Computation of $\int_{-\pi}^\pi \frac{\sin(x)}{1-2a\cos(x)+a^2}\sin(nx) \, dx$

Question

How can I show, by means of Fourier analysis, that for every $a>1$ and $n\in \mathbb{N}$

$$ \int_{-\pi}^\pi \frac{\sin(x)}{1-2a\cos(x)+a^2}\sin(nx)\,dx=\frac{\pi}{a^{n+1}}? $$

Answer 1

There are four key ingredients: $$ \sin(x)\sin(nx) = \frac{1}{2}\left[\cos((n-1)x)-\cos((n+1) x)\right]\tag{1} $$

$$ 1-2a\cos(x)+a^2 = (a-e^{ix})(a-e^{-ix}) \tag{2}$$ $$ \forall n,m\in\mathbb{Z},\qquad \int_{-\pi}^{\pi} e^{nix} e^{-mix}\,dx = 2\pi\cdot\delta(m,n)\tag{3}$$ $$ \frac{1}{a-e^{ix}} = \sum_{n\geq 0} \frac{e^{nix}}{a^{n+1}}\tag{4} $$ An a (hopefully) illuminating example: $$\begin{eqnarray*} \int_{-\pi}^{\pi}\frac{e^{-10ix}}{(a-e^{ix})(a-e^{-ix})}\,dx&=&\frac{1}{a^2}\int_{-\pi}^{\pi}e^{-10ix}\sum_{n\geq 0}\frac{e^{nix}}{a^n}\sum_{m\geq 0}\frac{e^{-mix}}{a^m}\,dx\\&=&\frac{2\pi}{a^2}\sum_{m\geq 0}\frac{1}{a^m}\cdot\frac{1}{a^{m+10}}\\&=&\frac{2\pi}{a^{12}}\cdot\frac{1}{1-\frac{1}{a^2}}.\tag{5} \end{eqnarray*}$$ Have a look at the Poisson kernel, too.