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.

  • $\begingroup$ Why is it not a tempered distribution? $\endgroup$
    – Fan Zheng
    Dec 6, 2015 at 19:47
  • $\begingroup$ 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...) $\endgroup$ Dec 6, 2015 at 21:53
  • $\begingroup$ 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\}$ $\endgroup$
    – reuns
    Jan 19, 2016 at 13:13

2 Answers 2


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.

  • $\begingroup$ What do you mean by the first statement? $\zeta(\frac12+it)$ is certainly not real for most real $t$ $\endgroup$ Dec 6, 2015 at 21:29
  • $\begingroup$ Probably you want to say that the imaginary part is an odd function? $\endgroup$ Dec 6, 2015 at 21:50
  • 1
    $\begingroup$ Thanks, I have corrected this. I had in mind the fact that $\zeta(\bar z)=\overline{\zeta(z)}$. $\endgroup$
    – Bazin
    Dec 7, 2015 at 10:45
  • $\begingroup$ 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. $\endgroup$ May 12, 2016 at 6:52
  • $\begingroup$ FYI, posted a question about this on MSE here: math.stackexchange.com/q/1782031/52694. Has gotten no responses yet... $\endgroup$ May 13, 2016 at 2:35

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