Start with the explicit formula
$$\sum_{n \le x}\Lambda(n) =\frac1{2i\pi} \int_{2-i\infty}^{2+i\infty} \frac{-\zeta'(s)}{\zeta(s)}\frac{x^s}s ds=1_{x > 1}\sum Res(\frac{-\zeta'(s)}{\zeta(s)}\frac{x^s}s)$$ $$=1_{x > 1}( x - \sum_\rho \frac{x^\rho}{\rho} - \frac12 \log 2\pi - \sum_{k=1}^\infty \frac{x^{-2k}}{-2k})$$
Since $\sum_\rho \frac1{|\rho|^2}<\infty$ the RHS converges in $L^1_{loc}$ thus we can differentiate both sides in the sense of distributions, the RHS being continuous at $1$ we get
$$\sum_n \Lambda(n) \delta(x-n) =1_{x > 1} - 1_{x > 1}\sum_\rho x^{\rho-1} +\frac{d}{dx} 1_{x > 1}\log(1-x^{-2})$$
If the RH is true, letting $x =e^u$ and multiplying both side by $e^{u/2}$, since $e^{u/2}\delta(e^u-n) = \frac{\delta(u-\log n)}{n^{1/2}}$
$$\sum_n \frac{\Lambda(n)}{n^{1/2}}\delta(u-\log n) =1_{u > 0}e^{u/2} - 1_{u > 0}\sum_t e^{itu} +e^{-u/2}\frac{d}{du}1_{u > 0} \log(1-e^{-2u}))$$
Making both side even
$$\sum_n \frac{\Lambda(n)}{n^{1/2}} (\delta(u-\log n)+\delta(u+\log n))=e^{|u|/2} - \sum_t e^{itu} -e^{-|u|/2}(\frac{d}{d|u|} \log(1-e^{-2|u|})$$
Which means that we have the Fourier transform in the sense of distributions $$\mathcal{F}^{-1}\left[2 \pi \sum_t \delta(\omega-t)\right] = e^{|u|/2}-e^{-|u|/2}(\frac{d}{d|u|} \log(1-e^{-2|u|})-\sum_n \frac{\Lambda(n)}{n^{1/2}} (\delta(u-\log n)+\delta(u+\log n)) $$
If the RH is not true then those things are true only in the sense of analytic functionals, for example they hold when using Gaussians as test functions.
