(4n−1)π2(4 n-1) \frac{\pi}{2}(4n−1)2π
nπ\mathrm{n} \pinπ
(4n+1)π2(4 n+1) \frac{\pi}{2}(4n+1)2π
(2n+1)π(2 n+1) \pi(2n+1)π