Bounds for Prime-Counting Function in terms of Second Chebyshev Function

Theorem
There exists a real function $R : \hointr 2 \infty \to \R$ such that:


 * $\ds \frac {\map \psi x} {\ln x} + \map R x \le \map \pi x \le \frac {2 \map \psi x} {\ln x} + \sqrt x$

for all real numbers $x \ge 2$, where:
 * $\pi$ is the prime counting function
 * $\psi$ is the second Chebyshev function
 * $R = \map \OO {\sqrt x \ln x}$ where $\OO$ is big-O notation.

Proof
We have, from the definition of the prime counting function:


 * $\ds \map \pi x = \sum_{p \le x} 1$

We can write:


 * $\ds \sum_{p \le x} 1 = \sum_{p \le \sqrt x} 1 + \sum_{\sqrt x < p \le x} 1$

We have that:

For $1 < \sqrt x < p$, we have from Logarithm is Strictly Increasing:


 * $0 < \map \ln {\sqrt x} < \ln p$

So:


 * $\ds \frac {\ln p} {\map \ln {\sqrt x} } > 1$

Then:

So:


 * $\ds \map \psi x \le \sqrt x + \frac 2 {\ln x} \map \psi x$

Now, from Order of Second Chebyshev Function, there exists a real function $r : \hointr 2 \infty \to \R$ such that:


 * $\ds \map \psi x = \sum_{p \le x} \ln p + \map r x$

where:


 * $r = \map \OO {\sqrt x \paren {\ln x}^2}$

For $p \le x$, we have from Logarithm is Strictly Increasing:


 * $0 < \ln p \le \ln x$

so:


 * $\ds \frac {\ln p} {\ln x} < 1$

Then:

From Product of Big-O Estimates, we have:


 * $\dfrac {\map r x} {\ln x} = \map \OO {\sqrt x \ln x}$

So if we take:


 * $\map R x = \dfrac {\map r x} {\ln x}$

for each $x \ge 2$, we have:


 * $\ds \map \pi x = \sum_{p \le x} 1 \ge \frac {\map \psi x} {\ln x} + \map R x$

with $R = \map \OO {\sqrt x \ln x}$.

So we obtain:


 * $\ds \frac {\map \psi x} {\ln x} + \map R x \le \map \pi x \le \frac {2 \map \psi x} {\ln x} + \sqrt x$

with $R = \map \OO {\sqrt x \ln x}$ as required.