TITLE: Evaluation of $\lim_{n\rightarrow\infty} \int_{-\pi}^\pi x^2 \frac{\sin(2nx)}{\sin x} dx$
QUESTION [9 upvotes]: 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!

REPLY [2 votes]: 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 $$\sin x\cos \left( {2k - 1} \right)x = \frac{{\sin 2kx - \sin \left( {2k - 2} \right)x}}{2}$$ $$\sum\limits_{k = 1}^n {\cos \left( {2k - 1} \right)x}  = \frac{{\sin 2nx}}{{2\sin x}}$$