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You seem to be over complicating things. For $\alpha$ fixed, this is relatively easy to prove. (But note, not true if $\alpha$ varies with $x$.) I leave it to you deduce the desired result as a consequence of the following theorem:

Theorem: Let $x>0$ and fix $0<\alpha<1$. Define $\omega_{x,\alpha}(n)=\sum_{p|n,\ p>x^\alpha} 1$. Then this function has average value $$\frac{1}{x}\sum_{n\leq x} \omega_{x,\alpha}(n)=-\log \alpha +O\left(\frac{1}{\log x}\right).$$

Proof: Notice that

$$\frac{1}{x}\sum_{n\leq x} \omega_{x,\alpha}(n)=\frac{1}{x}\sum_{x^{\alpha}<p<\leq x}\left[\frac{x}{p}\right] =\sum_{x^{\alpha}<p<\leq x}\frac{1}{p}-\frac{1}{x}\sum_{x^{\alpha}<p<\leq x}\left\{\frac{x}{p}\right\}$$

$$= \sum_{x^\alpha<p<x} \frac{1}{p} +O\left(\frac{1}{\log x}\right)=\log \log (x)-\log \log x^\alpha +O\left(\frac{1}{\log x}\right)$$ $$=-\log \alpha +O\left(\frac{1}{\log x}\right).$$

1

For $\alpha$ fixed, this is relatively easy to prove. (But note, not true if $\alpha$ varies with $x$.) I leave it to you deduce the desired result as a consequence of the following theorem:

Theorem: Let $x>0$ and fix $0<\alpha<1$. Define $\omega_{x,\alpha}(n)=\sum_{p|n,\ p>x^\alpha} 1$. Then this function has average value $$\frac{1}{x}\sum_{n\leq x} \omega_{x,\alpha}(n)=-\log \alpha +O\left(\frac{1}{\log x}\right).$$

Proof: Notice that

$$\frac{1}{x}\sum_{n\leq x} \omega_{x,\alpha}(n)=\frac{1}{x}\sum_{x^{\alpha}<p<\leq x}\left[\frac{x}{p}\right] =\sum_{x^{\alpha}<p<\leq x}\frac{1}{p}-\frac{1}{x}\sum_{x^{\alpha}<p<\leq x}\left\{\frac{x}{p}\right\}$$

$$= \sum_{x^\alpha<p<x} \frac{1}{p} +O\left(\frac{1}{\log x}\right)=\log \log (x)-\log \log x^\alpha +O\left(\frac{1}{\log x}\right)$$ $$=-\log \alpha +O\left(\frac{1}{\log x}\right).$$