First, we'll review the main function. Think of the prime counting function \(\pi\) as a so-called step function, where every time you hit a new prime you add 1.

Let's define a new function. Instead of adding 1 each time \(x\) hits a prime, we will add \(\log(p)\) (recall that this is the natural logarithm) each time we hit a prime \(p\). Of course, this value we add will get bigger as \(p\) gets bigger.

Earlier in this chapter we noted that the Prime Number Theorem is logically equivalent to the limit \(\lim_{x\to\infty}\frac{\pi(x)}{x/\log(x)}=1\). There are actually many such logical equivalences. One of them involves \(\Theta\): \begin{equation*}\lim_{x\to\infty}\frac{\Theta(x)}{x}=1\end{equation*} This is certainly numerically plausible. Here is a plot of both limits, along with the constant function 1.

As usual, proving such things completely is beyond the level of this course, but we can prove the following partial implication.

In order to prove this implication, we will first need a formula telling us more about \(\Theta(x)\). Our strategy will be to first turn \(\Theta(x)\) into an even more hopelessly complicated sum, but then use calculus to trickily get something usable by summing up integrals.

In order to do this, we need two subsidiary functions. First recall the notation \(m=\lfloor x\rfloor\) for the greatest integer less than \(x\). Secondly:

Then we can rewrite these step functions as weighted sums of \(a(n)\): \begin{equation*}\pi(x)=\sum_{n=1}^{m}a(n)\text{ and }\Theta(x)=\sum_{n=1}^{m}a(n)\log(n)\, .\end{equation*}

Our goal is to rearrange \(\Theta\) to be a sum of something involved \(\pi\). First we turn it into a difference of sums by rearranging (and using \(\log(1)=0\)): \begin{equation*}\Theta(x)=\sum_{1\leq n\leq x}a(n)\log(n)=\sum_{n=1}^{m}a(n)\log(n)=\end{equation*} \begin{equation*}\sum_{n=2}^{m}[\pi(n)-\pi(n-1)]\log(n)=\sum_{n=2}^{m}\pi(n)\log(n)-\sum_{n=1}^{m-1}\pi(n)\log(n+1)\end{equation*} This difference of sums can be combined into a single sum, with just two left over terms, the second of which is equal to \(0\). \begin{equation*}\Theta(x)=\sum_{n=2}^{m-1}\pi(n)[\log(n)-\log(n+1)]+\pi(m)\log(m)-\pi(1)\log(1)\, .\end{equation*}

To continue, we will rewrite this as an integral. We use a few key facts:

• The difference which appears in the last \(\Theta\) formula is an integral, \(\log(n+1)-\log(n)=\int_{n}^{n+1}\frac{dt}{t}\).
• We have that \(\pi(x)=\pi(m)\) is constant on \([m,x]\), so it may be factored out of any integral of a unit distance.
• We can rearrange and add sums and integrals as usual.

This yields the following rewrite. \begin{equation*}\Theta(x) = -\sum_{n=2}^{m-1}\Big[\pi(n)\int_{n}^{n+1}\frac{dt}{t}\Big]+\pi(m)\log(m)\end{equation*} \begin{equation*}=-\sum_{n=2}^{m-1}\Big[\pi(n)\int_{n}^{n+1}\frac{dt}{t}\Big]+\pi(m)\log(m)-\pi(x)\log(x)+\pi(x)\log(x)\end{equation*} \begin{equation*}=-\int_{2}^{m}\frac{\pi(t)dt}{t}+\pi(x)\log(x)-\int_{m}^{x}\frac{\pi(t)dt}{t} =\pi(x)\log(x)-\int_{2}^{x}\frac{\pi(t)dt}{t}\, .\end{equation*}

Now we have a formula for \(\Theta\) which will allow us to prove something.

We can divide the formula \(\Theta(x)=\pi(x)\log(x)-\int_{2}^{x}\frac{\pi(t)dt}{t}\) by \(x\): \begin{equation*}\frac{\Theta(x)}{x}=\frac{\pi(x)\log(x)}{x}-\frac{\int_{2}^{x}\frac{\pi(t)}{t}dt}{x}\, ,\end{equation*} Given that the Prime Number Theorem says that \(\lim_{x\to\infty}\) of the fraction with \(\pi(x)\) in it is 1, proving that \(\lim_{x\to\infty}\) of \(\frac{\Theta(x)}{x}\) is also 1 is equivalent to proving \begin{equation*}\lim_{x\to\infty}\frac{1}{x}\int_{2}^{x}\frac{\pi(t)}{t}dt=0\, .\end{equation*}

Now, the Prime Number Theorem also implies that \(\frac{\pi(t)}{t}\) and \(\frac{1}{\log(t)}\) are asymptotic, so that their integrals also are, \begin{equation*}\frac{1}{x}\int_{2}^{x}\frac{\pi(t)}{t}dt\text{ and } \frac{1}{x}\int_{2}^{x}\frac{dt}{\log(t)}\, .\end{equation*}

This reduces our proof to showing that the average value of \(1/\log(t)\) tends to zero. Since integral have a graphical interpretation, we now use the following graph of the integral limit to finish the proof!

Consider that one possible upper sum for the integral of \(1/\log(t)\) between 2 and 9 is the area of the two rectangles shown below, one with area \(\frac{1}{\log(2)}(\sqrt{9}-2)\) and the other with area \(\frac{1}{\log(\sqrt{9})}(9-\sqrt{9})\). (Of course \(\sqrt{9}=3\) but this form is more useful here.)

In general, the same argument should hold, so a possible overestimate of \(\int_2^x dt/\log(t)\) is \begin{equation*}\frac{1}{\log(2)}(\sqrt{x}-2)+\frac{1}{\log(\sqrt{x})}(x-\sqrt{x})\end{equation*} and we want the limit as \(x\to\infty\) of \(\frac{1}{x}\) times that quantity.

Now is the time to recklessly use logarithmic identities: \begin{equation*}\frac{1}{x}\left(\frac{1}{\log(2)}(\sqrt{x}-2)+\frac{1}{\log(\sqrt{x})}(x-\sqrt{x})\right)=\frac{1}{\log(2)x^{1/2}}-\frac{2}{x\log(2)}+\frac{1}{\log(\sqrt{x})}-\frac{1}{\log(\sqrt{x})x^{1/2}}\end{equation*} \begin{equation*}=\frac{1}{\log(2)x^{1/2}}-\frac{2}{x\log(2)}+\frac{2}{\log(x)}-\frac{2}{\log(x)x^{1/2}}\end{equation*} This last expression has positive powers of \(x\) and their logs in the denominators, so it pretty clearly goes to zero as \(x\to \infty\).

If the algebra doesn't convince you, perhaps the graphs will. Below, black is the overestimate to the integral and red is \(1/x\) times the integral.

The picture confirms our analytic proof that the limit of \(\frac{\theta(x)}{x}\) is the same as that of \(\frac{\pi(x)}{x/\log(x)}\), which is what we desired!