(4n+1)π2(4 n+1) \frac{\pi}{2}(4n+1)2π
2nπ2 n \pi2nπ
(4n−1)π2(4 n-1) \frac{\pi}{2}(4n−1)2π
(2n−1)π2(2 n-1) \frac{\pi}{2}(2n−1)2π