Let $\varphi(x) := \sum_{n\leq x} \log(n) $.
Show that
\begin{align}
\varphi(x) = x \cdot \log(x) + O(x) \quad \forall\,\, x > 1
\end{align}
I think I must show both inequalities, and if it’s the case, the first one $\leq$ holds.
But I have no idea how to do the second one, or maybe there’s another approach to solve it.
Any suggestions? Thanks in advance
Best Answer
A simple proof can be obtained using the Abel's summation formula. We have $$\sum_{n\leq x}\log\left(n\right)=x\log\left(x\right)-\int_{1}^{x}\frac{\left\lfloor t\right\rfloor }{t}dt$$ where $\left\lfloor t\right\rfloor $ is the floor function. Using the bound $\left\lfloor t\right\rfloor =O\left(t\right)$ the result follows. If you use $\left\lfloor t\right\rfloor =t+O\left(1\right)$ you can get an extra term in the formula.