17. Bernoulli sequence and Riemann zeta function

From the previous article¹, considering that the integration section starts from $1$, it appears that $\Delta e^{-px}$ should be corrected to a form such as $\Delta ((x-1)/(e^{p(x-1)}-1))$. Reference point² must has the finite value. Logarithmic function, $\ln x$, has the value when $x=1$, not $x=0$. This also explains that the Riemann zeta function is also related to the Bernoulli sequence. Therefore, $\zeta(s)$ can be modified as follows.³

 

(1) $\zeta(s)=\int x^{s-2}(\frac{1-g(x)}{e^{g(x)}-1} - \frac{g(x)}{(e^{g(x)}-1)^2})dx, g(x)=p(1/x-1), |x| \leq 1$

 

The reason the integration section is made smaller than $1$ is to make integration easier.

 

If $\zeta(s)=\sum a_m(s-s_0)^m$ ($m$: natural number), $a_m$ can be expressed by the polynomials of $x$ before the integration from $0$ to $1$ like $a_m(x)=\sum b_n x^n$. Then, integral form of $\zeta(s)$ is $\int{a_m(x)(s-s_0)^m}dx$ and though the integral value is different, the root of $\zeta(s)=0$ is same as $s_0$ in both equations. 

 

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Riemann_zeta_function_and_quantum_mechanics_20231221.pdf

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[Comparison with a well-known integral form of Riemann zeta function. 20204.02.21]

 

There is another well-known form of Riemann zeta function as

 

$$\zeta(s)\Gamma(s)=\int_{0}^{\infty}\frac{x^{s-1}}{e^x-1}dx$$

 

An example of the derivation of this equation is as follows.

 

$$\int_{0}^{\infty}\frac{x^{s-1}}{e^x-1}dx= \int_{0}^{\infty}x^{s-1}(e^{-x}+e^{-2x}+ e^{-3x}+...)dx$$

$$=\zeta(s)\int_{0}^{\infty}x^{s-1}e^{-x}dx=\zeta(s)\Gamma(s)$$

 

The integration interval of the Riemann zeta function must be from 1 to $\infty$, or 0 to 1, not from 0 to $\infty$. And Considering $x^{s-1}$ represents $1/n^s$, $x^{s-1}$ must be replaced with $x^{-s+1}$. Based on these two facts, we can infer that $x/(e^x-1)$ should be changed to $(x-1)/(e^{p(x-1)}-1)$. If the $p$ value is not assumed, the exponential form of $\pi$, which is the sum of the zeta function, does not appear.

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^{1. Refer to Riemann zeta function and quantum mechanics.}

^{2. The more precise mathematical definition of this term seems necessary. This also implies that it is possible that if $\sum (1/n) = \int(f(x)/\ln x)dx$, $\sum (1/n^s) =\int(f(x)/(\ln x)^s)dx$ regardless of $s$. The significance of the Bernoulli sequence is that the Riemann hypothesis can be approached more analytically by expressing seemingly irregular prime numbers as known functions.}