# 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.

• did you try using Abels Partial Summation formula and using the fact that $\displaystyle \sum_{p \leq x} \frac{\log p}{p}=\log x$ – happymath Jan 27 '15 at 4:28
• @happymath, the equality in your comment is not really an equality. – KCd Jan 27 '15 at 4:47
• @KCd sorry I just mentioned the dominant term – happymath Jan 27 '15 at 4:48
• $p_n\approx n\ln n$. – Lucian Jan 27 '15 at 6:34

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

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

• Isn't the error term too big? – happymath Jan 27 '15 at 4:42
• @happymath that's a little o. – Adam Hughes Jan 27 '15 at 4:42
• sorry my bad I am used to big O so i assumed it was a typo – happymath Jan 27 '15 at 4:46