## The Log of Zeta

Last time (too long ago, sorry), I finally got around to talking about the function $\zeta(s)$. Today, I think I’d like to talk about $\log \zeta(s)$.

First, though, I should mention Euler’s product formula. The goal is to re-write the infinite series $\zeta(s)=\sum_n n^{-s}$ as a product. For each prime $p$, we can tease out of the series above those terms where $n$ is $p$ to some power. That is, we can think about the sub-series $\sum_{k\geq 0} (p^k)^{-s}$. This is a geometric series that converges to $(1-p^{-s})^{-1}$. If we take two such series and multiply them together, which terms from our starting series for $\zeta(s)$ will we recover?

Suppose $p$ and $q$ are two primes, and consider the product $\displaystyle\left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\frac{1}{p^{3s}}+\cdots\right)\cdot \left(1+ \frac{1}{q^s}+\frac{1}{q^{2s}}+\frac{1}{q^{3s}}+\cdots\right)$.

The terms in $\sum_n n^{-s}$ that we obtain from this product are those where $n$ is a product of some power (possibly the 0 power) of $p$ and some power (again, possibly 0) of $q$. We could then think about another prime, say $r$, and it’s geometric series, and multiply it by the above, and obtain all the terms in $\sum_n n^{-s}$ where $n=p^aq^br^c$ for some $0\leq a,b,c$.

Continuing on, and recalling that every positive integer has a unique factorization as a product of primes, we obtain the magical formula (isn’t most of Euler’s work magical?): $\displaystyle \sum_n \frac{1}{n^s}=\prod_p (1-p^{-s})^{-1}$,

where the product is indexed by the primes. What’s nice about this, for us, at the moment, is that logarithms work well with products and powers, but not soo well for sums. Recalling the Taylor polynomial $\ln(1-x)=-1-x-x^2-x^3-\cdots=-\sum\limits_{n=0}^{\infty} x^n$

we find $\displaystyle \ln \zeta(s)=\sum_p\sum_n \frac{1}{n}p^{-ns}$.

That was fun. Incidentally, it’s not much of a jump from here to show that the series of prime reciprocals diverges. I mentioned it in a post a while ago.

Let’s switch gears for a little bit. I’m going to define a function $J(x)$ on the positive reals. If you’re following Edwards’ book with me (as I jump all over the place), this will be almost exactly his $J(x)$, differing only at the integers (for today anyway :)). Let me start by setting $J(0)=0$. The function $J(x)$ will be a step function, which I’ll define to mean piecewise linear with slope 0 on each piece. So to define $J(x)$ I only need to tell you when to jump, and by how much. The rule is: when you get to an integer $n$, if $n=p^k$ for some prime $p$, then you jump up by $1/k$. So at $2,3,5,7,11,\ldots$ you jump by 1, at $4,9,25,49,\ldots$ you jump by 1/2, at $8, 27, \ldots$ you jump by 1/3, etc. Here’s a little picture of $J(x)$, with a few values highlighted: Slightly more precisely, we can write $\displaystyle J(x)=\sum_{p^n\leq x}\frac{1}{n}$.

Now, let me see if I can convince you that there is some justification in writing $\displaystyle \ln \zeta(s)=s\int_0^{\infty} J(x)x^{-s-1}\ dx$.

Let’s work with the right-hand side. Since $J(x)=0$ near $x=0$, I’ll actually start my integral at $x=1$. I think the way to go about it is as follows: $\begin{array}{rcl} \displaystyle s\int_1^{\infty}J(x)x^{-s-1}\ dx &=& \displaystyle s\sum_{m=1}^{\infty}\int_m^{m+1}J(x)x^{-s-1}\ dx \\ {} & = & \displaystyle \sum_{m=1}^{\infty}J(m)\left(\frac{1}{m^s}-\frac{1}{(m+1)^s}\right).\end{array}$

Now suppose that $J(m)=J(m+1)$ for some $m$. Then the terms corresponding to $m$ and $m+1$ telescope as follows: $\begin{array}{c} \displaystyle J(m)\left(\frac{1}{m^s}-\frac{1}{(m+1)^s}\right)+J(m+1)\left(\frac{1}{(m+1)^s}-\frac{1}{(m+2)^s}\right)\\ =\displaystyle J(m)\left(\frac{1}{m^s}-\frac{1}{(m+2)^s}\right)\end{array}$.

If, also, $J(m)=J(m+2)$, this we can telescope another term into this one, and so on. So, really, the important $m$ in this sum are those where $J(x)$ jumps, which are the prime powers. Let $m_i=p_i^{n_i}$ be the $i$-th prime power (i.e., the point where $J(x)$ makes its $i$-th jump), starting with $m_1=2$. Then we can write $\begin{array}{rcl} \displaystyle s\int_0^{\infty}J(x)x^{-s-1}\ dx & = & \displaystyle \sum_i J(m_i)\left(\frac{1}{m_i^s}-\frac{1}{m_{i+1}^s}\right) \\ & = & \displaystyle \frac{1}{2}J(2)+\sum_{i=2}^{\infty} -J(m_{i-1})\frac{1}{m_i^s}+J(m_i)\frac{1}{m_i^s} \\ & = & \displaystyle \frac{1}{2}+\sum_{i=2}^{\infty} \frac{1}{m_i^s}\left(J(m_i)-J(m_{i-1})\right) \\ & = & \displaystyle \frac{1}{2}+\sum_{i=2}^{\infty} \frac{1}{m_i^s}\cdot \frac{1}{n_i} \\ & = & \displaystyle \sum_{p}\sum_n \frac{1}{n}p^{-ns} \\ & = & \ln \zeta(s).\end{array}$

What good is writing $\ln \zeta(s)$ this way? I guess if you know something about Fourier inversion (which I don’t) then you get to say that $\displaystyle J(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\log \zeta(s)x^s\frac{ds}{s},$

for $a=\text{Re}(s)>1$. What good is that? I think I’ll have to read some more and tell you about it tomorrow, but it’ll turn out to be useful once we have yet another description of $\ln \zeta(s)$, in terms of the 0s of $\zeta(s)$ (finally getting to those 0s everybody cares so much about).

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### One Response to “The Log of Zeta”

1. Another Formula for J(x) « ∑idiot's Blog Says:

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