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This question is from the MIT Integration Bee 2023 Quarterfinal #2. I'm doing these integrals to improve my quick math skills, and was wondering if there's a way to solve this within two minutes. The goal is to show that $$\lim_{n\to\infty} \frac{1}{n} \int_{0}^{n} \cos^2\left(\frac{\pi x^2}{\sqrt{2}}\right) \,\textrm{d}x = \frac{1}{2} $$

I've tried L'Hospital's Rule then FTC1 but I don't think the first part would even apply here. I'm not sure how to approach such a problem, especially if it's to be solved within two minutes.

Stamp
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    I think that the limit is 0 and not $1/2$, because the Fresnel integral $\int_0^\infty\cos(t^2),dt$ does converge, and hence $\int_0^n\cos\left(\frac{\pi x^2}{\sqrt 2}\right),dx$ has finite limit as $n\to\infty$. – Adren Feb 13 '23 at 18:48
  • @Adren I did notice that, and I realized I mis-typed the question itself. The cosine should be actually squared. I am aware, however, of the convergence of the Fresnel cosine integral. – Stamp Feb 13 '23 at 18:53
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    Ok, now it's sufficent to linearize $\cos^2(t)$ into $\frac12(1+\cos(2t))$ and to combine this with the previous argument. – Adren Feb 13 '23 at 18:57
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    I don't know how much work is supposed to be shown in these integration bees, but intuitively one might "guess" that the answer is $\frac 12$ based on the fact that we're looking at the average value of a function that oscillates between 0 and 1 without any systematic bias towards either bound. – Ben Grossmann Feb 13 '23 at 18:59
  • @BenGrossmann Actually I hadn't thought of it in that way! Thanks for the new insight. – Stamp Feb 13 '23 at 19:01

3 Answers3

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Let $x=nt$ and then \begin{eqnarray} &&\frac{1}{n} \int_{0}^{n} \cos^2\left(\frac{\pi x^2}{\sqrt{2}}\right) \,\textrm{d}x \\ &=&\frac12 + \frac12\int_0^1\cos\left(\sqrt2\pi n^2t^2\right) \textrm{d}t\\ &=&\frac12 + \frac12\int_0^1\frac{1}{2\sqrt2\pi n^2t}\textrm{d}\sin\left(\sqrt2\pi n^2t^2\right)\\ &=&\frac12 + \frac{1}{4\sqrt2\pi n^2t}\sin\left(\sqrt2\pi n^2t^2\right)\bigg|_0^1+\frac{1}{4\sqrt2\pi n^2}\int_0^1\frac{\sin\left(\sqrt2\pi n^2t^2\right)}{t^2}\textrm{d}x. \end{eqnarray} Note $$ \frac{1}{4\sqrt2\pi n^2t}\sin\left(\sqrt2\pi n^2t^2\right)\bigg|_0^1=\frac{1}{4\sqrt2\pi n^2}\sin\left(\sqrt2\pi n^2\right). $$ Since $$\frac{1}{4\sqrt2\pi n^2}\bigg|\frac{\sin\left(\sqrt2\pi n^2t^2\right)}{t^2}\bigg|\le\frac14,$$ one has, by DCT, $$ \lim_{n\to\infty}\frac{1}{4\sqrt2\pi n^2}\int_0^1\frac{\sin\left(\sqrt2\pi n^2t^2\right)}{t^2}\textrm{d}x=\int_0^1\lim_{n\to\infty}\frac{1}{4\sqrt2\pi n^2}\frac{\sin\left(\sqrt2\pi n^2t^2\right)}{t^2}\textrm{d}x=0.$$ So $$\lim_{n\to\infty} \frac{1}{n} \int_{0}^{n} \cos^2\left(\frac{\pi x^2}{\sqrt{2}}\right) \,\textrm{d}x = \frac{1}{2}. $$

xpaul
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Let's denote $$I(n)=\int_0^{n} \cos^2 \left( \frac{\pi x^2}{\sqrt 2}\right) dx$$ then $$I(n)=\frac{1}{2}\int_0^{n} \Big(1+\cos \left( \pi \sqrt2 x^2\right)\Big) dx=\frac{n}{2}+\frac{1}{4}\frac{1}{\sqrt{\pi\sqrt2}}\int_0^{n^2\pi\sqrt2}\frac{\cos x}{\sqrt x}dx$$ integrating by part $$=\frac{n}{2}+\frac{1}{4}\frac{1}{\sqrt{\pi\sqrt2}}\int_0^\infty\frac{\cos x}{\sqrt x}dx-\frac{1}{4}\frac{1}{\sqrt{\pi\sqrt2}}\int_{n^2\pi\sqrt 2}^\infty\frac{\cos x}{\sqrt x}dx$$ $$=\frac{n}{2}+\frac{1}{4}\frac{1}{\sqrt{\pi\sqrt2}}\int_0^\infty\frac{\cos x}{\sqrt x}dx+\frac{\sin(n^2\pi\sqrt 2)}{4n\pi\sqrt2}-\frac{1}{8}\frac{1}{\sqrt{\pi\sqrt2}}\int_{n^2\pi\sqrt 2}^\infty\frac{\sin x}{(\sqrt x)^3}dx$$ Using a closed contour in the complex plane (quarter-circle), we can show that $$\,\int_0^\infty\frac{\cos x}{\sqrt x}dx=\Re\int_0^\infty\frac{e^{ix}}{\sqrt x}dx=\Re e^{\frac{\pi i}{4}}\int_0^\infty\frac{e^{-t}}{\sqrt t}dt=\sqrt\frac{\pi}{2}$$ then $$I(n)=\frac{n}{2}\,+\,\frac{2^\frac{1}{4}}{8}\,+\,\frac{\sin(n^2\pi\sqrt 2)}{4n\pi\sqrt2}\,-\,\frac{1}{8}\frac{1}{\sqrt{\pi\sqrt2}}\int_{n^2\pi\sqrt 2}^\infty\frac{\sin x}{(\sqrt x)^3}dx=\frac{n}{2}\,+\,\frac{2^\frac{1}{4}}{8}\,+\,O\left(\frac{1}{n}\right)$$ $$\boxed{\,\,\frac{1}{n}I(n)=\frac{1}{2}\,+\,\frac{2^\frac{1}{4}}{8\,n}\,+\,O\left(\frac{1}{n^2}\right)\,\,}$$ The limit follows.

Svyatoslav
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    I wonder if there's a way to do this without contour integration, but the method here looks really neat, actually! – Stamp Feb 13 '23 at 19:06
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Transforming the integral by cosine half-angle formula yields $$ \begin{aligned} I(n) :& =\int_0^n \cos ^2\left(\frac{\pi x^2}{\sqrt{2}}\right) d x \\ & =\frac{1}{2} \int_0^n\left[1+\cos \left(\pi \sqrt{2} x^2\right)\right] d x \\ & =\frac{n}{2}+\frac{1}{2} \int_0^n \cos \left(\pi \sqrt{2} x^2\right) d x \end{aligned} $$Letting $\sqrt{\pi \sqrt{2}} x \rightarrow x$, then $$ \begin{aligned} I(n)&=\frac{n}{2} +\frac{1}{2 \sqrt{\pi \sqrt{2}}} \int_0^{n \sqrt{\pi \sqrt{2}}} \cos \left(x^2\right) d x \\ \lim _{n \rightarrow \infty}\left[\frac{1}{n} I(n)\right] & =\frac{1}{2}+\frac{1}{2 \sqrt{\pi \sqrt{2}}}\left(\lim _{n \rightarrow \infty} \frac{1}{n}\right)\left(\lim _{n \rightarrow \infty} \int_0^{n \sqrt{\pi \sqrt{2}}} \cos \left(x^2\right) d x\right)\\ & =\frac{1}{2}+\frac{1}{2 \sqrt{\pi \sqrt{2}}} \cdot 0 \cdot \sqrt{\frac{\pi}{8}}\quad (*) \\ & =\frac{1}{2} \end{aligned} $$ where (*) comes from the post.

Lai
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