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As a follow-up to my previous question. What is the integral of $\sin (2\pi x^2)$ as $x \in [\sqrt{n}, \sqrt{n+1}]$ we get this:

$$ \int_\sqrt{n}^{\sqrt{n+1}} \sin (2\pi x^2) \; dx \tag{$\ast$}$$

Since $\sin 2\pi n = 0$ for all $n \in \mathbb{Z}$ this integral should be close to zero. I could approximate by adding over the four corners:

$$ \int_\sqrt{n}^{\sqrt{n+1}} \sin (2\pi x^2) \; dx \approx \frac{1}{4} \left[ \sin (2\pi n) + \sin 2\pi (\sqrt{n}+ \frac{1}{4})^2 + \sin 2\pi (\sqrt{n}+ \frac{1}{2})^2 + \sin 2\pi (\sqrt{n}+ \frac{3}{4})^2\right] $$ This might be hard to estimate since we get unpredictible terms like. And I may have misaplied the trapezoid rule $$ \sin 2\pi (n + \frac{1}{2} \sqrt{n} + \frac{1}{4})$$ if we place the mesh and $\sqrt{n} < \sqrt{n + \frac{1}{4}} < \sqrt{n + \frac{1}{2}}< \sqrt{n + \frac{3}{4}}< \sqrt{n + 1}$ the trapeoid rule is: $$\tiny \sin(2\pi n ) \frac{ \sqrt{n+\tfrac{1}{4}} - \sqrt{n}}{2} + \sin (2\pi \sqrt{n + \frac{1}{4}})\left[ \sqrt{n+ \frac{1}{2} }- \sqrt{n}\right] + \sin (2\pi \sqrt{n + \frac{1}{2}})\left[ \sqrt{n+ \frac{3}{4} }- \sqrt{n+ \frac{1}{4}} \right] + \sin (2\pi \sqrt{n + \frac{3}{4}})\left[ \sqrt{n+ \frac{1}{2} }- \sqrt{n+ 1}\right] + \sin (2\pi \sqrt{n + 1})\left[ \frac{\sqrt{n+ \frac{3}{4} }- \sqrt{n+ 1}}{2} \right] $$ I think there is a mistake what should the trapezoid rule be? Maybe if I said: $$ \sqrt{n+ \frac{1}{4}} - \sqrt{n} = \sqrt{n} \left( \sqrt{1 + \frac{1}{4} } - 1 \right)= \frac{\sqrt{n}}{8} $$

Then after fixing all my mistakes I get something of order $\frac{1}{\sqrt{n}}$. How does this look?

\begin{eqnarray}\small \int \dots dx &\approx& \frac{1}{8} \left[ \sqrt{n }\cdot 0 + 2\sqrt{n + \tfrac{1}{4}}\cdot 1 + 2\sqrt{n + \tfrac{1}{2}}\cdot 0 + 2\sqrt{n + \tfrac{3}{4}}\cdot (-1) + \sqrt{n+1}\cdot 0\right] \\ \\ &=& \frac{1}{4} \left[ \sqrt{n+ \frac{1}{2}} - \sqrt{n + \frac{3}{4}}\right] = O(\frac{1}{\sqrt{n}})\end{eqnarray}

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  • $\begingroup$ Why do you want to use the trapezoidal rule here? $\endgroup$ – Mark Viola Sep 27 '16 at 18:34
  • $\begingroup$ you may also integrate by parts $$ I(n)=\int_{n}^{n+1}\frac{\sin(2\pi t)}{2\sqrt{t}}=-\frac{\cos(2\pi t)}{4\pi\sqrt{t}}\big|_{n}^{n+1}-\int_{n}^{n+1}\frac{\sin(2\pi t)}{8\pi t^{3/2}} $$ now expand the difference and show the remaining integral is dominated by this contribution $\endgroup$ – tired Sep 27 '16 at 18:53
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Setting $t=x^2$, we obtain the integral to be $$I = \int_n^{n+1} \sin(2\pi t) \dfrac{dt}{2\sqrt{t}}$$ We then have $$\vert 2I \vert = \left \vert \int_n^{n+1} \sin(2\pi t) \dfrac{dt}{\sqrt{t}}\right \vert \leq \int_n^{n+1} \dfrac{dt}{\sqrt{t}} = 2\left.\sqrt{t} \right \vert_{n}^{n+1} = 2\left(\sqrt{n+1}-\sqrt{n}\right)$$ Hence, we obtain that $$\vert I \vert \leq \dfrac1{\sqrt{n}+\sqrt{n+1}}$$

A better estimate can be determined as follows. We have $$I = \int_0^1 \dfrac{\sin\left(2\pi(n+t)\right)}{2\sqrt{n+t}}dt = \int_0^1 \dfrac{\sin(2\pi t)}{2\sqrt{n+t}}dt$$ Hence, we obtain $$\sqrt{n}I = \int_0^1 \dfrac{\sin(2\pi t)}{2\sqrt{1+t/n}}dt = \dfrac12\sum_{k=0}^{\infty} \dbinom{2k}k \left(\dfrac{-1}{4n}\right)^k \int_0^1 t^k \sin(2\pi t) dt = \dfrac1{8n\pi} - \dfrac3{32n^2 \pi} + \cdots$$ Hence, infact $I \sim \dfrac1{8\pi n^{3/2}} \left(1-\dfrac3{4n}\right)$.

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  • $\begingroup$ nice answer (+1) $\endgroup$ – tired Sep 27 '16 at 18:37
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One approach, following @Leg's answer:

$$I = \int_0^{1} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t}}=\int_0^{1/2} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t}}+\int_{1/2}^{1} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t}}$$

$$=\int_0^{1/2} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t}}-\int_{0}^{1/2} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t+1/2}}$$

$$=\int_0^{1/2}\sin{(2\pi t)}\frac{1}{2}\left(\dfrac{1}{\sqrt{n+t}}-\dfrac{1}{\sqrt{n+t+1/2}}\right)\,dt$$

Now, because:

$$\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+h}}=\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x}\sqrt{x+h}}=\frac{h}{\sqrt{x}\sqrt{x+h}(\sqrt{x+h}+\sqrt{x})}$$

Thus

$$I=\int_0^{1/2}\sin{(2\pi t)}\frac{1}{2}\frac{1}{2\sqrt{n+t}\sqrt{n+t+1/2}(\sqrt{n+t}+\sqrt{n+t+1/2})}$$

and taking simple bounds on the non-$\sin$ part of the integrand, we have:

$$\int_0^{1/2}\frac{\sin{(2\pi t)}}{8(n+1)^{3/2}}<I<\int_0^{1/2}\frac{\sin{(2\pi t)}}{8n^{3/2}}$$

Observing that $\int_0^{1/2}\sin(2\pi t)\,dt=\frac{1}{\pi}$, we see that:

$$\frac{1}{8\pi (n+1)^{3/2}}<I<\frac{1}{8\pi n^{3/2}}$$

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