Prime Number Theorem in Eulerian Logarithmic Integral Form

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Theorem

The Prime Number Theorem is equivalent to:

$\ds \lim_{x \mathop \to \infty} \frac {\map \pi x} {\map \Li x} = 1$

where:

$\map \pi x$ is the prime-counting function
$\map \Li x$ is the Eulerian logarithmic integral:
$\ds \map \Li x := \int_2^x \dfrac {\d t} {\ln t}$


Proof

Using Integration by Parts:

\(\ds \map \Li x\) \(=\) \(\ds \int_2^x \dfrac {\d t} {\ln t}\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac x {\ln x} - \dfrac 2 {\ln 2} + \int_2^x \dfrac {\d t} {\paren {\ln t}^2}\)

We have that $\dfrac 1 {\paren {\ln t}^2}$ is positive and decreasing for $t > 1$.

Let $x \ge 4$.

Then:

\(\ds 0\) \(<\) \(\ds \int_2^x \dfrac {\d t} {\paren {\ln t}^2}\)
\(\ds \) \(=\) \(\ds \int_2^{\sqrt x} \dfrac {\d t} {\paren {\ln t}^2} + \int_{\sqrt x}^x \dfrac {\d t} {\paren {\ln t}^2}\)
\(\ds \) \(<\) \(\ds \dfrac {\sqrt x - 2} {\paren {\ln 2}^2} + \dfrac {x - \sqrt x} {\paren {\ln \sqrt x}^2}\)
\(\ds \) \(<\) \(\ds \dfrac {\sqrt x} {\paren {\ln 2}^2} + \dfrac {4 x} {\paren {\ln \sqrt x}^2}\)
\(\ds \leadsto \ \ \) \(\ds 0\) \(<\) \(\ds \dfrac {\ds \int_2^x \frac {\d t} {\paren {\ln t}^2} } {x / \ln x}\)
\(\ds \) \(<\) \(\ds \dfrac {\ln x} {\sqrt x \paren {\ln 2}^2} + \frac 4 {\ln x}\)
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \lim_{x \mathop \to \infty} \dfrac {\ds \int_2^x \frac {\d t} {\paren {\ln t}^2} } {x / \ln x}\) \(=\) \(\ds 0\)

Then:

\(\ds \frac {\map \Li x} {x / \ln x}\) \(=\) \(\ds \dfrac {\dfrac x {\ln x} - \dfrac 2 {\ln 2} + \ds \int_2^x \dfrac {\d t} {\paren {\ln t}^2} } {x / \ln x}\) from $(1)$
\(\ds \) \(=\) \(\ds 1 - \dfrac {2 \ln x} {x \ln 2} + \dfrac {\ds \int_2^x \frac {\d t} {\paren {\ln t}^2} } {x / \ln x}\)
\(\ds \leadsto \ \ \) \(\ds \lim_{x \mathop \to \infty} \frac {\map \Li x} {x / \ln x}\) \(=\) \(\ds 1 - \lim_{x \mathop \to \infty} \dfrac 2 {\ln 2} \dfrac x {\ln x} - 0\) from $(2)$
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \lim_{x \mathop \to \infty} \frac {\map \Li x} {x / \ln x}\) \(=\) \(\ds 1\) as $\ds \lim_{x \mathop \to \infty} \dfrac x {\ln x} = 0$


Then:

\(\ds \lim_{x \mathop \to \infty} \frac {\map \pi x} {x / \ln x}\) \(=\) \(\ds \lim_{x \mathop \to \infty} \frac {\map \pi x} {\map \Li x} \frac {\map \Li x} {x / \ln x}\)
\(\ds \) \(=\) \(\ds \lim_{x \mathop \to \infty} \frac {\map \pi x} {\map \Li x}\) from $(3)$

$\blacksquare$


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