Order of Second Chebyshev Function

Theorem

Let $x \ge 2$ be a real number.

Then:

$\ds \map \psi x = \sum_{p \mathop \le x} \ln p + \map \OO {\sqrt x \paren {\ln x}^2}$

where:

$\OO$ is big-O notation

$\psi$ is the Second Chebyshev Function

the sum runs over the primes less than or equal to $x$.

Proof

From the definition of the Second Chebyshev Function, we have:

$\ds \map \psi x = \sum_{k \mathop = 1}^\infty \paren {\sum_{p^k \mathop \le x} \ln p}$

where the inner sum runs over the primes $p$ with $p^k \le x$.

That is, the primes $p$ with $p \le x^{1/k}$, so we can write:

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

Note that there are no primes with $p < 2$, so if:

$x^{1/k} < 2$

we have:

$\ds \sum_{p^k \mathop \le x} \ln p = 0$

That is, if:

$\ds k > \frac {\ln x} {\ln 2}$

So, we have:

\(\ds \map \psi x\)

\(=\)

\(\ds \sum_{1 \mathop \le k \mathop \le \frac {\ln x} {\ln 2} } \paren {\sum_{p \mathop \le x^{1/k} } \ln p}\)

\(\ds \)

\(=\)

\(\ds \sum_{p \mathop \le x} \ln p + \sum_{2 \mathop \le k \mathop \le \frac {\ln x} {\ln 2} } \paren {\sum_{p \mathop \le x^{1/k} } \ln p}\)

From Logarithm is Strictly Increasing, for $2 \le p \le x^{1/k}$ we have:

$\ds 0 < \ln p \le \map \ln {x^{1/k} }$

So:

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

\(\le\)

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

\(\ds \)

\(\le\)

\(\ds x^{1/k} \map \ln {x^{1/k} }\)

\(\ds \)

\(=\)

\(\ds \frac 1 k x^{1/k} \ln x\)

Logarithm of Power

Then, for $k \ge 2$, we have:

$\ds \frac 1 k x^{1/k} \ln x \le \frac 1 2 \sqrt x \ln x$

so:

\(\ds \sum_{2 \mathop \le k \mathop \le \frac {\ln x} {\ln 2} } \paren {\sum_{p \mathop \le x^{1/k} } \ln p}\)

\(\le\)

\(\ds \sum_{2 \mathop \le k \mathop \le \frac {\ln x} {\ln 2} } \paren {\frac 1 2 \sqrt x \ln x}\)

\(\ds \)

\(\le\)

\(\ds \frac 1 {2 \ln 2} \sqrt x \paren {\ln x}^2\)

From the definition of big-O notation, we have:

$\ds \sum_{2 \mathop \le k \mathop \le \frac {\ln x} {\ln 2} } \paren {\sum_{p \mathop \le x^{1/k} } \ln p} = \map \OO {\sqrt x \paren {\ln x}^2}$

so:

$\ds \map \psi x = \sum_{p \mathop \le x} \ln p + \map \OO {\sqrt x \paren {\ln x}^2}$

$\blacksquare$