Suppose we have $\alpha \in \mathbb{R}$. Then we know that

$$\sum_{1 \leq n \leq X} e(n \alpha) \ll \min \{ X, \|\alpha\|^{-1} \}$$

where $\| \cdot \|$ is the distance to the nearest integer. I was wondering if we put von Mangoldt function as a weight and consider the exponential sum

$$S(\alpha) = \sum_{1 \leq n \leq X} \Lambda(n) e(n \alpha)$$

can we obtain an estimate similar to above in terms of $\| \alpha \|$ and $X$?

Thank you very much!

PS Here $\Lambda$ is the von Mangoldt function, $e(m) = e^{2 \pi i m}$.


1 Answer 1


It's a classical sum going back to Vinogradov, but the estimate is more complicated. Usually, it is phrased in terms of an approximation $a/q$ of $\alpha$, with $a$ coprime to $q$, such that $|\alpha-a/q|\leq 1/q^2$ as $$S(\alpha)\ll ((Xq)^{1/2}+Xq^{-1/2}+X^{4/5})$$ (up to powers of $\log X$). This is explained in many places, e.g. in Vaughan's "Hardy--Littlewood method" if I remember right.

  • 1
    $\begingroup$ Yes. And the proof relies on "Vaughan's identity". $\endgroup$
    – GH from MO
    Jan 19, 2017 at 0:45

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