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

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SUMMARY

The discussion focuses on expressing the square of the Riemann zeta-function, \(\zeta^{2}(s)\), in terms of the divisor summatory function, D(x). The integral representation of D(x) is given by \(D(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty}\zeta^{2}(w)\frac{x^{w}}{w}dw\). A successful approach involves the relationship \(d(n)=D(n)-D(n-1)\) and the use of the Mellin inversion formula. A reference to M. Lukkarinen's doctoral dissertation from 2005 is provided for further insights into the Mellin transform of the square of the Riemann zeta-function.

PREREQUISITES
  • Understanding of the Riemann zeta-function and its properties
  • Familiarity with the divisor summatory function, D(x)
  • Knowledge of the Mellin transform and its applications
  • Basic calculus, particularly integration techniques
NEXT STEPS
  • Study the Mellin inversion formula and its applications in analytic number theory
  • Explore the properties of the divisor summatory function, D(x)
  • Read M. Lukkarinen's dissertation on the Mellin transform of the square of the Riemann zeta-function
  • Investigate the relationship between the roots of the Riemann zeta-function and divisor functions
USEFUL FOR

Mathematicians, number theorists, and researchers interested in analytic number theory, particularly those studying the properties of the Riemann zeta-function and divisor summatory 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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