Asymptotic formula for sums related to primes Suppose $0 < \alpha < 1$. What is the asymptotic formula for the sum
$$\displaystyle \sum_{p \leq x} \frac{\log p}{p^\alpha}?$$
Thanks for any insights.

We write this as
$$\sum_{p\le x}{\log n\over n^\alpha}\cdot 1_p$$
From here we use partial summation to get
$$\pi(x){\log x\over x^\alpha}-\int_1^x\pi(t){1-\alpha \log t\over t^{1+\alpha}}\,dt$$
using the PNT and monotonicity of the integral, this is asymptotic to
$$x^{1-\alpha}-\int_1^x {1-\alpha\log t\over t^\alpha\log t}\,dt$$
This gives
$$x^{1-\alpha}-\int_1^x{dt\over t^\alpha\log t}+\alpha\int_1^xt^{-\alpha}\,dt$$
which is easily computed to give

$$\left(1+{\alpha\over 1-\alpha}\right)x^{1-\alpha}+o(x^{1-\alpha})$$

