Arithmetic Average of Second Chebyshev Function

== Theorem ==
Let $x \ge 1$ be a real number.
Then:
:$\ds \sum_{n \mathop \le x} \map \psi {x/n} = x \ln x - x + \map \OO {\map \ln {x + 1} }$
where:
:the summation extends over all strictly positive integers $n$ such that $n \le x$
:$\OO$ is big-O notation
:$\psi$ is the second Chebyshev function.
== Proof ==
We have, by the definition of the second Chebyshev function:
:$\ds \sum_{n \mathop \le x} \map \psi {x/n} = \sum_{n \mathop \le x} \sum_{m \mathop \le x/n} \map \Lambda m$
where $\Lambda$ is the Von Mangoldt function. 
Consider the sum: 
:$\ds \sum_{n \mathop \le x} \sum_{m \mathop \le x/n} \map \Lambda m$
The sum runs over the pairs of natural numbers $\tuple {n, m}$ such that $n m \le x$.
This is equivalent to running over the divisors of each natural number $t$ such that $t \le x$. 
That is: 
:$\ds \sum_{n \mathop \le x} \sum_{m \mathop \le x/n} \map \Lambda m = \sum_{t \mathop \le x} \sum_{m \mathop \divides t} \map \Lambda m$
Then, from Sum Over Divisors of von Mangoldt is Logarithm, we have: 
:$\ds \sum_{t \mathop \le x} \sum_{m \mathop \divides t} \map \Lambda m = \sum_{t \mathop \le x} \ln t$
With a view to bound this sum, note that: 
:$\ds \sum_{t \mathop \le x} \ln t = \ln 1 + \sum_{2 \mathop \le t \mathop \le x} \ln t = \sum_{2 \mathop \le t \mathop \le x} \ln t$
We have, from Sum of Integrals on Adjacent Intervals for Integrable Functions: 
:$\ds \sum_{2 \mathop \le t \mathop \le x} \ln t = \sum_{2 \mathop \le t \mathop \le x} \paren {\int_{t - 1}^t \ln t \rd u}$
From Logarithm is Strictly Increasing, we have: 
:$\ln u \le \ln t \le \map \ln {u + 1}$
for $t - 1 \le u \le t$.
So, from Relative Sizes of Definite Integrals:
:$\ds \sum_{2 \mathop \le t \mathop \le x} \paren {\int_{t - 1}^t \ln u \rd u} \le \sum_{2 \mathop \le t \mathop \le x} \paren {\int_{t - 1}^t \ln t \rd u} \le \sum_{2 \mathop \le t \mathop \le x} \paren {\int_{t - 1}^t \map \ln {u + 1} \rd u}$
Now, from Sum of Integrals on Adjacent Intervals for Integrable Functions, we have: 
:$\ds \sum_{2 \mathop \le t \mathop \le x} \paren {\int_{t - 1}^t \map \ln {u + 1} \rd u} = \int_1^{\floor x} \map \ln {u + 1} \rd u$
and:
:$\ds \sum_{2 \mathop \le t \mathop \le x} \paren {\int_{t - 1}^t \ln u \rd u} = \int_1^{\floor x} \ln u \rd u$
We can compute: 
and:
:$\ds \int_1^{\floor x} \ln u \rd u = \floor x \ln \floor x - \floor x$
Since $x - 1 < \floor x \le x$, and $1 - 2 \ln 2 < 0$, we have: 
:$\paren {\floor x + 1} \map \ln {\floor x + 1} - \floor x - 2 \ln 2 + 1 \le \paren {x + 1} \map \ln {x + 1} - x$
and:
:$\paren {x - 1} \map \ln {x - 1} - \paren {x - 1} \le \floor x \ln \floor x - \floor x$
from Logarithm is Strictly Increasing.
We have therefore obtained that: 
:$\ds \paren {x - 1} \map \ln {x - 1} - \paren {x - 1} \le \sum_{n \mathop \le x} \map \psi {x/n} \le \paren {x + 1} \map \ln {x + 1} - x$
We will use this inequality to show that: 
:$\ds \sum_{n \mathop \le x} \map \psi {x/n} - \paren {x \ln x - x} = \map \OO {\map \ln {x + 1} }$
We have: 
:$\ds \paren {x - 1} \map \ln {x - 1} - x \ln x \le \sum_{n \mathop \le x} \map \psi {x/n} - \paren {x \ln x - x} \le \paren {x + 1} \map \ln {x + 1} - x \ln x$
We will show that for sufficiently large $x$ we have: 
:$\paren {x + 1} \map \ln {x + 1} - x \ln x \le 2 \map \ln {x + 1}$
and:
:$-2 \map \ln {x + 1} \le \paren {x - 1} \map \ln {x - 1} - x \ln x$
at which point we have the claim.
=== Lemma 1 ===
=== Lemma 2 ===
Putting these two inequalities together, for $x \ge 3$ we have: 
:$\ds -2 \map \ln {x + 1} \le \sum_{n \mathop \le x} \map \psi {x/n} - \paren {x \ln x - x} \le 2 \map \ln {x + 1}$
so, by the definition of big-O notation:
:$\ds \sum_{n \mathop \le x} \map \psi {x/n} - \paren {x \ln x - x} = \map \OO {\map \ln {x + 1} }$
so:
:$\ds \sum_{n \mathop \le x} \map \psi {x/n} = x \ln x - x + \map \OO {\map \ln {x + 1} }$
Category:Second Chebyshev Function
Category:Arithmetic Average of Second Chebyshev Function