What is $s\int_1^\infty\sin(2\,\pi\,n\,x)\,x^{-s-1}\,dx$? 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 $\Im(\zeta(1/2+i t)$
 A: What do you mean with "the result of the integral" ? For $\Re(s)> -1$ $$\int_1^\infty \sin(2\pi nx)x^{-s-1}dx=(2\pi n)^{s}\int_{2\pi n}^\infty \sin(x)x^{-s-1}dx$$ $$ = \lim_{b\to 0} (2\pi n)^{s}\int_{2\pi n}^\infty \frac{e^{-(i+b) x}-e^{-(b-i)x}}{2i}x^{-s-1}dx$$ $$=\lim_{b\to 0} (2\pi n)^{s}\int_{-2\pi (b+i) n}^\infty \frac{(i+b)^{s}}{2i}e^{-x}x^{-s-1}dx-(2\pi n)^{s}\int_{-2\pi (b-i) n}^\infty \frac{(b-i)^{s}}{2i}e^{-x}x^{-s-1}dx$$ $$=(2\pi n)^{s}\frac{i^s \Gamma(-s,-2i\pi n)-(-i)^s \Gamma(-s,2i\pi n)}{2i} $$
where $\Gamma(-s,2i\pi n)$ is the incomplete gamma function.
The gamma function is a special function whose almost every properties are well-understood, the incomplete gamma function is much more complicated.
The point is that from $\zeta(s)=s\int_1^\infty \lfloor x\rfloor x^{-s-1}dx$ we get two expressions for $\zeta(s)$ valid for $\Re(s)\in(-1,0)$ $$\zeta(s)=-s\int_0^\infty ( \{x\}-1/2)x^{-s-1}dx,\qquad \zeta(s)=\frac{s}{s-1}+\frac12 -s\int_1^\infty (\{x\}-1/2)x^{-s-1}dx$$
From the Fourier series $$\{x\}-1/2=-\sum_{n=1}^\infty \frac{\sin(2\pi nx)}{\pi n}$$ and the first integral we get the functional equation which is valid for $\Re(s) < 0$
$$\zeta(s)=s \int_0^\infty\sum_{n=1}^\infty \frac{\sin(2\pi nx)}{\pi n} x^{-s-1}dx=s\sum_{n=1}^\infty \int_0^\infty \frac{\sin(2\pi nx)}{\pi n} x^{-s-1}dx$$ $$=s \sum_{n=1}^\infty (2\pi)^s \pi^{-1} n^{s-1}\sin(\pi s/2)\Gamma(-s)=2^s \pi^{s-1} \zeta(1-s)\sin(\pi s/2)\Gamma(1-s)$$
whereas the second integral, which is valid for all $s$, gives
$$\zeta(s)=s \int_1^\infty \sum_{n=1}^\infty \frac{\sin(2\pi nx)}{\pi n} x^{-s-1}dx$$ $$=\frac{s}{s-1}+\frac12+ s \sum_{n=1}^\infty \pi^{-1} n^{s-1}\frac{i^s \Gamma(-s,-2i\pi n)-(-i)^s \Gamma(-s,2i\pi n)}{2i}$$
which is valid for all $s$.

As you see there is absolutely no point to look at ${}_2 F_1$ in this setting. The usefulness of ${}_2 F_1$ is to give : a contour integral representation of $\Gamma(-s,2\pi n)$, a power series representation, and a general expression that CAS can easily deal with (differentiation, integration, summation..)

A: NOTE.
Renus result can simplified into the form (after correcting some typos in its answer):
$$
\zeta(s)=\frac{s}{s-1}-\frac{1}{2}+\sum_{n\in\textbf{Z}^{*}}(2\pi i n)^{s-1}\Gamma(1-s,2\pi i n)\textrm{, }\forall s\in\textbf{C}-\{1\}
$$
Is this result known? Actualy is a representation of Riemann's zeta function in the whole plane!!!
