The Riemann Zeta Function is defined as the Dirichlet series of the form which is valid in the half-plane $\{s\in \mathbb{C}:\mathfrak{Re}(s)>1\}$.

The Riemann Zeta Function satisfies the following equation. This result is known as the Euler product formula

The Riemann Zeta function has no zeros for$\mathfrak{Re}\left(s\right)>1$.

Both sides of equation (1.2) converge for $\mathfrak{Re}\left(s\right)>1$.As the right side of this expression never becomes zero as it is a product of prime numbers, then the left side of this equation never becomes zero for $\mathfrak{Re}\left(s\right)>1$. □

The Riemann Zeta function, equation (1.1), can be continued analytically through the following functional equation, which is meromorphic for the whole complex plane $\{s\in \mathbb{C}:s\ne 1\}$ , where $s=1$ constitutes a simple pole with residue 1, of the form

The sine term in equation (1.3) vanishes at $s=-2,-4,-6,...$ while for the same values, $2^s{\pi }^{s-1}\ne 0$ . If Gamma function is expressed in terms of the Euler reflection formula, $\Gamma \left(s\right)\Gamma (1-s)=\pi /sin\left(\pi s\right)$, it never becomes zero and only the poles at $\Gamma (-2k)$, $k=0,1,2,...$ are present. $\zeta (1-s)\ne 0$ at $s=-2,-4,-6,...$ as per corollary 1.1. □

Let $\rho$ be a complex number such as $\mathfrak{Re}\left(\rho \right)=\sigma$ and $\mathfrak{Im}\left(\rho \right)=t$$\{\rho \in \mathbb{C}:0<\mathfrak{Re}(\rho )<1,0<\mathfrak{Im}(\rho )<\infty \}$. If $\zeta (\sigma +it)=0$, as $2^{\rho }{\pi }^{\rho -1}sin\left(\frac{\pi \rho }{2}\right)\Gamma (1-\rho )\ne 0$ in equation (1.3), then $\zeta (1-\sigma -it)=0$. □

As $\zeta :\mathbb{R}\to \mathbb{R}$ for $\{s\in \mathbb{C}:\mathfrak{Re}(s)\ne 1,\mathfrak{Im}(s)=0\}$ is well defined and is holomorphic on the upper half-plane $\{s\in \mathbb{C}:\mathfrak{Re}(s)\ne 1,\mathfrak{Im}(s)>0\}$, by the Schwarz reflection principle[2], $\overline{\zeta }\left(s\right)=\zeta \left(\overline{s}\right)$, and it implies that $\mathfrak{Re}\left(\zeta \right(\sigma +it\left)\right)=\mathfrak{Re}\left(\zeta \right(\sigma -it\left)\right)$ and $\mathfrak{Im}\left(\zeta \right(\sigma +it\left)\right)=-\mathfrak{Im}\left(\zeta \right(\sigma -it\left)\right)$ . □

Riemann Hypothesis (R.H.) [3] states that the real part of the non-trivial zeros of the Riemann Zeta function, that is to say, those zeros that are not “trivial”, equals $1/2$. Let be ${\rho }_n$ the n-th non-trivial zero of the Riemann Zeta function, then R.H. states that the non-trivialzeros1are the set $\{{\rho }_n\in \mathbb{C}:{\rho }_n=\sigma +i{\gamma }_n,\sigma =1/2,{\gamma }_n\in {\mathbb{R}}^{+},n\in \mathbb{N}\}$.

The Gamma function is defined, for $\mathfrak{Re}\left(s\right)>0$, as

The Dirichlet eta function can be written is terms of the Gamma function, for $\mathfrak{Re}\left(s\right)>0$, as

Let $S_{\infty }$ be an infinity geometric series of the form, converging for $-1<r<1$

As it is well know, this geometric series can be expressed in terms of a closed from

And let $r=-e^{-x}$, so

If I multiply both sides of the last expression I get

and

On one hand, if I use Mellin transform, I get2

On the other hand, I have

so,

and we get to equation (2.2)

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Riemann Zeta function can be continued analytically, for $\mathfrak{Re}\left(s\right)>0$, in terms of the Dirichlet eta function as

Let $\zeta \left(s\right)$ be expressed as, valid for $\mathfrak{Re}\left(s\right)>1$, that can be obtained from equation (1.1) by following an analogous procedure for proof of theorem (2.1) which is valid for $\mathfrak{Re}\left(s\right)>1$. Then I multiply both sides of the previous expression by $2^{-s}$ to obtain or more conveniently, I perform a variable change $(1/2)x=\alpha$, so I perform a fractional separation over the term $1/(e^{2\alpha }-1)$ as follows: so, I obtain Recalling equation (2.2) and the integral expression of the Riemann Zeta function used above, I have and, reorganizing terms, I come to equation (2.3)

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Conjecture 2.1 will be demonstrated from now on up to the end of the document.

I substitute equation (2.2) in equation (2.3) to obtain, I can multiply both sides of equation (2.4) by s to obtain Now, it is interesting to see that the integral of the last expression can be intelligently manipulated so that can yield a more useful expression. In effect, I realize that $(d/dx)x^s=s{x}^{s-1}$. On one hand, I perform the following variable changes, being $u=1/(e^x+1)$, then $du=-e^x/{(e^x+1)}^{2}dx$ and $dv=(d/dx)\left(x^s\right)dx$, then $v=x^s$. Given this, we can write the following I conveniently name $A\left(s\right)=s\zeta \left(s\right)$, $B\left(s\right)=1/\left((1-2^{1-s})\right)\Gamma \left(s\right)$, $C\left(s\right)$ will be the first term in the brackets, and ${\mathcal{I}}_1\left(s\right)$ will be the integral which belongs to the second term in brackets. Then, the domain of definition $\mathcal{D}$ of $A\left(s\right)$ can be expressed as ${\mathcal{D}}_A={\mathcal{D}}_B\bigcap \left({\mathcal{D}}_C\bigcup {\mathcal{D}}_{{\mathcal{I}}_1}\right)$. The function $1/(1-2^{1-s})$ is defined in the complex plane $\mathbb{C}\setminus \left\{1\right\}$ and the reciprocal of the Gamma function, $1/\Gamma \left(s\right)$ is defined, according to definition 2.2, in the half-plane $\mathfrak{Re}\left(s\right)>0$, so ${\mathcal{D}}_B=\{s\in \mathbb{C}:\sigma >0,-\infty <t<\infty \}\setminus \left\{1\right\}$. □

The following limits can be expressed as The limit when $x\to \infty$, regardless of the oscillatory behavior of the trigonometric functions, the exponential function in the denominator grows much faster than the polynomial term $x^{\sigma }$, so that For limit when $x\to 0$, both real and imaginary parts vanish if and only if $\sigma >0$, so that

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For ${\mathcal{I}}_1\left(s\right)$, when $x\gg 1$, then ${(e^x+1)}^2\approx e^{2x}$. Then I can approximate the integral as Recalling definition 2.2 and performing $s\mapsto s+1$, I get By comparing the two integrals above, and taking into account the fact that, as $\Gamma \left(s\right)$ is valid for $\mathfrak{Re}\left(s\right)>0$, $\Gamma (s+1)$ is valid for $\mathfrak{Re}\left(s\right)>-1$ according to definition 2.2, I come to the conclusion that the domain for which the integral converges is ${\mathcal{D}}_{{\mathcal{I}}_1}=\{s\in \mathbb{C}:\sigma >-1,0<t<\infty \}$. □

Now, let ${\mathcal{I}}_2\left(s\right)$ be the integral of the second term of the right side of the last expression: Proceeding analogously as I did with ${\mathcal{I}}_1\left(s\right)$, when $x\gg 1$, then ${(e^x+1)}^2\approx e^{2x}$, so I get which can be conveniently re-arranged with a variable change $u=2x$, giving as a result or, valid for $\mathfrak{Re}\left(s\right)>-1$. □

Equation (2.4) is valid for $\sigma >0$, so, according to the variable change in equation3 (2.8), $1-\sigma >0\Rightarrow \sigma <1$. Performing integration by parts over (2.8), I use the following variables, $u=1/(e^x+1)$, $du=-e^x/{(e^x+1)}^{2}dx$, $dv=(d/dx)\left(x^{1-s}\right)dx$, $v=x^{1-s}$, I obtain which remains valid for $\mathfrak{Re}\left(s\right)<1$.

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By following the same reasoning as per proof of proposition (2.1), the limits tend to zero for all $\mathfrak{Re}\left(s\right)<1$. □

When $x\gg 1$, then ${(e^x+1)}^2\approx e^{2x}$, so According to definition 2.2 and performing a variable change $s\mapsto 2-s$, I have As $\Gamma (2-s)$ is valid for $\mathfrak{Re}\left(s\right)<2$, this implies that the integral ${\mathcal{I}}_3\left(s\right)$ converges for $\mathfrak{Re}\left(s\right)<2$. Performing once more fractional separation over the exponential term in the integral, I can also write which can also be expanded as

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Let ${\mathcal{I}}_4\left(s\right)$ be the integral of the second term on the right side of the last expression: Proceeding analogously as before, when $x\gg 1$, then ${(e^x+1)}^2\approx e^{2x}$, so I get which can be conveniently re-arranged with a variable change $u=2x$, giving as a result or valid for $\mathfrak{Re}\left(s\right)<2$. □

See proof of proposition 2.6 for $1/\Gamma (2-s)$, proofs of propositions 2.3 and 2.6 for integral convergence of the term $x^{1-s}/{(e^x+1)}^2$.

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Let $f\left(s\right)$ be the numerator in (2.13), which can be expressed as $f\left(s\right)=1-e^{(-\sigma -it)ln2}$, with real part $\mathfrak{Re}\left(f\left(s\right)\right)=1-2^{-\sigma }cos\left(tln2\right)$ and imaginary part $\mathfrak{Im}\left(f\left(s\right)\right)=2^{-\sigma }sin\left(tln2\right)$. $\mathfrak{Re}\left(f\right(\rho \left)\right)=0$ ⇔$cos\left(\gamma ln2\right)=2^{\sigma }$ and $\mathfrak{Im}\left(f\right(\rho \left)\right)=0$ ⇔$\gamma =k\pi /ln2$$\forall \sigma \in (0,1)$ with $k=0,\pm 1,\pm 2,\pm 3,...$. As the values of $\gamma$ have to be the same for both real and imaginary parts, then $cos\left(k\pi \right)=2^{\sigma }$. For $k=2j+1$, with $j=0,\pm 1,\pm 2,\pm 3,...$, $cos\left(\right(2j+1\left)\pi \right)=-1$, and $2^{\sigma }>0$$\forall \sigma \in \mathbb{R}$, so $cos\left((2j+1)\pi \right)\ne 2^{\sigma }$$\forall \sigma \in \mathbb{R}$. For $k=2l$, with $l=0,\pm 1,\pm 2,\pm 3,...$, $cos\left(2l\pi \right)=1$, and $2^{\sigma }\ne 1$$\forall \sigma \in (0,1)$, so $cos\left(2l\pi \right)\ne 2^{\sigma }$ $\forall \sigma \in (0,1)$. I can conclude that $\nexists \sigma \in (0,1):cos\left(k\pi \right)=2^{\sigma }⟹f\left(s\right)\ne 0$ $\forall \sigma \in (0,1)$.

As $f\left(s\right)\ne 0$$\forall \sigma \in (0,1)$⟹$\Lambda \left(\rho \right)\ne 0$. □

Let $g\left(s\right)$ be the denominator in equation (2.13), which can be expressed as $g\left(s\right)=1-e^{(1-\sigma -it)ln2}$, with real part $\mathfrak{Re}\left(g\left(s\right)\right)=1-2^{1-\sigma }cos\left(tln2\right)$ and imaginary part $\mathfrak{Im}\left(g\left(s\right)\right)=2^{1-\sigma }sin\left(tln2\right)$. $\mathfrak{Re}\left(g\right(\rho \left)\right)=0$ ⇔$cos\left(\gamma ln2\right)=2^{\sigma -1}$ and $\mathfrak{Im}\left(g\right(\rho \left)\right)=0$ ⇔$\gamma =k\pi /ln2$$\forall \sigma \in (0,1)$ with $k=0,\pm 1,\pm 2,\pm 3,...$. As the values of $\gamma$ have to be the same for both real and imaginary parts, then $cos\left(k\pi \right)=2^{\sigma -1}$. For $k=2j+1$, with $j=0,\pm 1,\pm 2,\pm 3,...$, $cos\left(\right(2j+1\left)\pi \right)=-1$, and $2^{\sigma -1}>0$$\forall \sigma \in \mathbb{R}$, so $cos\left((2j+1)\pi \right)\ne 2^{\sigma -1}$$\forall \sigma \in (0,1)$. For $k=2l$, with $l=0,\pm 1,\pm 2,\pm 3,...$, $cos\left(2l\pi \right)=1$, and $2^{\sigma -1}\ne 1$ $\forall \sigma \in (0,1)$, so $cos\left(2l\pi \right)\ne 2^{\sigma -1}$$\forall \sigma \in (0,1)$. I can conclude that $\nexists \sigma \in (0,1):cos\left(k\pi \right)=2^{\sigma -1}⟹g\left(s\right)\ne 0$ $\forall \sigma \in (0,1)$.

As $g\left(s\right)$ are well defined for $\forall \sigma \in (0,1)$⟹ $\Lambda \left(\rho \right)$ is well defined. □

The Riemann Zeta function satisfies the following equation for $s=\rho$, which has no zeros and is well defined

If I set $\zeta \left(\rho \right)=0$ in equation (2.7) and I recall lemmas 2.1 and 2.2 to ensure $\Lambda \left(\rho \right)$ has no zeros and is well defined, respectively, and I recall corollary 1.1 to state that $\zeta (\rho +1)\ne 0$, in consequence, ${\mathcal{I}}_2\left(\rho \right)\ne 0$, and I multiply both sides of equation (2.7) by $1-2^{1-\rho }$, I obtain 2.14. □

Let $\Delta \left(s\right)$ be a complex-valued function of the form

Let $F\left(s\right)$ be the numerator in equation 2.15, which can be expressed as $F\left(s\right)=1-e^{(\sigma +it-1)ln2}$, with real part $\mathfrak{Re}\left(F\left(s\right)\right)=1-2^{\sigma -1}cos\left(tln2\right)$ and imaginary part $\mathfrak{Im}\left(F\left(s\right)\right)=-2^{\sigma -1}sin\left(tln2\right)$. $\mathfrak{Re}\left(F\right(\rho \left)\right)=0$⇔$cos\left(\gamma ln2\right)=2^{1-\sigma }$ and $\mathfrak{Im}\left(F\right(\rho \left)\right)=0$⇔$\gamma =k\pi /ln2$$\forall \sigma \in (0,1)$ with $k=0,\pm 1,\pm 2,\pm 3,...$. Again, as the values of $\gamma$ have to be the same for both real and imaginary parts, then $cos\left(k\pi \right)=2^{1-\sigma }$. For $k=2j+1$, with $j=0,\pm 1,\pm 2,\pm 3,...$, $cos\left(\right(2j+1\left)\pi \right)=-1$, and $2^{1-\sigma }>0$$\forall \sigma \in \mathbb{R}$, so $cos\left((2j+1)\pi \right)\ne 2^{1-\sigma }$ $\forall \sigma \in \mathbb{R}$. For $k=2l$, with $l=0,\pm 1,\pm 2,\pm 3,...$, $cos\left(2l\pi \right)=1$, and $2^{\sigma }\ne 1$$\forall \sigma \in (0,1)$, so $cos\left(2l\pi \right)\ne 2^{\sigma }$$\forall \sigma \in (0,1)$. I can conclude that $\nexists \sigma \in (0,1):cos\left(k\pi \right)=2^{\sigma }⟹F\left(s\right)\ne 0$ $\forall \sigma \in (0,1)$.

As $F\left(s\right)\ne 0$$\forall \sigma \in (0,1)$⟹$\Delta \left(\rho \right)\ne 0$. □

Let $G\left(s\right)$ be the denominator in equation (2.15), which can be expressed as $G\left(s\right)=1-e^{(\sigma +it)ln2}$, with real part $\mathfrak{Re}\left(G\left(s\right)\right)=1-2^{\sigma }cos\left(tln2\right)$ and imaginary part $\mathfrak{Im}\left(G\left(s\right)\right)=-2^{\sigma }sin\left(tln2\right)$. $\mathfrak{Re}\left(G\right(\rho \left)\right)=0$ ⇔$cos\left(\gamma ln2\right)=2^{-\sigma }$ and $\mathfrak{Im}\left(G\right(\rho \left)\right)=0$ ⇔$\gamma =k\pi /ln2$$\forall \sigma \in (0,1)$ with $k=0,\pm 1,\pm 2,\pm 3,...$. Following an identical reasoning, as the values of $\gamma$ have to be the same for both real and imaginary parts, then $cos\left(k\pi \right)=2^{-\sigma }$. For $k=2j+1$, with $j=0,\pm 1,\pm 2,\pm 3,...$, $cos\left(\right(2j+1\left)\pi \right)=-1$, and $2^{-\sigma }>0$$\forall \sigma \in \mathbb{R}$, so $cos\left((2j+1)\pi \right)\ne 2^{-\sigma }$ $\forall \sigma \in (0,1)$. For $k=2l$, with $l=0,\pm 1,\pm 2,\pm 3,...$, $cos\left(2l\pi \right)=1$, and $2^{-\sigma }\ne 1$$\forall \sigma \in (0,1)$, so $cos\left(2l\pi \right)\ne 2^{-\sigma }$$\forall \sigma \in (0,1)$. I can conclude that $\nexists \sigma \in (0,1):cos\left(k\pi \right)=2^{-\sigma }⟹G\left(s\right)\ne 0$ $\forall \sigma \in (0,1)$.

As $G\left(s\right)$ are well defined for $\forall \sigma \in (0,1)$⟹ $\Delta \left(\rho \right)$ is well defined. □

The Riemann Zeta function satisfies the following equation for $s=\rho$, which has no zeros and is well defined

If I set $\zeta \left(\rho \right)=0$ in equation (2.12) and I recall lemmas 2.3 and 2.4 to ensure $\Delta \left(\rho \right)$ has no zeros and is well defined, respectively, and I recall again corollary 1.1 to state that $\zeta (2-\rho )\ne 0$, in consequence, ${\mathcal{I}}_4\left(\rho \right)\ne 0$, and I multiplying both sides of (15) by $1-2^{\rho }$, I get to (2.16). □