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__

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