33. Verification of Riemann zeta function

From the previous article¹, $\zeta_e(s, x)$ can be wirtten like belows.

 

(1)    $\zeta_e(s, x)=\frac{q(x)(1-x^{s})}{(e^{i\pi p(x)}- 1)(1+ g_{s}(x))}$

 

$=\frac{x^s-1}{e^{i\pi p(x)}- 1}h_s(x)$

 

To make the integral form of zeta function more generally, assuming $i\pi p(x)$->$i\pi(p(x)-n), n:$ integer, this equation can be modified like belows.

 

(2)   $\zeta_e(s, x)=\frac{(x^s-1)h_s(x)}{e^{i\pi(p(x)-n)} - 1}$

 

Because $\zeta(s, y)=\int_0^y t^{s-2}b_0(t)dt$, it must be $\frac{\partial}{\partial x}\zeta_e(s, x)= \frac{\partial y}{\partial x} \frac{\partial}{\partial y} \zeta(s, y)$  when $x=e^{2\pi i/s}$.

 

(3) $\frac{((x^s-1)h_s(x))'}{e^{i\pi(p(x)-n)} - 1}-\frac{(x^s-1)h_s(x) i\pi p'(x) e^{i\pi (p(x)-n)}}{(e^{i\pi(p(x)-n)} - 1)^2}$

 

$= \frac{\partial y}{\partial x} y^{s-2}b_0(y)= \frac{\partial y}{\partial x} y^{s-2}( \frac{1}{e^{p(1/y-1)}-1}-\frac{p(1/y-1)e^{p(1/y-1)}}{(e^{p(1/y-1)}-1)^2})$

 

If $y=\frac{1}{i\pi(p(x)-n)/p+1}=1/P(x)$,

 

$= \frac{\partial y}{\partial x} (\frac{1}{P(x)})^{s-2}( \frac{1}{e^{i\pi(p(x)-n)}-1}-\frac{i\pi(p(x)-n)e^{i\pi(p(x)-n)}}{(e^{i\pi(p(x)-n)}-1)^2})$

 

Comparing both sides, 

 

$\frac{((x^s-1)h_s(x))'}{(x^s-1)h_s(x)}=\frac{p'(x)}{p(x)-n}$

 

$(x^s-1)h_s(x)=k(s)(p(x)-n), x \neq 1, k(s)$: the function of $s$.

 

(4)   $\zeta_e(s,x)=\frac{(x^s-1)h_s(x)}{e^{i\pi(p(x)-n)} - 1}=k(s)\frac{p(x)-n}{e^{i\pi(p(x)-n)} - 1}$

 

And

 

(5)²   $((x^s-1)h_s(x))'=(k(s)(p(x)-n))'=k(s)p'(x)$

 

$= \frac{\partial y}{\partial x} (\frac{1}{P(x)})^{s-2}=y'y^{s-2}$ when $x=e^{2\pi i/s}$.

 

Therefore, equation (4) holds true when $\zeta_e(s, x)$ converges. What's left is to verify the equation (4) when $x=e^{2\pi i/s}$.

 

If you quote this article, please cite the source.

Analytic_extension_of_Zeta_function_20240324.pdf

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Riemann_Zeta_Function_and_Quantum_Theory_20240319.pdf

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[Derivative or Division operator of unit vector. 2024.07.10]

 

Traditionally, differentiation and division are different, but if $p(x)$ is a function that is not differentiable, they can be treated like division or multiplication in equation (5). The reason the subscript $u$ is added next to the differential symbol in the equation below is to distinguish it from traditional differentiation.

 

(6)   $\frac{d}{dx_u}(e^{2\pi nxi})= \lim_{x \rightarrow 1}\frac{f(x) - f(1) }{x_u - 1}=\lim_{x \rightarrow 1}\frac{ e^{2\pi nxi} - e^{2\pi ni} }{e^{2\pi xi}- e^{2\pi i} }=n\frac{ e^{2\pi nxi} }{ e^{2\pi xi} }= ne^{2\pi(n-1)xi}$

($x_u:$ unit vector)

 

If equation (6) expresses derivative near $x=1$ and

 

(7)   $y=f(t)= e^{2\pi t}, \frac{\partial y}{\partial t}|_{t=ni}=2\pi $,

 

From (6), (7),

 

$e^{2\pi(n-1)xi}= 2\pi/n, e^{2\pi(n-2)xi}= 2\pi/(n-1), ...$

 

Because $e^{2\pi nxi} \approx e^{2\pi (n-k)xi}$, $y^n=(e^{2\pi nxi})^n=( 2\pi)^n/n!$ near $x=1$.

 

Here, the difference between $e^{2\pi nxi}$and $e^{2\pi t}$ is that while $nx, 0 \leq x \leq 1$ represents a quantized value, $t$ represents a continuous value. This can be treated as a way to express differential and integral values at discontinuous points of the zeta function.

Here, it is treated as

 

$1/P(x) \approx e^{-2\pi kxi}$.

 

If $p=\pi$,

 

$p(x)=((x-1)!+1)/x=-(e^{-2\pi kxi}-1)i+n$

 

$=(1-\cos2\pi kx)i+\sin2\pi kx+n$

 

Therefore, $n$ is an integer part of $p(x)$. $k$ is assumed as integer, but may be not in reality. 

 

It seems that $kx$ is replaced with $x+m$. $m$: integer.

 

When $x+m$ is a prime number, both sides of $p(x+m)$ are exactly the same as integers, but when $x+m$ is not a prime number, both sides are not the same and differ by $1/(x+m)$.³ That is,

 

$p(x+m)=((x+m-1)!+1)/(x+m)$

 

$=-(e^{-2\pi kxi}-1)i+n+(\frac{1}{x+m})$

 

Another thing to consider is that when $m$ becomes larger than a certain value, there may be two or more integers between $p(x+m)$ and $p(x+m+1)$. At this time, the value of $n$ can be changed accordingly as $x$ changes, but in that case $p(x)$ is discontinuous and cannot be differentiated. If $n$ represents all natural numbers in the integral expression of the zeta function, this part should be handled appropriately.

 

A way to extend to the entire complex range is to use a polynomial like

 

(8)   $f(z)=\sum a_k(z-z_k)^k, z=e^{2\pi xi}$. ($k$: integer)

 

Then, equation (6) can be modified like this.

 

$\frac{d}{dx_u}(e^{2\pi nxi})=\frac{d}{dz}z^n$. 

 

[Gap between infinitesimal and quantization. 2024.07.10]

 

From the equation (5), $k(s)p'(x)=y'y^{s-2}$,

 

$k(s)=y^{s-2}+k_2(s)$ assuming $p(x)=-(e^{-2\pi kxi}-1)i+n$ and can be differentiable.

 

In the zeta function of integral type, the integration interval is from 0 to 1, but this is a trigonometric function value, which indicates that the internal variable of this trigonometric function continues to increase. Therefore, it is not appropriate to compare using differential equations here when $x^s \neq 1$. $k_2(s)$ is the term that compensates the difference between derivative and quantization. If $y =1$ when $x$ approaches to 1, considering the results so far, $k(s)$ has the term $+1$.

 

 

[complex integral and trigonometric identities]

 

In the complex integral equation,

 

$\int z^ndz=\frac{z^{n+1}}{n+1}$

 

If $z=e^{ix}, dz=ie^{ix}dx$,

 

$\int z^ndz=\int e^{inx}\times ie^{ix}dx= \frac{e^{(n+1)xi}}{n+1}= \frac{z^{n+1}}{n+1}$

 

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^{1. Refer to About the roots of Riemann zeta function (final).}

^{2. Using $x^s=1$, $x=e^{2\pi i/s} \approx e^{2ip(s)} $. Then equation (5) becomes the differerntial equation of $s$, and equation (4) must be Riemann zeta function.} 

^{3. One way to solve this is to define $p(x)=((x-1)!+a(x))/x$, where $a(x)$ is the certain function that makes $a(x)=1$ when $x$ is a prime number.}