Integrate $\int_{-\pi}^{\pi} \frac{\sin \theta}{a - \sin\theta}$ where $a > 1$

Question

This is a question from the textbook Mathematical Methods for Physics and Engineering, Ex 24.13. I am having some issues to solve the definite integral by complex contour in the title. I tried to use a complex unit circle contour and convert the integral to its complex form \begin{align*} \int_{C} \frac{-\frac{1}{2}(z-z^{-1})}{az + \frac{1}{2}\imath (z^2-1)} dz \end{align*} and get $z = \imath a − \imath (a^2 − 1)^{1/2}$ as a simple pole suggested by the hint at the back. But I can't obtain the residue for the pole as given in the hint which is \begin{align*} -\frac{\imath}{2}(a^2-1)^{-1/2} \end{align*} I also think that in the unit circle the origin is a pole as well due to the numerator... Can anyone shed some lights for me? Any help would be appreciated! The answer should be \begin{align*} 2\pi (a(a^2-1)^{-1/2}-1) \end{align*}

Answer 1

Let's clear the integrand after the change of variable $z = e^{i\theta}$, which can be rewritten as $$ \int_{-\pi}^\pi \frac{\sin\theta}{a-\sin\theta} \,\mathrm{d}\theta = - \oint_C \frac{z^2-1}{z^2-2iaz-1} \frac{\mathrm{d}z}{iz}, $$ where $C$ is the unit circle centered at the origin. A simple pole is located at $z_0 = 0$ because of the last factor, whose residue is given by $$ \mathrm{Res}_{z_0}(f) = \displaystyle\lim_{z\rightarrow0} zf(z) = i, $$ where $f(z) := -\displaystyle\frac{z^2-1}{iz(z^2-2iaz-1)}$.

The quadratic factor in the denominator itself produces two simple poles at $z_\pm = i(a\pm\sqrt{a^2-1})$, whose residues are given by $$ \mathrm{Res}_{z_\pm}(f) = \displaystyle\lim_{z\rightarrow z_\pm} (z-z_\pm)f(z) = -\frac{z_\pm^2-1}{z_\pm-z_\mp}\frac{1}{iz_\pm} = -\frac{ia}{\sqrt{a^2-1}}, $$ where we used the factorization $z^2 - 2iaz - 1 = (z-z_+)(z-z_-)$ and the fact that $z_\pm^2-1 = 2iaz_\pm$.

Now, the crucial question is : which pole(s) lie(s) inside the unit circle $C$. For sure, $z_0$ has to be taken into account. But what for $z_\pm$ ? Actually, only $z_-$ is contained inside the unit disk when $a > 1$. In consequence, the residue formula leads to $$ \int_{-\pi}^\pi \frac{\sin\theta}{a-\sin\theta} \,\mathrm{d}\theta = 2\pi i \left(\mathrm{Res}_{z_0}(f) + \mathrm{Res}_{z_-}(f)\right) = -2\pi\left(1 - \frac{a}{\sqrt{a^2-1}}\right) $$