{-\kappa ct}\left(e^{iqct} \pm e^{-iqct}\right)\\ &= \frac{1}{i(\omega - qc) + \kappa} \pm \frac{1}{i(\omega + qc) + \kappa}\\ &= \frac{-i(\omega - qc) + \kappa}{(\omega - qc)^2 + \kappa^2} \pm \frac{-i(\omega + qc) + \kappa}{(\omega + qc)^2 + \kappa^2}\\ &=\frac{\left(-i(\omega - qc) + \kappa\right)\left((\omega + qc)^2 + \kappa^2\right) \pm \left(-i(\omega + qc) + \kappa\right)\left((\omega - qc)^2 + \kappa^2\right)}{\left((\omega - qc)^2 + \kappa^2\right) \left((\omega + qc)^2 + \kappa^2\right)}. \end{aligned} $$ The denominator is pretty much where we want to end up in (3.2), except for a factor of $c^2$. I admit I didn't dive into simplifying this expression, but I believe that it is impossible that all contributions of $\omega$ up to cubic order should cancel (especially with the different prefactors to the $\cos$ and the $\sin$ contribution in $S(q,t)$) to leave but the simple $\kappa^2$ numerator in (3.2).
Of course, one could also try the backward route, starting from $${\cal F}\left[e^{-a |t|}\right] = \frac{a}{a^2 + \omega^2}, $$ (Tables list an additional factor 2 in the numerator, but I think that only applies if we use the Fourier transform from $-\infty$ to $\infty$, the one for positive $t$ should not have the factor 2, but this isn't essential here). With standard manipulations, then, we have $${\cal F}\left[e^{-\kappa |c t|} e^{\pm i qc t} \right] = \frac{1}{c}\frac{\kappa}{\kappa^2 + \left(\omega/c \mp q\right)^2}, $$ With the convolution theorem, we have $$ \begin{aligned} S(q,t)&={\cal F}\left[\frac{8\kappa^2/c}{\left(\kappa^2 + \left(\frac{\omega}{c} - q\right)^2\right) \left(\kappa^2 + \left(\frac{\omega}{c} + q\right)^2\right)}\right]\\ &=8c \left[\left(e^{-\kappa |c\ \cdot\ |} e^{+ i qc\ \cdot\ } \right) * \left(e^{-\kappa |c\ \cdot\ |} e^{- i qc\ \cdot\ } \right)\right](t)\\ &= 8c \int_0^\infty du\, e^{-\kappa c (| t - u| + u) + iqc(t-2u)}\\ &= 8c e^{-\kappa c t} e^{iqct} \int_0^t du\, e^{- 2iqcu} + 8c e^{\kappa c t}e^{iqct}\int_t^\infty du\, e^{-2 \kappa cu - 2iqcu}\\ &= 8c \frac{1}{-2iqc} e^{-\kappa c t} e^{iqct} \left[e^{- 2iqct} - 1\right] - 8c e^{\kappa c t}e^{iqct}\frac{1}{-2\kappa c - 2iqc} e^{-2 \kappa ct - 2iqct}\\ &= \frac{8}{q} e^{-\kappa c t} \sin(qct) + \frac{4}{\kappa + iq} e^{-\kappa c t}e^{-iqct}. \end{aligned} $$ This again looks closer in form to $S(q,t)$, but it's not quite the same. If the difference lay in the fact that the time-domain Fourier transform should also stretch from $-\infty$ to $\infty$, the equation would change (including factors of 2) to $$ \begin{aligned} S(q,t)&= \left(\frac{i}{q} + \frac{1}{iq+\kappa}\right)e^{-\kappa c t}e^{-iqct}. \end{aligned} $$ This looks nice and slim, but I don't see where the derivation of $S(q,t)$ starting from $(x,t)$ is wrong.