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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.

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

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

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

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