approximating a geometric sum of exponentials by an integral

Question

Let $e(x) = e^{2 \pi i x}$. Let $I = [a,b]$ be a closed interval of real numbers. I am interested in the sum $$ S = \sum_{n \in I \cap \mathbb{Z}} e(\alpha n) $$ for some real number $\alpha.$ Do there exists constants $c_1, c_2>0$ such that we can approximate $S$ by integral, $$ c_2 |\int_I e(\alpha x) dx| \leq |S| \leq c_1 |\int_I e(\alpha x) dx| ? $$ The constants are independent of $I$ and $\alpha$. Do such constants exist? Thank you.

Answer 1

There are no such constants independent of $I$ and $\alpha$. For every $\alpha \neq 0$ we have $$\int_I e(\alpha x)\,dx = 0$$ whenever the length of $I$ is an integer multiple of $\frac{1}{\lvert\alpha\rvert}$, and not for all such intervals does the sum vanish too: By moving the interval left or right one can pick up or drop one term of the sum [if the length of the interval is an integer, this depends on the interval not being half-open], and since $e(x)$ never vanishes, at most one of the two intervals can lead to a zero sum.

Thus for every fixed $c$ the inequality $$\lvert S\rvert \leqslant c \biggl\lvert \int_I e(\alpha x)\,dx \biggr\rvert$$ is violated for infinitely many $I$ and $\alpha$.

Conversely, if $\alpha$ is not an integer, then $\lvert S\rvert$ can be made arbitrarily small while keeping the modulus of the integral above a fixed positive threshold: For a given run of integers $k \leqslant n \leqslant m$, you can choose $a$ freely in $(k-1,k]$ (and/or $b$ freely in $[m,m+1)$) without changing the sum, while the value of the integral varies. With $$K := \sup_{y \in [0,1)}\; \biggl\lvert \int_0^y e(\alpha x)\,dx\biggr\rvert$$ one can always choose the interval such that $$\biggl\lvert \int_I e(\alpha x)\,dx \biggr\rvert \geqslant \frac{K}{2}\,.$$ Hence for every fixed $c > 0$ also the inequality $$\lvert S\rvert \geqslant c \biggl\lvert \int_I e(\alpha x)\,dx \biggr\rvert$$ is violated for infinitely many $I$ and $\alpha$.