Letting $N\to\infty$ in an integral containing $\sin Nq$ I'm stuck in a path on a paper about thermal conductivity. There is a identity involving an integral that a I can't realize how they've perfomed it. Here is it:
$$\lim_{N\to \infty} \int_{0}^{\pi}\frac{g_1(q)dq}{1+g_2(q)\sin(Nq)}
=\int_{0}^{\pi}\frac{g_1(q)dq}{[1-g_2(q)^{2}]^{1/2}}$$
for "reasonable" functions $g_1$, $g_2$.
If you have any ideas to solve it or even the solution I'll be very grateful for help.
 A: $$ \int_0^{\pi} \! d\theta \frac{f(\theta)}{1-g(\theta) \sin N\theta} ~=~\int_0^{\pi} \! d\theta f(\theta) \sum_{n=0}^{\infty} g(\theta)^n \sin^n N\theta$$
$$ ~\longrightarrow~\int_0^{\pi} \! d\theta f(\theta) \sum_{p=0}^{\infty} \left(\begin{array}{c}  2p \\ p \end{array}\right) \left(\frac{g(\theta)}{2}\right)^{2p}
~=~\int_0^{\pi} \! d\theta  \frac{f(\theta)}{\sqrt{1-g(\theta)^{2}}}\quad\text{for}\quad N\to \infty,$$
because of (among other things) Riemann-Lebesgue Lemma. We use the fact that the $N$-independent middle term in the expansion of the sine
in terms of exponentials
$$\sin^n N\theta~=~\left(\frac{ie^{-iN\theta}-ie^{iN\theta}}{2}\right)^n
~=~\frac{i^ne^{-inN\theta}+\ldots +(-i)^ne^{inN\theta}}{2^n}$$ 
is 
$$0\qquad\text{for}\quad n\text{ odd},$$ 
and 
$$\frac{1}{2^{n}}\left(\begin{array}{c}  n \\ n/2 \end{array}\right)
~=~\frac{(1/2)_p}{p!}
~=~\left(\begin{array}{c}  p-1/2 \\ p \end{array}\right)
~=~ (-1)^p\left(\begin{array}{c}  -1/2 \\ p \end{array}\right)$$
$$\qquad\text{for}\quad n=2p\text{ even},$$ 
by a combinatorial argument from the binomial theorem. Finally, we use that 
$$ \frac{1}{(1-x)^s}~=~\sum_{p=0}^{\infty} \frac{(s)_px^p}{p!},$$
where $(s)_p=\frac{\Gamma(s+p)}{\Gamma(s)}$ is the Pochhammer symbol, cf. Abramowitz and Stegun, eq. (6.1.22), p. 256.
A: As $N$ gets large, the sine changes very rapidly, so that $g(q)$ is essentially constant.  So we want to know the average value of $1/(1+a\sin(Nq))$ as $Nq$ goes through a cycle of $2\pi$, and $g(q)$ remains roughly constant at the value $a$.
Let $Nq=x$.  During the following, make the substitution $z=e^{ix}$.  I use $\cos x$ instead of $\sin x$ because it makes the calculation a little easier.
$$\frac{1}{2\pi}\int_0^{2\pi}\frac{dx}{1+a\cos x}\\
=\frac{1}{2\pi}\int_0^{2\pi}\frac{2dx}{2+a(e^{ix}+e^{-ix})}\\
=\frac{1}{i\pi}\int_{|z|=1}\frac{dz}{2z+az^2+a}\\
=\frac{1}{ia\pi}\int_{|z|=1}\frac{dz}{(z+1/a)^2+1-(1/a)^2}
$$
The two roots of the quadratic have product equal 1, so one is inside the unit circle and the other outside.  The residue for the one inside the circle is $2\pi i/(2\sqrt{1/a^2-1})$, and the final answer is $1/\sqrt{1-a^2}$.
