7
$\begingroup$

I have seen the Fourier and Mellin transform for Riemann

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

where:

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

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

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

Where:

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

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.

$\endgroup$
0

1 Answer 1

8
$\begingroup$

$\zeta(s)=\sum_{n=1}^\infty n^{-s}$ 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 $f(u)= e^{-\sigma u}(S(u)-\sum_{k=0}^K \frac{B_k}{k!}e^{(1-k)u})$.

$\endgroup$
6
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.