For hours I have been trying to determine whether or not the following limit exists:

$$\displaystyle{ \lim_{n \to \infty} }\displaystyle\int_{0}^{1}\sin^2\left(\dfrac{1}{ny^2}\right)\mathrm{d}y$$

My first attempt was to try and solve it as an indefinite integral, hoping a nice closed form would result:

Starting with integration by parts gave $${\displaystyle\int_{0}^{1}}\sin^2\left(\dfrac{1}{ny^2}\right)\mathrm{d}y = y\cdot \sin^2\left( \dfrac{1}{ny^2} \right) + 2\displaystyle\int_{0}^{1} \dfrac{\sin\left(\dfrac{2}{ny^2}\right)}{ny^3}\mathrm{d}y$$

Which was not much help. Thus, I tried to see how far I could get with a series of substitutions, treating it as an indefinite integral:

$$ v=\dfrac{1}{y} \implies {\displaystyle\int_{}^{}}\sin^2\left(\dfrac{1}{ny^2}\right)\mathrm{d}y =-{\displaystyle\int}\dfrac{\sin^2\left(\frac{v^2}{n}\right)}{v^2} \space \mathrm{d}v$$

Then, integrating by parts:

$$ = -\dfrac{\sin^2\left(\frac{v^2}{n}\right)}{v}-{\displaystyle\int}-\dfrac{4\cos\left(\frac{v^2}{n}\right)\sin\left(\frac{v^2}{n}\right)}{n}\,\mathrm{d}v = -\dfrac{\sin^2\left(\frac{v^2}{n}\right)}{v} + \dfrac{4}{n}{\displaystyle\int}\cos\left(\dfrac{v^2}{n}\right)\sin\left(\dfrac{v^2}{n}\right)\space \mathrm{d}v $$

Which simplifies to

$$-\dfrac{\sin^2\left(\frac{v^2}{n}\right)}{v} + \dfrac4n{\displaystyle\int}\dfrac{\sin\left(\frac{2v^2}{n}\right)}{2}\space\mathrm{d}v \tag{$\ast$}$$

At this point, I realised that the initial substitution $v = 1/y$ will lead to problems at zero when determining the new limits so I modified the problem like this:

$$ v= \dfrac{1}{y} \implies \lim_{n \to \infty} {\displaystyle\int_{0}^{1}}\sin^2\left(\dfrac{1}{ny^2}\right)\mathrm{d}y = \lim_{n \to \infty} \left( \lim_{c \to 0}{\displaystyle\int_{c}^{1}}\dfrac{\sin^2\left(\frac{v^2}{n}\right)}{v^2} \space \mathrm{d}v \right) $$

I am still stuck at this point. However, referring back to ($\ast$), I have a few conjectures about the convergence of the individual terms:

Firstly, for a fixed $v$ $$\lim_{n \to \infty} -\dfrac{\sin^2\left(\frac{v^2}{n}\right)}{v} = 0$$

And secondly,

$${\displaystyle\int}\dfrac{\sin\left(\frac{2v^2}{n}\right)}{2}\space\mathrm{d}v$$ is bounded above thus

$$\lim_{n \to \infty}\dfrac4n{\displaystyle\int_{0}^{1}}\dfrac{\sin\left(\frac{2v^2}{n}\right)}{2}\space\mathrm{d}v = 0$$

Therefore the initial integral is indeed convergent. Right now I am trying to find the limit but no success yet. Any thoughts and ideas will be appreciated.

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    $\begingroup$ Do you have dominated convergence at your disposal? $\endgroup$ – Cameron Williams Jan 27 '20 at 13:09
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    $\begingroup$ I think you have $\int_0^1 {\sin ^2 \left( {\frac{1}{{ny^2 }}} \right)dy} = \int_1^{ + \infty } {\frac{1}{{v^2 }}\sin ^2 \left( {\frac{{v^2 }}{n}} \right)dv}$. $\endgroup$ – Gary Jan 27 '20 at 13:12
  • $\begingroup$ @Cameroon Williams Kindly explain, that is a new concept to me. $\endgroup$ – E.Nole Jan 27 '20 at 13:13
  • $\begingroup$ @E.Nole Dominated convergence allows you to pass limits around given certain conditions holding true. If you don't know what it is, you don't have it at your disposal. $\endgroup$ – Cameron Williams Jan 27 '20 at 13:14
  • $\begingroup$ @CameronWilliams since we are currently studying measure theory, I think we can use monotone convergence theorem for this problem. Would you elaborate? $\endgroup$ – E.Nole Jan 27 '20 at 13:19

After a change of variables the integral becomes $$I_{n}=\frac{1}{2}\int_{1}^{+\infty}\sin^2 \bigg(\frac{x}{n}\bigg) x^{-\frac{3}{2}}dx$$ Substituting $s=\frac{x}{n}$ we get $$I_{n}=\frac{1}{2\sqrt{n}}\int_{\frac{1}{n}}^{+\infty}\frac{\sin^2(s)}{s^{\frac{3}{2}}}ds \leq \frac{1}{2\sqrt{n}}\int_{0}^{+\infty}\frac{\sin^2(s)}{s^{\frac{3}{2}}}ds$$ And since $I_{n} \geq 0$ and the last integral is just a constant we conclude by the squeeze theorem $$\lim_{n \to +\infty} I_{n}=0$$

  • $\begingroup$ In the last integral, since $s$ is defined in terms of $n$ doesn't that make the last integral still dependent on $n$? $\endgroup$ – E.Nole Jan 27 '20 at 18:27
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    $\begingroup$ In the second to last yes, the left boundary depends on $n$, hence why I've bounded it with the integral in the interval $[0,+\infty]$. In the last one no: actually, it equals $\sqrt\pi$ $\endgroup$ – Gennaro Marco Devincenzis Jan 27 '20 at 18:36

Probably too complex

For the antiderivative first $$I_n=\int\sin^2\left(\dfrac{1}{ny^2}\right)\,dy$$ One integration by parts gives $$I_n=y \sin^2\left(\dfrac{1}{ny^2}\right)+\int\frac{2 }{n y^2}\sin \left(\frac{2}{n y^2}\right)\,dy$$ The remaining integral is computable in terms fo Fresnel sine integral.All of that makes $$I_n=y \sin ^2\left(\frac{1}{n y^2}\right)-\frac{\sqrt{\pi } S\left(\frac{2}{ \sqrt{n\pi } y}\right)}{\sqrt{n}}$$

Using the bounds $$J_n=\int_0^1\sin^2\left(\dfrac{1}{ny^2}\right)\,dy=\frac{-2 \sqrt{\pi } S\left(\frac{2}{ \sqrt{n\pi }}\right)+\sqrt{n}-\sqrt{n} \cos \left(\frac{2}{n}\right)+\sqrt{\pi }}{2 \sqrt{n}}$$

Expanding for large values of $n$ $$J_n=\frac{1}{2} \sqrt{\frac{\pi}{n}}-\frac{1}{3 n^2}+O\left(\frac{1}{n^4}\right)$$

Checking for $n=10$ using numerical integration $I_{10}=0.276921$ while the above truncated expansion gives $0.276916$.


Let the integral under limit be denoted by $I_n$. Since the integrand is non-negative we have $I_n\geq 0$.

Let's take an $\epsilon$ with $0<\epsilon<1$ and split the interval of integration into $[0, \epsilon] $ and $[\epsilon, 1]$ so that the integral $I_n$ is split as sum of two integrals. Since the integrand is bounded above by $1$ the first integral does not exceed $\epsilon $. Since $\sin^2x\leq x^2$ the second integral does not exceed $$\int_{\epsilon} ^{1}\frac{dy}{n^2y^4}=\frac{1}{3n^2}\left(\frac{1}{\epsilon^3}-1\right)$$ Therefore we have $$0\leq I_n\leq \epsilon+\frac{1}{3n^2}\left(\frac{1}{\epsilon^3}-1\right)\tag{1}$$ for all $n$ and all $\epsilon\in(0,1)$. Letting $n\to\infty$ we can see that $$0\leq \liminf_{n\to\infty} I_n\leq \limsup_{n\to\infty} I_n\leq \epsilon$$ Since $\epsilon\in(0,1)$ is arbitrary it follows that the desired limit is $0$.

As mentioned in comments, you can put $\epsilon =1/\sqrt{n}$ in the inequality $(1)$ and apply usual Squeeze theorem to get the desired result.

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    $\begingroup$ Nice, +1. You could just look at $[0,1/\sqrt n],[1/\sqrt n,1]$ and shorten the proof a bit. $\endgroup$ – zhw. Jan 27 '20 at 16:44
  • $\begingroup$ @zhw.: yeah you are right, I didn't notice it at first glance. $\endgroup$ – Paramanand Singh Jan 27 '20 at 20:26

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