We see all the zeros for \(\sigma=1/2\) between \(0\) and \(100\); there are 29 of them.

Our next goal is to see how this connection \[\ln(\zeta(s))=s\int_1^\infty J(x)x^{-s-1}dx\] relates to the zeros of the \(\zeta\) function (and hence the Riemann Hypothesis).

We will do this by a very powerful analogy, which Euler used to prove \(\zeta(2)=\frac{\pi^2}{6}\) and which, correctly done, does yield the right answer.

Recall basic algebra. The Fundamental Theorem of Algebra states that every polynomial factors over the complex numbers. For instance, \[f(x)=5x^3-5x=5(x-0)(x-1)(x+1)\; .\] So we could then say that \[\ln(f(x))=\ln(5)+\ln(x-0)+\ln(x-1)+\ln(x+1)\] Then if it turned out that \(\ln(f(x))\) was useful to us for some other reason, it would be reasonable to say that we can get information about that other material from adding up information about the zeros of \(f\) (and the constant \(5\)), because of the addition of \(\ln(x-r)\) for all the roots \(r\).

You can't really do this with arbitrary functions, of course, and \(\zeta\) doesn't work. This is true, for instance, because \(\zeta(1)\) diverges badly, no matter how you define the complex version of \(\zeta\).

But it happens that \(\zeta\) is very close to a function you can do that to, \((s-1)\zeta(s)\). Applying the idea above to \((s-1)\zeta(s)\) (and doing lots of relatively hard complex integrals, or some other formal business with difficult convergence considerations) allows us to essentially invert \[\ln(\zeta(s))=s\int_1^\infty J(x)x^{-s-1}dx\] to \[J(x)=Li(x)-\sum_{\rho}Li(x^\rho)-\ln(2)+\int_x^\infty\frac{dt}{t(t^2-1)\ln(t)}\]