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

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. – Bombyx mori Nov 10 '12 at 18:36
I like your question (+1). – user 1618033 Nov 10 '12 at 20:11
The Riemann-Lebesgue lemma does not apply, at least not directly, since $f(x) = \frac{x^2}{\sin x}$ is not integrable over $[-\pi,\pi]$. – Lukas Geyer Nov 10 '12 at 22:48

One can show$^{(1)}$ that $$\sum_{k=1}^n\cos(2k-1)x=\frac{\sin 2nx}{2\sin x}$$

So it suffices to evaluate $$\int_{-\pi}^{\pi}x^2\cos(2k-1)xdx$$ Can you do this?

SPOILER You can check that

$$\int_{ - \pi }^\pi {{x^2}} \cos \left( {2k - 1} \right)xdx = - \frac{{4\pi }}{{{{\left( {2k - 1} \right)}^2}}}$$

so it follows

$$\mathop {\lim }\limits_{n \to \infty } \int_{ - \pi }^\pi {{x^2}} \frac{{\sin \left( {2nx} \right)}}{{\sin x}}xdx = - 8\pi \sum\limits_{k \geqslant 1} {\frac{1}{{{{\left( {2k - 1} \right)}^2}}}} = - 8\pi \frac{{{\pi ^2}}}{8} = - {\pi ^3}$$

$(1)$: Since $$\frac{{\sin \left( {a + b} \right) - \sin \left( {a - b} \right)}}{2} = \sin b\cos a$$ we get $$\sin x\cos \left( {2k - 1} \right)x = \frac{{\sin 2kx - \sin \left( {2k - 2} \right)x}}{2}$$ $$\sum\limits_{k = 1}^n {\cos \left( {2k - 1} \right)x} = \frac{{\sin 2nx}}{{2\sin x}}$$

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Here is an (admittedly somewhat horrible) solution using both the Riemann-Lebesgue lemma and the residue theorem. Let $$f(x) = \frac{x^2}{\sin x} \quad \text{ and } \quad g(x) = \frac{\pi^2}{2} \frac{1-\cos x}{\sin x}$$ on $(-\pi,\pi)$. Then $$h(x) = f(x)-g(x) = \frac{x^2 - \frac{\pi^2}2(1-\cos x)}{\sin x}$$ is actually continuous on the closed interval $[-\pi,\pi]$, since both the numerator and denominator have simple zeros at $\pm \pi$ and at $0$, so that the quotient has removable singularities there. Then the Riemann-Lebesgue lemma implies that $$\lim_{n\to\infty} \int_{-\pi}^\pi h(x) \sin (2nx) \, dx = 0,$$ so that $$\lim_{n\to\infty} \int_{-\pi}^\pi f(x) \sin (2nx) \, dx = \lim_{n\to\infty} \int_{-\pi}^\pi g(x) \sin (2nx) \, dx$$ if either of those limits exists. The integral on the right-hand side for fixed $n$ can now be evaluated by standard residue theorem techniques, using the substitutions $z=e^{ix}$, $\cos x = \frac12(z+z^{-1})$, $dx = \frac{dz}{iz}$, $\sin x = \frac{1}{2i}(z-z^{-1})$, $\sin (2nx) = \frac{1}{2i} (z^{2n}-z^{-2n})$, which yields \begin{align*} \frac{\pi^2}{2} \int_{|z|=1|} & \frac{1-\frac12 (z+z^{-1})}{\frac1{2i}(z-z^{-1})}\frac{z^{2n}-z^{-2n}}{2i} \frac{dz}{iz} = \frac{\pi^2}{4i} \int_{|z|=1} \frac{2-z-z^{-1}}{z^{2n}} \frac{z^{4n}-1}{z^2-1}\, dz \\ &=\frac{\pi^2}{4i} \int_{|z|=1} \frac{2-z-z^{-1}}{z^{2n}} (1+z^2+z^4\ldots + z^{4n-2}) \, dz \\ &=\frac{\pi^2}{4i} \int_{|z|=1} (2-z-z^{-1}) (z^{-2n}+z^{-2n+2}+z^{-2n+4}\ldots + z^{2n-2}) \, dz \\ &= \frac{\pi^2}{4i} 2\pi i a_{-1} = \frac{\pi^3}{2} a_{-1} = \frac{\pi^3}{2}(-2) = -\pi^3 \end{align*} where $a_{-1}=-2$ is the coefficient of $z^{-1}$ in the integrand which is easily found by multiplying the terms in parentheses.

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