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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 $x=1$.


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


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


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?


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.


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


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


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 $\zeta(s)$.


Illustration of formula (3) evaluated at f=20

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


Illustration of formula (4) evaluated at f=20

Figure (2): Illustration of formula (4) 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 absolute value of formula (3) evaluated along the critical line at f=20

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


Illustration of real part of formula (3) evaluated along the critical line at f=20

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


Illustration of imaginary part of formula (3) evaluated along the critical line at f=20

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


Illustration of argument of formula (3) evaluated along the critical line at f=20

Figure (6): Illustration of formula (3) for $\text{Arg}\left(\zeta\left(\frac{1}{2}+i\,t\right)\right)$ evaluated at $f=20$


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.


Illustration of absolute value of formula (4) evaluated along the critical line at f=20

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


Illustration of real part of formula (4) evaluated along the critical line at f=20

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


Illustration of imaginary part of formula (4) evaluated along the critical line at f=20

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


Illustration of argument of formula (4) evaluated along the critical line at f=20

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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  • $\begingroup$ Replace your ${}_1F_2$ by something readable and meaningful if you want an answer. $\endgroup$
    – reuns
    Commented Apr 30, 2019 at 16:23
  • $\begingroup$ @reuns I updated formulas (3) and (4) in an attempt to make them more readable and easier to compare. I'll expand the $_1F_2$ functions when I find more time. $\endgroup$ Commented Apr 30, 2019 at 17:30

1 Answer 1

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  • 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 $s$ $$(1-2^{s})\zeta(s) = (1-2^{s}) (\frac{s}{s-1}-\frac12)+ \lim_{N \to \infty} \sum_{n=1}^N(-1)^{n+1}g_{n,0}(s)$$

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  • $\begingroup$ The integral $g_{n,0}(s)=s\int_1^{\infty}\frac{\sin(2\,\pi\,n\,x)}{\pi\,k}\,x^{-s-1}\,dx$ evaluates to $g_{n,0}(s)=\frac{2\,s}{s-1}\,_1F_2\left(\frac{1}{2}-\frac{s}{2};\frac{3}{2},\frac{3}{2}-\frac{s}{2};-n^2 \pi ^2\right)+2^s\,\pi^{s-1} \sin\left(\frac{\pi\,s}{2}\right)\,\Gamma(1-s)\,n^{s-1}$ which is conditionally valid for $\Re(s)>-1$, so it still involves a hypergeometric $_1F_2$ function. Also $\zeta(s)=\frac{s}{s-1}-\frac{1}{2}+\frac{1}{1-2^s}\sum _{n=1}^N (-1)^{n+1}g_{n,0}(s)$ doesn't seem to converge to $\zeta(s)$. $\endgroup$ Commented Apr 30, 2019 at 22:31
  • $\begingroup$ I also tried $g_{n,0}(s)=s\int_1^{\infty}\sin(2\,\pi\,n\,x)\,x^{-s-1}\,dx$ the way you've written it with somewhat similar results. $\endgroup$ Commented Apr 30, 2019 at 22:35
  • $\begingroup$ Please avoid hypergeometric functions, nobody cares that you use them to plug formulas in mathematica. Then do you understand what I did ? I integrated by parts $\int_1^\infty \sin(2 \pi nx)x^{-s-1}dx$ ($\ \ 2m$ times) to obtain something close to $\int_1^\infty \frac{\\sin(2 \pi nx)}{(2i\pi n)^{2m}}x^{-s-1-2m}dx$ that we can sum over $n$ for $\Re(s) > -2m$, giving an analytic continuation of $\zeta(s)$. When integrating by parts $\{x\}$ instead of $\sin(2\pi nx)$ this is the well-known Bernouilli numbers&polynomials / EMSF analytic continuation of $\zeta(s)$. $\endgroup$
    – reuns
    Commented Apr 30, 2019 at 22:36
  • $\begingroup$ I can expand the $_1F_2$ function as I did in my answer above if desired, but what is the result of the integral associated with $g_{n,0}(s)$ if it doesn't involve a $_1F_2$ function (or its equivalent expansion)? $\endgroup$ Commented Apr 30, 2019 at 22:46
  • $\begingroup$ Per your suggestion, I eliminated the hypergeometric $_1F_2$ function in my follow-on question at math.stackexchange.com/q/3447098. $\endgroup$ Commented Nov 22, 2019 at 20:59

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