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

  1. Feb 10, 2010 #1
    Hi,

    The divisor summatory function, [tex]D(x)[/tex], can be obtained from [tex]\zeta^{2}(s)[/tex] by [tex]D(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty}\zeta^{2}(w)\frac{x^{w}}{w}dw[/tex] and I was trying to express [tex]\zeta^{2}(s)[/tex] in terms of [tex]D(x)[/tex] but I didnt succeed, could someone help?
     
  2. jcsd
  3. Apr 6, 2010 #2
    Use the Mellin inversion formula.
     
  4. Apr 7, 2010 #3
    Hi,

    I've done this by a different aproach considering that [tex]d(n)=D(n)-D(n-1)[/tex] and [tex] D(0)=0[/tex] it follows that
    [tex]
    \begin{align}
    \zeta^{2}(s)&=\sum_{n=1}^{\infty} \frac{\sigma_{0}(n)}{n^{s}}=\sum_{n=1}^{\infty} \frac{D(n)-D(n-1)}{n^{s}} \nonumber\\
    &=\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\\
    &=\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\\
    &=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
    \end{align}
    [/tex]

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

    --
    Seven
     
  5. Apr 11, 2010 #4
    from the properties of Mellin transform i would bet that

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

    since the Mellin transform of [tex] \sum_{n=1}^{\infty}f(xn) [/tex] is [tex] \zeta (s) F(s) [/tex]


    here [x] means the floor function
     
  6. Dec 15, 2010 #5
    Hy everyone,

    I think that some time ago I've seen [tex]D(x)[/tex] expressed in terms of the roots of the [tex]\zeta(s)[/tex] function. Does anyone knows of references about this?
     
  7. Dec 15, 2010 #6
    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.
     
  8. Feb 2, 2011 #7
    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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