- Prove that
\[ \int_ {0} ^ {2 \pi} \frac {d \theta}{a + b \cos \theta} = \frac {2 \pi}{\sqrt {a ^ {2} - b ^ {2}}} \]
if \(a > | b |\) and \(a , b \in \mathbb { R }\)
Solution
1. If \(b = 0\), the integral is \(\frac{2\pi}{a}=\frac{2\pi}{\sqrt{a^2-b^2}}\).
The integrand is the constant \(1/a\).
2. If \(b\ne0\), then \(\displaystyle\int_0^{2\pi} \frac{d\theta}{a + b\cos\theta} = \frac{2}{i} \oint_{|z|=1} \frac{dz}{b z^2 + 2az + b}\).
With \(z = e^{i\theta}\) on the positively oriented unit circle, \(d\theta = \frac{dz}{iz}\) and \(\cos \theta = \frac{z + z^{-1}}{2}\), so the integrand becomes \(\frac{1}{a + b(z + z^{-1})/2}\cdot\frac{1}{iz}=\frac2i\cdot\frac{1}{bz^2+2az+b}\).
3. For \(b\ne0\), \(bz^2+2az+b=b(z-z_1)(z-z_2)\) with \(z_1 = \frac{-a + \sqrt{a^2 - b^2}}{b}\) inside and \(z_2 = \frac{-a - \sqrt{a^2 - b^2}}{b}\) outside the unit circle, and the residue of \(\frac{1}{b(z - z_1)(z - z_2)}\) at \(z_1\) is \(\frac{1}{2\sqrt{a^2 - b^2}}\).
The roots satisfy \(z_1 z_2 = 1\), and \(|z_2| = \frac{a + \sqrt{a^2 - b^2}}{|b|} > \frac{a}{|b|} > 1\), so \(|z_1| < 1\). The pole at \(z_1\) is simple with residue \(\frac{1}{b(z_1 - z_2)} = \frac{1}{2\sqrt{a^2 - b^2}}\).
4. Q.E.D.
For \(b\ne0\), the residue theorem and steps 2 and 3 give \[\int_0^{2\pi} \frac{d\theta}{a + b\cos\theta} = \frac{2}{i}\cdot 2\pi i\cdot\frac{1}{2\sqrt{a^2 - b^2}} = \frac{2\pi}{\sqrt{a^2 - b^2}},\] and step 1 covers \(b=0\).