I have seen the Fourier and Mellin transform for Riemann

$\Xi (t)=\xi ({\frac 12}+it)$


$\xi (s)={\tfrac {1}{2}}s(s-1)\pi ^{{-s/2}}\Gamma \left({\tfrac {1}{2}}s\right)\zeta (s)$

Fourier transform of $\Xi(t)$ is:

$\Xi (t) = \int_{-\infty}^\infty\Phi(u)e^{iut}\,du$


$\Phi(u) = \sum_{n=1}^\infty (4\pi^2n^4e^{9u/2} - 6n^2\pi e^{5u/2} ) exp(-n^2\pi e^{2u})$

But I had not seen something similar for Fourier or Mellin transform for Riemann $\zeta(s)$ itself.

Is there a "closed form" formula for Fourier or Mellin transform for Riemann $\zeta(s)$ itself ?

Thank you.


1 Answer 1


$\zeta(s)=\sum_{n=1}^\infty n^{-s}$ is the Laplace transform of the distribution $S(u) = \sum_{n=1}^\infty \delta(u-\log n)$.

  • $\zeta(s) = \mathcal{L}[S(u)](s) = \int_{-\infty}^\infty S(u) e^{-su}du$ converges for $\Re(s) > 1$.

  • For $\Re(s) \in (0,1)$ it becomes the bilateral Laplace transform $\zeta(s) = \mathcal{L}[S(u)-e^u](s) =\int_{-\infty}^\infty (S(u)-e^u) e^{-su}du$

  • For $\Re(s) \in (-1,0)$ it is $\zeta(s) = \mathcal{L}[S(u)-e^u+\frac{1}{2}](s) =\int_{-\infty}^\infty (S(u)-e^u+\frac{1}{2}) e^{-su}du$

  • For $\Re(s) \in (-K,-K+1)$ it is $\zeta(s) = \mathcal{L}[S(u)-\sum_{k=0}^K \frac{B_k}{k!}e^{(1-k)u}](s) =\int_{-\infty}^\infty (S(u)-\sum_{k=0}^K \frac{B_k}{k!}e^{(1-k)u}) e^{-su}du$ where $B_k$ are the Bernouilli numbers.

Thus for $\sigma \in (-K,-K+1)$, the inverse Fourier transform (in the sense of distributions) of $\hat{f}(\xi)=\zeta(\sigma+2i \pi \xi)$ is $f(u)= e^{-\sigma u}(S(u)-\sum_{k=0}^K \frac{B_k}{k!}e^{(1-k)u})$.

  • $\begingroup$ It is a consequence of the EMSF $\endgroup$
    – reuns
    Jul 26, 2017 at 4:25
  • $\begingroup$ Thank you ! Is there something similar for "closed form" Mellin transform of Riemann $\zeta(s)$ ? $\endgroup$
    – david
    Jul 26, 2017 at 15:25
  • 5
    $\begingroup$ @david Come on.. $\int_0^\infty f(x) x^{-s-1}dx=\int_{-\infty}^\infty f(e^u) e^{-su}du$ $\endgroup$
    – reuns
    Jul 27, 2017 at 3:53
  • $\begingroup$ Very instructive comment. Could you elaborate a little bit more on the meaning of the regularization terms for $0<|\Re(s)|<1$ and how to get them? $\endgroup$
    – Alexandre
    Jun 27, 2018 at 7:17
  • 1
    $\begingroup$ @Alexandre My answer really follows the same process as in $\frac{\zeta(s)}{s} = \int_1^\infty \lfloor x \rfloor x^{-s-1}dx$ for $\Re(s) > 1$ and $\frac{\zeta(s)}{s} = \frac{1}{s-1}+\int_1^\infty (\lfloor x \rfloor-x) x^{-s-1}dx=\int_0^\infty (\lfloor x \rfloor-x) x^{-s-1}dx$ for $\Re(s) \in (0,1)$ and $S(u)$ is the derivative of $\lfloor e^u \rfloor$ $\endgroup$
    – reuns
    Sep 8, 2018 at 21:55

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