问题及解答

令 $I(\alpha)=\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{1}{1+\tan^{\alpha}x}\mathrm{d}x$, 求 $I'(\alpha)$.

根据费因曼(Feynman)的方法, 

\[
\begin{split}
I'(\alpha)&=\int_0^{\frac{\pi}{2}}\Bigl(\dfrac{1}{1+\tan^{\alpha}x}\Bigr)'_{\alpha}\mathrm{d}x\\
&=\int_0^{\frac{\pi}{2}}\frac{-1}{(1+\tan^{\alpha}x)^2}\cdot\tan^{\alpha}x\cdot\ln\tan x\mathrm{d}x\\
&=\int_0^{\frac{\pi}{2}}\frac{-\alpha\cdot\tan^{\alpha-1}x\cdot\sec^2 x}{(1+\tan^{\alpha}x)^2}\cdot\frac{\tan x}{\alpha\sec^2 x}\cdot\ln\tan x\mathrm{d}x\\
&=\int_0^{\frac{\pi}{2}}\frac{\tan x}{\alpha\sec^2 x}\cdot\ln(\tan x)\mathrm{d}\frac{1}{1+\tan^{\alpha}x}\\
&=\int_0^{\frac{\pi}{2}}\frac{1}{\alpha}\cdot\tan x\cdot\cos^2 x\cdot\ln(\tan x)\mathrm{d}\frac{1}{1+\tan^{\alpha}x}\\
&=\frac{1}{\alpha}\int_0^{\frac{\pi}{2}}\sin x\cdot\cos x\cdot\ln(\tan x)\mathrm{d}\frac{1}{1+\tan^{\alpha}x}\\
\end{split}
\]

注意,

\[
\Bigl(\dfrac{1}{1+\tan^{\alpha}x}\Bigr)'_x=\frac{-1}{(1+\tan^{\alpha}x)^2}\cdot\alpha\tan^{\alpha-1}x\cdot\sec^2 x
\]

不要搞错了.