TITLE: Reference for the expected number of prime factors of n larger than n^alpha is -log alpha
QUESTION [6 upvotes]: Let $0 < \alpha < 1$ be a constant.  The expected number of prime factors of a "random" integer near $n$ which are greater than $n^\alpha$ is $-\log \alpha$.
It's my understanding that (properly formulated) this is a well-known fact in analytic number theory but I cannot find a reference for it.  Can anybody provide a reference?
Edited to add (March 28):: The asymptotic density of positive integers $n$ with $k$th largest factor smaller than $n^{1/\alpha}$ is $\rho_k(\alpha)$, where we have $L_0(\alpha) = [\alpha > 0]$ and 
$$ L_k(\alpha) = [\alpha \ge k] \int_k^\alpha L_{k-1}(t-1) \: {dt \over t}, $$
and $1-\rho_k(\alpha) = \sum_{n=0}^\infty {-k \choose n} L_{n+k}(\alpha)$.  (See Riesel, p. 162.)  The density of positive integers with $k$th largest factor larger than $n^{1/\alpha}$ is therefore $1-\rho_k(\alpha)$, and so the expected number of factors larger than $n^{1/\alpha}$ is $\sum_{k \ge 1} (1-\rho_k(\alpha))$.  Therefore the expected number of such factors is
$$ \sum_{k \ge 1} \sum_{n \ge 0} {-k \choose n} L_{n+k}(\alpha). $$
Letting $n+k = j$ we can rewrite this sum as
$$ \sum_{j \ge 1} \sum_{n=0}^{j-1} {n-j \choose n} L_j = \sum_{j \ge 1} L_j \left( \sum_{n=-0}^{j-1} (-1)^n {j-1 \choose n} \right) $$
and the inner sum is $0$ except when $j=1$, when it is $1$.  So the expected number of factors larger than $n^{1/\alpha}$ is $L_1(\alpha)$; this is $\log \alpha$.

REPLY [6 votes]: You seem to be over complicating things.  For $\alpha$ fixed, this is relatively easy to prove.  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 
$$\frac{1}{x}\sum_{n\leq x} \omega_{x,\alpha}(n)=\frac{1}{x}\sum_{x^{\alpha}<p<\leq x}\left[\frac{x}{p}\right] =\sum_{x^{\alpha}<p<\leq x}\frac{1}{p}-\frac{1}{x}\sum_{x^{\alpha}<p<\leq x}\left\{\frac{x}{p}\right\}$$  
$$= \sum_{x^\alpha<p<x} \frac{1}{p} +O\left(\frac{1}{\log x}\right)=\log \log (x)-\log \log x^\alpha +O\left(\frac{1}{\log x}\right)$$ 
$$=-\log \alpha +O\left(\frac{1}{\log x}\right).$$