# Can a functional equation of the form: $\zeta (s)=f(s) \zeta (s+1)$ exist?

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.

• Well you may intend some conditions on $f$. Easily $f(s)=1$ or $f(s)=1/(s+1)$ is possible to solve. Nov 6, 2017 at 13:40
• It certainly holds for $f(s)=\frac{\zeta(s)}{\zeta(s+1)}$. Welcome to the tautology club. Nov 6, 2017 at 13:41
• @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 Nov 7, 2017 at 10:21
• @reuns Shouldn't the $\zeta(s+\rho)$ in your last formula be $\zeta(s+k-\rho)$ ?
– Agno
Dec 2, 2017 at 19:19
– Agno
Dec 2, 2017 at 19:22

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 $$\frac{\partial\,\log\zeta(s)}{\partial s}=\frac{\zeta'(s)}{\zeta(s)}$$.

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

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

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

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

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

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

$Arg(\frac{\partial\,\log\zeta(s)}{\partial s})$ 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 $$log\,\zeta(s)$$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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