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Is it known if a functional equation of the form:

$$\zeta (s)=f(s) \zeta (s+1)$$

can exist?

If it is possible for such a functional equation to exist then I believe lots of wonderful things would happen. In particular one could solve this integral:

$$\int \log (\zeta (s)) \, ds$$

on the critical line by extending the validity of the Euler product formula to it.

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  • $\begingroup$ Well you may intend some conditions on $f$. Easily $f(s)=1$ or $f(s)=1/(s+1)$ is possible to solve. $\endgroup$
    – Macavity
    Nov 6, 2017 at 13:40
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    $\begingroup$ It certainly holds for $f(s)=\frac{\zeta(s)}{\zeta(s+1)}$. Welcome to the tautology club. $\endgroup$ Nov 6, 2017 at 13:41
  • $\begingroup$ @JackD'Aurizio Did you know the Riemann explicit formula gives for $\Re(s) > \sigma_{RH}$ : $\frac{-\zeta'(s)}{\zeta(s)}= \sum_{n=2}^\infty (\psi(n+1/2)-\psi(n-1/2)) n^{-s} = \sum_{n=2}^\infty (1-\sum_\rho \frac{(n+1/2)^\rho-(n-1/2)^\rho}{\rho})n^{-s}$ $=(\zeta(s)-1)-\sum_\rho \frac{1}{\rho}\sum_{k \ge 0} (\zeta(s+\rho)-1) {\rho \choose k} (2^{-k}-(-2)^{-k})$ where $\rho$ are the trivial and non-trivial zeros $\endgroup$
    – reuns
    Nov 7, 2017 at 10:21
  • $\begingroup$ @reuns Shouldn't the $\zeta(s+\rho)$ in your last formula be $\zeta(s+k-\rho)$ ? $\endgroup$
    – Agno
    Dec 2, 2017 at 19:19
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    $\begingroup$ @MatsGranvik Although it is the reciprocal of your question, maybe this link is helpful mathoverflow.net/questions/23378/zetas1-zetas $\endgroup$
    – Agno
    Dec 2, 2017 at 19:22

1 Answer 1

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Substituting $s=s+1$ into the relationship $\frac{\zeta (s-1)}{\zeta (s)}=\sum _{n=1}^{\infty } \frac{\phi (n)}{n^s}$ leads to $\zeta (s)=\zeta (s+1) \sum _{n=1}^{\infty } \frac{\phi (n)}{n^{s+1}}$ which is valid for $\Re(s)>1$.

Also, $\int\log(\zeta(s))\,ds=-\sum\limits_{n=2}^\infty\frac{\Lambda(n)}{\log(n)^2}\,n^{-s}$, but again this relationship only converges for $\Re(s)>1$.

I'm not sure where you're headed with this, but I suspect you'll be interested in the paper at the following link:

DISCRETE MELLIN CONVOLUTION AND ITS EXTENSIONS, PERRON FORMULA AND EXPLICIT FORMULAE


April 3, 2018 Update:


I've attempted to derive a formula for $\int\log\zeta(s)\,ds$ that assuming the Riemann hypothesis converges for $\Re(s)>\frac{1}{2}$ starting with the following relationship.

(1) $\quad\frac{\partial\,\log\zeta(s)}{\partial s}=\frac{\zeta'(s)}{\zeta(s)}=-s\left(\frac{1}{s-1}+\int_1^N x^{-s-1}(\psi(x)-x)\,dx\right),\quad N\to\infty\land\Re(s)>\frac{1}{2}$


The following formula for $\frac{\partial\,\log\zeta(s)}{\partial s}=\frac{\zeta'(s)}{\zeta(s)}$ was derived from relationship (1) above.

(2) $\quad\frac{\partial\,\log\zeta(s)}{\partial s}=\frac{\zeta'(s)}{\zeta(s)}=\frac{s\,N^{1-s}}{1-s}+\sum\limits_{n=1}^N\Lambda(n)\left(N^{-s}-n^{-s}\right),\quad N\to\infty\land\Re(s)>\frac{1}{2}$


Integrating (2) above leads to the following where I added the $-sgn(\Im(s))\,\pi\,i$ term to adjust for the branch point of the $Ei$ function at $0$.

(3) $\quad\log\zeta(s)=-sgn(\Im(s))\pi\,i-Ei\,((1-s) \log (N))+\frac{N^{1-s}}{\log (N)}+\sum\limits_{n=2}^N\Lambda(n)\left(\frac{n^{-s}}{\log(n)}-\frac{N^{-s}}{\log (N)}\right),\\$ $\qquad\qquad\qquad\qquad\qquad\qquad\qquad N\to\infty\land\Re(s)>\frac{1}{2}$


Integrating (3) above leads to the following where this time I added the $sgn(\Im(s))\,\pi\,i$ term to adjust for the branch point of the $Ei$ function at $0$.

(4) $\quad\int\log\zeta(s)\,ds=sgn(\Im(s))\,\pi\,i+(1-s)\,Ei\,((1-s)\log(N))-\frac{(\log(N)+1)\,N^{1-s}}{\log^2(N)}+\\$ $\qquad\qquad\quad\left( \begin{array}{cc} \{ & \begin{array}{cc} -i \pi s & \Im(s)>0 \\ i \pi s & \Im(s)<0 \\ \end{array} \\ \end{array} \right)+\sum\limits_{n=2}^N\Lambda(n)\left(\frac{N^{-s}}{\log^2(N)}-\frac{n^{-s}}{\log^2(n)}\right),\quad N\to\infty\land\Re(s)>\frac{1}{2}$


The formulas for $\frac{\partial\,\log\zeta(s)}{\partial s}$, $log\,\zeta(s)$, and $\int\log\zeta(s)\,ds$ in (2), (3), and (4) above are illustrated below where all plots are evaluated along the critical line $s=\frac{1}{2}+i\,t$ using the limit $N=1000$. The reference functions are illustrated in orange and the right-side of the formulas are illustrated in blue, The red discrete portions of the plots illustrate the evaluations of the right-side of the formulas at the non-trivial zeta zeros.

Note how the diverging oscillation in the plots below seems to decrease in magnitude as successive integrals are taken first from $\frac{\partial\,\log\zeta(s)}{\partial s}$ to $log\,\zeta(s)$ and second from $log\,\zeta(s)$ to $\int\log\zeta(s)\,ds$.


The following four plots illustrate the real, imaginary, absolute, and argument of formula (2) above for $\frac{\partial\,\log\zeta(s)}{\partial s}=\frac{\zeta'(s)}{\zeta(s)}$.


Illustration of <span class=$\Re(\frac{\partial\,\log\zeta(s)}{\partial s})$ associated with Formula (2) for $s=\frac{1}{2}+i\,t$">

Figure (1): $\text{Illustration of $\Re(\frac{\partial\,\log\zeta(s)}{\partial s})$ associated with Formula (2) for $s=\frac{1}{2}+i\,t$}$


Illustration of <span class=$\Im(\frac{\partial\,\log\zeta(s)}{\partial s})$ associated with Formula (2) for $s=\frac{1}{2}+i\,t$">

Figure (2): $\text{Illustration of $\Im(\frac{\partial\,\log\zeta(s)}{\partial s})$ associated with Formula (2) for $s=\frac{1}{2}+i\,t$}$


Illustration of <span class=$Abs(\frac{\partial\,\log\zeta(s)}{\partial s})$ associated with Formula (2) for $s=\frac{1}{2}+i\,t$">

Figure (3): $\text{Illustration of $Abs(\frac{\partial\,\log\zeta(s)}{\partial s})$ associated with Formula (2) for $s=\frac{1}{2}+i\,t$}$


Illustration of <span class=$Arg(\frac{\partial\,\log\zeta(s)}{\partial s})$ associated with Formula (2) for $s=\frac{1}{2}+i\,t$">

Figure (4): $\text{Illustration of $Arg(\frac{\partial\,\log\zeta(s)}{\partial s})$ associated with Formula (2) for $s=\frac{1}{2}+i\,t$}$


The following four plots illustrate the real, imaginary, absolute, and argument of formula (3) above for $log\,\zeta(s)$.


Illustration of <span class=$\Re(\log\zeta(s))$ associated with Formula (3) for $s=\frac{1}{2}+i\,t$">

Figure (5): $\text{Illustration of $\Re(\log\zeta(s))$ associated with Formula (3) for $s=\frac{1}{2}+i\,t$}$


Illustration of <span class=$\Im(\log\zeta(s))$ associated with Formula (3) for $s=\frac{1}{2}+i\,t$">

Figure (6): $\text{Illustration of $\Im(\log\zeta(s))$ associated with Formula (3) for $s=\frac{1}{2}+i\,t$}$


Illustration of <span class=$Abs(\log\zeta(s))$ associated with Formula (3) for $s=\frac{1}{2}+i\,t$">

Figure (7): $\text{Illustration of $Abs(\log\zeta(s))$ associated with Formula (3) for $s=\frac{1}{2}+i\,t$}$


Illustration of <span class=$Arg(\log\zeta(s))$ associated with Formula (3) for $s=\frac{1}{2}+i\,t$">

Figure (8): $\text{Illustration of $Arg(\log\zeta(s))$ associated with Formula (3) for $s=\frac{1}{2}+i\,t$}$


The following four plots illustrate the real, imaginary, absolute, and argument of formula (4) above for $\int\log\zeta(s)\,ds$.


Illustration of <span class=$\Re(\int\log\zeta(s)\,ds)$ associated with Formula (4) for $s=\frac{1}{2}+i\,t$]">

Figure (9): $\text{Illustration of $\Re(\int\log\zeta(s)\,ds)$ associated with Formula (4) for $s=\frac{1}{2}+i\,t$}$


Illustration of <span class=$\Im(\int\log\zeta(s)\,ds)$ associated with Formula (4) for $s=\frac{1}{2}+i\,t$">

Figure (10): $\text{Illustration of $\Im(\int\log\zeta(s)\,ds)$ associated with Formula (4) for $s=\frac{1}{2}+i\,t$}$


Illustration of <span class=$Abs(\int\log\zeta(s)\,ds)$ associated with Formula (4) for $s=\frac{1}{2}+i\,t$">

Figure (11): $\text{Illustration of $Abs(\int\log\zeta(s)\,ds)$ associated with Formula (4) for $s=\frac{1}{2}+i\,t$}$


Illustration of <span class=$Arg(\int\log\zeta(s)\,ds)$ associated with Formula (4) for $s=\frac{1}{2}+i\,t$">

Figure (12): $\text{Illustration of $Arg(\int\log\zeta(s)\,ds)$ associated with Formula (4) for $s=\frac{1}{2}+i\,t$}$


12/04/2018 Update:


Formula (7) below for $\zeta(s)=f(s)\,\zeta(s+1)$ was derived from the relationships illustrated in (5) and (6) below. Formula (7) below assumes the Riemann Hypothesis (RH).


(5) $\quad\zeta(s)=\frac{2\,\pi^{-\frac{1}{2}}\,\Gamma\left(\frac{s+3}{2}\right)}{(s-1)\,\Gamma\left(\frac{s}{2}\right)}\frac{\xi(s)}{\xi(s+1)}\zeta(s+1)$

(6) $\quad \xi(s)=\xi(0)\prod\limits_{k=1}^\infty\left(1-\frac{s}{\rho_k}\right)\left(1-\frac{s}{\rho_{-k}}\right)$

(7) $\quad \zeta(s)=\frac{2\,\pi^{-\frac{1}{2}}\Gamma\left(\frac{s+3}{2}\right)}{(s-1)\Gamma\left(\frac{s}{2}\right)}\left(\prod\limits_{k=1}^\infty\left(1-\frac{8\,s}{4\,\Im\left(\rho_k\right){}^2+(2\,s+1)^2}\right)\right)\zeta(s+1)\qquad\text{(assuming RH)}$


The following four figures illustrate formula (7) for $\zeta(s)$ above evaluated along the critical line $s=\frac{1}{2}+i\,t$ where formula (7) is evaluated over the first $200$ non-trivial zeta zeros in the upper half plane. Formula (7) is illustrated in orange, and $\zeta(s)$ is illustrated in blue as a reference.


Illustration of formula (7) for Abs(zeta(1/2+i t))

Figure (13): Illustration of formula (7) for $\left|\zeta(\frac{1}{2}+i\,t)\right|$


Illustration of formula (7) for Re(zeta(1/2+i t))

Figure (14): Illustration of formula (7) for $\Re\left(\zeta(\frac{1}{2}+i\,t)\right)$


Illustration of formula (7) for Im(\zeta(1/2+i t))

Figure (15): Illustration of formula (7) for $\Im\left(\zeta(\frac{1}{2}+i\,t)\right)$


Illustration of formula (7) for Arg(zeta(1/2+i t))

Figure (16): Illustration of formula (7) for $Arg\left(\zeta(\frac{1}{2}+i\,t)\right)$


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    $\begingroup$ What I am asking for is probably an impossibility. What you posted here is probably as close as one can get. $\endgroup$ Feb 20, 2018 at 18:16

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