Exercise 1.3.53* ($\varepsilon^2$ of Prime Number Theorem.)

Proof. a) $u\mapsto \pi (u)∕u$ is piecewise continuous on $[2,x]$, so is Riemann integrable on this interval.

Write $p_n$ the $n$-th prime number : $p_1=2,p_2=3,p_3=5,\dots$. Let $x\ge 2$ a real number. Define $n\in {\mathbb{N}}^{\text{∗}}$ by

Then

For $u\in [p_i,p_{i+1}[$, $\pi (u)$ is a constant, and equal to $i$, because the primes not exceeding $u$ are $p_1,p_2,\dots ,p_i$:

Therefore

and similarly

Replacing these values in (1), we obtain

Using $\pi (x)=n$, this gives

Since $p_1,\dots ,p_n$ are the prime numbers $p\le x$, we conclude

b) The limit superior of a function $f\ge 0$, defined on a neighborhood of $+\infty$, is defined by

Since $x\mapsto \underset{u\ge x}{\sup <!--\; FUNCTION\; APPLICATION\; -->}f(u)$ is a decreasing function, this limit exists in ${ℝ}^{\text{+}}\cup \{+\infty \}$.

Assume for contradiction that

Then, taking $𝜀=\frac{1-\rho }{2}>0$, there exists $x_0\ge 2$ such that

so

Writing $k=\frac{1+\rho }{2}$, then $k<1$, and for all $u\ge x_0$,

that is

By Theorem 1.9.1,

Then, by the equality (2), for every $x\ge x_0$,

Writing $c$ the constant $c={\int }_2^{x_0}\pi (u)∕u^{2}du$, this gives, using $\pi (x)\le x$,

The inequality (3) gives

where $C=2+c-k\log <!--\; FUNCTION\; APPLICATION\; -->(\log <!--\; FUNCTION\; APPLICATION\; -->(x_0))$ is a constant.

Dividing by $\log <!--\; FUNCTION\; APPLICATION\; -->(\log <!--\; FUNCTION\; APPLICATION\; -->(x))$, we obtain, for all $x\ge x_0$,

therefore

This shows $k\ge 1$: this is a constradiction, since $k<1$, so

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