Are these formulas for the Riemann zeta function $\zeta(s)$ globally convergent?

Question

This question assumes the following definitions. With respect to the integrals in (3) and (4) 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/delta functions of $S(x)$/$S'(x)$ at $x=0$ while simultaneously maximizing the desirable contribution of the step/delta functions of $S(x)$/$S'(x)$ at $x=1$.

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(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) $\quad S'(x)=1+2\sum\limits_{k=1}^f\cos(2\,\pi\,k\,x)\,,\qquad\quad f\to\infty$

(3) $\quad\zeta(s)=s\int\limits_{1/2}^\infty S(x)\,x^{-s-1}\,dx\\$ $\qquad\quad=\frac{2^{\,s-1}\,s}{s-1}-\left(2^{\,s-1}-2^{\,s} s\sum\limits_{k=1}^f\left(\frac{\,_1F_2\left(\frac{1}{2}-\frac{s}{2};\frac{3}{2},\frac{3}{2}-\frac{s}{2};-\frac{1}{4} k^2 \pi^2\right)}{s-1}-\pi^{\,s-1} k^{\,s-1}\sin\left(\frac{\pi\,s}{2}\right)\,\Gamma(-s)\right)\right)\\$ $\qquad\quad=2^s\left(\frac{1}{2\,(s-1)}+\sum\limits_{k=1}^f\left(\pi^{s-1}\sin\left(\frac{\pi\,s}{2}\right)\,\Gamma(1-s)\,k^{s-1}+\frac{s}{s-1}\,_1F_2\left(\frac{1}{2}-\frac{s}{2};\frac{3}{2},\frac{3}{2}-\frac{s}{2};-\frac{1}{4} k^2 \pi^2\right)\right)\right)\\$ $\qquad\quad=2^s\left(\frac{1}{2 (s-1)}+\sum\limits_{k=1}^f\left(\pi^{s-1}\sin\left(\frac{\pi\,s}{2}\right)\,\Gamma (1-s)\,k^{s-1}+s\sum\limits_{j=0}^{\infty}\frac{(\pi\,i\,k)^{2\,j}}{(s-2\,j-1)\,\Gamma(2\,j+2)}\right)\right)$

(4) $\quad\zeta(s)=\int\limits_{1/2}^\infty S'(x)\,x^{-s}\,dx\\$ $\qquad\quad=\frac{2^{s-1}}{s-1}+2^s\sum\limits_{k=1}^f\left(\frac{\,_1F_2\left(\frac{1}{2}-\frac{s}{2};\frac{1}{2},\frac{3}{2}-\frac{s}{2};-\frac{1}{4} k^2 \pi^2\right)}{s-1}+\pi^{s-1} \sin\left(\frac{\pi\,s}{2}\right)\,\Gamma(1-s)\,k^{s-1}\right)\\$ $\qquad\quad=2^s\left(\frac{1}{2\,(s-1)}+\sum\limits_{k=1}^f\left(\pi^{s-1}\sin\left(\frac{\pi\,s}{2}\right)\,\Gamma(1-s)\,k^{s-1}+\frac{1}{s-1}\,_1F_2\left(\frac{1}{2}-\frac{s}{2};\frac{1}{2},\frac{3}{2}-\frac{s}{2};-\frac{1}{4} k^2 \pi^2\right)\right)\right)\\$ $\qquad\quad=2^s\left(\frac{1}{2\,(s-1)}+\sum\limits_{k=1}^f\left(\pi^{s-1}\sin\left(\frac{\pi\,s}{2}\right)\,\Gamma (1-s)\,k^{s-1}+\sum\limits_{j=0}^{\infty}\frac{(\pi\,i\,k)^{2\,j}}{(s-2\,j-1)\,\Gamma(2\,j+1)}\right)\right)$

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Formulas (3) and (4) above for $\zeta(s)$ are illustrated following the questions below.

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Question (1): Are formulas (3) and/or (4) for $\zeta(s)$ above globally convergent as $f\to\infty$?

Question (2): If so, does global convergence of formulas (3) and/or (4) for $\zeta(s)$ have any implications with respect to the Riemann Hypothesis?

Question (3): If not, what are the convergence ranges of these two formulas?

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I believe formulas (3) and (4) above can be shown to be equivalent if either of the two equalities below can be shown to be true. Note (6) below would follow from (5) below but not necessarily vice-versa.

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(5) $\quad \frac{s}{s-1}\,_1F_2\left(\frac{1}{2}-\frac{s}{2};\frac{3}{2},\frac{3}{2}-\frac{s}{2};-\frac{1}{4} k^2 \pi ^2\right)=\frac{1}{s-1}\,_1F_2\left(\frac{1}{2}-\frac{s}{2};\frac{1}{2},\frac{3}{2}-\frac{s}{2};-\frac{1}{4} k^2 \pi ^2\right)\\$ $$s\sum\limits_{j=0}^{\infty}\frac{(\pi\,i\,k)^{2\,j}}{(s-2\,j-1)\,\Gamma(2\,j+2)}=\sum\limits_{j=0}^{\infty}\frac{(\pi\,i\,k)^{2\,j}}{(s-2\,j-1)\,\Gamma(2\,j+1)}$$

(6) $\quad\frac{s}{s-1}\sum\limits_{k=1}^{\infty}\,_1F_2\left(\frac{1}{2}-\frac{s}{2};\frac{3}{2},\frac{3}{2}-\frac{s}{2};-\frac{1}{4} k^2 \pi ^2\right)=\frac{1}{s-1}\sum\limits_{k=1}^{\infty}\,_1F_2\left(\frac{1}{2}-\frac{s}{2};\frac{1}{2},\frac{3}{2}-\frac{s}{2};-\frac{1}{4} k^2 \pi ^2\right)\\$ $$s\sum\limits_{k=1}^\infty\sum\limits_{j=0}^{\infty}\frac{(\pi\,i\,k)^{2\,j}}{(s-2\,j-1)\,\Gamma(2\,j+2)}=\sum\limits_{k=1}^\infty\sum\limits_{j=0}^{\infty}\frac{(\pi\,i\,k)^{2\,j}}{(s-2\,j-1)\,\Gamma(2\,j+1)}$$

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Question (4): Can either of the equalities illustrated in (5) and (6) above be proven to be true?

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The following two figures illustrate formulas (3) and (4) for $\zeta(s)$ in orange where both formulas are evaluated with the upper limit $f=20$. The underlying blue reference function is $\zeta(s)$.

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

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Figure (2): Illustration of formula (4) for $\zeta(s)$ evaluated at $f=20$

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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.

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

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

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

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Figure (6): 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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The following four figures illustrate the absolute value, real part, imaginary part, and argument of formula (4) for $\zeta(s)$ evaluated along the critical line $s=\frac{1}{2}+i\,t$ in orange where formula (4) 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 (4) at the first $10$ non-trivial zeta-zeros in the upper half-plane.

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Figure (7): Illustration of formula (4) for $\left|\zeta\left(\frac{1}{2}+i\,t\right)\right|$ evaluated at $f=20$

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Figure (8): Illustration of formula (4) for $\Re\left(\zeta\left(\frac{1}{2}+i\,t\right)\right)$ evaluated at $f=20$

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Figure (9): Illustration of formula (4) for $\Im\left(\zeta\left(\frac{1}{2}+i\,t\right)\right)$ evaluated at $f=20$

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Figure (10): Illustration of formula (4) for $\text{Arg}\left(\zeta\left(\frac{1}{2}+i\,t\right)\right)$ evaluated at $f=20$

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Answer 1

• Avoid ${}_2F_1$. You don't need any hypergeometric function to talk about $g_{n,0}(s) = \int_1^\infty \sin(2\pi nx) s x^{-s-1}dx, \Re(s) > 0$ or its analytic continuation $g_{n,1}(s) =\int_1^\infty \frac{\cos(2\pi nx)-1}{2\pi n} s(s+1)x^{-s-2}dx,\Re(s) > -1$. Continuing this way we'll have

$$g_{n,2m}(s) = \prod_{l=0}^{2m-1} (s+l) \int_1^\infty \frac{\sin(2\pi nx)}{(2i\pi n)^{2m}} x^{-s-1-2m}dx + \sum_{k=1}^m \frac{\prod_{l=0}^{2k-1} (s+l)}{(2i\pi n)^{2k-1}},\Re(s) > -2m$$

• Let $h_{N,2m}(x) = \sum_{n=1}^N (-1)^{n+1} \frac{\sin(2\pi nx)}{(2i\pi n)^{2m}}$ then $\lim_{N \to \infty} h_{N,2m}=h_{\infty,2m}$ converges in $L^1(\Bbb{R/Z})$ thus $\lim_{N \to \infty} \sum_{n=1}^N (-1)^{n+1} g_{n,2m}(s)$ converges to an analytic function for $\Re(s) > -2m$ which must be the analytic continuation of $\lim_{N \to \infty} \sum_{n=1}^N (-1)^{n+1} g_{n,0}(s)=\int_1^\infty (2\{2x\}-\{x\}-\frac12)sx^{-s-1}dx$.

• Since $\zeta(s) = \frac{s}{s-1}-\frac12+\int_1^\infty (\frac12-\{x\})s x^{-s-1}dx$ for $\Re(s) > 0$ then for every $s$ $$(1-2^{s})\zeta(s) = (1-2^{s}) (\frac{s}{s-1}-\frac12)+ \lim_{N \to \infty} \sum_{n=1}^N(-1)^{n+1}g_{n,0}(s)$$