Equivalence of Definitions of Second Chebyshev Function

Theorem

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


 * $(2): \quad \displaystyle \map \psi x = \sum_{1 \mathop \le n \mathop \le x} \map \Lambda n$


 * $(3): \quad \displaystyle \map \psi x = \sum_{p \mathop \le x} \floor {\log_p x} \ln p$

where:
 * $p$ is a prime number
 * $\map \Lambda n$ is the von Mangoldt function
 * $\floor {\, \cdot \, }$ denotes the floor function.

Proof
The equivalence:
 * $\displaystyle \sum_{p^k \mathop \le x} \ln p \equiv \sum_{1 \mathop \le n \mathop \le x} \map \Lambda n$

follows directly from the definition of the von Mangoldt function.

Let $N = \floor x$.

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:

Notice this sum will have:
 * as many $\map \ln 2$ terms as there are powers of $2$ less than or equal to $N$
 * as many $\map \ln 3$ 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 $\floor {\log_p N}$ times.

Hence:
 * $\displaystyle \sum_{1 \mathop \le n \mathop \le x} \map \Lambda n \equiv \sum_{p \mathop \le x} \floor {\log_p x} \ln p$