I understand that for $\nu(n)$, the number of divisors of $n$, let me denote $\log_2 x=\log\log x$. Then $$\sum_{n\le x}\Big(\nu(n)-\log_2x\Big)^2=O(x\log_2x)$$
The book told me about the theorem (due to Erdos and Kac), for $S(x;\alpha,\beta)=\#\left\{ n\le x : \alpha\le\dfrac{\nu(n)-\log_2n}{\sqrt{\log_2(n)}}\le\beta\right\}:=\#S$, then $$\lim_{x\rightarrow\infty}\dfrac{S(x;\alpha,\beta)}{x}=\dfrac{1}{\sqrt{2\pi}}\int_{\alpha}^\beta e^{-t^2/2} dt$$
I want to prove that $\displaystyle\sum_{n\le x}\Big(\nu(n)-\log_2n\Big)^2\ll x\log_2 x$ by using the theorem stated above.
What I am confused is that if $n\in S$ then $\displaystyle \delta(n):=\left|\dfrac{\nu(n)-\log_2 n}{\sqrt{\log_2 (n)}}\right|\le \max(|\alpha|,|\beta|)$ and if we choose $|\alpha|,|\beta|\le \sqrt x$ then the statement I want follows for $n\in S$, but what about $n\not\in S, n\le x$? I confuse about the phrase "normally distributed". What does it mean exactly in this sense?
As I guess, if we choose $\alpha,\beta$ that $\beta-\alpha$ is wide enough then $\delta(n)$ that satisfies the condition in $S$ would be wide enough, too. Therefore, the number of terms left from the distribution (the term on the left and right part of normal distribution) would be small.
How can I add the left terms that I mentioned to cover all $n\le x$?
I proved it by using the method that resembled Turan that \begin{equation} \begin{split} \sum_{1<n\le x}\nu(n) & =x\log_2x+O(x) \\ \sum_{n\le x} \nu(n)^2&= x(\log_2x)^2+O(x\log_2x) \end{split} \end{equation} and by partial summation \begin{equation} \begin{split} \sum_{1<n\le x}(\log_2 n)^2 & =\sum_{1<n\le x}1\cdot(\log_2n)^2 = (\left[\,x\right]-1)\log^2_2x-\int_2^x (\left[\,t\right]-1)\cdot 2\log_2 t\cdot\left(\dfrac{1}{t\log t}\right) dt\\ &= x(\log_2 x)^2+O(\log^2_2 x)+O(x\log_2 x)= x(\log_2 x)^2+O(x\log_2 x)\\ \sum_{1<n\le x} \nu(n)\log_2 n & = \sum_{n\le x} \nu(n)\cdot \log_2 n=\Big(x\log_2 x+O(x)\Big)(\log_2 x)-\int_2^x (t\log_2 t+O(t))\left(\dfrac{1}{t\log t}\right) dt \\ & = x(\log_2 x)^2+O(x\log_2 x) \end{split} \end{equation}
Hence, we obtain \begin{equation} \begin{split} \sum_{1<n\le x}(\nu(n)-\log_2 n)^2 & = \sum \nu(n)^2-2\sum\nu(n)\log_2n+\sum(\log_2 n)^2 = O(x\log_2 x) \end{split} \end{equation} as desired. Therefore, almost all numbers $n$ have $\log_2 n$ numbers of prime factors as Hardy and Ramanujan stated (I find their theorem indeed elegant).