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Section25.7The Riemann Explicit Formula

Now we are finally ready to see Riemann's result, by plugging in this formula for \(J\) into the Moebius inverted formula for \(\pi\) given by \begin{equation*}\pi(x)=J(x)-\frac{1}{2}J(\sqrt{x})-\frac{1}{3}J(\sqrt[3]{x})-\frac{1}{5}J(\sqrt[5]{x})+\frac{1}{6}J(\sqrt[6]{x})+\cdots\end{equation*} It is true that Riemann did not prove the following formula fully rigorously, and indeed one of the provers of the Prime Number Theorem mentioned taking decades as part of that effort just to prove all the statements Riemann made in this one paper. Nonetheless, it is certainly Riemann's formula for \(\pi(x)\), and an amazing one:

It is worth making two points about the transition to this formula. First, if you're wondering where the \(\log(2)\) at the end of the previous section went, it went to 0 because \(\sum_{n=1}^\infty \frac{\mu(n)}{n}=0\), though this is very hard to prove. (In fact, it is a consequence of the Prime Number Theorem; see Exercise 25.9.5.)

Secondly, each \(\rho\) is a zero above the real axis, and then \(\bar{\rho}\) is the corresponding one below the real axis. The summation is over every single zero not on the real axis. In particular, these \(\rho\) are conjectured by the Riemann Hypothesis to all have real part equal to \(1/2\), which would make things particularly tidy.

Now let's see this formula in action.

This graphic shows just how good it can get. Again, notice the waviness, which allows it to approximate \(\pi(x)\) not just once per “step” of the function, but along the steps.

We can also just check out some numerical values.

Many wonderful facts would follow from the truth of the Riemann Hypothesis, or from a natural generalization.

So can you prove that there are no other zeros other than those on the critical line to contribute to these approximations to \(\pi(x)\)? If so, welcome to the future of number theory!