计算 $\dps{\int_{-\infty}^{+\infty}\frac{\rd x}{(x^2+2x+2)^n}}$. (中国科学院)

解答: 设 $$\bex I_n=\int_{\bbR} \frac{\rd x}{(x^2+2x+2)^n} =\int_{\bbR} \frac{\rd (x+1)}{[(x+1)^2+1]^n} =\int_{\bbR}\frac{\rd t}{(t^2+1)^n}=2\int_0^\infty \frac{\rd t}{(t^2+1)^n}, \eex$$ 则由分部积分, $$\beex \bea I_n&=2\frac{t}{(t^2+1)^n}|_0^\infty +4n \int_0^\infty \frac{t^2}{(t^2+1)^{n+1}}\rd t\\ &=4n\int_0^\infty \frac{(t^2+1)-1}{(t^2+1)^{n+1}}\rd t\\ &=2n\sex{I_n-I_{n+1}}. \eea \eeex$$ 故 $$\bex I_{n+1}=\frac{2n-1}{2n}I_n,\quad I_n=\frac{2n-3}{2n-2}I_{n-1}=\cdots =\frac{(2n-3)!!}{(2n-2)!!}I_1=\frac{(2n-3)!!}{(2n-2)!!}\cdot \frac{\pi}{2}. \eex$$