Let $v(\beta) := \sum_{n\le X} e(n\beta)$ where $e(\alpha) := e^{2\pi i \alpha}$. It is not hard to show that $$\log X\ll \int_0^1 |v(\beta)| d\beta\ll \log X$$ and by considering the underlying Diophantine equation, we have that for $k\ge 0, $ $$\int_0^1 |v(\beta)|^{2k}d\beta\sim C_kX^{2k - 1}.$$

For general $p\ge 0$, is the order of magnitude of $\int_0^1 |v(\beta)|^pd\beta$ known?

Edit: corrected the actual exponent for the $2k$th moment.


1 Answer 1


Yes. Without loss of generality, $X\geq 2$ is an integer. Then we have explicitly $$ v(\beta)=e\left(\frac{(X+1)\beta}{2}\right)\frac{\sin(\pi X\beta)}{\sin(\pi\beta)},\qquad\beta\not\in\mathbb{Z}.$$ It follows for any $p>0$ that $$\int_{|\beta|\leq\frac{1}{4X}}|v(\beta)|^p\,d\beta\asymp_p\int_{|\beta|\leq \frac{1}{4X}}X^p\,d\beta\asymp_p X^{p-1}.$$ Now we estimate the rest of the integral carefully. On the one hand, $$\int_{\frac{1}{4X}<|\beta|\leq\frac{1}{2}}|v(\beta)|^p\,d\beta\ll_p\int_{\frac{1}{4X}<|\beta|\leq \frac{1}{2}}|\beta|^{-p}\,d\beta\asymp_p \begin{cases}X^{p-1},&p>1;\\\log X,&p=1;\\1,&p<1.\end{cases}$$ On the other hand, \begin{align*}\int_{\frac{1}{4X}<|\beta|\leq\frac{1}{2}}|v(\beta)|^p\,d\beta&\gg_p\sum_{1\leq m\leq\frac{X}{2}}\int_{\frac{1}{4}<m-|X\beta|<\frac{3}{4}}\frac{d\beta}{|\beta|^{p}}\\ &\asymp_p\sum_{1\leq m\leq\frac{X}{2}}\frac{1}{X}\left(\frac{X}{m}\right)^p\asymp_p\begin{cases}X^{p-1},&p>1;\\\log X,&p=1;\\1,&p<1.\end{cases}\end{align*} Putting together these three bounds, $$\int_{|\beta|\leq 1/2}|v(\beta)|^p\,d\beta\asymp_p \begin{cases}X^{p-1},&p>1;\\\log X,&p=1;\\1,&p<1.\end{cases}$$ This also means that in your second display the correct exponent of $X$ is $2k-1$, not $k$.

  • $\begingroup$ Is it possible to get asymptotics and not just estimates within a constant factor for general exponents $p$? $\endgroup$ Commented Mar 20, 2018 at 19:55

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