Fejer 积分

在分析中，Fejer 积分是一个常义积分，它的基本形式是 $$\int_0^\pi \left( \dfrac{\sin nx}{\sin x} \right)^p \mathrm{d}x.$$

基础

1. $\int_0^\pi \dfrac{\sin nx}{\sin x} \mathrm{d}x = \begin{cases} 0, & n = 2k, \\ \pi, & n = 2k-1, \end{cases} \quad k \in \N^+.$
2. $\int_0^\pi \left( \dfrac{\sin nx}{\sin x} \right)^2 \mathrm{d}x = n\pi.$
3. $\int_0^\frac{\pi}{2} \dfrac{\sin nx}{\sin x} \mathrm{d}x = \begin{cases} 2 \displaystyle{\sum_{i=1}^k} \dfrac{(-1)^{k-1}}{2k-1}, & n = 2k, \\ \dfrac{\pi}{2}, & n = 2k-1, \end{cases} \quad k \in \N^+.$
4. $\int_0^\frac{\pi}{2} \left( \dfrac{\sin nx}{\sin x} \right)^2 \mathrm{d}x = \dfrac{n\pi}{2}.$

递推结论

假设$m \leqslant k, m, k \in \N^+$，那么有

1. $\int_0^\pi \dfrac{\sin^{2k+1} nx}{\sin^{2m+1} x} \mathrm{d}x = \dfrac{2k(2k+1)n^2}{2m(2m-1)} \int_0^\pi \dfrac{\sin^{2k-1} nx}{\sin^{2m-1} x} \mathrm{d}x - \dfrac{(2k+1)^2 n^2 - (2m-1)^2}{2m(2m-1)} \int_0^\pi \dfrac{\sin^{2k+1} nx}{\sin^{2m-1} x} \mathrm{d}x.$
2. $\int_0^\pi \dfrac{\sin^{2k} nx}{\sin^{2m} x} \mathrm{d}x = \dfrac{2k(2k-1)n^2}{(2m-1)(2m-2)} \int_0^\pi \dfrac{\sin^{2k-2} nx}{\sin^{2m-2} x} \mathrm{d}x - \dfrac{(2k)^2 n^2 - (2m-2)^2}{(2m-1)(2m-2)} \int_0^\pi \dfrac{\sin^{2k} nx}{\sin^{2m-2} x} \mathrm{d}x.$
3. $\int_0^\pi \dfrac{\sin^{2k+1} nx}{\sin x} \mathrm{d}x = \begin{cases} 0, & n = 2k, \\ \dfrac{(2k-1)!!}{(2k)!!} \pi, & n = 2k-1, \end{cases} \quad k \in \N^+.$
4. $\int_0^\pi \dfrac{\sin^{2k} nx}{\sin^2 x} \mathrm{d}x = \begin{cases} n \pi, & k = 1, \\ \dfrac{(2k-3)!!}{(2k-2)!!} n \pi, & k > 1, \end{cases} \quad k \in \N^+.$

利用如上两个递推公式以及初始条件（上文的基础）可以求得 Fejer 积分的值。