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)
Best Answer
Expanding on Frank's answer: by partial summation, we have that $$\sum_{p \leq x} \frac{(\log p)^k}{p} = \frac{(\log x)^k}{x} \pi(x) - \int_{2}^{x} \pi(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt.$$ Using the fact that $\pi(x) = \mathrm{Li}(x) + E(x)$, where $E(x) = O(e^{-c\sqrt{\log x}})$, we have that the first term is equal to $$\frac{(\log x)^k}{x} \mathrm{Li}(x) + O((\log x)^k e^{-c\sqrt{\log x}}) = \frac{(\log x)^k}{x} \mathrm{Li}(x) + O(e^{-c_k \sqrt{\log x}}).$$ For the second term, $$- \int_{2}^{x} \pi(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt = R_1(x) + R_2 + R_3(x),$$ where $$R_1(x) = - \int_{2}^{x} \mathrm{Li}(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt,$$ $$R_2 = - \int_{2}^{\infty} E(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt,$$ $$R_3(x) = \int_{x}^{\infty} E(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt.$$ Let's look at $R_1(x)$ first. Using the fact that $\mathrm{Li}(t) = \int_{2}^{s} \frac{ds}{\log s}$ and interchanging the order of integration, then evaluating the integral with respect to $t$, we obtain $$R_1(x) = - \frac{(\log x)^k}{x} \mathrm{Li}(x) + \frac{(\log x)^k}{k} - \frac{(\log 2)^k}{k}.$$ For $R_3(x)$, a simple calculation shows that $$R_3(x) \ll - \int_{x}^{\infty}{(\log t)^{k-1} e^{-c\sqrt{\log t}} \left(\frac{kt - \log t}{t^2}\right) dt} = - \int_{\log x}^{\infty}{u^{k-1} e^{-c\sqrt{u}} \left(\frac{ke^{u} - u}{e^u}\right) du}.$$ We can rewrite this integral as $$- \int_{\log x}^{\infty}{\left(ku^{k-1} e^{-c\sqrt{u}} - \frac{c}{2} u^{k - 1/2} e^{-c\sqrt{u}}\right) + \left(\frac{c}{2} u^{k - 1/2} e^{-c\sqrt{u}} - u^k e^{-c\sqrt{u} - u}\right) du}$$ and from this it is clear that $$R_3(x) \ll_k - \int_{\log x}^{\infty}{\left(ku^{k-1} e^{-c\sqrt{u}} - \frac{c}{2} u^{k - 1/2} e^{-c\sqrt{u}}\right) du} = (\log x)^k e^{-c \sqrt{\log x}} \ll e^{-c_k \sqrt{\log x}}.$$ Finally, it's clear from the estimate for $R_3(x)$ that the integral defining $R_2$ converges, and so we obtain $$\sum_{p \leq x} \frac{(\log p)^k}{p} = \frac{(\log x)^k}{k} - \frac{(\log 2)^k}{k} + R_2 + O_k(e^{-c\sqrt{\log x}}).$$
Note that it's possible to make this valid for much more than just nonnegative integers $k$ - you could also have $k$ complex.