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.
Best Answer
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 $$\lim_{n \to +\infty} I_{n}=0$$