This was [asked on MSE](https://math.stackexchange.com/q/1546777/52694) and got a lot of upvotes but no answers, so I'm posting it here.


Is there a known expression for the (distributional) Fourier transform of the Riemann zeta function, taken along the critical line?


I'd love to say that it's a weighted sum of delta distributions, logarithmically spaced and decreasing in amplitude, as in


$\sum\_n \frac{\delta(\omega+\log(n))}{n^{1/2}}$


but this fails to be a tempered distribution, and fails in general when the exponent in the denominator is less than 1.



Fourier transform of the critical line of zeta?
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
$$\zeta(s)=s\int\_0^{+\infty}\frac{\lfloor
x\rfloor-x}{x^{s+1}}\,dx\qquad (0<\sigma<1).$$


