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I came across this problem which says:

Let $f:[-1,1]\rightarrow \ \mathbb{R}$ be continuous. Assume that $\int_{-1}^{1}f(t)\, dt =2$. Then $$\lim_{n\to\infty} \int_{-1}^{1}f(t)\sin^2(nt)\,dt$$ equals to

a) $0$

b) $1$

c) $f(1) - f(-1)$

d) Does not exist

I have taken $f(t)=1$ so that it satisfies the given definite integral. Then I see the solution to be $1$. Am I correct? I am looking for a better way to solve it. Please help.

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I don't get it. We know that $\int_{-1}^{1}f(t)dt =2$ and we want to find the limit of $\int_{-1}^{1}f(t)dt$. The limit of a constant? The thing under the $\lim$ sign should be a function or a sequence, otherwise I don't understand what the problem asks for. –  Dan Shved Nov 20 '12 at 9:47
2  
@budha: Did you learn Riemann-Lebesgue Lemma? –  P.. Nov 20 '12 at 10:03
1  
For a starting, the constant $f=1$ is an easy choice, and if the statement is true in this form, (i.e. doesn't depend too much on $f$), then it must give the right result. –  Berci Nov 20 '12 at 10:08
1  
Then you did the best. I personally dislike these quizes with given possible answers, but if that happens, just pick the easiest example. Anyway, the Riemann-Lebesgue lemma could help you, write $\sin^2$ as $1-\cos^2$ then use $2\cos^2x-1=\cos(2x)$.. –  Berci Nov 20 '12 at 10:22
1  
@budha: In my opinion the best approach is what Berci suggested but uses a theorem (lemma) you didn't learn. In what course did you take the exams? –  P.. Nov 20 '12 at 10:35

1 Answer 1

up vote 1 down vote accepted

Write $\sin^2(nx)=\dfrac{1-\cos{(2nx)}}{2}$.
Then $$\displaystyle{\int_{-1}^{1}f(t)\sin^2(nt)\,dt=\frac{1}{2}\int_{-1}^{1}f(t)\,dt-\frac{1}{2}\int_{-1}^{1}f(t)\cos(2nt)\,dt=1-\frac{1}{2}\int_{-1}^{1}f(t)\cos(2nt)\,dt}.$$ From Riemann-Lebesgue Lemma ( proof )we have $$\lim_{n\to\infty} \int_{-1}^{1}f(t)\cos(2nt)\,dt=0.$$ Therefore $\displaystyle{\lim_{n\to\infty} \int_{-1}^{1}f(t)\sin^2(nt)\,dt=1.}$

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