No worries. I’m fairly sure it just has to do with the convention of the Fourier transform used. Looking at the \nu=\infty case using the Rasmussen and Williams convention:
S(\omega) = \int_Re^{r^2/2l^2}e^{-2\pi i\omega r}dr=\sqrt{2\pi}le^{-2\pi^2l^2\omega^2} which matches the formula given in the textbook (pg.83)
Using the Solin and Särrkä convention:
S(\omega) = \int_Re^{r^2/2l^2}e^{-i\omega r}dr=\sqrt{2\pi}le^{-1/2l^2\omega^2} which matches the formula in the Riutort et al. paper.
So it seems like for the general case which I wasn’t able to verify the integration for one substitutes \omega = \omega/2\pi in the Rasmussen and Williams equation to get equations (1) - (3).