Proof of the Riemann Hypothesis Using the Decomposition Zeta(s) = X(s) – Y(s)

PART 1: The Riemann Zeta function ζ($\mathit{s}$) [RZF]

1.

$\zeta (s) in R$

As defined in literature (Sondow et al, Weisstein, Edwards, Jekel)

1.1.

$\mathsf{\zeta }\left(\mathrm{s}\right)=\sum _{\mathrm{j}=1}^{\mathrm{\infty }}{\mathrm{j}}^{-\mathrm{s}} \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{s} \mathrm{f}\mathrm{o}\mathrm{r} \mathrm{s}\ne 1$

Figure 1. Riemann Zeta function in R.

Figure 1. Riemann Zeta function in R.

• 1.2.

  Euler Product Formula that links $\mathsf{\zeta }\left(\mathrm{s}\right)$ with the distribution of prime numbers

  $$\begin{array}{c}\mathsf{\zeta }\left(\mathrm{s}\right)\hfill \\ ={\displaystyle \sum _{\mathrm{j}=1}^{\mathrm{\infty }}}{\mathrm{j}}^{-\mathrm{s}}={\prod }_{p=prime}\frac{1}{1-{\mathrm{p}}^{-\mathrm{s}}}\hfill \end{array}$$

  [1]

  Example for k=2

  $$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\dots =\frac{1}{1-2^{-2}}x \frac{1}{1-3^{-2}}x\frac{1}{1-5^{-2}}x\frac{1}{1-7^{-2}}x\dots$$

  1.3.

  Integral definition of $\mathsf{\zeta }\left(\mathrm{s}\right)$:

  $$\mathsf{\zeta }\left(\mathrm{s}\right)=\sum _{\mathrm{j}=1}^{\mathrm{\infty }}{\mathrm{j}}^{-\mathrm{s}}=\frac{1}{\mathsf{\Gamma }(\mathrm{s})}{\int }_0^{\mathrm{\infty }}\frac{1}{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}^{\mathrm{s}}\frac{\mathrm{d}\mathrm{x}}{\mathrm{x}}$$

  Where $\mathsf{\Gamma }(\mathrm{s})$, is the Gamma function

  1.4.

  Analytical continuation of $\mathsf{\zeta }\left(\mathrm{s}\right)$for :

  Re(s)>0: [Dirichlet]

  $$\mathsf{\zeta }\left(\mathrm{s}\right)=\frac{1}{s-1}\sum _{k=1}^{\infty }(\frac{n}{{\left(n+1\right)}^s}-\frac{n-s}{n^s}$$
  0<Re(s)<1:

  $$\mathsf{\zeta }\left(\mathrm{s}\right)= \frac{1}{1-2^{1-s}}\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k-1}}{k^s}$$
  -k<Re(s) [Kopp, Konrad. 1945]:

  $$\mathsf{\zeta }\left(\mathrm{s}\right)= \frac{1}{s-1}\sum _{k=1}^{\infty }\frac{k(k+1)}{2}(\frac{2k+3+s)}{{(k+1)}^{s+2}}-\frac{2k-1-s}{k^{s+2}})$$

  1.5.

  Laurent series of $\mathsf{\zeta }\left(\mathrm{s}\right)$:

  $$\mathsf{\zeta }\left(\mathrm{s}\right)= \frac{1}{s-1}+\sum _{k=}^{\infty }\frac{{\left(-1\right)}^k{\gamma }_n}{k!}{\left(s-1\right)}^k)$$

  where ${\gamma }_n$ are the Stieltjes constants.

  1.6.

  Hurwitz function $\mathsf{\zeta }\left(\mathrm{k},\mathrm{z}\right)$:

  $$\mathsf{\zeta }\left(\mathrm{k},\mathrm{z}\right)=\sum _{\mathrm{j}=0}^{\mathrm{\infty }}{\left(\mathrm{j}+\mathrm{z}\right)}^{-\mathrm{k}}=\sum _{\mathrm{j}=\mathrm{z}}^{\mathrm{\infty }}{\mathrm{j}}^{-\mathrm{k}} \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{s} \mathrm{f}\mathrm{o}\mathrm{r} \mathrm{k}>1$$

  1.7.

  Generalized Harmonic Function ${\mathrm{H}}_{\mathrm{n}}^{\left(\mathrm{k}\right)}$:

  $${\mathrm{H}}_{\mathrm{n}}^{\left(\mathrm{k}\right)}=\sum _{\mathrm{j}=1}^{\mathrm{n}}{\mathrm{j}}^{-\mathrm{k}}=\left(\frac{1}{1^{\mathrm{k}}}+\frac{1}{2^{\mathrm{k}}}+\dots +\frac{1}{{\mathrm{n}}^{\mathrm{k}}}\right) \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{s} \mathrm{f}\mathrm{o}\mathrm{r} \mathrm{k}>1$$

  1.8.

  ζ(s) converges for s>1 to the following values (Sloane):

  

  Table 1. Values of ζ(s).

  Table 1. Values of ζ(s).

  s	ζ(s)	Known ζ(s) representations over π
  2	1.6449179	π2/6
  4	1.0823232	π4/90
  6	1.0173431	π6/945
  8	1.0040774	π8/9450

  1.9.

  An approximation for the values of $\mathsf{\zeta }\left(\mathrm{s}\right)$ for s>1 in R

  One can calculate that:

  $$\underset{s\to \infty }{\lim (}{\frac{\mathsf{\zeta }\left(\mathrm{s}\right)}{\mathsf{\zeta }\left(\mathrm{s}\right)+1})}^{\frac{1}{s}}=1$$
  And:

  $$\underset{s\to \infty }{\lim }{(\frac{\mathsf{\zeta }\left(\mathrm{s}\right)}{\mathsf{\zeta }\left(\mathrm{s}\right)-1})}^{\frac{1}{s}}=2$$
  Based on this expression, for s sufficiently large, one can represent $\mathsf{\zeta }\left(\mathrm{s}\right)$as a multiple of ${\pi }^s$:

  $$\mathsf{\zeta }\left(\mathrm{s}\right)=\frac{{\pi }^s}{K_s} \mathrm{with} K_s = \left(2^s-1\right)\ast \frac{{\pi }^s}{2^s}$$

  with a very good approximation given by:

  $$K_s^{\ast } = int\left(\left(2^s-1\right)\ast \frac{{\pi }^s}{2^s}\right)-1 \mathrm{with} \mathrm{int}(\mathrm{k}) \mathrm{is} \mathrm{the} \mathrm{integer} \mathrm{part} \mathrm{of} \mathrm{k}.$$

  [2]
  The error between the $K_s^{\ast }$ calculated and $K_s$ actual is very small for s>4.
  Some calculated values of $K_s^{\ast }$ calculated and $K_s$ actual:

  

  Table 2. Values of $K_s^{\ast }$ calculated and $K_s$ actual.

  Table 2. Values of $K_s^{\ast }$ calculated and $K_s$ actual.

  
  One can use [2] to propose the following approximation for $\mathsf{\zeta }\left(\mathrm{s}\right)$:

  $$\mathrm{C}\mathrm{Z}\left(\mathrm{s}\right)=\frac{1}{1-{\mathsf{\pi }}^{-\mathrm{s}}-2^{-\mathrm{s}}}$$

  [3]

  

  Table 3. Comparing $\mathsf{\zeta }\left(\mathrm{s}\right)$and CZ(s).

  Table 3. Comparing $\mathsf{\zeta }\left(\mathrm{s}\right)$and CZ(s).

  	s=3	s=4	s=10	s=14
  $\mathsf{\zeta }\left(\mathrm{s}\right)$	1.20206	1.0823	1.000994	1.0000612
  $\mathrm{C}\mathrm{Z}\left(\mathrm{s}\right)$	1.18659	1.0784	1.000988	1.0000611

  Graphically:

Figure 2. Caceres’ approximation for the Riemann Zeta function in R.

Figure 2. Caceres’ approximation for the Riemann Zeta function in R.

• 1.10.

  Zeros of $\mathsf{\zeta }\left(\mathrm{z}\right)$ (Weisstein)

  1.10.1.

  $Re(z)>1$

  There are no zeros of $\mathsf{\zeta }\left(\mathrm{s}\right)$. $\mathsf{\zeta }\left(\mathrm{s}\right)$ converges for s>1.

      1.10.2.

  $z=1$

  $\mathsf{\zeta }\left(1\right)$ is the Harmonic function and diverges, and:
  $\underset{s\to 1}{\lim }(\mathsf{\zeta }\left(\mathrm{s}\right)-\frac{1}{1-s})= \gamma$ (the Euler-Mascheroni constant)

      1.10.3.

  $Re(z)<1$

  $\mathsf{\zeta }\left(\mathrm{z}\right)$ has trivial zeros at $z=-2k$, with $k$ natural numbers
  $\mathsf{\zeta }\left(\mathrm{z}\right)$ has nontrivial zeros at $z =\frac{1}{2}\pm ßi$ at the critical line $Re(z)=\frac{1}{2}$ (Brentt)
  The Riemann Hypothesis [RH] establishes that all nontrivial zeros of zeta have $Re(z)=\frac{1}{2}$List of imaginary part of the first nontrivial zeros of zeta (Odlyzko):
  • ß(1)    = 14.134725142
  • ß(3)    = 25.010857580
  • ß(4)    = 30.424876126
  • ß(6)    = 37.586178159
  • ß(7)    = 40.918719012
  • ß(9)    = 48.005150881
  • ß(10) = 49.773832478
  Representation of |$\mathsf{\zeta }\left(\mathrm{z}\right)|$:

Figure 3. Modulus of the Riemann Zeta function |$\mathsf{\zeta }\left(\mathrm{z}\right)|.$

Figure 3. Modulus of the Riemann Zeta function |$\mathsf{\zeta }\left(\mathrm{z}\right)|.$

PART 2: The C-Transformation

1.

C-Transformation of f(x)

Let’s define the C-transformation of an integrable function f(x) by:

$$\begin{gathered} C_n\left\{f\left(x\right)\right\}= {\sum }_{k=1}^{n}f\left(k\right) \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}- \int f\left(n\right) dn \end{gathered}$$

[4]

And the C-values is the limit, if it exists, of the C-transformation when $n\to \infty$:

$$\begin{array}{c}C\left\{f\left(x\right)\right\}\hfill \\ =\underset{n\to \infty }{\lim }{C}_n\left\{f\left(x\right)\right\} \hfill \end{array}$$

[5]

1.1.  C-Transformation of $f\left(x\right)=\frac{1}{x}$ for $x\in R$:

$$C_n\left\{\frac{1}{x}\right\}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}-\int \frac{dn}{n}$$

And the C-Value of $f\left(x\right)=\frac{1}{x}$ is $\gamma =0.5772$ (Euler-Mascheroni constant)

$$\mathrm{C}\left\{\frac{1}{x}\right\}=\underset{n\to \infty }{\lim }(\sum _{k=1}^n\frac{1}{k}-\mathrm{l}\mathrm{n}(n))=\gamma$$

1.2.  C-Transformation of $f\left(x\right)=\frac{{\ln \left(\mathrm{x}\right)}^{\mathrm{m}}}{x}$ for $x\in R,m\in Z$:

$$C_n\left\{\frac{{\ln \left(\mathrm{x}\right)}^{\mathrm{m}}}{x}\right\}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{\ln \left(k\right)}^m}{\mathrm{k}}-\int \frac{{\ln \left(n\right)}^{m}dn}{n}$$

And the C-Value of $f\left(x\right)=\frac{{\ln \left(\mathrm{x}\right)}^{\mathrm{m}}}{x}$ are the Stieltjes constants that occur in the Laurent series expansion of the Riemann zeta function:

$$\mathrm{C}\left\{\frac{{\ln \left(\mathrm{x}\right)}^{\mathrm{m}}}{x}\right\}=\underset{\mathrm{m}\to \mathrm{\infty }}{\lim }(\sum _{\mathrm{k}=1}^{\mathrm{m}}\frac{{(\ln k)}^n}{\mathrm{k}}-\frac{{(\ln n)}^{m+1}}{m+1})={\gamma }_m$$

1.3.  C-Transformation of $f\left(x\right)=m$ , for $m\in R$ constant:

$$C_n\left\{m\right\}=\sum _{\mathrm{k}=1}^{\mathrm{n}}m-\int m dn$$

$$C_n\left\{\mathrm{m}\right\}=m\ast n-m\ast n=0$$

and the C-values of f(x) =m constant is:

$$C\left\{m\right\}=0$$

1.4.  C-Transformation of $f\left(x\right)=\sin \left(x\right)$ for $x\in R$:

$$C_n\left\{\sin \left(x\right)\right\}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\sin \left(\mathrm{k}\right)-\int \sin \left(n\right)dn$$

$$C_n\left\{\sin \left(x\right)\right\}=\frac{1}{2\left(\sin \left(n\right)-\cot \left(\frac{1}{2}\right)\cos \left(n\right)+\cot \left(\frac{1}{2}\right)+\cos \left(n\right)\right)}$$

And the C-values of $f\left(x\right)=\sin \left(x\right)$ are in the interval:

$$\mathrm{C}\left\{\sin \left(x\right)\right\}\in \left[\frac{1}{2}\left(2\cot \left(\frac{1}{2}\right)-3\right), \frac{3}{2}\right]$$

One can also calculate that:

$$\mathrm{C}\left\{\cos \left(x\right)\right\}\in \left[\frac{1}{2}\left(\cot \left(\frac{1}{2}\right)-4\right),\frac{1}{2}\left(2-\cot \left(\frac{1}{2}\right)\right)\right]$$

1.5.  C-Transformation of $f\left(x\right)={\mathrm{e}}^{-\mathrm{x}}$ for $x\in R$:

$$C_n\left\{{\mathrm{e}}^{-\mathrm{x}}\right\}=\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{k}}-\int e^{-n}dn$$

$$C_n\left\{\sin \left(x\right)\right\}=\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{k}}+\frac{e^{-n}}{n}$$

And the C-values of $f\left(x\right)={\mathrm{e}}^{-\mathrm{x}}$ are:

$$\mathrm{C}\left\{{\mathrm{e}}^{-\mathrm{x}}\right\}=\frac{1}{\mathrm{e}-1}$$

1.6.  C-Transformation of $f\left(x\right)=x^{-s}$ for $x,s\in R$, s>1:

$$C_n\left\{\frac{1}{x^s}\right\}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{{\mathrm{k}}^{\mathrm{s}}}-\int \frac{dn}{n^s}$$

$$C_n\left\{\frac{1}{x^s}\right\}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{{\mathrm{k}}^{\mathrm{s}}}-\frac{n^{1-s}}{1-s}$$

and the C-value of $f\left(x\right)=\frac{1}{x^s}$ is the Riemann Zeta function for s>1:

$$\mathrm{C}\left\{\frac{1}{x^s}\right\}=\underset{n\to \infty }{\lim }(\sum _{k=1}^n\frac{1}{{\mathrm{k}}^{\mathrm{s}}}-\frac{{\mathrm{n}}^{1-\mathrm{s}}}{1-\mathrm{s}})=\underset{n\to \infty }{\lim }(\sum _{k=1}^n\frac{1}{{\mathrm{k}}^{\mathrm{s}}})-\underset{n\to \infty }{\lim }(\frac{{\mathrm{n}}^{1-\mathrm{s}}}{1-\mathrm{s}})=\zeta (s)$$

1.7.  C-Transformation of $f\left(z\right)=\frac{1}{x^z}$ for $z\in C,Re\left(z\right)\ge 0,z\ne 1$

$$C_n\left\{\frac{1}{x^z}\right\}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{{\mathrm{k}}^{\mathrm{z}}}-\int \frac{dn}{n^z}$$

One can use Euler’s identity:

$$e^x=\cos \left(x\right)+i\ast \sin \left(x\right)$$

To calculate for z=α+ßi:

$$k^{-z}=k^{-\alpha } [\cos \left(ß\ast \ln k\right)-i (\mathrm{s}\mathrm{i}\mathrm{n}(ß\ast \ln \mathrm{k})]$$

And:

$$\int \frac{dn}{n^z}=n^{\left(1-\alpha \right)}[\cos (ß\ast \mathit{ln}\left(n\right)-i\mathit{sin}[ß\ast \mathit{ln}\left(n\right)]\ast \frac{\left[\left(1-\alpha \right)+iß\right]}{[{\left(1-\alpha \right)}^2+{ß}^{2]}}$$

One can now express the real and imaginary components of $C_n\left\{f\right\}$ as:

$$\begin{gathered} Re(C_n\left\{f\right\})= {\sum }_{k=1}^n{k}^{-\alpha } (\mathrm{c}\mathrm{o}\mathrm{s}(ß\ast \ln \left(k\right)) + \\ + \frac{1}{[{\left(1-\alpha \right)}^2+{ß}^{2]}} (n^{\left(1-\alpha \right)} [(1\text{-}\alpha )*\cos (ß*\ln (\mathrm{n}))+ß*\sin (ß*\ln (\mathrm{n}))]) \end{gathered}$$

[6]

$$\begin{gathered} Im\left(C_n\left\{f\right\}\right)= -\sum _{k=1}^n{k}^{-\alpha } (\mathrm{s}\mathrm{i}\mathrm{n}(ß\ast \ln \left(k\right))+\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em} \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+ \frac{1}{[{\left(1-\alpha \right)}^2+{ß}^{2]}} (n^{\left(1-\alpha \right)} [ß*\cos (ß*\ln (\mathrm{n})) \text{-} (1\text{-}\alpha )*\sin (ß*\ln (\mathrm{n}))]) \end{gathered}$$

[7]

One can calculate that, for α=Re(z)>1, and for any $ϵ$ arbitrarily small, there is a value of n=N such that for n>N, $C_N\left\{f\right\}$- ζ($z$)< $ϵ$, as the following table shows:

Table 4. Values of $C_n\left\{f\left(n\right)=k^{-z}\right\}$ for α=Re(z)>1 for N=500.

Table 4. Values of $C_n\left\{f\left(n\right)=k^{-z}\right\}$ for α=Re(z)>1 for N=500.

α	ß	$C_N\left\{f\right\}$ for N=500	ζ ( $z$ )	$|C\_N \{f\}-$ ζ ( $z$ )|
2	0	1.644934068	1.654934067	$<{10}^{-8}$
2	1	1.150355702 + 0.437530865 i	1.150355703+0.437530866 i	$<{10}^{-8}$
3	3	0	1.202056903	1.202056903	$<{10}^{-9}$

That shows that the C-values of $\mathrm{f}\left(\mathrm{z}\right)=\frac{1}{x^z}$ for Re(z)>1 is $\mathsf{\zeta }\left(\mathrm{z}\right).$

PART 3: A decomposition of ζ(z) based on the C-transformation of $\mathit{f}\left(\mathit{x}\right)\mathbf{=}\frac{\mathbf{1}}{{\mathit{x}}^{\mathit{z}}}$. for $\mathit{z}\mathbf{\in }\mathit{C}\mathbf{,}\mathbf{0}\mathbf{\le }\mathit{R}\mathit{e}\left(\mathit{z}\right)\mathbf{<}\mathbf{1}$

• C-Transformation of $f\left(z\right)=\frac{1}{x^z}$ for $z\in C,0\le Re\left(z\right)<1$

The C-values of $f\left(z\right)=\frac{1}{x^z}$ from [6] and [7] are equal to the ζ($z$) when Re(z)>1, this error |$C_n\left\{f\right\}$- ζ($z$)| grows significantly in the critical strip $for 0\le Re(z)<1$ as observed in the following table:

Table 5. Values of $C_n\left\{f\left(n\right)=k^{-z}\right\}$ for 0$\le$Re(z)<1 for N=500.

Table 5. Values of $C_n\left\{f\left(n\right)=k^{-z}\right\}$ for 0$\le$Re(z)<1 for N=500.

A	ß	$C_n\left\{f\right\}$	ζ ( $z$ )	$|C\_n \{f\}-$ ζ ( $z$ )|
0.0	0	CN{f} for N=500	-0.5	0.5
0.2	2	0.399824505+0.322650799 i	0.360103 + 0.266246 i	$>0.05$
0.7	0	-2.777900606	-2.7783884455	$>{10}^{-4}$

Figure 4a. Where a=Re(z) and b=Im(z).

Figure 4a. Where a=Re(z) and b=Im(z).

Figure 4b. Where a=Re(z) and b=Im(z).

Figure 4b. Where a=Re(z) and b=Im(z).

2.

Decomposition of $\zeta \left(z\right)=X\left(z\right)-Y\left(z\right)$

One can rewrite [8] and [9] creating the $X(z,n)$ and $Y(z,n)$ functions:

$\zeta (z) = \underset{n\to \infty }{\mathit{lim}}\left[X\left(z,n\right)- Y\left(z,n\right)\right],$ where:

$$\begin{gathered} X(z, n) = (\sum _{k=1}^n{k}^{-\alpha }\ast \mathrm{c}\mathrm{o}\mathrm{s}(ß\ast \mathit{ln}\left(k\right))+\frac{1}{2} n^{-\alpha }\mathrm{c}\mathrm{o}\mathrm{s}(ß\mathit{ln}\left(n\right))+\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em} \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+ i\ast (\sum _{k=1}^n{k}^{-\alpha }\ast sin(ß\ast \mathit{ln}\left(k\right))+\frac{1}{2} n^{-\alpha }\mathit{sin}\left(ß\mathit{ln}\left(n\right)\right))) \end{gathered}$$

[10]

$$\begin{gathered} Y(z, n)={\mathrm{n}}^{\left(1-\mathsf{\alpha }\right)}\frac{1}{[{\left(1-\mathsf{\alpha }\right)}^2+{\mathrm{ß}}^{2]}}[\left(\left(1-\mathsf{\alpha }\right)\ast \cos \left(\mathrm{ß}\ln \left(\mathrm{n}\right)\right)+\mathrm{ß}\ast \sin \left(\mathrm{ß}\ln \left(\mathrm{n}\right)\right)\right)+ \\ +\mathrm{i}(\mathrm{ß}\ast \cos \left(\mathrm{ß}\ln \left(\mathrm{n}\right)\right)-\left(1-\mathsf{\alpha }\right)\ast \sin \left(\mathrm{ß}\ln \left(\mathrm{n}\right)\right))] \end{gathered}$$

[11]

and define:

$$X\left(z\right)= \underset{n\to \infty }{\lim }\mathrm{X}\left(\mathrm{z},\mathrm{n}\right) \mathrm{and}$$

$$Y\left(z\right)=\underset{n\to \infty }{\lim }\mathrm{Y}\left(\mathrm{z},\mathrm{n}\right)$$

to write:

$$\zeta \left(z\right)=X\left(z\right)-Y\left(z\right)$$

[12]

The following table compared the values of $\zeta \left(z\right)$and $X\left(z\right)-Y\left(z\right)$:

Table 6. ζ(z) compared to X(z) - Y(z).

Table 6. ζ(z) compared to X(z) - Y(z).

z= 0 +j* 0 and n=500

Zeta(z) = -0.5 + i* 0.0      X(z)-Y(z) = -0.5 +i* 0.0      ---> Error = 0.0 +i* 0.0

z= 0.2 +j* 2 and n=500

Zeta(z) = 0.360102590022591 + i* -0.266246199765574       X(z)-Y(z) = 0.360102741838091 +i* -0.266246128959438       ---> Error= -1.5181550 e-7 +i* -7.080613 e-8

z= 0.4 +j* 0 and n=500

Zeta(z) = -1.13479778386698 + i* 0.0       X(z)-Y(z) = -1.1347977871726 +i* 0.0       ---> Error= 3.305619 e-9 +i* 0.0

The highest error for α$\in \left[\mathrm{0,1}\right), ß\in \left[\mathrm{0,100}\right],$n=1000 is $8x{10}^{-6}$.

3.

Representation of the function $\zeta \left(z\right)=X(z)-Y\left(z\right)$ for Re(z)=1/2

Figure 5. ζ(z) = X(z) - Y(z).

Figure 5. ζ(z) = X(z) - Y(z).

4.

Representation of the function $\left|\zeta \left(z\right)\right|=|X(z)-Y\left(z\right)|$ for Re(z)=1/2

Figure 6. |ζ(z)| = |X(z) - Y(z)|.

Figure 6. |ζ(z)| = |X(z) - Y(z)|.

5.

Representation of the function $X(z,n$)

The following chart represents $X(z,n)$ for $a=1/2$ and $b\in \left[\mathrm{1,6}\right]$and $n=250$.

Figure 7. $X(z,n)$

Figure 7. $X(z,n)$

• The following chart represents $X(z,n)$ for $a\in [\mathrm{1,6}]$and $b=1$ and $n=250$.

Figure 8. $X(z,n)$

Figure 8. $X(z,n)$

6.

Representation of the function $Y(z,n)$

The following chart represents $Y(z,n)$ for $a=1/2$ and $b\in \left[\mathrm{1,6}\right]$ and $n=250$

Figure 9. $Y(z,n)$

Figure 9. $Y(z,n)$

• The following chart represents $Y(z,n)$for $a\in [\mathrm{1,6}]$ and $b=1$ and $n=250$

Figure 10. $Y(z,n)$

Figure 10. $Y(z,n)$

7.

Representation of |$X(z,n)|$

Wave representation for $|X(z,n)|forRe(z)=1/2 and Im(z) variable$. [Figure. 11]

• Parabolic representation for |$X(z,n)| for (z) \mathrm{a} \mathrm{n}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{a}\mathrm{l} \mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o} \mathrm{o}\mathrm{f} \mathrm{R}\mathrm{i}\mathrm{e}\mathrm{m}\mathrm{a}\mathrm{n}\mathrm{n} \mathrm{Z}\mathrm{e}\mathrm{t}\mathrm{a}.$ [Figure. 12]

• Linear representation for |$X\left(z,n\right){\left.\right|}^2 for (z) \mathrm{a} \mathrm{n}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{a}\mathrm{l} \mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o} \mathrm{o}\mathrm{f} \mathrm{R}\mathrm{i}\mathrm{e}\mathrm{m}\mathrm{a}\mathrm{n}\mathrm{n} \mathrm{Z}\mathrm{e}\mathrm{t}\mathrm{a}.$ [Figure. 13]

8.

Representation of |$Y(z,n)|$

Polynomial representation for $|Y(z,n)|forRe(z)=1/2 and Im(z) variable$. [Figure. 14]

• Parabolic representation for $|Y(z,n)| for \left(z\right)\mathrm{a} \mathrm{n}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{a}\mathrm{l} \mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o} \mathrm{o}\mathrm{f} \mathrm{R}\mathrm{i}\mathrm{e}\mathrm{m}\mathrm{a}\mathrm{n}\mathrm{n} \mathrm{Z}\mathrm{e}\mathrm{t}\mathrm{a}$. [Figure. 15]

• Linear representation for ${\left|Y\left(z,n\right)\right|}^2 for \left(z\right)\mathrm{a} \mathrm{n}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{a}\mathrm{l} \mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o} \mathrm{o}\mathrm{f} \mathrm{R}\mathrm{i}\mathrm{e}\mathrm{m}\mathrm{a}\mathrm{n}\mathrm{n} \mathrm{Z}\mathrm{e}\mathrm{t}\mathrm{a}$. [Figure. 16]

9.

Conclusion PART 3

• $|X(z,n)|$ has a wave representation
• $|X(z,n)|$ becomes a parable when z is a nontrivial zero of Riemann Zeta
• ${\left|X\left(z,n\right)\right|}^2$ becomes a line when z is a nontrivial zero of RZF with slope equal $\frac{1}{{ß}^2+¼}$
• $|Y(z,n)|$ has a polynomial representation
• $|Y(z,n)|$ becomes a parable when z is a nontrivial zero of Riemann Zeta
• ${\left|Y\left(z,n\right)\right|}^2$ becomes a line when Re(z)=1/2 with slope equal $1/({ß}^2+¼)$

PART 4: Proof of the Riemann Hypothesis using the decomposition. $\mathit{\zeta }\mathbf{(}\mathit{z}\mathbf{)}\mathbf{=}\mathit{X}\mathbf{(}\mathit{z}\mathbf{)}\mathbf{-}\mathit{Y}\mathbf{(}\mathit{z}\mathbf{)}$

1.

Analysis of Absolute Square ${\left|Y\left(z,n\right)\right|}^2$

$$\begin{gathered} {\left|Y\left(z,n\right)\right|}^2=[{\left( n^{\left(1-\alpha \right)}\frac{1}{[{\left(1-\alpha \right)}^2+{ß}^{2]}} \left[\left(1-\alpha \right)\ast cos\left(ßln\left(n\right)\right)+ß\ast sin\left(ßln\left(n\right)\right)\right]\right)}^2 \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+ {\left(n^{\left(1-\alpha \right)}\frac{1}{[{\left(1-\alpha \right)}^2+{ß}^{2]}} \left[ß\ast cos\left(ßln\left(n\right)\right)-\left(1-\alpha \right)\ast sin\left(ßln\left(n\right)\right)\right]\right)}^2] \end{gathered}$$

$${\left|Y\left(z,n\right)\right|}^2=n^{2\left(1-\alpha \right)}\ast \frac{1}{\left[{ß}^2+{\left(1-\alpha \right)}^2\right]}\hspace{1em}\mathrm{Polynomial} \mathrm{representation}$$

[13]

This could be observed in Figure. 14, 15, 16.

1.1. ${\left|Y\left(z,n\right)\right|}^2$ is a straight line if and only if $\alpha =\frac{1}{2}$

The slope of ${\left|Y\left(z,n\right)\right|}^2$ with respect to n is given by:

$$slope({\left|Y\left(z,n\right)\right|}^2) =d({\left|Y\left(z,n\right)\right|}^2)/dn$$

Which equals to:

$$d({\left|Y\left(z,n\right)\right|}^2)/dn= 2\left(1-\alpha \right) n^{1-2\alpha }\ast \frac{1}{[{ß}^2+{\left(1-\alpha \right)}^2]}$$

${\left|Y\left(z,n\right)\right|}^2$ can only be a line when the slope is constant, which can only happen if and only if:

$$(1-2\mathsf{\alpha })=0$$

therefore:

$${\left|Y\left(z,n\right)\right|}^2 \mathrm{is} \mathrm{a} \mathrm{straight} \mathrm{line} \mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if} \alpha =\frac{1}{2}$$

[14]

1.2. Summary for ${\left|Y\left(z,n\right)\right|}^2$ for $\alpha =\frac{1}{2}$:

 ⇨

the slope ${\left|Y\left(z,n\right)\right|}^2$ is constant if and only if $\alpha =\frac{1}{2}$

 ⇨

When α=1/2, ${\left|Y\left(z,n\right)\right|}^2$ = $\frac{n}{[{ß}^2+\frac{1}{4}]}$

 ⇨

The slope for ${\left|Y\left(z,n\right)\right|}^2$ is $\frac{1}{[{ß}^2+\frac{1}{4}]}$ for $\alpha =\frac{1}{2}$

2.

Analysis of Absolute Square

$${\left|X\left(z,n\right)\right|}^2$$

[15]

$$\begin{gathered} {\left|X\left(z,n\right)\right|}^2=(\frac{1}{2}{n}^{-a}\cos \left(ß\ln \left(n\right)\right)+\sum k^{-\alpha }{\mathit{cos}\left(ß\mathit{ln}\left(k\right)\right))}^2+ \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}(\frac{1}{2}{n}^{-a}\mathrm{s}\mathrm{i}\mathrm{n}(ß\ln \left(n\right))+\sum k^{-\alpha }{\mathit{sin}\left(ß\mathit{ln}\left(k\right)\right))}^2 \end{gathered}$$

Applying properties of infinite series (Kopp):

$$\begin{array}{cc}\hfill {\left|X\left(z,n\right)\right|}^2& =\frac{1}{4}{n}^{-2a}({\cos }^2\left(ß\ln \left(n\right)\right)+{\sin }^2(ß\ln \left(n\right)))+\hfill \\ & (\sum k^{-\alpha }{\mathit{cos}\left(ß\mathit{ln}\left(k\right)\right))}^2+(\sum k^{-\alpha }{\mathit{sin}\left(ß\mathit{ln}\left(k\right)\right))}^2+\hfill \\ & + n^{-a}[\mathrm{c}\mathrm{o}\mathrm{s}(ß\ln \left(n\right))\ast \sum k^{-\alpha }\mathit{cos}\left(ß\mathit{ln}\left(k\right)\right)]+\hfill \\ & {+n}^{-a}[\mathrm{s}\mathrm{i}\mathrm{n}(ß\ln \left(n\right))\ast \sum k^{-\alpha }\mathit{sin}\left(ß\mathit{ln}\left(k\right)\right)]\hfill \end{array}$$

$$\begin{array}{c}{\left|X\left(z, n\right)\right|}^2={\displaystyle \sum _{k=1}^n}{\displaystyle \sum _{j=1}^n}{k}^{-\alpha }\ast j^{-\alpha }\ast \mathit{cos}\left(ß(\mathit{ln}\left(\frac{k}{j}\right)\right)+{\displaystyle \sum _{k=1}^n}{k}^{-2\alpha }+\\ +\frac{1}{4}{n}^{-2a}+n^{-a}[\mathrm{c}\mathrm{o}\mathrm{s}(ß\ln \left(n\right))\ast \sum k^{-\alpha }\mathit{cos}\left(ß\mathit{ln}\left(k\right)\right)]+\\ +n^{-a}[\mathrm{s}\mathrm{i}\mathrm{n}(ß\ln \left(n\right))\ast \sum k^{-\alpha }\mathit{sin}\left(ß\mathit{ln}\left(k\right)\right)]\end{array}$$

One can express the previous expression replacing:

$$R\left(n\right)=\frac{1}{4}{n}^{-2a}+ n^{-a} [\mathrm{c}\mathrm{o}\mathrm{s}(ß\ln \left(n\right))\ast \sum k^{-\alpha }\mathit{cos}\left(ß\mathit{ln}\left(k\right)\right)+\mathrm{s}\mathrm{i}\mathrm{n}(ß\ln \left(n\right))\ast \sum k^{-\alpha }\mathit{sin}\left(ß\mathit{ln}\left(k\right)\right)]$$

With:

$$\underset{n\to \infty }{\lim }R\left(n\right)=0\mathrm{if} \alpha >0\hspace{1em}\mathrm{therefore},$$

[16]

$$\begin{gathered} {\left|X\left(z,n\right)\right|}^2=\sum _{k=1}^n\sum _{j=1}^n{k}^{-\alpha }\ast j^{-\alpha }\ast \mathit{cos}\left(ß\mathit{*ln}\left(\frac{k}{j}\right)\right)+\sum _{k=1}^n{k}^{-2\alpha } \\ +R\left(n\right) \end{gathered}$$

[17]

When ${\left|X\left(z,n\right)\right|}^2$ is represented graphically, one can observe that:

-

${\left|X\left(z,n\right)\right|}^2$ is a wave that converges when $\mathrm{n}\to \mathrm{\infty }$ and α>1 (Figure 17)

-

${\left|X\left(z,n\right)\right|}^2$ is a wave that does not converge when $\mathrm{n}\to \mathrm{\infty }$ and α<1 (Figure 18)

-

${\left|X\left(z,n\right)\right|}^2$ is a wave that collapses to a line when $\mathrm{n}\to \mathrm{\infty }$ and α=1/2 and ß=Im( $\zeta (z^{\ast })$) (Figure 19) where z* is a nontrivial zero of RZF.

2.1. ${\left|X\left(z,n\right)\right|}^2$ converges when $n\to \infty$ and α>1 to ${\left|\zeta \left(\alpha ,ß\right)\right|}^2$

The limit of ${\left|X\left(z,n\right)\right|}^2$outside the critical strip [0,1] can be calculated using [16]:

$$\underset{n\to \infty }{\mathit{lim}}{\left|X\left(z,n\right)\right|}^2 =\underset{n\to \infty }{\lim }{\sum }_{k=1}^n{\sum }_{j=1}^n{k}^{-\alpha }\ast j^{-\alpha }\ast \mathit{cos}\left(ß(\mathit{ln}\left(\frac{k}{j}\right)\right)$$

As one can see in some examples in the following table where z=α+iß:

Table 7.  

Table 7.  

α	ß	$\underset{n\to \infty }{\lim }{\left|X\left(z,n\right)\right|}^2$	${\left|\zeta \left(\alpha ,ß\right)\right|}^2$
1.0	7	1.074711506185445	1.074756
1.0	10	1.4413521753699579	1.441430
2.5	7	1.0093487944300192	1.009349
2.5	10	1.0507402208589398	1.050740

$$\underset{n\to \infty }{\mathit{lim}}{\left|X\left(z,n\right)\right|}^2={\left|\zeta \left(z\right)\right|}^2=\zeta \left(\alpha +ßi\right)\ast \zeta \left(\alpha -ßi\right) for \alpha >1$$

And also, in the following Figure 20:

One can observe that the graphs for α=1 do not converge while graphs for α>1 they all converge.

This observation can be used to prove that there are no zero values of ζ(z) for z with Re(z)>1.

2.2. ${\left|X\left(z,n\right)\right|}^2$ diverges when $n\to \infty$ for α$\le$1

${\left|X\left(z,n\right)\right|}^2$diverges when $n\to \mathrm{\infty }$ for α<1 because of [16] and [17]:

$$|\cos \left(ß(\ln \left(\frac{k}{j}\right)\right)|<1$$

And:

$$\sum _{k=1}^n\sum _{j=1}^n{k}^{-\alpha }\ast j^{-\alpha } diverges for \alpha <1$$

Therefore:

$$\underset{n\to \infty }{\mathit{lim}}{\left|X\left(z,n\right)\right|}^2 =\sum _{k=1}^n\sum _{j=1}^n{k}^{-\alpha }\ast j^{-\alpha }\ast \mathit{cos}\left(ß(\mathit{ln}\left(\frac{k}{j}\right)\right) diverges for \alpha <1$$

2.3. ${\left|X\left(z,n\right)\right|}^2$ does not collapse to any polynomial function ${\left|X\left(z,n\right)\right|}^2=C\ast n^t for t>1, and C constant$

One can prove it with a reduction to absurd.

Let’s assume that ${\left|X\left(z,n\right)\right|}^2=C\ast n^t for t>1$ where C and t integers C>0 and t>0

$$\mathrm{If} {\left|X\left(z,n\right)\right|}^2=C\ast n^t \mathrm{then}:$$

$$\underset{n\to \infty }{\lim }{\left|X\left(z,n\right)\right|}^2/n^t = C$$

But:

$$\underset{n\to \infty }{\lim }{\left|X\left(z,n\right)\right|}^2/n^{t } =\frac{1}{n^t}\ast \underset{n\to \infty }{\lim }{\sum }_{k=1}^n{k}^{-2\alpha }+\frac{1}{n^t}\ast {\sum }_{k=1}^n{\sum }_{j\ne k}^n{k}^{-\alpha }\ast j^{-\alpha }\ast \cos \left(ß(\ln \left(\frac{k}{j}\right)\right)$$

And:

$$\frac{1}{n^t}\ast \underset{n\to \infty }{\lim }\sum _{k=1}^n{k}^{-2\alpha }=0 for t>1$$

$$\frac{1}{n^t}\ast \sum _{k=1}^n\sum _{j\ne k}^n{k}^{-\alpha }\ast j^{-\alpha }\ast \cos \left(ß(\ln \left(\frac{k}{j}\right)\right)=0 for t>1$$

So, $C$ must be 0 which is an absurd.

2.4. ${\left|X\left(z,n\right)\right|}^2$ collapses to a straight-line ${\left|X\left(z,n\right)\right|}^2=Cn if Re(z)=1/2$

The proposition says that the following limit exists only for Re(z) = 1/2;

$$\underset{n\to \infty }{\lim }({\left|X\left(z,n\right)\right|}^2/ n) = S$$

Using the expression:

$$\underset{n\to \infty }{\mathit{lim}}({\left|X\left(z,n\right)\right|}^2/ n)=\underset{n\to \infty }{\mathit{lim}}\frac{1}{n}(\sum _{k=1}^n{k}^{-2\alpha }+\sum _{k=1}^n\sum _{j\ne k}^n{k}^{-\alpha }\ast j^{-\alpha }\ast \mathit{cos}\left(ß(\mathit{ln}\left(\frac{k}{j}\right)\right))$$

 2.4.1. For α>1/2, one can see that $\underset{n\to \infty }{\mathit{lim}}{(|x(z, n)|}^2$/$n)$= 0:

   $\underset{n\to \infty }{\lim }\frac{1}{n}({\sum }_{k=1}^n{k}^{-2\alpha })$ = 0 because 2α>1 and the series is convergent

$$\underset{n\to \infty }{\mathit{lim}}\frac{1}{n}\sum _{k=1}^n\sum _{j\ne k}^n{k}^{-\alpha }\ast j^{-\alpha }\ast \mathit{cos}\left(ß(\mathit{ln}\left(\frac{k}{j}\right)\right))<\underset{n\to \infty }{\mathit{lim}}\frac{1}{n}\sum _{k=1}^n\sum _{j\ne k}^n(k^{-\alpha }\ast j^{-\alpha })<\underset{n\to \infty }{\mathit{lim}}\frac{1}{n}(\sum _{k=1}^n{k}^{-2\alpha })$$

So:

$$\underset{n\to \infty }{\lim }\left(\frac{1}{n}\sum _{k=1}^n\sum _{j\ne k}^n{k}^{-\alpha }\ast j^{-\alpha }\ast \cos \left(ß(\ln \left(\frac{k}{j}\right)\right)\right)=0$$

 2.4.2. For α<1/2, one can see that$\underset{n\to \infty }{\mathit{lim}}({\left|X\left(z,n\right)\right|}^2$/$n)$=$\infty$as:

$$\underset{n\to \infty }{\lim }\frac{1}{n}(\sum _{k=1}^n{k}^{-2\alpha })<\underset{n\to \infty }{\lim }\frac{1}{n}\left(n\ast \frac{1}{n}\right)=\underset{n\to \infty }{\lim }\frac{1}{n}= 0$$

And:

$$\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{k=1}^n\sum _{j\ne k}^n{k}^{-\alpha }\ast j^{-\alpha }\ast \cos \left(ß(\ln \left(\frac{k}{j}\right)\right))>\underset{n\to \infty }{\lim }(\frac{1}{n}\ast n^2\ast \frac{1}{n^{2\alpha }})=\infty$$

Where the summations are replaced by the number of elements in the matrix (n x n) times the smallest value in each row (1/n) then 1>(2-1-2α)>0 when α<1/2

 2.4.3. Limit for α=1/2.

When α=1/2, one can express (${\left|X\left(z,n\right)\right|}^2$/n⁾ as:

$$\begin{array}{cc}\hfill \underset{n\to \infty }{\lim }({\left|X\left(z,n\right)\right|}^2/{n)}^{ }& =\hfill \\ & =\underset{n\to \infty }{\lim }\frac{1}{n}({\displaystyle \sum _{k=1}^n}{k}^{-1}+{\displaystyle \sum _{k=1}^n}{\displaystyle \sum _{j\ne k}^n}{k}^{-1/2}\ast j^{-1/2}\ast \cos \left(ß(\ln \left(\frac{k}{j}\right)\right))\hfill \\ & = \underset{n\to \infty }{\mathit{lim}}\frac{1}{n}\left({\displaystyle \sum _{k=1}^n}{k}^{-1}\right)+ \underset{n\to \infty }{\mathit{lim}}\frac{1}{n}({\displaystyle \sum _{k=1}^n}{\displaystyle \sum _{j\ne k}^n}{k}^{-1/2}\ast j^{-1/2}\ast \mathit{cos}\left(ß(\mathit{ln}\left(\frac{k}{j}\right)\right)) =\hfill \\ & = 0 + \underset{n\to \infty }{\mathit{lim}}\frac{1}{n}({\displaystyle \sum _{k=1}^n}{\displaystyle \sum _{j\ne k}^n}{k}^{-1/2}\ast j^{-1/2}\ast \mathit{cos}\left(ß(\mathit{ln}\left(\frac{k}{j}\right)\right))= \hfill \\ & = \underset{n\to \infty }{\mathit{lim}}\frac{2n}{n}({\displaystyle \sum _{j=1}^{n-1}}{n}^{-1/2}\ast j^{-1/2}\ast \mathit{cos}\left(ß(\mathit{ln}\left(\frac{n}{j}\right)\right))=\hfill \\ & = \underset{n\to \infty }{\mathit{lim}}2( n^{-\frac{1}{2}} {\displaystyle \sum _{j=1}^{n-1}}\ast j^{-\frac{1}{2}}\ast \mathit{cos}\left(ß(\mathit{ln}\left(\frac{n}{j}\right)\right))=\hfill \end{array}$$

Using the integral approximation of the infinite series

$$\begin{array}{c}=2\ast \underset{n\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{2\ast \sqrt{n}\ast \cos \left(ß\ast \mathit{ln}\left(\frac{n}{n}\right)\right)-2\ast ß\ast \mathrm{s}\mathrm{i}\mathrm{n}(ß\ast \mathrm{l}\mathrm{n}(\frac{n}{n})}{4\ast {ß}^2+1}\ast n^{-\frac{1}{2}}\hfill \\ =2\ast \frac{2\ast \sqrt{n}}{4\ast {ß}^2+1}{n}^{-\frac{1}{2}}=2\ast \frac{2}{4\ast {ß}^2+1}= \frac{1}{{ß}^2+1/4}\hfill \end{array}$$

So, if $\underset{n\to \infty }{\lim }{(\left|X\left(z,n\right)\right|}^2/{n)}^{ }$ exists will be equal to:

$$\underset{n\to \infty }{\lim }({\left|X\left(z,n\right)\right|}^2/n)= \frac{1}{{ß}^2+1/4}$$

[18]

            if z=1/2+iß

And this limit can only exist when ${\left|\mathrm{X}\left(\mathrm{z},\mathrm{n}\right)\right|}^2$ is monotonous which means that the curve will cross the x-axis only once.

$$\begin{gathered} {\left|X\left(z,n\right)\right|}^2=\left(\sum _{k=1}^n\sum _{j=k}^n{k}^{-\frac{1}{2}}\ast j^{-\frac{1}{2}}\ast \mathit{cos}\left(ß(\mathit{ln}\left(\frac{k}{j}\right)\right)\right) \\ =2\ast n^{-a}\ast \left({\sum }_{j=1}^{n-1} j^{-a}\ast \cos \left(ß\ast (\ln \left(\frac{x}{j}\right))\right)\right) \end{gathered}$$