# 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.

• Why is it not a tempered distribution? Dec 6, 2015 at 19:47
• I might suggest looking at the Guinand Fourier-transform version of Riemann's explicit formula. (This is often referred-to as Weil's version, but Guinand's paper predates Weil's by 5 years...) Dec 6, 2015 at 21:53
• the inverse Fourier transform of $\zeta(1/2+2 i \pi f) e^{-\epsilon f^2}$ when $\epsilon \to 0$ will $\to$ to $-e^{-x/2}\frac{d}{dx} \{e^x\}$ Jan 19, 2016 at 13:13

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

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.

• What do you mean by the first statement? $\zeta(\frac12+it)$ is certainly not real for most real $t$ Dec 6, 2015 at 21:29
• Probably you want to say that the imaginary part is an odd function? Dec 6, 2015 at 21:50
• Thanks, I have corrected this. I had in mind the fact that $\zeta(\bar z)=\overline{\zeta(z)}$. Dec 7, 2015 at 10:45
• Thanks for this - how are you getting the Fourier transform of that second term? You end up with two nested integrals and I'm not seeing how you're rearranging things here. May 12, 2016 at 6:52
• FYI, posted a question about this on MSE here: math.stackexchange.com/q/1782031/52694. Has gotten no responses yet... May 13, 2016 at 2:35