沃利斯公式

$$I_n = \int_0^\frac{\pi}{2} \sin^n x dx = \int_0^\frac{\pi}{2} \cos^n x dx \\ = \begin{cases} \dfrac{(n-1)!!}{n!!} \cdot \dfrac{\pi}{2} &, n为偶数 \\ \dfrac{(n-1)!!}{n!!} &, n为奇数 \\ \end{cases}$$

注意

注意是双阶乘不是阶乘

证明:

$$I_n = -\int_0^\frac{\pi}{2} \sin^{n-1} x d(\cos x) \\ = - \sin^{n-1} x \cos x|_0^\frac{\pi}{2} + \int_0^\frac{\pi}{2} \cos x d(\sin^{n-1} x) \\ = (n-1) \int_0^\frac{\pi}{2} \cos^2 x \sin^{n-2} x dx \\ = (n-1) \int_0^\frac{\pi}{2} (1-\sin^2 x) \sin^{n-2} x dx \\ = (n-1)(I_{n-2} - I_n) \\ \\ \implies I_n = \dfrac{n-1}{n} I_{n-2} \space (n \geq 2)$$

推论: 因为 $\sin x \in [0, 1]$, 故 $I_{2k+1} \leq I_{2k} \leq I_{2k-1}$,

$$\dfrac{(2k)!!}{(2k+1)!!} \leq \dfrac{(2k-1)!!}{(2k)!!} \cdot \dfrac{\pi}{2} \leq \dfrac{(2k-2)!!}{(2k-1)!!} \\ 1 \leq [\dfrac{(2k-1)!!}{(2k)!!}]^2 \cdot (2k+1) \cdot \dfrac{\pi}{2} \leq \dfrac{2k+1}{2k}$$

因为 $\lim_{k \to \infty} \dfrac{2k+1}{2k} = 1$ 由夹逼定理得:

$$\lim_{k \to \infty} [\dfrac{(2k)!!}{(2k-1)!!}]^2 \dfrac{1}{2k+1} = \dfrac{\pi}{2}$$