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

1. Feb 10, 2010

### AtomSeven

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?

2. Apr 6, 2010

### Eynstone

Use the Mellin inversion formula.

3. Apr 7, 2010

### AtomSeven

Hi,

I've done this by a different aproach considering that $$d(n)=D(n)-D(n-1)$$ and $$D(0)=0$$ it follows that
\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}

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

--
Seven

4. Apr 11, 2010

### zetafunction

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

5. Dec 15, 2010

### AtomSeven

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?

6. Dec 15, 2010

### jackmell

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.

7. Feb 2, 2011

### AtomSeven

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