问题及解答

令 $I(\alpha )={\displaystyle {\int }_0^{\frac{\pi }{2}}{\displaystyle \frac{1}{1+{\tan }^{\alpha }x}}\mathrm{d}x}$, 求 $I^{\prime }(\alpha )$.

根据费因曼(Feynman)的方法, 

$$\begin{array}{rl}{I}^{\prime }(\alpha )& ={\int }_0^{\frac{\pi }{2}}({\displaystyle \frac{1}{1+{\tan }^{\alpha }x}}{)}_{\alpha }^{\prime }\mathrm{d}x\\ & ={\int }_0^{\frac{\pi }{2}}\frac{-1}{(1+{\tan }^{\alpha }x{)}^2}\cdot {\tan }^{\alpha }x\cdot \ln \tan x\mathrm{d}x\\ & ={\int }_0^{\frac{\pi }{2}}\frac{-\alpha \cdot {\tan }^{\alpha -1}x\cdot {\sec }^{2}x}{(1+{\tan }^{\alpha }x{)}^2}\cdot \frac{\tan x}{\alpha {\sec }^{2}x}\cdot \ln \tan x\mathrm{d}x\\ & ={\int }_0^{\frac{\pi }{2}}\frac{\tan x}{\alpha {\sec }^{2}x}\cdot \ln (\tan x)\mathrm{d}\frac{1}{1+{\tan }^{\alpha }x}\\ & ={\int }_0^{\frac{\pi }{2}}\frac{1}{\alpha }\cdot \tan x\cdot {\cos }^{2}x\cdot \ln (\tan x)\mathrm{d}\frac{1}{1+{\tan }^{\alpha }x}\\ & =\frac{1}{\alpha }{\int }_0^{\frac{\pi }{2}}\sin x\cdot \cos x\cdot \ln (\tan x)\mathrm{d}\frac{1}{1+{\tan }^{\alpha }x}\end{array}$$

注意,

$$({\displaystyle \frac{1}{1+{\tan }^{\alpha }x}}{)}_x^{\prime }=\frac{-1}{(1+{\tan }^{\alpha }x{)}^2}\cdot \alpha {\tan }^{\alpha -1}x\cdot {\sec }^{2}x$$

不要搞错了.