In Davenport's Multiplicative Number Theory, I am reading the proof of Vinogradov's Theorem and it is estimating the sum
$$S(\alpha)=\sum_{\substack{k \leq N \\ (k,q)=1}}\Lambda(k) e(k \alpha)$$
Then it computes $$S(\alpha)^3$$ and uses this to get an estimate for
$$r(N)=\int_{0}^{1}S(\alpha)^3 e(-N\alpha) d\alpha$$
In the step, I am stuck in the following. I would appreciate if anyone can explain this step
$$ r(N) = \int_{0}^{1}S(\alpha)^3 e(-N\alpha) d\alpha $$ $$ = e\Big(\frac{-Na}{q}\Big)\int_{-\frac{1}{Q}}^{\frac{1}{Q}}S(\alpha)^3 e(-N\beta) d\beta $$ $$= e\Big(\frac{-Na}{q}\Big)\int_{-\frac{1}{Q}}^{\frac{1}{Q}}\Big(\frac{\mu(q)}{\phi(q)^3}T(\beta)^3+O\Big(N^3 \operatorname{exp}(-c_1\sqrt{\log N})\Big)\Big) d\beta $$ $$ \frac{\mu(q)}{\phi(q)^3} e\Big(\frac{-Na}{q}\Big)\int_{-\frac{1}{Q}}^{\frac{1}{Q}}T(\beta)^3 d\beta + e\Big(\frac{-Na}{q}\Big)\int_{-\frac{1}{Q}}^{\frac{1}{Q}}O\Big(N^3 \operatorname{exp}(-c_2\sqrt{\log N})\Big) d\beta $$ $$ \frac{\mu(q)}{\phi(q)^3} e\Big(\frac{-Na}{q}\Big)\int_{-\frac{1}{Q}}^{\frac{1}{Q}}T(\beta)^3 d\beta + O\Big(N^2 \operatorname{exp}(-c_2\sqrt{\log N})\Big) $$
I cannot understand how $$ e\Big(\frac{-Na}{q}\Big)\int_{-\frac{1}{Q}}^{\frac{1}{Q}}O\Big(N^3 \operatorname{exp}(-c_1\sqrt{\log N})\Big) d\beta= O\Big(N^2 \operatorname{exp}(-c_2\sqrt{\log N})\Big) $$
Here are my thoughts, $e\Big(\frac{-Na}{q}\Big)$ has modulus $1$, so it is none of our concern. Secondly considering the integrand, we have $$N^3 \operatorname{exp}(-c_1\sqrt{\log N}) \ll N^3 \operatorname{exp}(-c_3\sqrt{\log N})$$ for some $c_3$ and hence, we have an upperbound on the integrand for sufficently large $N$. And thus this integral is bounded by $$\frac{2}{Q}N^3 \operatorname{exp}(-c_3\sqrt{\log N})=\frac{2(\log N)^B}{1}N^2 \operatorname{exp}(-c_3\sqrt{\log N}) \ll O\Big(N^2 \operatorname{exp}(-c_2\sqrt{\log N})\Big)$$ for some $c_2$, we have the desired estimates. Is this correct?