# 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? – Cameron Williams Jan 27 '20 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 '20 at 13:12
• @Cameroon Williams Kindly explain, that is a new concept to me. – E.Nole Jan 27 '20 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. – Cameron Williams Jan 27 '20 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? – 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 $$\lim_{n \to +\infty} I_{n}=0$$

• In the last integral, since $s$ is defined in terms of $n$ doesn't that make the last integral still dependent on $n$? – E.Nole Jan 27 '20 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 $\sqrt\pi$ – 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.

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