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 $(3)$ |
$\blacksquare$
Sources
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.16$: The Sequence of Primes