If $\varphi$ 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
$$\zeta(s)=s\int_0^{+\infty}\frac{\lfloor
x\rfloor-x}{x^{s+1}}\,dx\qquad (0<\sigma<1).$$