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Section 25.6 Connecting to Zeros

Subsection 25.6.1 Where are the zeros?

Our next goal is to see how this connection
\begin{equation*} \log(\zeta(s))=s\int_1^\infty J(x)x^{-s-1}dx \end{equation*}
relates to the zeros of the \(\zeta\) function (and hence the Riemann Hypothesis).
We see all the zeros for \(\sigma=1/2\) between \(0\) and \(100\text{;}\) there are 29 of them.
We will connect to \(\zeta\) by means of a very powerful analogy, the one which Euler used to prove \(\zeta(2)=\frac{\pi^2}{6}\) (see the end of Subsection 20.4.2) and which, correctly done, does yield the right answer.
Begin the analogy by recalling basic algebra. The Fundamental Theorem of Algebra states that every polynomial factors over the complex numbers. For instance,
\begin{equation*} f(x)=5x^3-5x=5(x-0)(x-1)(x+1)\text{.} \end{equation*}
If we take the logarithm of such a factorization, we can say things like
\begin{equation*} \log(f(x))=\log(5)+\log(x-0)+\log(x-1)+\log(x+1) \end{equation*}
Then if it turned out that \(\log(f(x))\) was useful to us for some other reason \(R\text{,}\) it would be reasonable to say that we can get information about the otherwise-mysterious \(R\) from adding up information about the zeros of \(f\) (and the constant \(5\)), because of the addition of \(\log(x-r)\) for all the roots \(r\text{.}\)
You can’t really do this with arbitrary functions, of course. Disappointingly, \(\zeta\) is definitely a function where this doesn’t work, mostly because \(\zeta(1)\) diverges so badly, no matter how you define the complex version of \(\zeta\text{.}\)
But it so happens that \(\zeta\) is very close to a function you can analyze this way, \((s-1)\zeta(s)\text{.}\) Applying the logarithm factoring idea 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 the equation
\begin{equation*} \log(\zeta(s))=s\int_1^\infty J(x)x^{-s-1}dx \end{equation*}
to the even more surprising formula
\begin{equation} J(x)=Li(x)-\sum_{\rho}Li(x^\rho)-\log(2)+\int_x^\infty\frac{dt}{t(t^2-1)\log(t)}\tag{25.6.1} \end{equation}

Subsection 25.6.2 Analyzing the connection

It is hard to overestimate the importance of the formula (25.6.1). Each piece comes from something inside \(\zeta\) itself, inverted in this special way.
  • First, \(Li(x)\) comes from the fact that we needed \((s-1)\zeta(s)\) to apply this inversion, not just \(\zeta(s)\text{.}\) In fact, this particular inversion can be seen by integrating, as it’s true that
    \begin{equation*} s\int_1^\infty Li(x)x^{-s-1}dx=-\log(s-1) \end{equation*}
    so one can see that \(s-1\) and \(Li\) seem to correspond.
  • Second, each \(Li(x^\rho)\) comes from each of the zeros of \(\zeta\) on the line \(\sigma=1/2\) in the complex plane. This is the part which most closely corresponds to the factoring.
  • The constant term \(\log(2)\) comes from the constant when you do the factoring, similarly to the \(5\) in the example above using \(f(x)=5x^3-5x\text{.}\)
  • Finally, the integral in (25.6.1) comes from the zeros of \(\zeta\) at \(-2n\) we mentioned just before the statement of 25.3.7.
To give you a sense of how complicated (25.6.1) really is, here is a plot of just one small piece of it.
Plot of \(Li(20^{1/2+it})\)
Figure 25.6.1. Plot of \(Li(20^{1/2+it})\)
This is the plot of \(Li(20^{1/2+it})\) up through the first zero of \(\zeta\) above the real axis. It’s beautiful, but also forbidding. After all, if takes that much twisting and turning to get to \(Li\) of the first zero, what is in store if we have to add up over all infinitely many of them to calculate \(J(20)\text{?}\)
So at the very least, it would be helpful to know where all of those mysterious zeros live! This is why the Riemann Hypothesis is so important; it pins them down quite dramatically.