Why $\int_{Y^2}^{{Y'}^2} z^{-1/2} e^{2\pi i N z}dz = O(Y^{-1})$?

Question

For $Y'>Y>0$, $N$ being any positive integer, I need to show $$\int_{Y^2}^{{Y'}^2} z^{-1/2} e^{2\pi i N z}dz = O(Y^{-1})$$ as $Y$ goes to inifinity. I tried integration by parts with $u= e^{2 \pi i Nz}, dv = z^{-1/2}$ but could not proceed. I appreciate any help!

Note: This equation is from Davenport's book Multiplicative Number Theory (page 15, second edition). It helps one to conclude the sign of (quadratic) Gauss sum. In summary, calculating the Gauss sum simplifies to calculating the summation $\sum_{x=0}^{q-1} e^{{2 \pi x^2}/q}$ and Dirichlet's method evaluates even the more general sum $\sum_{n=0}^{N-1} e^{{2 \pi n^2}/N}$. For this we apply Poisson summation formula to $\cos 2 \pi x^2 / N$ and $\sin 2 \pi x^2 / N$ and combine the results, and my question justifies the type of Poisson summation that is used is applicable in this certain case.

Answer 1

Let $f: [1,\infty)\rightarrow \mathbb{C}$ be defined by $$ f(z) = z^{-1/2} e^{2\pi i N z} $$ where $N\geq1$.

Let's call a function $g: [1,\infty)\rightarrow \mathbb{C}$ almost antiperiodic if there exist positive constants $\alpha$ and $c$ such that $$ | g(x) + g(x+c)| < \frac{\alpha}{x^{(3/2)}} $$ for all $x>1$.

I claim that $f$ is almost antiperiodic with $c=\frac1{2 N}$ because $$\begin{aligned} \left|f(x) + f\left(x+c\right)\right| &= \left| \frac1{\sqrt{x}} - \frac1{\sqrt{x+c}} \right|\\&= \left| \frac{\sqrt{x+c} - \sqrt{x}}{\sqrt{x}\sqrt{x+c}}\right| \\&= \frac{c}{\sqrt{x}\sqrt{x+c}\cdot({\sqrt{x+c} + \sqrt{x}})} \\&\leq\frac{c}{x^{(3/2)}}.\end{aligned}$$

I also claim that if

1. a continuous function $g:[1,\infty)\rightarrow C$ is almost antiperiodic,
2. there exists a real number $\gamma$ such that $|g(x)|< \gamma/\sqrt{x}$ for all $x>1$, and
3. $b>a\geq1$,

then there exists a constant $\beta>0$ such that $$ \left|\int_{x=a}^b g(x) dx\right| < \frac{\beta}{\sqrt{a}}. $$

Proof of second claim: Let $m$ be the largest integer less than $\frac{b-a}{2 c}$. Then $$\left|\int_{x=a}^b g(x) dx\right| $$ $$=\left|\sum_{i=0}^{m-1}\left(\int_{x=a+2ic}^{a+2ic+c} g(x) dx +\int_{x=a+2ic+c}^{a+2ic+2c} g(x) dx \right) + \int_{x=a+2mc}^{b} g(x) dx \right|$$ $$=\left|\sum_{i=0}^{m-1}\left(\int_{x=a+2ic}^{a+2ic+c} (g(x)+ g(x+c)) dx \right) + \int_{x=a+2mc}^{b} g(x) dx \right|$$ $$\leq \left|\sum_{i=0}^{m-1}\left(\int_{x=a+2ic}^{a+2ic+c} (g(x)+ g(x+c)) dx \right)\right| + \left|\int_{x=a+2mc}^{b} g(x) dx \right|$$ $$\leq \sum_{i=0}^{m-1}\left|\int_{x=a+2ic}^{a+2ic+c} (g(x)+ g(x+c)) dx \right| + \frac{2 c \gamma}{\sqrt{a}} $$ $$\leq \sum_{i=0}^{m-1}\int_{x=a+2ic}^{a+2ic+c} \frac{\alpha}{x^{(3/2)}} dx + \frac{2 c \gamma}{\sqrt{a}} $$ $$\leq \int_{x=a}^{b} \frac{\alpha}{x^{(3/2)}} dx + \frac{2 c \gamma}{\sqrt{a}} $$ $$= \frac{2\alpha}{\sqrt{a}} - \frac{2\alpha}{\sqrt{b}} + \frac{2 c \gamma}{\sqrt{a}} $$ $$< \frac{2\alpha}{\sqrt{a}} + \frac{2 c \gamma}{\sqrt{a}} $$ $$= \frac{2(\alpha + c\gamma) }{\sqrt{a}}$$ proving the second claim.

It follows from the claims above that $$ \int_{z=Y^2}^{(Y')^2} z^{-1/2} e^{2\pi i N z} dz \in O(1/Y) $$ if $1<Y<Y'$.

Edited for grammar, algebra errors, added a bound on $g$ in the second claim in order to make it true, and added a proof of second claim as requested by Gary which was quite helpful because it was wrong as stated originally.