Section 6.1 Elementary results on distribution of primes
Proof.
We have \(\pi(1) = 0\text{,}\) \(\pi(2) = 1\text{,}\) \(\pi(3) = 2\text{,}\) \(\pi(4) = 2\text{,}\) and \(\pi(5) = 3\text{.}\) Meanwhile, \(\theta(1) = 0\text{,}\) \(\theta(2) = \log 2\text{,}\) \(\theta(3) = \theta(4) = \log 2 + \log 3 \approx 1.79\text{,}\) and \(\theta(5) = \log 2 + \log 3 + \log 5 \approx 3.40\text{.}\) So indeed \(\theta(5) \geq \pi(5)\text{.}\) For \(p \geq 5\) we have \(\log p \gt 1\text{.}\) So in the sums for \(\theta\) and \(\pi\text{,}\) all terms with \(p \geq 5\) are greater for \(\theta\) than for \(\pi\text{.}\) This shows that \(\theta(x) \geq \pi(x)\) for \(x \geq 5\text{.}\)
The difference between \(\theta\) and \(\psi\) is
\begin{equation*}
\psi(x) - \theta(x) = \sum_{p^k \leq x, k \geq 2} \log p\text{.}
\end{equation*}
This difference is always nonnegative because each \(\log p \gt 0\text{.}\)
Theorem 6.1.2.
For every \(x \geq 1\text{,}\) we have
\begin{equation}
0 \leq \psi(x) - \theta(x) \leq \frac{\sqrt{x}(\log x)^2}{2 \log 2}\text{.}\tag{6.1.1}
\end{equation}
In particular, \(\psi(x) = \theta(x) + O(\sqrt{x}(\log x)^2)\text{.}\)Proof.
We already proved the first statement. For the second one,
\begin{align*}
\psi(x) \amp = \sum_{n \leq x} \Lambda(n) \\
\amp = \sum_{p^k \leq x} \log p \\
\amp = \sum_{k=1}^{\infty} \sum_{p \leq x^{1/k}} \log p \\
\amp = \sum_{k=1}^{\infty} \theta(x^{1/k}) \text{.}
\end{align*}
Despite appearances, this is actually a finite sum: We have \(\theta(y) = 0\) for \(y \lt 2\text{.}\) So \(\theta(x^{1/k}) = 0\) if \(x^{1/k} \lt 2\text{,}\) in other words if \(k \gt \frac{\log x}{\log 2}\text{.}\) Therefore
\begin{equation}
\psi(x) - \theta(x) = \sum_{2 \leq k \leq \frac{\log x}{\log 2}} \theta(x^{1/k})\text{.}\tag{6.1.2}
\end{equation}
The largest term is \(\theta(\sqrt{x})\) (for \(k=2\)) and the number of terms is at most \(\frac{\log x}{\log 2}\text{.}\) Therefore the sum is bounded above,
\begin{equation*}
\psi(x) - \theta(x) \leq \theta(\sqrt{x}) \frac{\log x}{\log 2} \text{.}
\end{equation*}
Furthermore,
\begin{equation}
\theta(\sqrt{x}) = \sum_{p \leq \sqrt{x}} \log p
\leq \sum_{p \leq \sqrt{x}} \log \sqrt{x}
\leq \sqrt{x} \log \sqrt{x}\tag{6.1.3}
\end{equation}
since there are at at most \(\sqrt{x}\) terms. With this we get
\begin{equation}
\psi(x) - \theta(x) \leq \sqrt{x} (\log \sqrt{x}) \frac{\log x}{\log 2}\text{.}\tag{6.1.4}
\end{equation}
After simplifying with \(\log \sqrt{x} = \frac{1}{2} \log x\) we have the claimed result.
Remark 6.1.3.
Observe that twice in the above proof we bounded sums in the same way: bounding each term (every \(\theta(x^{1/k})\) is less than or equal to the largest one, which is \(\theta(x^{1/2})\text{;}\) every \(\log p\) is less than or equal to \(\log \sqrt{x}\text{,}\) when \(p \leq \sqrt{x}\)), and bounding the number of terms (in particular, a sum over primes \(p \leq y\) has at most \(y\) terms).
The previous theorem gives a comparison between \(\psi\) and \(\theta\text{.}\) Now we give similar comparisons between \(\theta\) and the prime counting function \(\pi\text{.}\)
Theorem 6.1.4.
For \(x \geq 2\) we have
\begin{equation}
\theta(x) = \pi(x) \log x - \int_2^x \frac{\pi(t)}{t} dt\tag{6.1.5}
\end{equation}
and
\begin{equation}
\pi(x) = \frac{\theta(x)}{\log x} + \int_2^x \frac{\theta(t)}{t (\log t)^2} dt\text{.}\tag{6.1.6}
\end{equation}
Proof.
The proof uses the Riemann-Stieltjes integral and integration by parts. Define a sequence by \(a_n = 1\) if \(n\) is a prime number and \(a_n = 0\) otherwise. Then \(\pi(x) = \sum_{n \leq x} a_n\text{,}\) and
\begin{align}
\theta(x) \amp = \sum_{p \leq x} \log p \tag{6.1.7}\\
\amp = \sum_{1 \lt n \leq x} a_n \log n \tag{6.1.8}\\
\amp = \int_1^x \log t \; d\pi(t) \tag{6.1.9}\\
\amp = \pi(x) \log x - \int_1^x \frac{\pi(t)}{t} dt \tag{6.1.10}\\
\amp = \pi(x) \log x - \int_2^x \frac{\pi(t)}{t} dt \text{.}\tag{6.1.11}
\end{align}
Here are the reasons for each step.
The second claim follows similarly. Let \(b_n = \log n\) if \(n\) is prime, \(b_n = 0\) otherwise. Then \(\theta(x) = \sum_{n \leq x} b_n\text{,}\) and
\begin{align}
\pi(x) \amp = \sum_{2 \leq n \leq x} \frac{1}{\log n} b_n \tag{6.1.12}\\
\amp = \int_{2^{-}}^x \frac{1}{\log t} d\theta(t) \tag{6.1.13}\\
\amp = \frac{\theta(x)}{\log x} + \int_2^x \frac{\theta(t)}{t (\log t)^2)} \tag{6.1.14}
\end{align}
where we once again used the Riemann-Stelejes integral, and integration by parts.
One reason to care about the Chebyshev functions is that they give equivalent statements to the Prime Number Theorem:
Theorem 6.1.5.
The following statements are equivalent:- \(\displaystyle \pi(x) \sim \frac{x}{\log x}\)
- \(\displaystyle \theta(x) \sim x\)
- \(\displaystyle \psi(x) \sim x\)
Proof.
We have
\begin{equation*}
\theta(x) = \pi(x) \log x - \int_2^x \frac{\pi(t)}{t} dt,\text{.}
\end{equation*}
Dividing by \(x\) gives
\begin{equation*}
\frac{\theta(x)}{x} = \frac{\pi(x)}{x/\log x} - \frac{1}{x} \int_2^x \frac{\pi(t)}{t} dt\text{.}
\end{equation*}
Assume that \(\pi(x) \sim x/\log x\text{.}\) We need to show that the integral term goes to zero. In this case we have \(\frac{\pi(t)}{t} \ll \frac{1}{\log t}\text{.}\) Therefore
\begin{equation*}
\int_2^x \frac{\pi(t)}{t} dt \ll \int_2^x \frac{1}{\log t} dt\text{,}
\end{equation*}
because there is a threshold \(T\) and constant \(C\) such that for \(t \gt T\text{,}\) \(\frac{\pi(t)}{t} \leq C \frac{1}{\log t}\text{;}\) then for \(x \gt T\text{,}\)
\begin{equation*}
\int_T^x \frac{\pi(t)}{t} dt \ll C \int_T^x \frac{1}{\log t} dt\text{.}
\end{equation*}
Notice that these integrals diverge to infinity as \(x \to \infty\text{.}\) The error (difference) between \(\int_2^T \frac{\pi(t)}{t} dt\) and \(C \int_2^T \frac{1}{\log t} dt\) is some constant amount; eventually it is insignificant compared to the integrals from \(T\) to \(x\text{,}\) for large \(x\) (and we can replace \(C\) with, say, \(2C\)).
So, we have:
\begin{equation*}
\frac{1}{x} \int_2^x \frac{\pi(t)}{t} dt \ll \frac{1}{x} \int_2^x \frac{1}{\log t} dt
\end{equation*}
We can bound the last integral in a similar way to previous bounds of sums: the maximum value of the integrand on this interval is \(\frac{1}{\log 2}\) (because the function \(\frac{1}{\log t}\) is monotone decreasing) and the length of the interval is less than \(x\text{.}\) So the value of the integral is less than or equal to \(\frac{x}{\log 2}\text{.}\) This gives us:
\begin{align*}
\frac{1}{x} \int_2^x \frac{\pi(t)}{t} dt \amp \ll \frac{1}{x} \int_2^x \frac{1}{\log t} dt \\
\amp \ll \frac{1}{x} \frac{x}{\log 2} \\
\amp = \frac{1}{\log 2} \text{.}
\end{align*}
So the “error” term (the integral) is bounded by a constant. But we want a stronger result: we want the error to go to zero.
We can get this by a trick that is slightly reminiscent of the hyperbola method: we split the interval of integration into two parts as follows.
\begin{align*}
\frac{1}{x} \int_2^x \frac{\pi(t)}{t} dt \amp \ll \frac{1}{x} \int_2^x \frac{1}{\log t} dt \\
\amp = \frac{1}{x} \int_2^{\sqrt{x}} \frac{1}{\log t} dt + \frac{1}{x} \int_{\sqrt{x}}^x \frac{1}{\log t} dt \\
\amp \ll \frac{1}{x} \frac{\sqrt{x}}{\log 2} + \frac{1}{x} \frac{x}{\log \sqrt{x}} \\
\amp = \frac{1}{\sqrt{x} \log 2} + \frac{1}{\log \sqrt{x}} \text{.}
\end{align*}
This does go to zero. Therefore, if \(\frac{\pi(x)}{x/\log x} \to 1\text{,}\) then \(\frac{\theta(x)}{x} \to 1\) as well.
Remark 6.1.6.
By splitting the interval of integration into two parts, we have one part where the bound for the integrand is the same (\(1/\log 2\)) but the interval is shorter (\(\sqrt{x}\) instead of \(x\)) and another part where the interval is the same length (or rather, we use the same bound \(x\) for its length) but the integrand is bounded by a smaller amount (\(1/\log \sqrt{x}\) instead of \(1/\log 2\)).
The converse (\(\frac{\theta(x)}{x} \to 1\) implies \(\frac{\pi(x)}{x/\log x} \to 1\)) is similar.
From our comparison of \(\psi\) and \(\theta\) we have
\begin{equation*}
0 \leq \frac{\psi(x)}{x} - \frac{\theta(x)}{x} \leq \frac{(\log x)^2}{2 \log 2 \sqrt{x}}\text{.}
\end{equation*}
Therefore
\begin{equation*}
\lim_{x \to \infty} \frac{\psi(x)}{x} - \frac{\theta(x)}{x} = 0\text{,}
\end{equation*}
so
\begin{equation*}
\frac{\psi(x)}{x} \to 1 \Leftrightarrow \frac{\theta(x)}{x} \to 1\text{.}
\end{equation*}
