This response to my question Are these formulas for the Riemann zeta function $\zeta(s)$ globally convergent? didn't answer my question, but rather proposed an alternate approach which was intended to eliminate the hypergeometric $_1F_2$ function from my formulas. The response claims a hypergeometric function is not needed to talk about the integral defined in (1) below, but Mathematica evaluates this integral as illustrated in (2) below.
(1) $\quad g_{n,0}(s)=s\int_1^\infty\sin(2\,\pi\,n\,x)\,x^{-s-1}\,dx\,,\,\Re(s)>0$
(2) $\quad g_{n,0}(s)=\frac{2\,s}{s-1}\,_1F_2\left(\frac{1}{2}-\frac{s}{2};\frac{3}{2},\frac{3}{2}-\frac{s}{2};-n^2 \pi ^2\right)+2^s\,\pi^{s-1} \sin\left(\frac{\pi\,s}{2}\right)\,\Gamma(1-s)\,n^{s-1}\,,\\$ $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\,\Re(s)>-1$
I realize the hypergeometric $_1F_2$ function can be expanded as I did in an update of my original question (which contained a slightly different $_1F_2$ function).
Question: What is the result of the integral associated with $g_{n,0}(s)$ defined in (1) above if it doesn't involve a hypergeometric $_1F_2$ function (or its equivalent expansion)?
Based on the definition in (3) below, the relationship illustrated in (4) below, my original derivation, and the answers below I believe all of the formulas for $\zeta(s)$ defined in (5) to (9) below are globally convergent.
(3) $\quad S(x)=x-\left(\frac{1}{2}-\frac{1}{\pi}\sum\limits_{k=1}^\infty\frac{\sin(2\,\pi\,k\,x)}{k}\right)$
(4) $\quad\zeta(s)=s\int\limits_1^\infty S(x)\,x^{-s-1}\,dx$
(5) $\quad\zeta(s)=\frac{s}{s-1}-\frac{1}{2}+\sum\limits_{k=1}^\infty\left(\frac{2 s\,_1F_2\left(\frac{1}{2}-\frac{s}{2};\frac{3}{2},\frac{3}{2}-\frac{s}{2};-k^2 \pi^2\right)}{s-1}+2^s \pi ^{s-1} \sin\left(\frac{\pi s}{2}\right)\,\Gamma(1-s)\,k^{s-1}\right)$
(6) $\quad\zeta(s)=\frac{s}{s-1}-\frac{1}{2}+i (2 \pi)^{s-1}\sum\limits_{k=1}^\infty k^{s-1}\left(e^{-\frac{i \pi s}{2}} \Gamma(1-s,-2 \pi i k)-e^{\frac{i \pi s}{2}} \Gamma(1-s,2 \pi i k)\right)$
(7) $\quad\zeta(s)=\frac{s}{s-1}-\frac{1}{2}+\sum\limits_{k=1}^\infty\left((-2 \pi i k)^{s-1} \Gamma(1-s,-2 \pi i k)+(2 \pi i k)^{s-1} \Gamma (1-s,2 \pi i k)\right)$
(8) $\quad\zeta(s)=\frac{s}{s-1}-\frac{1}{2}+\sum\limits_{k=1}^\infty (E_s(-2 \pi i k)+E_s(2 \pi i k))$
(9) $\quad\zeta(s)=\frac{s}{s-1}-\frac{1}{2}+\frac{i s}{2 \pi}\sum\limits_{k=1}^\infty\frac{E_{s+1}(2 \pi i k)-E_{s+1}(-2 \pi i k)}{k}$
Based on the definition in (3) above, the relationship illustrated in (10) below, my original derivation, and the answers below I believe the formulas for $\zeta(s)$ defined in (11) and (12) below are also globally convergent.
(10) $\quad\zeta(s)=s\int\limits_{1/2}^\infty S(x)\,x^{-s-1}\,dx$
(11) $\quad\zeta(s)=2^{s-1}\left(\frac{s}{s-1}-1+2 s \sum\limits_{k=1}^\infty \left(\frac{\, _1F_2\left(\frac{1}{2}-\frac{s}{2};\frac{3}{2},\frac{3}{2}-\frac{s}{2};-\frac{1}{4} k^2 \pi ^2\right)}{s-1}-\pi ^{s-1} \sin\left(\frac{\pi s}{2}\right)\,\Gamma(-s)\,k^{s-1}\right)\right)$
(12) $\quad\zeta(s)=2^{s-1}\left(\frac{s}{s-1}-1+\sum\limits_{k=1}^\infty (E_s(-i k \pi)+E_s(i k \pi))\right)$
The following two figures illustrate the relationship illustrated in (10) above seems to converge better than the relationship illustrated in (4) above. The figures below illustrate formulas (8) and (12) for $\zeta(s)$ above evaluated along the critical line $s=1/2+i t$ where both formulas are evaluated over the first 20 terms of their associated series. Formulas (8) and (12) are illustrated in orange, and the underlying blue reference function is $\zeta(s)$. The red discrete portions of the two figures below illustrate the evaluation of formulas (8) and (12) for $\zeta(s)$ above at the first ten non-trivial zeta-zeros in the upper-half plane.
Figure (1): Illustration of Formula (8) for $\Im(\zeta(1/2+i t)$
Figure (2): Illustration of Formula (12) for $\Im(\zeta(1/2+i t)$
...After some corrections
The integral you want to evaluate is $$ I(n,s)=\int^{\infty}_{1}\sin(2\pi n x)x^{-s-1}dx.\tag 1 $$ With change of variable $2\pi nx=y$, we get $$ I(n,s)=\int^{\infty}_{2\pi n}\sin(y)\left(2\pi n\right)^{s+1}y^{-s-1}(2\pi n)^{-1}dy=(2\pi n)^{s}\int^{\infty}_{2\pi n}\frac{\sin(y)}{y^{s+1}}dy= $$ $$ (2\pi n)^s\int^{\infty}_{-\infty}\frac{\sin(y)}{y}\frac{X_{[2\pi n,\infty)}(y)}{y^s}dy. $$ We also have the next Fourier pairs $$ \frac{\sin(t)}{t}\leftrightarrow \pi X_{[-1,1]}(\gamma)\textrm{ and }\frac{X_{[2\pi n,\infty)}(t)}{t^s}\leftrightarrow (i\gamma)^{s-1}\Gamma(1-s,2\pi i n \gamma), $$ where the Fourier transform has been considered as $$ \widehat{f}(\gamma)=\int^{\infty}_{-\infty}f(t)e^{-it\gamma}dt. $$ Hence $$ I(n,s)=\frac{(2\pi n)^s}{2\pi}\int^{1}_{-1}\pi(i\gamma)^{s-1}\Gamma(1-s,2\pi i n \gamma)d\gamma= $$ $$ =\frac{(2\pi n)^s}{2}\int^{1}_{-1}\Gamma(1-s,2\pi i n\gamma)(i\gamma)^{s-1}d\gamma=\frac{(2\pi n)^s}{2i}\int^{i}_{-i}\Gamma(1-s,2\pi n \gamma)\gamma^{s-1}d\gamma= $$ $$ =\ldots\textrm{ using Mathematica }\ldots= $$ $$ =\frac{i(2\pi n)^s}{2s}e^{-i\pi s/2}\left(\Gamma(1-s,-2i n\pi)-e^{i\pi s}\Gamma(1-s,2in\pi)\right)+\frac{\sin(2n\pi)}{s},\tag 2 $$ where $n\in\textbf{R}-\{0\}$ and $Re(s)>0$.
Set now $$ C(s,x)=e^x-\sum^{s}_{k=0}\frac{x^k}{k!},\tag 3 $$ in the sense that $s$ is in whole $\textbf{C}$, by using the analytic continuation: $$ \sum^{s}_{k=0}\frac{x^k}{k!}:=e^x-\sum^{\infty}_{k=0}\frac{x^{k+s+1}}{\Gamma(k+s+2)}\textrm{, }\forall s\in \textbf{C}\textrm{, when }x\neq 0.\tag 4 $$ Then $$ C(s,x)=e^x\left(1-\frac{\Gamma(s+1,x)}{\Gamma(s+1)}\right)\tag 5 $$ and $$ \frac{d}{dx}C(s,x)=C(s-1,x).\tag 6 $$ Then also $$ \Gamma(1+s,x)=\left(1-e^{-x}C(s,x)\right)\Gamma(1+s).\tag 7 $$ The function $\Gamma(1-s,z)$ can evaluated using (7) from the analytic continuation (4),(3): $$ \Gamma(1-s,z)=\left(1-e^{-z}\sum^{\infty}_{k=0}\frac{z^{k-s+1}}{\Gamma(k-s+2)}\right)\Gamma(1-s).\tag{10} $$ Actualy (10) is valid for all $s\in\textbf{C}$, when $z\neq 0$ and this agree with the analytic continuation used in Mathematica program. After all above $I(n,s)$ can analyticaly expanded in $\textbf{C}$, when $n\neq 0$.
I don't have proof about the Mathematica symbolic calculation right now for (2), but going the opposite direction as in comments it seems more convinient.
CONTINUING.
From the one hand we have to evaluate $$ I(n,s)=\int^{\infty}_{1}\frac{\sin(2\pi n t)}{t^{s+1}}dt $$ From the other hand set $$ E_s(z):=z^{s-1}\Gamma(1-s,z)\textrm{, }z\neq 0. $$ Set also $$ E^{*}_s(z):=\int^{\infty}_{1}\frac{e^{-tz}}{t^s}dt\textrm{, }Re(z)>0. $$ Hence $$ E_s(z)=E^{*}_s(z)\textrm{, }Re(z)>0. $$ Also $$ \partial_zE_s(z)=-E_{s-1}(z)\textrm{, }Re(z)>0. $$ Also with integration by parts $$ zE_{s}(z)=e^{-z}-sE_{s+1}(z)\Leftrightarrow s\frac{E_{s+1}(z)}{z}=\frac{e^{-z}}{z}-E_{s}(z)\textrm{, }Re(z)>0.\tag{11} $$ However if $Re(s)>-1$, then we can define $E^{*}_s(z)$, for $Re(z)\geq0$, $z\neq 0$. Hence for $n$ non zero integer, we have $$ I(n,s)=2^{-1}i\int^{\infty}_{1}\left(e^{-2\pi n i t}-e^{2\pi n i t}\right)t^{-s-1}dt= $$ $$ =2^{-1}iE_{s+1}(2\pi i n)-2^{-1}iE_{s+1}(-2\pi n i)\textrm{, }Re(s)>-1.\tag{12} $$ But (see [T] pages 13-15): $$ \zeta(s)=\frac{1}{s-1}+\frac{1}{2}+s\int^{\infty}_{1}\left(\frac{1}{2}-\{x\}\right)x^{-s-1}\textrm{, }Re(s)>-1\tag{13} $$ and $$ \frac{1}{2}-\{x\}=\sum^{\infty}_{n=1}\frac{\sin(2\pi n x)}{\pi n},\tag{14} $$ if $x$ is not integer. Hence $$ \zeta(s)=\frac{1}{s-1}+\frac{1}{2}+s\int^{\infty}_{1}\sum^{\infty}_{n=1}\frac{\sin(2\pi n x)}{\pi n}x^{-s-1}dx\textrm{, }Re(s)>-1.\tag{15} $$ But $$ \int^{\infty}_{1}\sum^{\infty}_{n=1}\frac{\sin(2\pi n x)}{\pi n}x^{-s-1}dx =\sum^{\infty}_{k=1}\int^{k+1}_{k}\sum^{\infty}_{n=1}\frac{\sin(2\pi n x)}{\pi n}x^{-s-1}dx= $$ $$ =\sum^{\infty}_{k,n=1}\int^{k+1}_{k}\frac{\sin(2\pi n x)}{\pi n}x^{-s-1}dx. $$ Assume now the integral $$ I_1(k,n,s):=\int^{k+1}_{k}\frac{\sin(2\pi n x)}{x^{s+1}}dx. $$ Using integration by parts we have $$ \left|I_1(k,n,s)\right|=\left|\frac{1}{2\pi n}\left(\frac{1}{k^{s+1}}-\frac{1}{(k+1)^s}\right) -\frac{s+1}{2 \pi n}\int^{k+1}_{k}\frac{\cos(2\pi n x)}{x^{s+2}}dx\right|\leq $$ $$ \leq\frac{1}{2\pi n}\left|\frac{1}{k^{s+1}}-\frac{1}{(k+1)^{s+1}}\right|+\frac{s+1}{2\pi n}\left|\int^{k+1}_{k}x^{-s-2}dx\right|= $$ $$ =\frac{1}{\pi n}\left|\frac{1}{k^{s+1}}-\frac{1}{(k+1)^{s+1}}\right|\leq\frac{(s+1)}{\pi n k^{s+2}} $$ Hence $$ \sum^{\infty}_{k,n=1}\int^{k+1}_{k}\frac{\sin(2\pi n x)}{\pi n}x^{-s-1}dx=\sum^{\infty}_{k,n=1}\frac{I_1(k,n,s)}{\pi n}. $$ But $$ \left|\frac{I_1(k,n,s)}{\pi n}\right|\leq \frac{(s+1)}{\pi^2 n^2 k^{s+2}}\textrm{, }Re(s)>-1. $$ Hence the double sum $$ \sum^{\infty}_{k,n=1}\int^{k+1}_{k}\frac{\sin(2\pi n x)}{\pi n}x^{-s-1}dx $$ is absolutely convergent. Hence we can rearange the order of summation, to get $$ \sum^{\infty}_{n,k=1}\int^{k+1}_{k}\frac{\sin(2\pi n x)}{\pi n}x^{-s-1}dx=\sum^{\infty}_{n=1}\int^{\infty}_{1}\frac{\sin(2\pi n x)}{\pi n}x^{-s-1}dx. $$ Hence from (11),(12),(15): $$ \zeta(s)=\frac{1}{s-1}+\frac{1}{2}+\frac{is}{2\pi}\sum^{\infty}_{n=1}\left(\frac{E_{s+1}(2\pi i n)}{n}-\frac{E_{s+1}(-2\pi i n)}{n}\right)= $$ $$ =\frac{1}{s-1}+\frac{1}{2}-\sum^{\infty}_{n=1}\left(s\frac{E_{s+1}(2\pi i n)}{2\pi i n}+s\frac{E_{s+1}(-2\pi i n)}{-2\pi i n}\right)= $$ $$ =\frac{1}{s-1}+\frac{1}{2}-\sum^{\infty}_{n=1}\left(\frac{e^{-2\pi i n}}{2\pi i n}-E_{s}(2\pi i n)+\frac{e^{2\pi i n}}{-2\pi i n}-E_{s}(-2\pi i n)\right). $$ Hence we get $$ \zeta(s)=\frac{1}{s-1}+\frac{1}{2}+\sum_{n\in\textbf{Z}^{*}}E_s(2\pi i n)\textrm{, }Re(s)>-1.\tag{16} $$
REFERENCES.
[T] E.C. Titchmarsh. ''The Theorey of the Riemann zeta-function''. Oxford. At the Clarendon press. (1951).