# A Formulation of The Riemann Hypothesis

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?

• ? 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 ? – reuns Aug 31 '16 at 5:01
• 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. – hardmath Aug 31 '16 at 5:09
• 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 – Sergio Charles Aug 31 '16 at 15:50
• 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 $\int_{\tilde{C_1^+}} \frac{\zeta'(s)}{\zeta(s)} ds = 0$ – reuns Aug 31 '16 at 16:25
• 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 – reuns Sep 1 '16 at 1:13

• 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 $Re(s) = 1/2$ – reuns Sep 1 '16 at 1:49