Evaluating $\int_0^\pi \frac{\sin{mt}\sin{nt}}{a+b\cos{nt}}dt$

Question

How can the following integral be evaluated?

$$\int_0^\pi \frac{\sin{mt}\sin{nt}}{a+b\cos{nt}}dt$$

for $m,n \in \mathbb{N}$ and $a,b \in \mathbb{R}$.

I have tried some substitutions and developing that $\sin{mt}$ into a combination of powers of $\sin{t}$, but with no success.

Edit: We can assume $a>0$, $b<0$, $|a|>|b|$, $n>1$, $m>>1$

Any help will be welcomed.

Answer 1

Let $$I=\int_0^\pi \frac{\sin{mt}\sin{nt}}{a+b\cos{nt}}dt$$

By enforcing the substitution $t\mapsto-t$, $$I=\int^0_{-\pi} \frac{\sin{mt}\sin{nt}}{a+b\cos{nt}}dt$$

Then, we can see $$2I=\int^{\pi}_{-\pi} \frac{\sin{mt}\sin{nt}}{a+b\cos{nt}}dt$$

Let $z=e^{it}$, the integral equals $$-\oint_{|z|=1} \frac{(z^m-z^{-m})(z^n-z^{-n})}{2a+b(z^n+z^{-n})}\frac{dz}{iz}$$ $$=i\oint_{|z|=1} \frac{(z^{2m}-1)(z^{2n}-1)}{(z^{m+1})(2az^n+bz^{2n}+b)}dz$$

The poles are:

1. $z=0$, of order $m+1$, or, of order $m+n+1$ if $b=0$.
2. For $b\ne0$, $z^n=\frac{-a\pm\sqrt{a^2-b^2}}{b}$, which gives $2n$ roots.

I cannot further generalize it. Indeed, calculation of residue at each pole is tedious as well. Moreover, you need to do some work to determine if the pole lies inside the unit circle.

I am searching for some better solutions...

EDIT:

For simplicity, assume $n>1$.

Then, between $0$ and $\pi$, $\cos(nt)$ attained all possible values at least once.

Thus, if $\frac{\cos^{-1}(-a/b)}n$ is real, your integrand must have a singularity within the integration interval. Thus, your integral becomes improper, which I would not want to deal with at the moment.

Therefore, I assume $\vert\frac{a}{b}\vert>1$.

According to my analysis, for $\frac{a}{b}>0$, only the positive roots will lie within the unit circle(i.e. only consider $+$ in $\pm$). for $\frac{a}{b}<0$, only the negative roots will lie within the unit circle(i.e. only consider $-$ in $\pm$).