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

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 \(J\).