Let \(n\) be a positive integer. Compute \[ \int_0^{2\pi}\frac{1-\cos(n\theta)}{1-\cos\theta}\,d\theta. \]
Solution
1. For \(\theta\notin2\pi\ZZ\),\[ \frac{1-\cos(n\theta)}{1-\cos\theta} = \abs{\sum_{k=0}^{n-1}e^{ik\theta}}^2. \]
Using \[ \abs{1-e^{it}}^2=2(1-\cos t) \] and the geometric-sum identity, \[ \sum_{k=0}^{n-1}e^{ik\theta} = \frac{1-e^{in\theta}}{1-e^{i\theta}}, \] we obtain \[ \abs{\sum_{k=0}^{n-1}e^{ik\theta}}^2 = \frac{2(1-\cos(n\theta))}{2(1-\cos\theta)}. \] The quotient has a removable singularity at each point of \(2\pi\ZZ\), where both sides extend continuously with value \(n^2\).
2.\[ \int_0^{2\pi} \abs{\sum_{k=0}^{n-1}e^{ik\theta}}^2\,d\theta =2\pi n. \]
Expand: \[ \abs{\sum_{k=0}^{n-1}e^{ik\theta}}^2 = \sum_{j,k=0}^{n-1}e^{i(j-k)\theta}. \] For an integer \(m\), \[ \int_0^{2\pi}e^{im\theta}\,d\theta = \begin{cases} 2\pi,&m=0,\\ 0,&m\ne0. \end{cases} \] Thus only the \(n\) diagonal terms \(j=k\) survive the integration, giving \(2\pi n\).
3. The requested integral equals\[ \boxed{2\pi n}. \]
2. Changing the value at the removablesingularity does not affect the integral.
4. Q.E.D.
Step 3 gives the value of the integral.