Lower and upper bound for Chebyshev's psi function

In this post, I prove the inequalities \begin{equation}\label{eq:1} \log 2 \cdot x - 3 \log x \le \psi(x) \le 2 \log 2 \cdot x + \frac{3}{\log 2} \log^2 x \end{equation} for all real \( x \ge 2 \). Hereby is \( \psi(x) = \sum_{n \le x } \Lambda(n) \) with \( \Lambda \) the von Mangoldt function. This is a second post motivated by Chebyshev's paper "Mémoire sur les nombres premiers" from 1852. In this paper, Chebyshev uses a more complex variant of the calculation below to obtain stronger inequalities. Similar inequalities can also be found on page 50 in Murty's book.
I showed in a previous post that if \( T(x) = \sum_{ 1 \le n \le x} \log n\ \) then \begin{equation}\label{eq:2} \psi(x) - \psi\left(\frac{x}{2} \right)\le T(x) - 2\ T\left( \frac{x}{2} \right) \le \psi(x) \end{equation} For all real \( x \ge 1 \), one has \begin{equation}\label{eq:3} x \log x - x - \log x +1 \le T(x) \le x \log x - x + \log x +1 \end{equation} These inequalities are easily proved by comparing the sum in the expression \( T(x) = \sum_{ 1 \le n \le x} \log n \) with the corresponding integral. From \eqref{eq:3} one immediately deduces that for all real \( x \ge 2 \) \begin{equation}\label{eq:4} \log 2 \cdot x - 3 \log x + 2 \log 2 -1 \le T(x) - 2\ T\left( \frac{x}{2} \right) \end{equation} and \begin{equation}\label{eq:5} T(x) - 2\ T\left( \frac{x}{2} \right) \le \log 2 \cdot x + 3 \log x - 2 \log 2 -1 \end{equation} Combining \eqref{eq:2} and \eqref{eq:4} and using \( 2 \log 2 -1 > 0\ \) gives \begin{equation}\label{eq:6} \log 2 \cdot x - 3 \log x \le \psi(x) \end{equation} The lower bound of \eqref{eq:1} is thus proved. Combining \eqref{eq:2} and \eqref{eq:5} and using \( - 2 \log 2 -1 < 0 \), one has for all real \( x \ge 2 \) that \begin{equation}\label{eq:7} \psi(x) - \psi\left(\frac{x}{2} \right)\le \log 2 \cdot x + 3 \log x \end{equation} To prove the upper bound, take \( m \in \mathbb{N} \) such that \( 2 \le \frac{x}{2^m} < 4 \). Applying \eqref{eq:7} several times gives \begin{align*} \psi(x) - \psi\left(\frac{x}{2} \right) &\le \log 2 \cdot x + 3 \log x\\ \psi\left(\frac{x}{2} \right)- \psi\left(\frac{x}{4} \right) &\le \log 2 \cdot \frac{x}{2} + 3 \log \frac{x}{2}\\ \psi\left(\frac{x}{4} \right)- \psi\left(\frac{x}{8} \right) &\le \log 2 \cdot \frac{x}{4} + 3 \log \frac{x}{4}\\ \cdots\\ \psi\left(\frac{x}{2^m} \right)- \psi\left(\frac{x}{2^{m+1}} \right) &\le \log 2 \cdot \frac{x}{2^m} + 3 \log \frac{x}{2^m} \end{align*} Because \( 1 \le \frac{x}{2^{m+1}} < 2 \), it follows that \( \psi\left(\frac{x}{2^{m+1}} \right) = 0 \). Adding the inequalities thus gives \begin{equation*} \psi(x) \le 2 \log 2 \cdot x + 3 (m+1) \log x \end{equation*} The inequality \( 2^{m+1} \le x \) gives \( m + 1 \le \frac{\log x }{\log2 } \). The upper bound of \eqref{eq:1} is thus also proved.

Here is a graph that illustrates the inequality \eqref{eq:1}.

The red line is the upper bound, the black line is \( \psi(x) \), the green line is the lower bound.

Other posts about Chebyshev's mémoire

1. How not to prove Bertrand's postulate
2. Variations on a theme by Chebyshev