# Derive a bound for a general Gauss sum (from Iwaniec and Kowalski pg 199)

I am reading 'Analytic Number Theory' by Iwaniec and Kowalski and get stuck on this:

On page 199,they say

...we get $$\left| S_f(N) \right|^2\leq N+\sum_{ 1\leq \mathcal{l} < N} \min(2N,\left \| 2\alpha \mathcal{l} \right \| ^{-1}).$$ Hence one can deduce that $$\left| S_f(N) \right|\leq 2\sqrt[]{\alpha}N+\frac{1}{\sqrt[]{\alpha}}\log\frac{1}{\alpha}$$if $$0<\alpha \leq \frac{1}{2}$$,which restriction can always be arranged...

where $$\left \| \alpha \right \|$$ denotes the distance of $$\alpha$$ to the nearest integer,and $$S_f(N)=\sum_{1\leq \mathcal{n} \leq N}e(\alpha n^2+\beta n)$$. I suppose one needs to summon $$\sum_{1\leq \mathcal{l} \leq x}\frac{1}{\mathcal{l}}=\log x +\gamma +O(\frac{1}{x})$$ to fill the gap.

Can somebody help?

We can assume $$\alpha < 1/4$$ because otherwise we can apply the trivial bound $$|S_f(N)| \leq N$$.
Let's partition the sum as follows: \begin{align*} \sum_{ 1\leq \ell < N} \min(2N,\| 2\alpha \ell \| ^{-1}) &= \sum_{\substack{1\leq \ell < N \\ 2\alpha\ell < 1/2}} \min(2N,\| 2\alpha \ell \| ^{-1})\\ &+ \sum_{1 \leq m < 2\alpha(N-1)} \sum_{\substack{1\leq \ell < N \\ m-1/2 \leq 2\alpha\ell < m+1/2}} \min(2N,\| 2\alpha \ell \| ^{-1}) \end{align*} We have the inequalities \begin{align*} \sum_{\substack{1\leq \ell < N \\ 2\alpha\ell < 1/2}} \min(2N,\| 2\alpha \ell \| ^{-1}) &\leq \frac{1}{2\alpha} + \frac{1}{2\alpha} \int_{1/2\alpha}^{1/2} \frac{1}{x} \, dx\\ &= \frac{1}{2\alpha} + \frac{1}{2\alpha} \log \frac{1}{4\alpha} \end{align*} and \begin{align*} \sum_{\substack{1\leq \ell < N \\ m-1/2 \leq 2\alpha\ell < m+1/2}} \min(2N,\| 2\alpha \ell \| ^{-1})) &\leq 2N + 2 \left(\frac{1}{2\alpha} + \frac{1}{2\alpha} \int_{1/2\alpha}^{1/2} \frac{1}{x} \, dx \right)\\ &= 2N + \frac{1}{\alpha} + \frac{1}{\alpha} \log \frac{1}{4\alpha}. \end{align*}
Therefore we have \begin{align*} |S_f(N)|^2 &\leq N + \left(\frac{1}{2\alpha} + \frac{1}{2\alpha} \log \frac{1}{4\alpha} \right) + 2\alpha(N-1)\left(2N + \frac{1}{\alpha} + \frac{1}{\alpha} \log \frac{1}{4\alpha} \right)\\ &\leq 4 \alpha N^2 + 4N \log \frac{1}{\alpha} + \frac{1}{\alpha} \left( \log \frac{1}{\alpha} \right)^2\\ &= \left(2 \sqrt{\alpha} N + \frac{1}{\sqrt{\alpha}} \log \frac{1}{\alpha} \right)^2 \end{align*} where the second inequality can be shown after a bit of work (using $$\alpha < 1/4$$).
• Did you simply ignore the constraints $1\leq \ell <N$ when giving $\sum_{1\leq \ell <N \\ 2\alpha\ell < \frac{1}{2}}\min(2N,||2\alpha\ell||^{-1})\leq \frac{1}{2\alpha}+\frac{1}{2\alpha}\int^\frac{1}{2}_\frac{1}{2\alpha}\frac{1}{x} dx$ ? May 28, 2020 at 9:01
• Yes, I do. That inequality holds for any $N$. May 28, 2020 at 14:10