Is the following estimate correct?
Let $\tau (n)$ be the function that counts how many divisors of n are there. Then:
$$ \sum\limits_{n\leq x} \log(\tau(n))=\log 2 \log\log x + O(1) $$
I've been trying to prove it is, but I can't. If it is not correct, how could I arrive at an estimate similar to this one?
On
Using the bounds $$2^{\omega\left(n\right)}\leq\tau\left(n\right)\leq2^{\Omega\left(n\right)} $$ where $\omega\left(n\right) $ is the number of distinct prime factors of $n $ and $\Omega\left(n\right) $ is the number of the total prime factors of $n $, we get $$\log\left(2\right)\sum_{n\leq x}\omega\left(n\right)\leq\sum_{n\leq x}\log\left(\tau\left(n\right)\right)\leq\log\left(2\right)\sum_{n\leq x}\Omega\left(n\right) $$ and the asymptotic of these two function is well known $$\sum_{n\leq x}\Omega\left(n\right)=x\left(\log\left(\log\left(x\right)\right)+O\left(1\right)\right) $$ $$\sum_{n\leq x}\omega\left(n\right)=x\left(\log\left(\log\left(x\right)\right)+O\left(1\right)\right) $$ (see here and here) hence $$\sum_{n\leq x}\log\left(\tau\left(n\right)\right)=x\log\left(2\right)\left(\log\left(\log\left(x\right)\right)+O\left(1\right)\right). $$
One could try using the identity $$\sum_{n\leq x} \log \tau(n)=\sum_{p\leq x} a_p(x) \log p,$$ where $$a_p(x):= \sum_{1<n\leq x}\left(1+\nu_p(n)\right)=\sum_{k\geq 0}\Big[\frac{x}{p^k}\Big],$$ where $[y]$ is the integer part of a real number $y$ and $\nu_p(n)$ is the usual $p$-adic valuation of $n$, i.e. the largest integer $\nu$ satisfying $p^\nu|n$.