# Questions on convergence of formula for $\zeta(s)$

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?

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 $$\zeta(s)$$.

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.

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

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

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

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

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

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).

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