Fourier transform of the critical line of zeta? This was asked on MSE 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.
 A: 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
$$\zeta(s)=s\int_0^{+\infty}\frac{\lfloor
x\rfloor-x}{x^{s+1}}\,dx\qquad (0<\sigma<1).$$
A: The function $\mathbb R\ni t\mapsto\zeta(\frac12+it)$
is  analytic and smaller in absolute value than $C(1+\vert t\vert)^{1/6}$
(the $1/6$ may be replaced by $9/56$ and even by a slightly smaller number). It is thus a tempered distribution. We have, with $E(x)$ standing for the floor function,
\begin{multline}
\zeta(\frac 1 2 +it)=-it\int_{1}^{+\infty}\bigl(x-E(x)\bigr)x^{-\frac3 2 -it }dx
-\frac 1 2\int_{1}^{+\infty}\bigl(x-E(x)\bigr)x^{-\frac3 2 -it }dx\\-\frac{1+2it}{1-2it},
\tag{$\ast$}
\end{multline}
an identity which follows from the first step of the Euler-Maclaurin formula.
With the above formula, it is easy to find an explicit expression for the Fourier transform: in fact, we need only to calculate the Fourier transform of $t\mapsto e^{-it \ln x}$, which is $\delta_0(\tau+\frac{\ln x}{2π})$
and moreover, for $\phi$ in the Schwartz space,
the integral
$$
\int_{1}^{+\infty}\bigl(x-E(x)\bigr)x^{-\frac3 2 }\phi(-\frac{\ln x}{2π})dx,
$$
is absolutely converging. As a result, the Fourier transform of the second term in $(\ast)$ is given by
$$
\int_{1}^{+\infty}\bigl(x-E(x)\bigr)x^{-\frac3 2 }\delta_0(\tau+\frac{\ln x}{2π})dx,
$$
which makes sense as a Radon measure. The first term has a Fourier transform which is essentially the derivative of the above Radon measure, because of the $t$ in front, whereas the Fourier transform of the last term is easy to get explicitly.
