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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.


Illustration of Formula (8) for Im(zeta(1/2+i t)

Figure (1): Illustration of Formula (8) for $\Im(\zeta(1/2+i t)$


Illustration of Formula (12) for Im(zeta(1/2+i t)

Figure (2): Illustration of Formula (12) for $\Im(\zeta(1/2+i t)$

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3 Answers 3

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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..)

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  • $\begingroup$ There are problems in your last three formulas. In the first of these the left side should be preceded by a minus sign: "$=s\sum\limits_{n=1}^\infty(2\pi)^s...$" should be "$=-s\sum\limits_{n=1}^\infty(2\pi)^s...$". In the second of these "$\frac{s}{s-1}-\frac{1}{2}$" was omitted. Your final formula is also incorrect (see the corrected formula in the second answer posted as a NOTE by Nikos Bagis and the related comments). $\endgroup$ Nov 21, 2019 at 21:11
  • $\begingroup$ The $_1F_2$ function may not be useful in this context, but I believe it is in the context of my original question at math.stackexchange.com/q/3207348. You indicated "The gamma function is a special function whose almost every properties are well-understood, the incomplete gamma function is much more complicated". Note the expansions of the $_1F_2$ functions in the formulas in my original question only involve complete gamma functions, so it seems to me these formulas would perhaps be of some interest. $\endgroup$ Nov 21, 2019 at 21:22
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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!!!

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  • $\begingroup$ Some use $n\in\textbf{Z}^{*}$ for non-zero integers, others use it for non-negative integers. Assuming you meant the non-zero integers, the formula $\zeta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{s}{s-1}-\frac{1}{2}+\sum\limits_{n=1}^K\left((2 \pi i n)^{s-1} \Gamma (1-s,2 \pi i n)+(-2 \pi i n)^{s-1} \Gamma (1-s,-2 \pi i n)\right)\right)$ does seem to converge globally. Mathematica simplifies this formula to $\zeta (s)=\underset{K\to\infty}{\text{lim}}\left(\frac{s}{s-1}-\frac{1}{2}+\sum_\limits{n=1}^K\left(E_s(2 \pi i n)+E_s(-2 \pi i n)\right)\right)$. $\endgroup$ Nov 21, 2019 at 14:28
  • $\begingroup$ Yes but is it some known result? I search in Wikipedia and Wolfram. It seems to be something new. A right thing to do is to examine the rate of convergence in $\textbf{C}$. $\endgroup$ Nov 21, 2019 at 15:17
  • $\begingroup$ I don't know if it's a known result, but I suspect it may not be. I couldn't seem to find it at functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta/06. $\endgroup$ Nov 21, 2019 at 15:37
  • $\begingroup$ 1) I will search the matter more deep and if I find something I will post it in this or in the other post you have. I have a book of E.C. Titchmarsh. "The Theory of the Riemann Zeta-Function". Oxford. At the Clarendon Press. (1951). $\endgroup$ Nov 21, 2019 at 16:16
  • $\begingroup$ 2) I observe now that function $E_s(z):=\int^{\infty}_{1}\frac{e^{-tz}}{t^s}dt$, from its deffinition is related with $\int^{\infty}_{1}\sin(tz)t^{-s}$ of your question (using $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$). 3) Renus post is also very interesting. However I don't know much about the theories of Riemann zeta function. $\endgroup$ Nov 21, 2019 at 16:39
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...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).

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  • $\begingroup$ I haven't had time to thoroughly review your updated answer, but I derived a new formula for $\zeta(s)$ based on your answer which is illustrated at math.stackexchange.com/q/3447098. $\endgroup$ Nov 22, 2019 at 21:03
  • $\begingroup$ I should have mentioned the context for this question is a sum over the positive integers $n$. With the assumption $n>0$, Mathematica simplifies formula (2) to $I(n,s)=\frac{\pi n}{s}(E_s(-2 i n \pi )+E_s(2 i n \pi))+\frac{\sin(2 \pi n)}{s}$. As you pointed out in a comment on my follow-on question, when $n$ is an integer the $\frac{\sin(2 \pi n)}{s}$ term evaluates to zero and can be deleted further simplifying the result. $\endgroup$ Nov 24, 2019 at 18:24
  • $\begingroup$ Formula (4) still doesn't make sense to me. If $s$ has an imaginary component, what is the meaning of the sum from $k=0$ to $s$ on the left side? $\endgroup$ Nov 24, 2019 at 18:25
  • $\begingroup$ As you pointed out in a comment on your second answer posted as a NOTE, $\sin(2 \pi n x)=\frac{1}{2} i e^{-2 i \pi n x}-\frac{1}{2} i e^{2 i \pi n x}$, and the definition $E_m(y)=\int\limits_1^\infty\frac{e^{-y x}}{x^m} \, dx$ can be used to derive the expression $I(n,s)=\frac{i}{2}(E_{s+1}(2 i n \pi )-E_{s+1}(-2 i n \pi ))$ by setting $m=s+1$ and first $y=2 i \pi n x$ and second $y=-2 i \pi n x$. With the assumption $n\in \mathbb{Z}\land n>0$, this expression is equivalent to $I(n,s)=\frac{\pi n}{s}(E_s(-2 i n \pi )+E_s(2 i n \pi))$. $\endgroup$ Nov 24, 2019 at 19:35

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