# Prime Number Theorem in Eulerian Logarithmic Integral Form

## 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 $(3)$

$\blacksquare$