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To preface this inquiry, I am aware of the fastidious nature of the mathematics presented on this exchange. In addition to that, I apologize for my ineptness and the contention in this particular subject; however, this is of great interest to me. I ask for pardon with the great length of this text.

The following formulation invokes the Argument Principle due to Cauchy. In this sense, consider the closed rectifiable Jordan curve $\mathcal D^{+}= \mathcal D^{+}(T)$ with vertices $(\frac{1}{2}+\epsilon)-iT, 2-iT, 2+iT, (\frac{1}{2}+\epsilon)+iT$, traversed in that order, i.e., with positive orientation. Likewise, let $D^{-}=D^{-}(T)$ denote the closed contour with vertices $-1-iT, (\frac{1}{2}+\epsilon)-iT, (\frac{1}{2}+\epsilon)+iT, -1+iT$, and positive orientation. In both cases $\epsilon$ is chosen to be arbitrarily small. Furthermore, invoking the detail that the zeros of the function $\xi(s)$, where $s=\sigma+it$, are identical to the non-trivial zeros of the function $\zeta(s)$, and the fact that $\xi(s)$ has no poles; we obtain the number of zeros of $\zeta$ in the region $\mathcal D^{+}$ and $\mathcal D^{-}$ are $$N_+(T)=\frac{1}{2\pi i}\int_{\mathcal D^+} \frac{{\xi}^\prime (s)}{\xi(s)}ds=\frac{1}{2 \pi}Im\left(\int_{\mathcal D^+} \frac{{\xi}^\prime (s)}{\xi(s)}ds\right)$$ and $$N_-(T)=\frac{1}{2\pi i}\int_{\mathcal D^-} \frac{{\xi}^\prime (s)}{\xi(s)}ds=\frac{1}{2 \pi}Im\left(\int_{\mathcal D^-} \frac{{\xi}^\prime (s)}{\xi(s)}ds\right),$$ respectively. By the symmetry of both $\xi$ and $\mathcal D^{+}, \mathcal D^{-}$, $N_-(T)=N_+(T)$. Let $\mathcal C_0^{+}$ denote the contour with vertices $\frac{1}{2}+\epsilon, 2, 2+iT, (\frac{1}{2}+\epsilon)+iT$, traversed in the positive sense. Similarly, let $\mathcal C_1^{+}$ denote the contour with vertices $(\frac{1}{2}+\epsilon)-iT, 2-iT, 2, \frac{1}{2}+\epsilon$, also traversed in the positive sense. Decomposing $\mathcal D^{+}$ into $\mathcal C_0^{+}$ and $\mathcal C_1^{+}$ yields $\mathcal D^{+}:=\mathcal C_0^{+} \cup \mathcal C_1^{+}$, whereby $\mathcal D^{+}:=\mathcal C_0^{+} \cap \mathcal C_1^{+}=[\frac{1}{2}+\epsilon,2].$ Hence, the former integral $N_+(T)$ can be expressed as $$N_+(T)=\frac{1}{2 \pi}Im\left(\int_{\mathcal C_0^{+}} \frac{{\xi}^\prime (s)}{\xi(s)}ds\right)+\frac{1}{2 \pi}Im\left(\int_{\mathcal C_1^{+}} \frac{{\xi}^\prime (s)}{\xi(s)}ds\right).$$ Since $\xi$ is positive real on the portion of the real axis $\mathcal C_0^{+} \cap \mathcal C_1^{+}$, the argument of $\xi$ does not change there. Let $\tilde {\mathcal C_0^{+}}, \tilde {\mathcal C_1^{+}}$ denote $\mathcal C_0^{+}, \mathcal C_1^{+}$ without $\mathcal C_0^{+} \cap \mathcal C_1^{+}$. Therefore

$$\begin{align*} N_+(T)& =\frac{1}{2 \pi}Im\left(\int_\tilde {\mathcal C_0^{+}} \frac{{\xi}^\prime (s)}{\xi(s)}ds\right)+\frac{1}{2 \pi}Im\left(\int_\tilde {\mathcal C_1^{+}} \frac{{\xi}^\prime (s)}{\xi(s)}ds\right)\\ & \equiv \frac{1}{2 \pi}Im\left(\int_\tilde {\mathcal C_0^{+}} \frac{{\zeta}^\prime (s)}{\zeta(s)}ds\right)+\frac{1}{2 \pi}Im\left(\int_\tilde {\mathcal C_1^{+}} \frac{{\zeta}^\prime (s)}{\zeta(s)}ds\right)\end{align*}$$

Since the set of zeros of $\zeta$ is discrete, we may assume no zero of $\zeta$ has imaginary part $T$. This integral can be decomposed into the contribution over the vertical lines, and over the one horizontal line. In particular,

$$\begin{align*} \frac{1}{2 \pi}Im\left(\int_\tilde {\mathcal C_0^{+}} \frac{{\zeta}^\prime (s)}{\zeta(s)}ds\right) &= Im \left(i \int_0^{-T} \frac{{\zeta}^\prime ((\frac{1}{2}+\epsilon)+iy)}{\zeta(\frac{1}{2}+\epsilon)+iy)} dy+\int_{\frac{1}{2}+\epsilon}^{2} \frac{{\zeta}^\prime (\sigma-iT)}{\zeta(\sigma-iT)} d\sigma \right)\\ &=Im\left(\int_{\frac{1}{2}+\epsilon}^{2} \frac{{\zeta}^\prime (\sigma-iT)}{\zeta(\sigma-iT)} d\sigma \right)-Im \left(i \int_{-T}^{0} \frac{{\zeta}^\prime ((\frac{1}{2}+\epsilon)+iy)}{\zeta(\frac{1}{2}+\epsilon)+iy)} dy \right)\end{align*}$$

for which the integral over the vertical segment $[2-iT,2]$ vanishes, due to the boundedness of the total variation of $arg(\zeta(2+it))$ on the segment $[2-iT,2]$. It can be shown that $$\frac{\zeta^{\prime}(s)}{\zeta(s)}=-\frac{1}{s}-\frac{1}{s-1}+\frac{1}{2}log(\pi)+\sum_\rho \frac {1}{s-\rho}-\frac{1}{2}\frac{\Gamma^{\prime}(\frac{s}{2})}{\Gamma(\frac{s}{2})}=\sum_\rho \frac {1}{s-\rho}+O(log|t|),$$ for $\sigma \ge -1, |t| \ge 2.$ This, in turn, can be used to show $$\frac{\zeta^{\prime}(s)}{\zeta(s)}=\sum_{|\gamma_n-t|\le 1} \frac{1}{s-\rho_n}+O(log|t|),$$ with the same condition $\sigma \ge -1, |t| \ge 2.$ As such, it follows that

$$\begin{align*}Im\left(\int_{\frac{1}{2}+\epsilon}^{2} \frac{{\zeta}^\prime (\sigma-iT)}{\zeta(\sigma-iT)} d\sigma \right)&=Im\left(\sum_{|\gamma_n-T| \le 1}\int_{\frac{1}{2}+\epsilon}^{2} \frac{d\sigma}{\sigma-it-\rho_n} \right)+Im\left(\sum_{|\gamma_n-T| \le 1}\int_{\frac{1}{2}+\epsilon}^{2} O(log|T|)dT \right)\\ &\equiv \sum_{|\gamma_n-T| \le 1}Im\left(\int_{\frac{1}{2}+\epsilon}^{2} \frac{d\sigma}{\sigma-it-\rho_n} \right)\end{align*}$$

By the Argument Principle, each summand equals the net change of $arg(s-\rho_n)$ on $[\frac{1}{2}+\epsilon-iT,2-iT]$; thus, its absolute value is less than $\pi.$ In similar vein, by the von Mangoldt estimate of the vertical density of roots $\rho_n$: $N(T+1)-N(T) \le \sum_{T \le \gamma_n \le T+1}1 < 2logT,$ the number of summands is less than $4logT.$ The modulus of the latter integral is $$\left |\sum_{|\gamma_n-T| \le 1}Im\left(\int_{\frac{1}{2}+\epsilon}^{2} \frac{d\sigma}{\sigma-it-\rho_n} \right) \right|<4\pi logT,$$ for large $T.$ Consequently $$\sum_{|\gamma_n-T| \le 1}Im\left(\int_{\frac{1}{2}+\epsilon}^{2} \frac{d\sigma}{\sigma-it-\rho_n} \right)=O(logT).$$

Similarly $$\begin{align*}Im \left(i \int_{-T}^{0} \frac{{\zeta}^\prime ((\frac{1}{2}+\epsilon)+iy)}{\zeta(\frac{1}{2}+\epsilon)+iy)} dy \right)&=Im\left(i \int_{-T}^{0} \sum_{|\gamma_n-T|\le 1} \frac{dy}{(\frac{1}{2}+\epsilon)+iy-\rho_n}+i\int_{-T}^{0}O(log|T|)dy \right)\\ &=Im\left(i \sum_{|\gamma_n-T|\le 1} \int_{-T}^{0} \frac{dy}{(\frac{1}{2}+\epsilon)+iy-\rho_n}\right)+TO(log|T|)\end{align*}$$

As before, each summand equals the net change of $arg(s-\rho_n)$ on $[(\frac{1}{2}+\epsilon)-iT, \frac{1}{2}+\epsilon].$ Therefore, the absolute variation in the argument is less than $\pi.$ By the above estimate of $N(T+1)-N(T),$ the number of summands is less than $4logT.$ Thus $$Im\left(\sum_{|\gamma_n-T|\le 1} \int_{-T}^{0} \frac{dy}{(\frac{1}{2}+\epsilon)+iy-\rho_n}\right)+TO(log|T|)<4\pi logT $$ for large $T.$ Finally, $$\sum_{|\gamma_n-T|\le 1}Im\left(i \int_{-T}^{0} \frac{dy}{(\frac{1}{2}+\epsilon)+iy-\rho_n}\right)+TO(log|T|)=O(logT)+O(TlogT),$$ which implies that $$\frac{1}{2 \pi}Im\left(\int_\tilde {\mathcal C_0^{+}} \frac{{\zeta}^\prime (s)}{\zeta(s)}ds\right)=O(logT)-O(TlogT),$$ keeping the negative signature for convenience. Therefore $$N_+(T)=\frac{1}{2 \pi}Im\left(\int_\tilde {\mathcal C_1^{+}} \frac{{\zeta}^\prime (s)}{\zeta(s)}ds\right)+O(logT)-O(TlogT).$$

Using this formulation of Argument Principle, would the Riemann Hypothesis follow if it was shown that $$\frac{1}{2 \pi}Im\left(\int_\tilde {\mathcal C_1^{+}} \frac{{\zeta}^\prime (s)}{\zeta(s)}ds\right)$$ exactly cancels $O(logT)-O(TlogT)$ as $\epsilon$ tends to zero and as $T$ tends to infinity? In addition to this, is there a differential-geometric approach to the Riemann Hypothesis?

Thanks in advance.

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    $\begingroup$ ? the Riemann hypothesis is $\int_{\begin{array}{l}\sigma-i\epsilon \ \to \ \sigma-iT \\ \to \ 2-iT\ \to \ 2-i\epsilon \\ \to \sigma-i\epsilon\end{array}} \frac{{\zeta}^\prime (s)}{\zeta(s)}ds = 0$ for every $T, \sigma > 1/2,\epsilon > 0$. And did you read Titchmarsh's book ? $\endgroup$ – reuns Aug 31 '16 at 5:01
  • $\begingroup$ I split four of the equation chains (ones that had display problems) across multiple lines using the $\LaTeX$ align* environment. Please check my edit did not introduced unintended errors or unwanted effects. $\endgroup$ – hardmath Aug 31 '16 at 5:09
  • $\begingroup$ I have been introduced to the general theory of the Riemann Zeta Function via 'Rimann's Zeta Function' by Harold Edwards. However, I have yet to read Titchmarsh's book. Also, what does this particular notation mean?@user1952009 $\endgroup$ – Sergio Charles Aug 31 '16 at 15:50
  • $\begingroup$ the RH is that $\frac{\zeta'(s)}{\zeta(s)}$ is analytic on the rectangle I wrote (the $\epsilon > 0$ is for avoiding the pole at $s=1$), so if your contour $\tilde{C_1^+}$ means the rectangle I wrote, then the RH is $\int_{\tilde{C_1^+}} \frac{\zeta'(s)}{\zeta(s)} ds = 0$ $\endgroup$ – reuns Aug 31 '16 at 16:25
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    $\begingroup$ I don't get it. $F(s) = (s-1)\zeta(s)$ is entire, so integrating $\frac{1}{2i\pi}\frac{F'(s)}{F(s)}$ on the boundary of any finite region just counts (with multiplicity) the (finite) number of zeros it has in that region, and no cancelling occurs. Of course, $\frac{F'(s)}{F(s)} = \frac{\zeta'(s)}{\zeta(s)} + \frac{1}{s-1}$, so up to a constant $1$, the same is true for $\frac{\zeta'(s)}{\zeta(s)}$. And Titchmarsh's book is there, see p.210 $\endgroup$ – reuns Sep 1 '16 at 1:13
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As a place-holder "answer" by the specs of this site/forum: the baseline situation is that many people have thought about RH for 150+ years. Stietljes thought he had a proof a few decades after Riemann's 1859 paper, but it evaporated...

The wonderful fallout of Riemann's paper was much of the late 19th century, and early 20th, development of the function theory of one complex variable. Indeed, the things needed to make Riemann's epiphanies "provable" were not done until after Hadamard's results (1890s) on product expressions for entire functions, and, equally significantly, the Phragmen-Lindelof results c. 1908.

Landau, Hardy, Littlewood, ... bore down quite hard on complex-variables ideas. Cf Levinson.

But, no, ... we do not have useful information. And, equally, re-inventing wheels (while eminently interesting to oneself, at least), is not ... significant.

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    $\begingroup$ as far as I understand it now, there are only two analytic approaches to the RH : the one trying to show $\zeta(s) = 0 \implies \zeta(1-s) \ne 0$ when $Re(s) \ne 1/2$, and the other one trying to say $\xi(1/2+it)$ changes of sign (at least asymptotically) in the neighborhood of every zero of $\zeta(s)$. the two are by nature very complicated, because there are many functions close to $\zeta(s)$ having some zeros off $Re(s) = 1/2$ $\endgroup$ – reuns Sep 1 '16 at 1:49

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