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I need an approach to analytically evaluating this limit:

$$\lim_{n\rightarrow\infty} \int_{-\pi}^\pi x^2 \frac{\sin(2nx)}{\sin x} dx$$

Numerically, I see that the answer is $-\pi^3$. Similarly, if I replace $x^2$ with $x^4$, I get $-\pi^5$. I vaguely recall seeing this result obtained analytically and not necessarily using advanced ideas, but I just can't remember any details. I know the fraction in the integrand relates to Chebyshev polynomials of the second kind. Thoughts anyone? Thanks!

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    Hint: Riemann-Lesbegue Lemma.2012-11-10
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    I like your question (+1).2012-11-10
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    The Riemann-Lebesgue lemma does not apply, at least not directly, since $f(x) = \frac{x^2}{\sin x}$ is not integrable over $[-\pi,\pi]$.2012-11-10

2 Answers 2