Number theory

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Prime counting functions:
Prime counting function $\pi(x)$ = number of primes $\leq x$. It is approximately $\text{li}(x) \sim x/\ln x$
(and auxiliary big Pi function counting powers of primes by fractions. It is even more close to $\text{li}(x)$)

Logarithm of products of primes:
Prime theta function: $\vartheta(x) =$ logarithm of product of primes $\leq x$.
Prime psi function: $\psi(x)$ = logarithm of least common multiple of natural numbers $\leq x$.
Firstly let's define $\psi$ for $x\geq 1$:

$$\psi(x) = \sum_{\text{prime}\ p} \left\lfloor \log_p x \right\rfloor \ln p $$ it satisfies: $$\sum_{1\leq n\leq x} \psi(x/n) = \ln \lfloor x \rfloor !$$ therefore (using Moebius inversion): $$\psi(x) = \sum_{n\geq 1} \mu(n) \ln \lfloor x/n \rfloor !$$ Now using function $\theta$ defined using sum over primes $p$:

$$\theta(x) = \sum_{\text{prime}\ p\leq x} \ln p$$ We can express also $\psi$ as:

$$\psi(x) = \sum_{n\geq 1} \theta( \sqrt[n]{x} )$$ Therefore (using inversion)

$$\theta(x) = \sum_{n\geq 1} \mu(n) \psi( \sqrt[n]{x} )$$

Prime counting functions $\pi$ and $Pi$

For an arbitrary $a\in [0,2]$ $$\Pi(x) = \int_a \frac{\text{d}\psi(x)}{\ln x} = \frac{\psi(x)}{\ln x} - \int_a \frac{\psi(x)\ \text{d}x}{x \ln^2 x} $$ $$\pi(x) = \int_a \frac{\text{d} \theta(x)}{\ln x } = \frac{\theta(x)}{\ln x} - \int_a \frac{\theta(x)\ \text{d}x}{x \ln^2 x}$$ $$\text{li}(x) = \int_0 \frac{d x}{\ln x} = \frac{ x }{\ln x} - \int \frac{x \ \text{d} x}{x \ln^2 x}$$ And like $\psi$ and $\theta$ have small differences from $x$, similarly $\Pi$ and $\pi$ have small differences from $\text{li}$, because: $$\Pi(x)-\text{li}(x) = \int \frac{\text{d}(\psi(x) - x)}{\ln x} = \frac{\psi(x)-x}{\ln x} - \int \frac{\psi(x)-x}{x \ln^2 x} \text{d}x$$ $$\pi(x)-\text{li}(x) = \int \frac{\text{d}(\theta(x) - x)}{\ln x} = \frac{\theta(x)-x}{\ln x} - \int \frac{\theta(x)-x}{x \ln^2 x}\text{d}x $$ $$\vartheta(x) = \int_0 (\ln x) \text{d}\pi(x) = \pi(x) \ln x - \int_0 \frac{ \pi(x) }{x} \text{d}x$$