Let's begin by defining a new function.

Riemann called the function above \(f\). Following [C.3.4] and [C.3.1], we will call it \(J(x)\). It is very similar to \(\pi(x)\) in its definition, so it's not surprising that it looks similar.

This looks like it's an infinite sum, but for any given \(x\), it is finite. For instance, let's calculate \(J(20)\): \begin{equation*}J(20)=\pi(20)+\frac{1}{2}\pi(\sqrt{20})+\frac{1}{3}\pi(\sqrt[3]{20})+\frac{1}{4}\pi(\sqrt[4]{20})=8+\frac{2}{2}+\frac{1}{3}+\frac{1}{4}=9\frac{7}{12}\end{equation*} because \(\sqrt[5]{20}\approx 1.8\) and \(\pi(\sqrt[5]{20})\approx\pi(1.8)=0\), so the sum ends there, and we can see that on the graph.

Okay, so we have this new function. Yet another arithmetic function. So what?

Ah, but what have we been doing to all our arithmetic functions to see what they can do, to get formulas for them? We've been Moebius inverting them, naturally! (Recall Section 23.2.) In this case, Moebius inversion could be really great, since it would give us information about the thing being added, which is the all-important \(\pi(x)\).

The only thing standing in our way is that \begin{equation*}J(x)=\sum_{n=1}^\infty \frac{1}{n}\pi\left(x^{1/n}\right)\end{equation*} is not a sum over divisors. But it turns out that, just like when we took the limits of the sum over divisors \(\sum_{d\mid n}\frac{1}{d}\), we got \(\sum_{n=1}^\infty \frac{1}{n}\), we can do the same thing with Moebius inversion.

We can use this by setting \(g=J\) with \(f(x/n)=\frac{1}{n}\pi\left(x^{1/n}\right)\). Applying this, we achieve a very important result writing \(\pi(x)\) in terms of \(J\).\begin{equation}\pi(x)=\sum_{n=1}^\infty \mu(n)\frac{J(x^{1/n})}{n}=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\label{men-1}\tag{25.4.1}\end{equation}