ディリクレ積分の一般化みたいなやつ

$$\begin{array}{rl}I& ={\int }_0^{\mathrm{\infty }}\frac{{\sin }^{n}x}{x^m}dx\\ & =\frac{1}{m-1}{\left[\frac{{\sin }^{n}x}{x^{m-1}}\right]}_0^{\mathrm{\infty }}-\frac{n}{1-m}{\int }_0^{\mathrm{\infty }}\frac{{\sin }^{n-1}x\cos x}{x^{m-1}}dx\\ & =\frac{n}{m-1}{\int }_0^{\mathrm{\infty }}\frac{{\sin }^{n-1}x\cos x}{x^{m-1}}dx\end{array}$$
$$\begin{array}{rl}J& ={\int }_0^{\mathrm{\infty }}\frac{{\sin }^{n-1}x\cos x}{x^{m-1}}dx\\ & =\frac{1}{2-m}{\left[\frac{{\sin }^{n-1}x\cos x}{x^{m-2}}\right]}_0^{\mathrm{\infty }}-\frac{1}{2-m}{\int }_0^{\mathrm{\infty }}\frac{(n-1){\sin }^{n-2}x{\cos }^{2}x+{\sin }^{n}x}{x^{m-2}}dx\\ & =\frac{n-1}{m-2}{\int }_0^{\mathrm{\infty }}\frac{{\sin }^{n-2}x{\cos }^{2}x}{x^{m-2}}dx-\frac{1}{m-2}{\int }_0^{\mathrm{\infty }}\frac{{\sin }^{n}x}{x^{m-2}}dx\\ & =\frac{n-1}{m-2}({\int }_0^{\mathrm{\infty }}\frac{{\sin }^{n-2}x}{x^{m-2}}dx-{\int }_0^{\mathrm{\infty }}\frac{{\sin }^{n}x}{x^{m-2}}dx)-\frac{1}{m-2}{\int }_0^{\mathrm{\infty }}\frac{{\sin }^{n}x}{x^{m-2}}dx\\ & =\frac{n-1}{m-2}f(n-2,m-2)-\frac{n}{m-2}f(n,m-2)\end{array}$$
以上より、
$$f(n,m)=\frac{n(n-1)}{(m-1)(m-2)}f(n-2,m-2)-\frac{n^2}{(m-1)(m-2)}f(n,m-2)$$