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Section 25.5 Connecting to Zeta

Subsection 25.5.1 Turning the golden key

Now, this looks just as hopeless as before. How is \(J\) going to help us calculate \(\pi\text{,}\) if we can only calculate \(J\) in terms of \(\pi\) anyway?

Here is where Riemann “turns the Golden Key”, as Derbyshire puts it. Because \(\zeta\) has an Euler product over the set of primes, we can just possibly connect it to each prime. It turns out this will in fact connect \(\zeta\) to \(J\text{.}\) This is the goal of the rest of the current section.

In the next section, we will see how the zeros of \(\zeta\) give us an exact formula for \(J\text{;}\) then we will finally plug \(J\) back into the Moebius-inverted formula for \(\pi\) to get an exact formula for \(\pi\) in Section 25.7. Here is a plot of that formula, as a foretaste.

Prime pi, log integral, and a better approximation
Figure 25.5.1. \(\pi(x)\text{,}\) \(Li(x)\text{,}\) and something better than \(Li\)

We can see above that this has the potential to be a very good approximation, even given that I did limited calculations here. The most interesting thing is the gentle waves you should see; this is quite different from the other types of approximations we had, and seems to have the potential to mimic the more abrupt nature of the actual \(\pi(x)\) function much better in the long run. (See [E.4.3] for more details along these lines, connecting to Fourier series, which we will not pursue.)

Subsection 25.5.2 Detailing the connections

Now let's connect \(J\) and \(\zeta\text{.}\) Recall the Euler product for \(\zeta\) again:

\begin{equation*} \zeta(s)=\prod_{p}\frac{1}{1-p^{-s}} \end{equation*}

The trick to getting information about primes out of this, as well as connecting to \(J\text{,}\) is to take the logarithm of the whole thing. This will turn the product into a sum, something we can work with much more easily 3 :

\begin{equation*} \log(\zeta(s))=\sum_{p}\log\left(\frac{1}{1-p^{-s}}\right)=\sum_{p}-\log\left(1-p^{-s}\right) \end{equation*}
This reminds me of the old joke about Noah's ark and logarithms. So, after the ark lands, all the animals are … having baby animals, let's say. Except the snakes. No baby snakes. Noah asks what the problem is – they seem to be missing the point. Snakes say, no worries, just give us a wooden bench or sawhorse or something. Noah wonders what's up, but gives it to them. Next morning, tons of baby snakes! Naturally Noah has to ask where the magic was. “Simple; adders need a log table to multiply.”

Adding just fractions would have perhaps allowed using a geometric series to make this a sum, but what could we do with a sum of logarithms?

Question 25.5.2.

What can we do with \(-\log()\) of some sum, not a product?

Solution

We can use its Taylor series!

\begin{equation*} -\log(1-x)=\sum_{k=1}^\infty \frac{x^k}{k} \end{equation*}

So we plug it in:

\begin{equation*} \log(\zeta(s))=\sum_{p}\sum_{k=1}^\infty \frac{(p^{-s})^k}{k} \end{equation*}

Now we will manipulate this in two big steps. First we'll rewrite the fraction as an integral, and then we will try to somehow add up the integrals.

Standard improper integral work (Exercise 25.9.3) from second-semester calculus shows that we can rewrite the summands:

\begin{equation*} \frac{(p^{-s})^k}{k}=\frac{s}{k}\int_{p^k}^\infty x^{-s-1}dx\text{.} \end{equation*}

That means we can rewrite the logarithm of \(\zeta\) as

\begin{equation*} \log(\zeta(s))=\sum_{p}\sum_{k=1}^\infty \frac{(p^{-s})^k}{k} \end{equation*}
\begin{equation*} =\sum_{p}\sum_{k=1}^\infty\frac{s}{k}\int_{p^k}^\infty x^{-s-1}dx=s\sum_{p}\sum_{k=1}^\infty\int_{p^k}^\infty \frac{1}{k}x^{-s-1}dx\text{.} \end{equation*}

This is a very large sum of integrals. We can rewrite this as a single integral, but we will need to pay close attention.

First, we can unify all these integrals from \(p^k\) to \(\infty\) by making them all have the same endpoints. This is done somewhat artificially, by writing

\begin{equation*} \int_{p^k}^\infty \frac{1}{k}x^{-s-1}dx=\int_1^{p^k}\frac{1}{k}\cdot 0\cdot x^{-s-1}\; dx+\int_{p^k}^\infty \frac{1}{k}x^{-s-1}dx \end{equation*}

This yields the integral of a piecewise-defined function, but it for every \(k\) and \(p\) it is defined from \(1\) to \(\infty\text{.}\)

Now comes the most surprising part. What function would I get if I added up all those integrals in the double sum for \(\log(\zeta(s))\text{,}\) \(\sum_{p}\sum_{k=1}^\infty \frac{(p^{-s})^k}{k}\text{?}\) To see this, let us add up all of the piecewise integrands, organizing by the powers \(k\) for any given prime \(p\text{.}\)

  1. Whenever \(x\) reaches \(p^1=p\text{,}\) the sum of all those functions would add \(\frac{1}{1}x^{-s-1}\text{.}\) Adding up all of these for all \(p\) means the total function would include

    \begin{equation*} \pi(x)x^{-s-1}\ldots \end{equation*}

  2. Whenever \(x\) reaches \(p^2\text{,}\) the sum of all those functions would add \(\frac{1}{2}x^{-s-1}\text{.}\) This, however, is the same thing as when \(\sqrt{x}\) hits a prime, so we can add it to the previous point. The total function would include would include

    \begin{equation*} \frac{1}{2}\pi(\sqrt{x})x^{-s-1}\ldots \end{equation*}

  3. When \(x\) reaches a cube of a prime, the sum adds \(\frac{1}{3}x^{-s-1}\text{.}\) This is the same thing as adding a new part when \(\sqrt[3]{x}\) hits a prime, that is adding

    \begin{equation*} \frac{1}{3}\pi(\sqrt[3]{x})x^{-s-1} \end{equation*}

And so forth for each \(k\text{.}\) In short, adding up all these piecewise integrands seems to give a big integrand

\begin{equation*} \left(\pi(x)+\frac{1}{2}\pi(\sqrt{x})+\frac{1}{3}\pi(\sqrt[3]{x})+\cdots\right)x^{-s-1}\text{.} \end{equation*}

But this sum of all the piecewise integrands is \(J(x)\text{,}\) multiplied by \(x^{-s-1}\text{.}\) Hence

\begin{equation*} \log(\zeta(s))=s\sum_{p}\sum_{k=1}^\infty\int_{p^k}^\infty \frac{1}{k}x^{-s-1}dx=s\int_1^\infty J(x)x^{-s-1}dx\text{.} \end{equation*}

This completes our connection of \(\zeta\) and \(J\text{.}\)