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Suppose that $f$ is continuous on $[0, 1]$. Then calculate the following limit:
$$\lim_{n\to\infty} \frac{\displaystyle\int_{0}^{1} f(x) \sin^{2n} (2 \pi x) \space dx}{\displaystyle\int_{0}^{1} e^{x^2}\sin^{2n} (2 \pi x) \space dx}$$

What should i start with? (high school problem)

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Are there any restrictions on $f$? – nullUser Jul 7 '12 at 13:08
Well, what about starting to tell us what is $\,f\,$, what have you tried so far, where're you stuck...? – DonAntonio Jul 7 '12 at 13:09
Is this a high-school problem? – Siminore Jul 7 '12 at 13:55
I'd love to know what high school, where and under what syllabus and/or under what teacher is this a high school problem... – DonAntonio Jul 8 '12 at 2:29
A student, any student, has to construct his learning building in such a way that (s)he can study stuff based on what (s)he already knows. This question is no high school stuff, not even close, in any educational system I know, and that's what I asked what's your high school, where and what's your mathematics syllabus: it really interests me. – DonAntonio Jul 8 '12 at 18:24
up vote 2 down vote accepted

For every continuous function $g$ defined on $(-1,1)$ and every $\delta$ in $[0,1]$, define $$ I_n(g)=\int_{-1}^1g(x)\cos^{2n}\left(\frac\pi2 x\right)\,\mathrm dx,\qquad J_n(\delta)=\int_{-\delta}^{+\delta}\cos^{2n}\left(\frac\pi2 x\right)\,\mathrm dx. $$ Fix a continuous function $g$ defined on $(-1,1)$. Since $g$ is continuous, there exists some finite $c$ such that $|g|\leqslant c$ uniformly on $(-1,1)$. Since $g$ is continuous at $0$, for every $(a,b)$ such that $a\lt g(0)\lt b$, there exists some $\delta$ in $[0,1]$ such that $a\lt g\lt b$ uniformly on $(-\delta,\delta)$. Hence, $$ aJ_n(\delta)-c(J_n(1)-J_n(\delta))\leqslant I_n(g)\leqslant bJ_n(\delta)+c(J_n(1)-J_n(\delta)). $$ Here is a basic but crucial fact:

For every $\delta$ in $[0,1]$, $J_n(1)-J_n(\delta)\ll J_n(\delta)$ when $n\to\infty$.

Now, fix some continuous functions $g$ and $h$ defined on $(-1,1)$ such that $h(0)\ne0$, say, $h(0)\gt0$. Assume without loss of generality that $g\gt0$ everywhere (if necessary, add to $g$ a large multiple of $h(0)$). Then, for every positive $(a,b,a',b')$ such that $a\lt g(0)\lt b$ and $a'\lt h(0)\lt b'$, $$ \frac{a}{b'}\leqslant\liminf\limits_{n\to\infty}\frac{I_n(g)}{I_n(h)}\leqslant\limsup\limits_{n\to\infty}\frac{I_n(g)}{I_n(h)}\leqslant\frac{b}{a'}. $$ Hence, $$ \lim\limits_{n\to\infty}\frac{I_n(g)}{I_n(h)}=\frac{g(0)}{h(0)}. $$ To solve the question asked, apply the result above to the functions $g$ and $h$ defined on $(-1,1)$ by $g(x)=f(\frac14(x+1))+f(\frac14(x+3))$ and $h(x)=e(\frac14(x+1))+e(\frac14(x+3))$ with $e(t)=\mathrm e^{t^2}$.

Note: To prove the basic but crucial fact mentioned above, one can show that $J_n(\delta)\sim2/\sqrt{\pi n}$ when $n\to\infty$, for every $\delta$ in $(0,1]$.

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Thank you for your solution. – user 1618033 Jul 7 '12 at 20:13
how did you think of this approach? Did you meet similar problems before? – user 1618033 Jul 7 '12 at 20:26
Yes, as already mentioned in my comment (did you go and read the WP page?). The idea is that $\sin^{2n}(2\pi x)\to0$ except at $x=\frac14$ and $x=\frac34$ where $\sin^{2n}(2\pi x)=1$ identically, hence all that matters are the values of the function integrated, at these points. – Did Jul 7 '12 at 20:34

This isn't at all high-school level, but pursuing the delta function idea, if you let $I_n = \int_0^{\pi}\sin^{2n}(2\pi x)\,dx = \int_{\pi}^{2\pi}\sin^{2n}(2\pi x)\,dx$, then your limit is the same as $${\int_0^{\pi} f(x){\sin^{2n}(2\pi x) \over I_n} \, dx + \int_{\pi}^{2\pi} f(x){\sin^{2n}(2\pi x) \over I_n}\, dx \over \int_0^{\pi} e^{x^2} {\sin^{2n}(2\pi x) \over I_n} \, dx + \int_{\pi}^{2\pi} e^{x^2} {\sin^{2n}(2\pi x) \over I_n}\, dx }$$ The functions $\sin^{2n}(2\pi x) \over I_n$ converge in a distribution sense to $\delta(x - {1 \over 4})$ on $(0,\pi)$ and to $\delta(x - {3 \over 4})$ on $(\pi,2\pi)$ (Alternatively, you can talk about approximations to the identity). So as $n$ goes to infinity the above converges to ${f(1/4) + f(3/4) \over e^{1/16} + e^{9/16}}$.

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Thank you for your solution. – user 1618033 Jul 7 '12 at 20:14
the approach needed for this problem is totally different from the major part of the calculus problems i've seen so far. – user 1618033 Jul 7 '12 at 20:31
@Chris'sister - then you've learned something ... – Mark Bennet Jul 7 '12 at 21:09

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