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Sorry if this is a duplicate, but can't find it anywhere. I'm learning some additive number theory and am stuck on showing some bounds. Let $S(\alpha, X) = \displaystyle \sum_{x=1}^{X} e(\alpha x)$ (where $e(\beta)$ just means $e^{2 \pi i \beta}$). First, I need that $|S(\alpha, X)| << (2|| \alpha||)^{-1}$ where $|| \alpha|| = |\alpha \text{ mod } 1|$.

Here is what I've gathered: $S(\alpha, X) = \frac{e(\alpha)-e(\alpha(X+1))}{1-e(\alpha)}$ by the geometric series formula. Then, the magnitude of this is $$|S(\alpha, X)|^2 = \frac{2-e(\alpha X)-e(-\alpha X)}{2-e(\alpha)-e(-\alpha)} = \frac{2-2\cos(2 \pi \alpha X)}{2-2 \cos (2 \pi \alpha)}$$ I'm not sure where to go from here. I thought I remembered at one point getting an expression with a single $\sin (2\pi \alpha)$, which is easy to compare to $||\alpha||^{-1}$.

The other bound I'm stuck with is $$\int_{0}^{1} |S(\alpha, X)| \text{ d} \alpha << \log (2X)$$ Again, I'm not really sure how to proceed except with fiddling with the cosine expression I got above, but I have no idea how to relate that to something like $\log (2X)$.

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For the first bound note that $2-2 \cos (2 \pi \alpha)=4\sin^2{(\pi \alpha)}$ and $4\sin^2{(\pi \alpha)} \ge ||\alpha||^2$, while the numerator is bounded by $4$

For the second bound, we use the first bound and notice that if $\frac{1}{X} \le \alpha \le 1-\frac{1}{X}$ we can use the first bound in the integral but for $\alpha$ small or near $1$ it is better to use the trivial bound $X$, so the integral is majorized by:

$\int_0^{\frac{1}{X}}Xdu+\int_{\frac{1}{X}}^{\frac{1}{2}}\frac{4}{u}du+\int_{\frac{1}{2}}^{1-\frac{1}{X}}\frac{4}{1-u}du+\int_{1-\frac{1}{X}}^{1}Xdu=2+8\log X-8\log 2<< \log X$

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