Square of the Riemann zeta-function in terms of the divisor summatory function.

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Discussion Overview

The discussion centers on expressing the square of the Riemann zeta-function, \(\zeta^{2}(s)\), in terms of the divisor summatory function, \(D(x)\). Participants explore various approaches, including the use of the Mellin inversion formula and alternative methods for deriving relationships between these mathematical constructs.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using the Mellin inversion formula to express \(D(x)\) in terms of \(\zeta^{2}(s)\).
  • Another participant presents an alternative approach, deriving \(\zeta^{2}(s)\) using the relationship \(d(n) = D(n) - D(n-1)\) and provides a series of equations leading to a formulation involving \(D(x)\).
  • A different participant proposes a conjecture regarding the properties of the Mellin transform and its relation to \(D(x)\) and \(\sigma_{0}(n)\).
  • One participant recalls having seen \(D(x)\) expressed in terms of the roots of the \(\zeta(s)\) function and seeks references on this topic.
  • Another participant expresses uncertainty about applying the Mellin inversion formula and requests clarification on how to proceed.
  • A reference is provided by a participant, citing a doctoral dissertation that discusses the Mellin transform of the square of the Riemann zeta-function.

Areas of Agreement / Disagreement

Participants express differing approaches to the problem, with no consensus on the best method to express \(\zeta^{2}(s)\) in terms of \(D(x)\). Some participants agree on the potential of the Mellin inversion formula, while others explore alternative methods or express uncertainty.

Contextual Notes

Some participants note limitations in their approaches, such as unresolved mathematical steps or difficulties in representing integrands appropriately for inversion. The discussion reflects a range of assumptions and conditions that may affect the validity of proposed methods.

Who May Find This Useful

This discussion may be of interest to mathematicians and researchers focused on analytic number theory, particularly those studying the properties of the Riemann zeta-function and divisor functions.

AtomSeven
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Hi,

The divisor summatory function, D(x), can be obtained from \zeta^{2}(s) by D(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty}\zeta^{2}(w)\frac{x^{w}}{w}dw and I was trying to express \zeta^{2}(s) in terms of D(x) but I didnt succeed, could someone help?
 
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Use the Mellin inversion formula.
 
Hi,

I've done this by a different approach considering that d(n)=D(n)-D(n-1) and D(0)=0 it follows that
<br /> \begin{align}<br /> \zeta^{2}(s)&amp;=\sum_{n=1}^{\infty} \frac{\sigma_{0}(n)}{n^{s}}=\sum_{n=1}^{\infty} \frac{D(n)-D(n-1)}{n^{s}} \nonumber\\<br /> &amp;=\sum_{n=1}^{\infty} \frac{D(n)}{n^{s}}-\sum_{n=1}^{\infty} \frac{D(n-1)}{n^{s}}=\sum_{n=1}^{\infty} \frac{D(n)}{n^{s}}-\sum_{n=1}^{\infty} \frac{D(n)}{(n+1)^{s}} \nonumber\\<br /> &amp;=\sum_{n=1}^{\infty}D(n)\bigg\{ \frac{1}{n^{s}} - \frac{1}{(n+1)^{s}} \bigg\}=\sum_{n=1}^{\infty}D(n)\int_{n}^{n+1}\frac{s}{x^{s+1}} dx \nonumber\\<br /> &amp;=s\sum_{n=1}^{\infty}\int_{n}^{n+1}\frac{D(x)}{x^{s+1}} dx =s\int_{1}^{\infty}\frac{D(x)}{x^{s+1}} dx \nonumber<br /> \end{align}<br />

So it would be interesting to see if anyone could solve this using the Mellin inversion aproach.

--
Seven
 
from the properties of Mellin transform i would bet that

D= \sum_{n\le x}\sigma_{0} = \sum_{n\ge1}[(x/n)]

since the Mellin transform of \sum_{n=1}^{\infty}f(xn) is \zeta (s) F(s)here [x] means the floor function
 
Hy everyone,

I think that some time ago I've seen D(x) expressed in terms of the roots of the \zeta(s) function. Does anyone knows of references about this?
 
Eynstone said:
Use the Mellin inversion formula.

I don't see how to do that. Can you show me (and the OP I assume too) how to do that please?

I did try ok. If I need to show my work, I could but I got to a spot where I tried to represent the integrand in the form that I think I could have inverted it, the inversion didn't come out well.
 
For those interested here is a refference:

M. Lukkarinen, The Mellin transform of the square of Riemann’s zeta-function and Atkinson’s formula, Doctoral Dissertation, Annales Acad. Sci. Fennicae, No. 140, Helsinki, 2005
 

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