Contour integration of $\zeta(s)\zeta(2s)$ - MathOverflow

Contour integration of $\zeta(s)\zeta(2s)$

2011-03-16T02:59:32Z

I have been looking at this for days and I am going insane.

I need to show that for a dirichlet series equal to $\zeta(s)\zeta(2s)$ the sum of the coefficients less that x is $x\zeta(2)+O(x^(3/4))$ and then expand that to the $\Pi \zeta(ks)$ for all k in an effort to find the formula for the number of non-isomorphic abelian groups.

I know that using perron's formula there is a simple pole at $s=1$ that gives a residue of $X\zeta(2)$ but I can't find a contour that converges or the exact error term.

Answer by David Hansen for Contour integration of $\zeta(s)\zeta(2s)$

2011-03-16T03:10:38Z

This case, at least, you can do by hand; if $c(n)$ is the $n$th coefficient of $\zeta(s)\zeta(2s)$, then

$\sum_{n\leq X}c(n)=\sum_{nm^2\leq X}1=\sum_{m}\sum_{n\leq X/m^2}1 = \sum_{m < \sqrt{X}}(m^{-2} X+O(1))=\zeta(2)X+O(\sqrt{X}).$

Answer by Daniel Loughran for Contour integration of $\zeta(s)\zeta(2s)$

2011-03-16T10:37:08Z

If you havn't done so already, you might find it useful to look at the proof of theorem 12.2 on the divisor problem in Titchmarsh - The theory of the Riemann zeta function. Here, he goes through a detailed application of Perron's formula for the function $\zeta^k(s)$, which I believe to be very similar to your case.

Indeed, for $s=\sigma + it$ and $\sigma>1/2$, $\zeta(2s)$ is absolutley convergent and hence uniformly bounded with respect to $t$. So this will not contribute to the contours that you choose (as long as $\sigma>1/2$!). What you need then is good upper bounds for the order of the zeta function in the critical strip.

To get these, one normally finds the order of the function at two points, and then uses the Phragmén–Lindelöf principle for strips to get estimates for the function between these two points. For example, it is known that $\zeta(1/2 + it) = O(t^{1/4})$ (see The Lindelöf hypothesis), although there are much better bounds available than that. This is all done in Titchmarsh's book.

I hope this helps!