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 $\alpha < 1/4$).
