Suppose \(a>b>0\) and calculate \[ \int_0^{2\pi} {1 \over (a+b\cos(\theta))^2} \,d\theta .\]
Solution
1. Substitute \(z = e^{i\theta}\), so \(d\theta = \frac{dz}{iz}\) and \(\cos\theta = \frac{z + z^{-1}}{2}\).
standard substitution.
2. Then \(a + b\cos\theta = a + \frac{b}{2}(z + z^{-1}) = \frac{2az + bz^2 + b}{2z}\).
1.
3. Hence\[\int_0^{2\pi} \frac{d\theta}{(a + b\cos\theta)^2} = \oint_{|z|=1} \frac{1}{\left(\frac{2az + bz^2 + b}{2z}\right)^2} \frac{dz}{iz} = \oint_{|z|=1} \frac{4z}{i(2az + bz^2 + b)^2}\,dz.\]
1 and 2.
4. The integrand has poles where \(bz^2 + 2az + b = 0\), i.e. \(z = \frac{-a \pm \sqrt{a^2 - b^2}}{b}\).
solve the quadratic.
5. Since \(a > b > 0\), one root \(z_1 = \frac{-a + \sqrt{a^2 - b^2}}{b}\) has \(|z_1| < 1\) and the other \(z_2 = \frac{-a - \sqrt{a^2 - b^2}}{b}\) has \(|z_2| > 1\).
\(z_1 z_2 = 1\) (product of roots is \(b/b = 1\)), and \(z_1\) is the one inside the unit circle.
6. The integrand has a double pole at \(z_1\) (inside the unit circle), and the residue there is \(\frac{a}{i(a^2 - b^2)^{3/2}}\).
computing the residue of \(\frac{4z}{i b^2 (z - z_1)^2 (z - z_2)^2}\) at \(z = z_1\) (a double pole).
7. By the residue theorem, the integral is \(2\pi i \cdot \frac{a}{i(a^2 - b^2)^{3/2}} = \frac{2\pi a}{(a^2 - b^2)^{3/2}}\).
6.
8. Hence \(\int_0^{2\pi} \frac{d\theta}{(a + b\cos\theta)^2} = \frac{2\pi a}{(a^2 - b^2)^{3/2}}\).
7.
9. Q.E.D.
8.