Equivalence of Definitions of Second Chebyshev Function

Theorem

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

Definition 1

$\ds \forall x \in \R: \map \psi x := \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p$

where, for each $k$, the summation extends over all powers of prime numbers $p$ such that $p^k \le x$.

Definition 2

$\ds \forall x \in \R: \map \psi x := \sum_{1 \mathop \le n \mathop \le x} \map \Lambda n$

where $\Lambda$ is the von Mangoldt function.

Definition 3

$\ds \forall x \in \R: \map \psi x := \sum_{p \mathop \le x} \floor {\log_p x} \ln p$

where:

the summation extends over all prime numbers $p$ such that $p \le x$

$\floor {\, \cdot \,}$ denotes the floor function.

Proof

Definition 1 equivalent to Definition 2

The equivalence:

$\ds \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p \equiv \sum_{1 \mathop \le n \mathop \le x} \map \Lambda n$

follows directly from the definition of the von Mangoldt function.

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$\Box$

Definition 1 equivalent to Definition 3

If $x < 2$, then:

$\ds \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p = \sum_{p \mathop \le x} \floor {\log_p x} \ln p = 0$

since the sums are empty.

So take $x \ge 2$.

Consider the sum:

$\ds \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p$

This sum runs over the pairs of natural numbers $\tuple {k, p}$ such that $p$ is a prime number with $p^k \le x$.

Equivalently, for each prime $p$, we sum over the natural numbers $k$ with $p^k \le x$.

Since $p^k \le x$ implies that $p \le x$, such $k$ only exists if $p \le x$.

Note that for a fixed prime number $p$, we have $p^k \le x$ if and only if:

$k \ln p \le \ln x$

by Logarithm of Power and Logarithm is Strictly Increasing.

Since $p \ge 2$, this is equivalent to:

$\ds k \le \frac {\ln x} {\ln p} = \log_p x$

So, we can equivalently sum $\ln p$ over the pairs of natural numbers $\tuple {p, k}$ such that $p \le x$ is a prime number and $k \le \log_p x$.

That is:

\(\ds \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p\)

\(=\)

\(\ds \sum_{p \mathop \le x} \sum_{k \mathop \le \log_p x} \ln p\)

\(\ds \)

\(=\)

\(\ds \sum_{p \mathop \le x} \ln p \paren {\sum_{k \mathop \le \log_p x} 1}\)

\(\ds \)

\(=\)

\(\ds \sum_{p \mathop \le x} \floor {\log_p x} \ln p\)

for $x \ge 2$.

Hence:

$\ds \forall x \in \R: \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p = \sum_{p \mathop \le x} \floor {\log_p x} \ln p$

$\blacksquare$