- Prove that
\[ \int_ {0} ^ {2 \pi} \frac {d \theta}{(a + \cos \theta) ^ {2}} = \frac {2 \pi a}{(a ^ {2} - 1) ^ {3 / 2}}, \quad \text { whenever } a > 1. \]
Solution
Put \(z = e^{i\theta}\), so \(\cos\theta = \frac{z+z^{-1}}{2}\) and \(d\theta = \frac{dz}{iz}\), and let \(z_0 = -a+\sqrt{a^2-1}\), \(z_1 = -a-\sqrt{a^2-1}\) be the roots of \(z^2+2az+1\).
1. \(\displaystyle\int_0^{2\pi} \frac{d\theta}{(a+\cos\theta)^2} = \oint_{|z|=1} \frac{4z}{i(z-z_0)^2(z-z_1)^2}\,dz\), and \(z_0\) is the only pole inside \(\abs z=1\).
Since \(\bigl(a+\frac{z+z^{-1}}{2}\bigr)^2=\frac{(z^2+2az+1)^2}{4z^2}\), the substitution gives the integrand \(\frac{4z^2}{(z^2+2az+1)^2}\cdot\frac{1}{iz}\). For \(a>1\) the roots are real with \(z_0z_1=1\) and \(\abs{z_1}=a+\sqrt{a^2-1}>1\), so \(\abs{z_0}<1<\abs{z_1}\).
2. The residue at the double pole \(z_0\) is \(\dfrac{a}{i(a^2-1)^{3/2}}\).
The residue is the derivative of \(\frac{4z}{i(z-z_1)^2}\) at \(z_0\): \[\frac{4}{i}\cdot\frac{(z_0-z_1)-2z_0}{(z_0-z_1)^3} = \frac{4}{i}\cdot\frac{-(z_0+z_1)}{(z_0-z_1)^3} = \frac4i\cdot\frac{2a}{8(a^2-1)^{3/2}},\] using \(z_0+z_1=-2a\) and \(z_0-z_1=2\sqrt{a^2-1}\).
3. Q.E.D.
By the residue theorem and steps 1 and 2, the integral is \(2\pi i\cdot\frac{a}{i(a^2-1)^{3/2}}=\boxed{\dfrac{2\pi a}{(a^2-1)^{3/2}}}\).