Compute \[ \int_{\abs z = 1} \qty{z + {1\over z}}^{2n} {dz \over z} \] and use it to show that \[ \int_0^{2\pi} \cos^{2n}(\theta) \, d\theta = 2\pi \qty{1\cdot 3 \cdot 5 \cdots (2n-1) \over 2 \cdot 4 \cdot 6 \cdots (2n)} .\]
Solution
Goal: Compute \(\int_{\abs z = 1} \qty{z + \frac1z}^{2n} \frac{dz}{z}\) and use it to show \(\int_0^{2\pi} \cos^{2n}\theta\, d\theta = 2\pi \frac{1\cdot 3 \cdots (2n-1)}{2\cdot 4 \cdots (2n)}\).
1. Expand \(\qty{z + \frac1z}^{2n} = z^{-2n}(z^2 + 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} z^{2k - 2n}\).
\(z + 1/z = z^{-1}(z^2 + 1)\); raising to the power \(2n\) and applying the binomial theorem gives the stated expansion (valid for \(z \neq 0\), in particular on \(\abs z = 1\)).
2. The coefficient of \(z^{-1}\) in \(\qty{z + \frac1z}^{2n} \frac1z\) is \(\binom{2n}{n}\).
By 1, \(\qty{z + \frac1z}^{2n} \frac1z = \sum_{k=0}^{2n} \binom{2n}{k} z^{2k - 2n - 1}\). The power \(z^{-1}\) occurs when \(2k - 2n - 1 = -1\), i.e. \(k = n\), with coefficient \(\binom{2n}{n}\).
3. \(\int_{\abs z = 1} \qty{z + \frac1z}^{2n} \frac{dz}{z} = 2\pi i \binom{2n}{n}\).
By the residue theorem, the integral around the positively oriented unit circle is \(2\pi i\) times the residue at \(z = 0\), which by 2 is the coefficient of \(z^{-1}\), namely \(\binom{2n}{n}\) (there are no other singularities on or inside the circle).
4. Parametrize \(z = e^{i\theta}\), \(0 \leq \theta \leq 2\pi\): the integral becomes \(i\int_0^{2\pi} \qty(2\cos\theta)^{2n} d\theta\).
\(z + 1/z = e^{i\theta} + e^{-i\theta} = 2\cos\theta\), and \(dz/z = ie^{i\theta} d\theta / e^{i\theta} = i\, d\theta\).
5. Equate 3 and 4: \(\int_0^{2\pi} \cos^{2n}\theta\, d\theta = \frac{2\pi}{2^{2n}}\binom{2n}{n}\).
From 3 and 4, \(i\cdot 2^{2n}\int_0^{2\pi}\cos^{2n}\theta\, d\theta = 2\pi i \binom{2n}{n}\); dividing by \(i\cdot 2^{2n}\) gives the formula.
6. \(\frac{1}{2^{2n}}\binom{2n}{n} = \frac{1\cdot 3 \cdots (2n-1)}{2\cdot 4 \cdots (2n)}\).
\(\binom{2n}{n} = \frac{(2n)!}{(n!)^2}\). Write \((2n)! = (1\cdot 3 \cdots (2n-1))(2\cdot 4 \cdots (2n))\) and \(n! = \frac{2\cdot 4 \cdots (2n)}{2^n}\) (each factor \(k\) replaced by \(2k/2\)), so \(\frac{(2n)!}{(n!)^2} \frac1{2^{2n}} = \frac{(1\cdot 3 \cdots (2n-1))(2\cdot 4 \cdots (2n))}{(2\cdot 4 \cdots (2n))^2/2^{2n}} \cdot \frac1{2^{2n}} = \frac{1\cdot 3 \cdots (2n-1)}{2\cdot 4 \cdots (2n)}\).
7. Q.E.D.
5 and 6 give \(\int_0^{2\pi}\cos^{2n}\theta\, d\theta = 2\pi\frac{1\cdot 3 \cdots (2n-1)}{2\cdot 4 \cdots (2n)}\), as required.