\[I_k = \int_0^\infty e^{-nx} \sin^{2k}\! x {\rm d}x \quad(n,k\in \mathbb{N}^*) \]

Solution:

\[\begin{align*} I_k &= \int_0^\infty e^{-nx} \sin^{2k}\! x {\rm d}x \\ &=\left[-\frac{1}{n}e^{-n x}\sin^{2k}\! x\right]_0^\infty-\int_0^\infty -\frac{2k}{n}e^{-nx} \sin^{2k-1}\! x \cos x{\rm d}x \\ &=\frac{2k}{n} \int_0^\infty e^{-n x}\sin^{2k-1}\!x\cos x{\rm d}x \\ &=\frac{2k}{n}\left(\left[-\frac{1}{n}e^{-n x}\sin^{2k-1}\!x\cos x\right]_0^\infty-\int_0^\infty -\frac{1}{n}e^{-n x}\frac{{\rm d}}{{\rm d}x}(\sin^{2k-1}\!x\cos x){\rm d}x \right) \\ &=\frac{2k}{n^2}\int_0^\infty e^{-n x}\left((2k-1)\sin^{2k-2}\!x \cos^2 x-\sin^{2k}\!x\right){\rm d}x \\ &=\frac{2k}{n^2}\int_0^\infty e^{-n x}\left((2k-1)\sin^{2k-2}\!x (1-\sin^2 x)-\sin^{2k}\!x\right){\rm d}x \\ &=\frac{2k}{n^2}\int_0^\infty e^{-n x}\left((2k-1)\sin^{2k-2}\!x-2k\sin^{2k}\!x\right){\rm d}x \\ &=\frac{2k\times(2k-1)}{n^2}I_{k-1}-\frac{2k \times2k}{n^2}I_{k}\\ n^2 I_k &= 2k(2k-1)I_{k-1}-4k^2I_k \\ (n^2+4k^2)I_k&=2k(2k-1)I_{k-1} \\ I_k &= \frac{2k(2k-1)}{n^2+4k^2} I_{k-1} \end{align*} \]

接下来只需求出 \(I_0\) 即可

\[I_0 = \int_0^\infty e^{-nx}{\rm d}x= \left[-\frac{1}{n}e^{-nx}\right]_0^\infty = \frac{1}{n} \]

经过简单递推容易得到

\[I_k = \frac{(2k)!}{n \prod_{j=1}^k (n^2 + 4j^2)} \]

\(\square\)