Section25.5Connecting to zeta

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

Here is where Riemann “turns the Golden Key”, as Derbyshire puts it.

  • We will now use the Euler product for \(\zeta\) to connect \(\zeta\) to \(J\).
  • Then we will see how the zeros of \(\zeta\) give us an exact formula for \(J\).
  • Then we will finally plug \(J\) back into the Moebius-inverted formula for \(\pi\), giving an exact formula for \(\pi\)!

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.

Subsection25.5.1Detailing the connections

So let's see what this is. First, let's connect \(J\) and \(\zeta\).

Recall the Euler product for \(\zeta\) again: \[\zeta(s)=\prod_{p}\frac{1}{1-p^{-s}}\]

The trick to getting information about primes out of this, as well as connecting to \(J\), 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. \[\ln(\zeta(s))=\sum_{p}\ln\left(\frac{1}{1-p^{-s}}\right)=\sum_{p}-\ln\left(1-p^{-s}\right)\]

If we just had the fraction, we could use the geometric series to turn this into a sum (in fact, that is where it came from), but we got rid of it.

Question: So what can we do with \(-\ln()\) of some sum, not a product?

Answer: We can use its Taylor series! \[-\ln(1-x)=\sum_{k=1}^\infty \frac{x^k}{k}\]

Plug it in: \[\ln(\zeta(s))=\sum_{p}\sum_{k=1}^\infty \frac{(p^{-s})^k}{k}\]

Question: I don't see anything very useful here. Can I get anything good out of this?

Answer: Yes, in two big steps.

First, we rewrite \(\frac{(p^{-s})^k}{k}\) as an integral, so that we could add it up more easily.

  • Standard Calculus II improper integral work shows that \[\frac{(p^{-s})^k}{k}=\frac{s}{k}\int_{p^k}^\infty x^{-s-1}dx\]
  • That means \[\ln(\zeta(s))=\sum_{p}\sum_{k=1}^\infty \frac{(p^{-s})^k}{k}=\sum_{p}\sum_{k=1}^\infty\frac{s}{k}\int_{p^k}^\infty x^{-s-1}dx=s\sum_{p}\sum_{k=1}^\infty\int_{p^k}^\infty \frac{1}{k}x^{-s-1}dx\]

Next, it would be nice to get this as one big integral. But of what function, and with what endpoints?

We could unify these integrals from \(p^k\) to \(\infty\) somewhat artificially, by writing \[\int_{p^k}^\infty \frac{1}{k}x^{-s-1}dx=\int_1^{p^k}\frac{1}{k}\cdot 0\cdot x^{-s-1}\; dx+\int_{p^k}^\infty \frac{1}{k}x^{-s-1}dx\] That would give an integral from \(1\) to \(\infty\), though it would be defined with a piecewise integrand.

What function would I get if I added up all those piecewise integrands?

  • Well, it would be a function which added \(\frac{1}{1}x^{-s-1}\) when \(x\) reached a prime \(p\); so it would include \[\pi(x)x^{-s-1}\ldots\]
  • It would add \(\frac{1}{2}x^{-s-1}\) when it reached a square of a prime \(p^2\), which is the same as adding it when \(\sqrt{x}\) hits a prime; so it would include \[\frac{1}{2}\pi(\sqrt{x})x^{-s-1}\ldots\]
  • And it adds \(\frac{1}{3}x^{-s-1}\) when \(x\) reaches a cube of a prime, which is when \(\sqrt[3]{x}\) hits a prime; and that which adds \[\frac{1}{3}\pi(\sqrt[3]{x})x^{-s-1}\]
In short, adding up all these piecewise integrands seems to give a big integrand \[\left(\pi(x)+\frac{1}{2}\pi(\sqrt{x})+\frac{1}{3}\pi(\sqrt[3]{x})+\cdots\right)x^{-s-1}\]

But this is \(J(x)\), of course (multiplied by \(x^{-s-1}\)). Hence \[\ln(\zeta(s))=s\sum_{p}\sum_{k=1}^\infty\int_{p^k}^\infty \frac{1}{k}x^{-s-1}dx=s\int_1^\infty J(x)x^{-s-1}dx\] This completes our connection of \(\zeta\) and \(J\).