## The first Chebyshev function:

$$\vartheta(x) = \sum_{p \leq x} \ln p$$

where $$p \in \mathbb{P}$$.

## The second Chebyshev function:

If $$\Lambda$$ denotes the Von-Mangoldt function:

$$\psi(x) = \sum_{n \leq x} \Lambda(n) = \sum_{p \leq x} \lfloor \log_p x \rfloor \cdot \ln p$$

## Relation between the Chebyshev functions:

The second Chebyshev function may be expressed as:

$$\psi(x) = \sum_{p \leq x} k \cdot \ln p$$

where $$p^k \leq x < p^{k+1}$$.

This allows us to deduce:

$$\psi(x) = \sum_{n=1}^\infty \vartheta(x^{\frac{1}{n}}) = \sum_{n=1}^{\lfloor \log_2 x \rfloor} \vartheta(x^{\frac{1}{n}})$$

since $$\vartheta(x^{\frac{1}{n}}) = 0$$ for $$n > \log_2 x$$.

## $$e^{\psi(n)} = \text{lcm}([1,n])$$

### Proof:

Given the formulas:

$$\text{lcm}([1,n]) = \prod_{i=1}^{\lfloor \log_2 n \rfloor} p_i^{\alpha_i}$$

$$\psi(n) = \sum_{i=1}^{\lfloor \log_2 n \rfloor} k_i \cdot \ln p_i$$

we’ll note that $$\alpha_i > 0 \implies \alpha_i \cdot \ln p_i = k_i \cdot \ln p_i$$ for the reason that:

$$p_i^{k_i} \leq p_i^{\alpha_i} < p_i^{k_i + 1}$$

QED.

## $$e^{\vartheta(n)} \leq 4^n$$

### Proof by induction:

We’ll start by observing that:

$$e^{\vartheta (n)} = \prod_{p \leq n} p$$

For the base case $$n=2$$,

$$2 \leq 4^2$$

For the inductive hypothesis we set $$n = 2m -1$$,

$$\prod_{p \leq 2m-1} p \leq 4^{2m-1}$$

Now, we may consider the cases of even and odd numbers separately. For the even case,

$$\prod_{p \leq 2m} p \leq 4^{2m-1} < 4^{2m}$$

and for the odd case $$n = 2m+1$$, given that $$2m-1 > m+1$$ we have:

$$\prod_{p \leq 2m+1} p = \prod_{p \leq m+1} p \cdot \prod_{p \leq m+2}^{2m+1} p \leq 4^{m+1} \cdot {2m+1 \choose m}$$

so we may deduce that:

$$\prod_{p \leq 2m+1} p \leq 4^{m+1} \cdot 4^m = 4^{2m+1}$$

QED.

## $$\vartheta(n) \sim n$$

### Proof:

Given that the prime numbers are distributed according to,

$$\pi(n) \sim \frac{n}{\ln n}$$

we may expect that for any random variable $$X_n \sim \pi(n)$$

$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \ln X_k = \lim_{n \to \infty} \frac{1}{n} \sum_{X \leq n}\mathbb{E}[\ln X]$$

where $$\mathbb{E}[\sum_{k=1}^n \text{Bool} \circ (\ln X_k > 0)] \sim \frac{n}{\ln n}$$ and therefore:

$$\sum_{X \leq n}\mathbb{E}[\ln X] \approx \int_{2}^n \ln x \cdot P(x \in \mathbb{P}) dx \approx n$$

since $$P(x \in \mathbb{P}) \approx \frac{1}{\ln x}$$.

It follows that:

$$\lim_{n \to \infty} \frac{\vartheta(n)}{n} = 1$$

QED.

## References:

1. Chebyshev, P. L. “Mémoir sur les nombres premiers.” J. math. pures appl. 17, 366-390, 1852

2. Hardy, G. H. and Wright, E. M. “The Functions theta(x) and psi(x)” and “Proof that theta(x) and psi(x) are of Order x.” §22.1-22.2 in An Introduction to the Theory of Numbers, 5th ed. Oxford, England: Clarendon Press, pp. 340-342, 1979.