TITLE: Asymptotic Formula for a Mertens Style Sum
QUESTION [7 upvotes]: Hello,
I am wondering if there is a simple asymptotic formula for
$$\sum_{p\leq x}\frac{\left(\log p\right)^{k}}{p},$$
where $k\geq0$ is some integer. If $k$ is $0,$ by using the Prime Number Theorem we have 
$$\sum_{p\leq x}\frac{1}{p}=\log \log x+b+O\left(e^{-c\sqrt{\log x}}\right).$$
Similarly, the prime number theorem and integration by parts solves the case $k=1$ and gives 
$$\sum_{p\leq x}\frac{\left(\log p\right)}{p}=\log x+C+O\left(e^{-c\sqrt{\log x}}\right).$$ 
My question is do these integrals have a nice asymptotic formula for every $k$? Specifically, I mean with an error term of the form $O\left(e^{-c\sqrt{\log x}}\right).$
Thanks!
Remark:  This question is related, and in particular if it is solved with a nice enough asymptotic, then so is this.  (but not vice versa)

REPLY [6 votes]: Here is an answer that is similar in spirit to Frank and Peter's answers, but possibly simpler.
Summing by parts, we see that
$$ \sum_{p\le x} \frac{\log^k p}{p} = (\log x)^{k-1} \sum_{p\leq x} \frac{\log p}{p} -(k-1)\int_{2^-}^x (\log u)^{k-2}\sum_{p\le u} \frac{\log p}{p} \frac{du}{u}.$$
Now use the formula
$$ \sum_{p\leq x} \frac{\log p}{p} = \log x + c_1 + O(\exp(-c_2\sqrt{\log x}))$$
and it is not hard to derive that
$$ \sum_{p\le x} \frac{\log^k p}{p} = \frac{\log^k x}{k} + c_3 + O(\exp(-c_4\sqrt{\log x})).$$
This seems easier than dealing with $Li(x)$.