Suppose \(a > b > 0\) . Evaluate \(\begin{array} { r } { \int _ { 0 } ^ { 2 \pi } \frac { d \theta } { ( a + b \cos \theta ) ^ { 2 } } } \end{array}\)
Solution
Put \[ d=\sqrt{a^2-b^2}>0. \] First consider the auxiliary integral \[ J(a) = \int_0^{2\pi} \frac{d\theta}{a+b\cos\theta}. \]
1. Under the substitution \(z=e^{i\theta}\),\[ J(a) = \frac2i \int_{\abs{z}=1} \frac{dz}{bz^2+2az+b}. \]
On the unit circle, \[ d\theta=\frac{dz}{iz} \] and \[ \cos\theta=\frac12\left(z+z^{-1}\right). \] Therefore \[ \begin{aligned} \frac{d\theta}{a+b\cos\theta} &= \frac{1}{a+\frac b2(z+z^{-1})} \frac{dz}{iz}\\ &= \frac{2\,dz} {i(bz^2+2az+b)}. \end{aligned} \] Integrating once counterclockwise around the unit circle gives the claim.
2. The roots of\[ bz^2+2az+b \]are\[ \alpha=\frac{-a+d}{b}, \qquad \beta=\frac{-a-d}{b}, \]with\[ \abs{\alpha}<1<\abs{\beta}. \]
The quadratic formula gives the two roots. Their product is \[ \alpha\beta=1. \] Moreover, \[ \abs{\alpha} = \frac{a-d}{b} = \frac{b}{a+d} <1, \] because \(a+d>b\). Therefore \[ \abs{\beta}=\frac1{\abs{\alpha}}>1. \]
3. The auxiliary integral is\[ J(a) = \frac{2\pi}{\sqrt{a^2-b^2}}. \]
By step 2, only the simple pole \(z=\alpha\) lies inside the unit circle. For \[ G(z) = \frac2{i(bz^2+2az+b)}, \] the residue at \(\alpha\) is \[ \begin{aligned} \Res(G;\alpha) &= \frac2{i(2b\alpha+2a)}\\ &= \frac1{i(b\alpha+a)}. \end{aligned} \] Since \[ b\alpha+a=d, \] one has \[ \Res(G;\alpha) = \frac1{id} = -\frac{i}{d}. \] The residue theorem and step 1 therefore give \[ J(a) = 2\pi i \left(-\frac{i}{d}\right) = \frac{2\pi}{d}. \] Substitute \(d=\sqrt{a^2-b^2}\).
4. Differentiation with respect to \(a\) may be passed through theintegral defining \(J(a)\), and\[ J'(a) = -\int_0^{2\pi} \frac{d\theta}{(a+b\cos\theta)^2}. \]
Since \(a>b\), choose \[ 0<\delta<a-b. \] For \(s\) with \(\abs{s-a}<\delta\) and every real \(\theta\), \[ s+b\cos\theta \geq s-b > a-b-\delta > 0. \] Thus the function \[ (s,\theta) \longmapsto \frac1{s+b\cos\theta} \] and its \(s\)-derivative are continuous on a compact parameter rectangle around \(a\) and \([0,2\pi]\). The standard differentiation-under-the-integral theorem applies, giving \[ \begin{aligned} J'(a) &= \int_0^{2\pi} \frac{\partial}{\partial a} \left( \frac1{a+b\cos\theta} \right) d\theta\\ &= -\int_0^{2\pi} \frac{d\theta}{(a+b\cos\theta)^2}. \end{aligned} \]
5. The requested integral is\[ \boxed{ \int_0^{2\pi} \frac{d\theta}{(a+b\cos\theta)^2} = \frac{2\pi a}{(a^2-b^2)^{3/2}}. } \]
Differentiate the formula in step 3: \[ J'(a) = -\frac{2\pi a}{(a^2-b^2)^{3/2}}. \] Combine this with step 4 and multiply by \(-1\).
6. Q.E.D.
Step 5 is the requested evaluation.