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This question assumes definition (1) below and relationship (2) below. With respect to the integral in (2) below, I selected $\frac{1}{2}$ as the lower integration bound because this is the ideal location for minimizing the undesirable contribution of the step of $S(x)$ at $x=0$ while simultaneously maximizing the desirable contribution of the step of $S(x)$ at $x=1$.


(1) $\quad S(x)=x-\left(\frac{1}{2}-\frac{1}{\pi}\sum\limits_{k=1}^f\frac{\sin(2\,\pi\,k\,x)}{k}\right),\quad f\to\infty$

(2) $\quad\zeta(s)=s\int\limits_{1/2}^\infty S(x)\,x^{-s-1}\,dx$


I originally illustrated a couple of formulas for $\zeta(s)$ based on definition (1) and relationship (2) above in my earlier question Are these formulas for the Riemann zeta function $\zeta(s)$ globally convergent? which involved the hypergeometric $_1F_2$ function.


The question here is about formula (3) below which was also derived from definition (1) and relationship (2) above but is also based on this answer to my follow-on question What is $s\int_1^\infty\sin(2\,\pi\,n\,x)\,x^{-s-1}\,dx$?


(3) $\quad\zeta(s)=\underset{f\to\infty}{\text{lim}}\quad 2^{\,s-1}\left(\frac{s}{s-1}-1+\sum\limits_{n=1}^f\left(E_s(i n \pi)+E_s(-i n \pi)\right)\right)$


Formula (3) above for $\zeta(s)$ is illustrated following the questions below.


Question (1): Is formula (3) for $\zeta(s)$ above globally convergent as $f\to\infty$?

Question (2): If so, does global convergence of formula (3) for $\zeta(s)$ have any implications with respect to the Riemann Hypothesis?

Question (3): If not, what is the convergence range of this formula?


Formula (5) below defines another globally convergent formula for $\zeta(s)$ based on relationship (4) below and this second answer to my question What is $s\int_1^\infty\sin(2\,\pi\,n\,x)\,x^{-s-1}\,dx$?.

(4) $\quad\zeta(s)=s\int\limits_1^\infty S(x)\,x^{-s-1}\,dx$

(5) $\quad\zeta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{s}{s-1}-\frac{1}{2}+\sum\limits_{n=1}^K\left((2 \pi i n)^{s-1} \Gamma (1-s,2 \pi i n)+(-2 \pi i n)^{s-1} \Gamma (1-s,-2 \pi i n)\right)\right)\\$ $\qquad\qquad\quad=\underset{K\to\infty}{\text{lim}}\left(\frac{s}{s-1}-\frac{1}{2}+\sum_\limits{n=1}^K\left(E_s(2 \pi i n)+E_s(-2 \pi i n)\right)\right)$


Note formula (3) for $\zeta(s)$ above was derived from the relationship $\zeta(s)=s\int\limits_{1/2}^\infty S(x)\,x^{-s-1}\,dx$ and formula (5) for $\zeta(s)$ above was derived from the relationship $\zeta(s)=s\int\limits_1^\infty S(x)\,x^{-s-1}\,dx$.


Question (4): Can a globally convergent formula for $\zeta(s)$ be derived from the more general integral $\zeta(s)=s\int\limits_a^\infty S(x)\,x^{-s-1}\,dx$ for any $0<a\le 1$?


The following figure illustrate formula (3) for $\zeta(s)$ in orange where formula (3) is evaluated with the upper limit $f=20$. The underlying blue reference function is $\zeta(s)$.


Illustration of formula (3) for zeta(s)

Figure (1): Illustration of formula (3) for $\zeta(s)$ evaluated at $f=20$


The following four figures illustrate the absolute value, real part, imaginary part, and argument of formula (3) for $\zeta(s)$ evaluated along the critical line $s=\frac{1}{2}+i\,t$ in orange where formula (3) is evaluated with the upper limit $f=20$. The underlying blue reference function is $\zeta(\frac{1}{2}+i\,t)$. The red discrete portion of the plot illustrates the evaluation of formula (3) at the first $10$ non-trivial zeta-zeros in the upper half-plane.


Illustration of formula (3) for |zeta(1/2+it)|

Figure (2): Illustration of formula (3) for $\left|\zeta\left(\frac{1}{2}+i\,t\right)\right|$ evaluated at $f=20$


Illustration of formula (3) for Re(zeta(1/2+it))

Figure (3): Illustration of formula (3) for $\Re\left(\zeta\left(\frac{1}{2}+i\,t\right)\right)$ evaluated at $f=20$


Illustration of formula (3) for Im(zeta(1/2+it))

Figure (4): Illustration of formula (3) for $\Im\left(\zeta\left(\frac{1}{2}+i\,t\right)\right)$ evaluated at $f=20$


Illustration of formula (3) for Arg(zeta(1/2+it))

Figure (5): Illustration of formula (3) for $\text{Arg}\left(\zeta\left(\frac{1}{2}+i\,t\right)\right)$ evaluated at $f=20$

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  • $\begingroup$ If $n$ is integer then $\sin(n\pi)=0$ in (3). $\endgroup$ Commented Nov 22, 2019 at 21:10
  • $\begingroup$ @NikosBagis Good point. I deleted the sin term from formula (3). $\endgroup$ Commented Nov 22, 2019 at 21:21

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You already asked the same question two times.

Look at the analytic continuations of $$g(s,2\pi n)=\int_{2\pi n}^\infty \sin(x)x^{-s-1}dx, \int_1^\infty \sin(2\pi nx)x^{-s-1}dx= (2\pi n)^s g(s,2\pi n), \Re(s) > 1$$

  • For $\Re(s) > 0$, $\zeta(s) = \frac{s}{s-1}-\frac12+s\int_1^\infty (\frac12-\{x\})x^{-s-1}dx$

  • Look at the Fourier series $\frac12-\{x\}=\sum_{n=1}^\infty \frac{\sin(2\pi nx)}{\pi n}$. Theorem : since $\frac12-\{x\}\in L^2(\Bbb{R/Z})$ the Fourier series converges in the $L^2(\Bbb{R/Z})$ thus in the $L^1(\Bbb{R/Z})$ norm.

  • Thus for $\Re(s) > 0$ $$\zeta(s) = \frac{s}{s-1}-\frac12+s\lim_{N \to \infty}\int_1^\infty \sum_{n=1}^N \frac{\sin(2\pi nx)}{\pi n}x^{-s-1}dx$$ $$=\frac{s}{s-1}-\frac12+s\sum_{n=1}^\infty \frac{(2\pi n)^s g(s,2\pi n)}{\pi n}$$

The question now is if $\sum_{n=1}^\infty \frac{(2\pi n)^s g(s,2\pi n)}{\pi n}$ converges on a larger domain. The answer is yes, by integrating by parts two times $$ g(s,2\pi n) = (s+1)( (2\pi n)^{-s-2} - (s+2) g(s+2,2\pi n) )$$

This proves

$$\sum_{n=1}^\infty \frac{(2\pi n)^s g(s,2\pi n)}{\pi n}$$ converges and is analytic for all $s$.

Qed. $$\zeta(s) = \frac{s}{s-1}-\frac12+s\sum_{n=1}^\infty \frac{(2\pi n)^s g(s,2\pi n)}{\pi n}$$ is valid for all $s$.

It works exactly the same way when starting with $\zeta(s)= \frac{s 2^{s-1}}{s-1}-\frac14 +s\int_{1/2}^\infty (\frac12-\{x\})x^{-s-1}dx$ obtaining $\zeta(s) = \frac{s 2^{s-1}}{s-1}-\frac14 +s\sum_{n=1}^\infty \frac{(2\pi n)^s g(s,\pi n)}{\pi n}$

It has nothing to do with the Riemann hypothesis, you need to look at the linear combinations of Dirichlet L-functions keeping exactly the same properties as $\zeta(s)$ for the analytic continuation, functional equation, series and integral representations, only loosing the Euler product and thus having infinitely many zeros in $\Re(s)\in (1-\epsilon,1+\epsilon)$.

The series representations in the critical strip have to do with the Lindelöf hypothesis (which stays true for linear combinations of Dirichlet series satisfying it).

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  • $\begingroup$ So a globally convergent formula be derived from the more general integral $\int_a^\infty$ for any $0<a\le 1$? $\endgroup$ Commented Nov 23, 2019 at 4:19

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