I am trying to show $\{\sqrt{p}: p \text{ primes }\}$ is equidistributed modulo 1. Using Weyl's criterion, it is sufficient to show for each nonzero integer $k$,
\begin{equation} \sum_{n \leq x}e(k\sqrt{n}) \Lambda(n) =o(x), \text{ as } x \to \infty. \end{equation}
I found an exercise on chapter 13 "sum over primes" in Iwaniec H., Kowalski E. - Analytic number theory regarding this: it's the exercise 2 on page 348: for any real number $\alpha \neq 0$, $$ \sum_{n\le x} e(\alpha \sqrt{n}) \Lambda(n) \ll_\alpha x^{\frac{5}{6}} (\log x)^4. $$ The idea is to use the following Vaughan's identity and estimate each sum separately: Let $y,z \geq 1$, then $$ \Lambda(n)=\sum_{\substack{b \mid n\\ b \leq y}} \mu(b) \log \frac{n}{b}-\sum_{\substack{bc \mid n\\ b \leq y, c \leq z}} \mu(b) \Lambda(c)+\sum_{\substack{bc \mid n\\ b > y, c >z}} \mu(b) \Lambda(c)+R(n), $$ where $R(n)=0$ when $n>z$, and $R(n)=\Lambda(n)$ when $n \leq z$.
I came up with the following lemmas as some basic preparation:
- For any real number $\alpha> 0$, $$ \sum_{n\le x} e(\alpha \sqrt{n}) \ll x^{\frac{1}{2}} \alpha^{-1}+\alpha^2, $$ where the implicit constant does not depend on $\alpha$. (Using Van der Corput Method)
- For any real number $\alpha > 0$, $N \leq x$, and any complex numbers $\beta_m,\gamma_n$ with $|\beta_m| \leq 1$ and $|\gamma_n| \leq 1$, we have $$ A(M,N)=\sum_{M<m \leq 2M} \sum_{N<n \leq 2N} \beta_m \gamma_n e(\alpha \sqrt{mn}) \ll MN^{1/2} + M^{1/4} N^{3/4}\alpha^{-1/2}(\log N)^{1/2}+N^{5/4} \alpha. $$
- For any real number $\alpha > 0$, $N \leq x$, and any complex numbers $\beta_m$ with $|\beta_m| \leq 1$, we have $$ B(x; N)=\sum_{N<n \leq 2N} |\sum_{mn \leq x} \beta_m e(\alpha \sqrt{mn})| \ll N^{1/2} x^{1/2}+x^{3/4} \alpha^{-1/2} (\log x)^{1/2}+x^{5/4} N^{-3/4} \alpha. $$
- (Use lemma 2 and lemma3, Estimate for bilinear form) For any real number $\alpha \neq 0$, and any complex numbers $\beta_m, \gamma_n$ with $|\beta_m| \leq 1, |\gamma_n| \leq 1$, $M,N<\sqrt{x}$, we have $$ \sum_{\substack{mn \leq x\\ m>M,n>N}} \beta_m \gamma_n e(\alpha \sqrt{mn}) \ll x^{3/4}+x^{3/4} \alpha^{-1/2} (\log x)^{3/2}+x^{5/4} (N^{-3/4}+M^{-3/4})\alpha +x^{5/8}\alpha \log x +MN. $$
By Vaughan's idenity, we have \begin{align*} \sum_{n\le x} e(\alpha \sqrt{n}) \Lambda(n) &= \sum_{\substack{n\le x\\ b \mid n\\ b \leq y}} \mu(b) e(\alpha \sqrt{n}) \log \frac{n}{b}-\sum_{\substack{n\le x \\ bc \mid n\\ b \leq y, c \leq z}} \mu(b) \Lambda(c)e(\alpha \sqrt{n})\\ & \quad +\sum_{\substack{n\le x \\ bc \mid n\\ b > y, c >z}} \mu(b) \Lambda(c)e(\alpha \sqrt{n})+O(\sum_{n \leq z} \Lambda(n)). \end{align*}
For the first sum, using lemma1 and partial summation, we get \begin{align*} \sum_{\substack{n\le x\\ b \mid n\\ b \leq y}} \mu(b) e(\alpha \sqrt{n}) \log \frac{n}{b} &=\sum_{\substack{bl\le x\\ b \leq y}} \mu(b) e(\alpha \sqrt{bl}) \log l \\ &\ll \sum_{b \leq y} |\sum_{l \leq x/b} e(\alpha \sqrt{bl}) \log l |\\ &\ll_{\alpha} x^{1/2} \log x \log y+y^2 \log x. \end{align*}
For the second sum, using lemma1 and partial summation, we get \begin{align*} \sum_{\substack{n\le x \\ bc \mid n\\ b \leq y, c \leq z}} \mu(b) \Lambda(c)e(\alpha \sqrt{n}) &=\sum_{\substack{bcl\le x \\ b \leq y, c \leq z}} \mu(b) \Lambda(c)e(\alpha \sqrt{bcl}) \\ &\ll \sum_{b \leq y, c \leq z} |\mu(b) \Lambda(c)| |\sum_{bcl \leq x} e(\alpha \sqrt{bcl})|\\ &\ll \sum_{b \leq y, c \leq z} |\sum_{bcl \leq x} e(\alpha \sqrt{bcl})| \Lambda(c)\\ &\ll_{\alpha} \sum_{b \leq y, c \leq z} (x^{1/2} (bc)^{-1}+bc) \Lambda(c)\\ &\ll x^{1/2} \log y\log z+y^2z^2. \end{align*}
For the third sum, using lemma 4, it is easy to show that \begin{align*} \sum_{\substack{n\le x \\ bc \mid n\\ b > y, c >z}} \mu(b) \Lambda(c)e(\alpha \sqrt{n}) &=\log x \sum_{\substack{bk\le x \\ b > y, k >z}} \mu(b) \gamma_k e(\alpha \sqrt{bk}). \ll_{\alpha} x^{3/4} (\log x)^{5/2}+x^{5/4} (y^{-3/4}+z^{-3/4})\log x+yz\log x. \end{align*}
However, it seems some of my estimates could be improved, because based on the above estimate, there is no way to choose y,z such that $$ \sum_{n\le x} e(\alpha \sqrt{n}) \Lambda(n) =o(x). $$ And the main issue here is that the $y^2z^2$ term from the second estimate seems to be too large, and this comes from the $\alpha^2$ term from lemma 1( the estimate for standard exponential sum). I had some ideas that could improve the $\alpha^2$ term to $\alpha^{3/2}$, but I guess $\alpha^{3/2}$ is still not good enough.
Any suggestions would be appreciated. Thank you very much！
Updated 1: I guess I figured out how to do this, for lemma 2 and lemma 3, we can take advantage that $n>N$ and use an improved version of lemma1.
Updated 2: Could someone point out what is the best known upper bound for $$ \sum_{n\le x} e(\alpha \sqrt{n}) \Lambda(n) $$
