Equivalence of Definitions of Second Chebyshev Function

Theorem
These definitions of the second Chebyshev function are equivalent:


 * $$\psi \left({x}\right) = \sum_{p^k \le x} \ln p$$


 * $$\psi \left({x}\right) = \sum_{1 \le n \le x} \Lambda \left({n}\right)$$


 * $$\psi \left({x}\right) = \sum_{p \le x} \left \lfloor {\log_p x} \right \rfloor \ln p$$

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 $$\sum_{p^k \le x} \ln p \equiv \sum_{1 \le n \le x} \Lambda \left({n}\right)$$ follows directly from the definition of the von Mangoldt function.


 * 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:

$$ $$

Notice this sum will have:
 * as many $$\ln(2) \ $$ terms as there are powers of $$2 \ $$ less than or equal to $$N \ $$,
 * as many $$\log(3) \ $$ 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 $$\left \lfloor {\log_p N} \right \rfloor \ $$ times.

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