# Equivalence of Definitions of Second Chebyshev Function

## Theorem

The following definitions of the concept of **Second Chebyshev Function** are equivalent:

- $(1): \quad \displaystyle \psi \left({x}\right) = \sum_{p^k \mathop \le x} \ln p$

- $(2): \quad \displaystyle \psi \left({x}\right) = \sum_{1 \mathop \le n \mathop \le x} \Lambda \left({n}\right)$

- $(3): \quad \displaystyle \psi \left({x}\right) = \sum_{p \mathop \le x} \left \lfloor {\log_p x} \right \rfloor \ln p$

where:

- $p$ is a prime number
- $\Lambda \left({n}\right)$ is the von Mangoldt function
- $\left \lfloor {\ldots} \right \rfloor$ denotes the floor function.

## Proof

The equivalence:

- $\displaystyle \sum_{p^k \mathop \le x} \ln p \equiv \sum_{1 \mathop \le n \mathop \le x} \Lambda \left({n}\right)$

follows directly from the definition of the von Mangoldt function.

$\Box$

Let $N = \left \lfloor {x} \right \rfloor$.

It can be seen directly that all the above summations are exactly the same whether performed on $N$ or $x$.

Hence we need only to prove the equivalence for integral arguments.

First we expand the von Mangoldt function:

\(\displaystyle \sum_{n \mathop = 1}^N \Lambda \left({n}\right)\) | \(=\) | \(\displaystyle \Lambda \left({1}\right) + \Lambda \left({2}\right) + \cdots + \Lambda \left({N}\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0 + \ln \left({2}\right) + \ln \left({3}\right) + \ln \left({2}\right) + \ln \left({5}\right) + 0 + \ln \left({7}\right) + \ln \left({2}\right) + \ln \left({3}\right) + 0 + \cdots\) |

Notice this sum will have:

- as many $\ln \left({2}\right)$ terms as there are powers of $2$ less than or equal to $N$,
- as many $\ln \left({3}\right)$ terms as there are powers of $3$ less than or equal to $N$

and in general, if $p$ is a prime less than $N$, $\ln p$ will occur in this sum $\left \lfloor {\log_p N} \right \rfloor$ times.

Hence:

- $\displaystyle \sum_{1 \mathop \le n \mathop \le x} \Lambda \left({n}\right) \equiv \sum_{p \mathop \le x} \left \lfloor {\log_p x} \right \rfloor \ln p$

$\blacksquare$