$$ I(r)=\int_{0}^{\pi} \ln (1\pm2r\cos\theta+r^{2}) , \mathrm{d}\theta =\left{ \begin{align} & 0, & \left| r \right| < 1 \ & 2\pi \ln \left| r \right| , & \left| r \right| > 1 \end{align} \right. $$
证明思路 | 递推法
- 证明 $I(r)=I(-r)$
- 证明 $I(r)=\frac{1}{2}I(r^{2})$
- 递推证明 $I(r)=\frac{1}{2^{n}}I(r^{2^{n}})$
- 求值
- $\left| r \right|<1$ 时
- $r^{2^{n}}\to 0$
- $I(r^{2^{n}})\to 0$
- $I(r)=0$
- $\left| r \right| >1$ 时
- $I(r)=I\left( \frac{1}{r} \right)+\int_{0}^{\pi} \ln r^{2} , \mathrm{d}\theta=0+2\pi \ln \left| r \right|$
证明思路 | 一致收敛法
- 糊弄证明一致收敛
- 求 $\frac{ \partial }{ \partial r }I(r)$
- 积回来