# Bounding the prime counting function

How can I get inequalities that bound the prime counting function if I have the following inequalities for some functions $f(x)$ and $g(x)$: $$g(x)<\psi(x)<f(x),$$ where $\psi(x)$ is second Chebyshev function (the summatory function for the von Mangoldt function)?

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Chapter 5 of this book contains a pretty nice walkthrough of Chebyshev’s theorem (and the whole prime number theorem) that uses nothing but elementary number theory and combinatorics techniques to achieve such bounds.

Define:
$$\psi(x) = \sum_{n \le x} \Lambda(n) = \sum_{p^k \le x} \ln p$$
$$\theta(x) = \sum_{p \le x} \ln p$$
$$\pi(x) = \sum_{p \le x} 1$$

We’ll compare $\psi$ to $\theta$ and $\theta$ to $\pi$.

• $\psi(x) – \theta(x) = \sum_{p^k \le x, k\ge 2} \ln p$. So in the sum we care only about primes that are at most $\sqrt{x}$. Each such prime $p$ appears in the sum $\lfloor \frac{\ln x}{\ln p} \rfloor -1$ times, so we can write it as follows:
$$\psi(x) -\theta(x) = \sum_{p \le \sqrt{x}} \ln p (\lfloor \frac{\ln x}{\ln p} \rfloor -1)$$

Now there are a few options:

i. We can bound each term from above by $\ln x$, so $$0 \le \psi(x) – \theta(x) \le \pi(\sqrt{x}) \ln x$$ And then use the trivial bound $\pi(\sqrt{x}) \le \sqrt{x}$ bound, to find $\psi(x) – \theta(x) \le \sqrt{x}\ln x$.

ii. Similar to i, but use the Chebyshev bound $\pi(\sqrt{x}) = O(\frac{\sqrt{x}}{\ln x})$ to prove $\psi(x) – \theta(x) = O(\sqrt{x})$ (with constant around 2).

iii. A more involved argument shows that the real constant is around 1: Bound each term by $\ln x – \ln p$, so the difference becomes $\ln x \pi(\sqrt{x}) – \theta(\sqrt{x})$, which behaves like $\ln x \frac{\sqrt{x}}{\ln \sqrt{x}} – \sqrt{x} \sim \sqrt{x}$.

• To understand the relation between $\pi$ and $\theta$ we need Abel Summation applied to $a_n = 1_{n \text{ is prime}}, b_n = \log n$. As $\theta(x) = \sum_{n \le x} a_n b_n$, we find:
$$\theta(m) = (\sum_{n \le m} a_n) b_{m} – \sum_{n < m} (b_{n+1}-b_{n}) (\sum_{i \le n} a_n) =$$
$$\pi(m) \ln m – \sum_{n < m} \ln(1 + \frac{1}{n}) \pi(n)$$
(I worked with $m$ integer but it is true in general, up to some tiny error term.)

To bound the last sum, note that $0 \le \ln(1+\frac{1}{n}) \le \frac{1}{n}$ (it’s pretty tight for large $n$), so $0 \le \sum_{n < m} \ln(1 + \frac{1}{n}) \pi(n) \le \sum_{n < m} \frac{\pi(n)}{n}$.

i. The trivial bound is $m$, which follows form $\pi(x) \le x$.

ii. A more complex one is $O(\sum_{n < m} \frac{1}{\ln n}) = O(\frac{m}{\ln m})$ (constant around 1), based on $\pi(x) = O(\frac{x}{\ln x})$, which is based of Chebyshev’s estimates.

All in all, $\psi(x) = \pi(x) \ln x + O(\frac{x}{\ln x})$, so $g(x) < \psi(x) < f(x) \implies \frac{g(x)+O(\frac{x}{\ln x})}{\ln x} < \pi(x) < \frac{f(x)+O(\frac{x}{\ln x})}{\ln x}$. Precise error terms depend on what you’re assumeing, but I believe the constant of the $O()$ is around $1$.