# How to find $\lim_{n \to \infty}\int_{0}^{1}\sin^2\left(\frac{1}{ny^2}\right)\,\mathrm{d}y$ if it exists?

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.

• Do you have dominated convergence at your disposal? Jan 27, 2020 at 13:09
• I think you have $\int_0^1 {\sin ^2 \left( {\frac{1}{{ny^2 }}} \right)dy} = \int_1^{ + \infty } {\frac{1}{{v^2 }}\sin ^2 \left( {\frac{{v^2 }}{n}} \right)dv}$.
– Gary
Jan 27, 2020 at 13:12
• @Cameroon Williams Kindly explain, that is a new concept to me. Jan 27, 2020 at 13:13
• @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. Jan 27, 2020 at 13:14
• @CameronWilliams since we are currently studying measure theory, I think we can use monotone convergence theorem for this problem. Would you elaborate? Jan 27, 2020 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 $$\lim_{n \to +\infty} I_{n}=0$$

• In the last integral, since $s$ is defined in terms of $n$ doesn't that make the last integral still dependent on $n$? Jan 27, 2020 at 18:27
• 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 $\sqrt\pi$ Jan 27, 2020 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.

• Nice, +1. You could just look at $[0,1/\sqrt n],[1/\sqrt n,1]$ and shorten the proof a bit.
– zhw.
Jan 27, 2020 at 16:44
• @zhw.: yeah you are right, I didn't notice it at first glance. Jan 27, 2020 at 20:26