In 2008 I thought on this same problem. IfIf $\varphi$ is in the class of Schwartz we have $$\int_{-\infty}^{+\infty}\varphi(t)\zeta(\frac12+it)\,dt= \sum_{n=0}^\infty\Bigl\{ \frac{1}{\sqrt{n+1}}\widehat{\varphi}\Bigl(\frac{1}{2\pi}\log (n+1)\Bigr)- 2\pi\int_{x_n}^{x_{n+1}}e^{\pi y}\widehat{\varphi}(y)\,dy\Bigr\}$$ where $x_0=-\infty$ and $x_n=\frac{1}{2\pi}\log n$.
So that $\zeta(\frac12+it)$ is the Fourier transform of the tempered distribution defined by $$\varphi\in{\mathcal S}\mapsto \sum_{n=0}^\infty\Bigl\{ \frac{1}{\sqrt{n+1}}\varphi\Bigl(\frac{1}{2\pi}\log (n+1)\Bigr)- 2\pi\int_{x_n}^{x_{n+1}}e^{\pi y}\varphi(y)\,dy\Bigr\}$$
In general we can not separate the sum in two, but if $\varphi$ is such that $$\int_{-\infty}^{+\infty} e^{\pi y}|\widehat{\varphi}(y)|\,dy<+\infty$$ we can simplify and put $$\int_{-\infty}^{+\infty}\varphi(t)\zeta(\frac12+it)\,dt=\sum_{n=1}^\infty \frac{1}{\sqrt{n}}\widehat{\varphi}\Bigl(\frac{1}{2\pi}\log n\Bigr)-2\pi \int_{-\infty}^{+\infty} e^{\pi y}\widehat{\varphi}(y)\,dy.$$
We can say that $\zeta(\frac12+it)$ is the Fourier transform of a tempered distribution that can be obtained extending the measure $$\mu=\sum_{n=1}^\infty \frac{1}{\sqrt{n}}\delta_{\frac{1}{2\pi}\log n}-\nu,\quad \text{where} \quad \nu(dx)=2\pi e^{\pi x}\,dx,$$ in the indicated way.
To prove this I started from the formula (2.1.5) of Titchmarsh $$\zeta(s)=s\int_0^{+\infty}\frac{\lfloor x\rfloor-x}{x^{s+1}}\,dx\qquad (0<\sigma<1).$$