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:

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:

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:

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:


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