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.

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    $\begingroup$ What is the relationship between $Y$ and $Y'$? Also, why do you call this a number-theory problem? $\endgroup$
    – davidlowryduda
    Sep 27, 2022 at 15:34
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    $\begingroup$ I suggest you give the source and context of the problem. It's good to give enough context to make it obvious to potential answerers what the problem is, and also enough context so that future people with the same problem will find this problem. Also: if you integrate by parts but switch $u$ and $dv$, you get the claimed result by noting that the remaining integral can be bounded absolutely. $\endgroup$
    – davidlowryduda
    Sep 27, 2022 at 15:51
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    $\begingroup$ Oh I see, and one also need to note that $|e^{ix}|=1$. Thank you for guidance, and I will edit my question as you said! I was just trying to ask as shortly as possible so that is why I did not give a source, but I will do it for my future questions. $\endgroup$
    – CCCC
    Sep 27, 2022 at 16:22
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    $\begingroup$ Let $f(z) = z^{-1/2} e^{2\pi i N z}$. Notice that $f(z) + f(z + 1/(2N)) = z^{-1/2} - (z+1/(2N))^{-1/2}$ which is in $O(z^{(-3/2)})$, so $\int_{z=a}^b f(z) dz \in O(1/\sqrt{a})$. $\endgroup$
    – irchans
    Sep 27, 2022 at 17:35
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    $\begingroup$ Given that $a,b\in\Bbb R_{+}$ with $a<b$, we can apply the triangle inequality, $$\left|\int_a^b x^{-1/2}\exp(2\pi \mathrm i nx)\mathrm dx\right|\leq\int_a^b\big|x^{-1/2}\exp(2\pi\mathrm inx)\mathrm dx\big| \\ =\int_a^b x^{-1/2}\mathrm dx=2(\sqrt{b}-\sqrt{a})=\mathrm O(\sqrt{a})$$ $\endgroup$
    – K.defaoite
    Sep 27, 2022 at 18:37

1 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.

  • $\begingroup$ What is the proof of your second claim? (also, you can use \mathbb{C} to produce $\mathbb{C}$) $\endgroup$
    – Gary
    Sep 27, 2022 at 21:30
  • $\begingroup$ Alright Gary, give me some time to add a "proof" of claim 2. (I can see, I also need a bound on $g$.) $\endgroup$
    – irchans
    Sep 27, 2022 at 21:47
  • $\begingroup$ @Gary Could you check my proof of the second claim? $\endgroup$
    – irchans
    Sep 28, 2022 at 11:06

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