📝 题目
例 2 利用 $\displaystyle{\frac{1}{2\pi }{\int }_{0}^{2\pi }\frac{1 - {r}^{2}}{1 - {2r}\cos x + {r}^{2}}\mathrm{\;d}x = 1}$ ,试计算积分
$$ I\left( r\right) = {\int }_{0}^{2\pi }\ln \left( {1 - {2r}\cos x + {r}^{2}}\right) \mathrm{d}x\;\left( {r > 0,r \neq 1}\right) . $$
💡 答案与解析
解 在矩形 $0 \leq x \leq {2\pi },0 \leq r \leq {r}_{0}\left( { < 1}\right)$ 上,函数 $\ln (1 - {2r}\cos x +$ $\left. {r}^{2}\right)$ 及其对 $r$ 的偏导数在矩形上连续,故 $I\left( r\right) \in C\left\lbrack {0,{r}_{0}}\right\rbrack$ ,及
$$ {I}^{\prime }\left( r\right) = {\int }_{0}^{2\pi }\frac{2\left( {r - \cos x}\right) }{1 - {2r}\cos x + {r}^{2}}\mathrm{\;d}x\;\left( {0 \leq r \leq {r}_{0}}\right) . $$
由 ${r}_{0}$ 的任意性,得出 $I\left( r\right) \in C\lbrack 0,1)$ 且上述积分在 $\lbrack 0,1)$ 上成立. 当 $r > 0$ 时,
$$ {I}^{\prime }\left( r\right) = \frac{1}{r}{\int }_{0}^{2\pi }\left\lbrack {1 - \frac{1 - {r}^{2}}{1 - {2r}\cos x + {r}^{2}}}\right\rbrack \mathrm{d}x = 0, $$
故
$$ I\left( r\right) = c\;\left( {0 < r < 1}\right) . $$
由 $I\left( r\right) \in C\lbrack 0,1)$ ,所以 $I\left( r\right) = c\left( {0 \leq r < 1}\right)$ ,并令 $r = 0$ ,定出任意常数 $c = I\left( 0\right) = 0$ ,最后得 $I\left( r\right) = 0\left( {0 \leq r < 1}\right)$ . 当 $r > 1$ 时,
$$ I\left( r\right) = {\int }_{0}^{2\pi }\ln \left( {1 - {2r}\cos x + {r}^{2}}\right) \mathrm{d}x $$
$$ = {\int }_{0}^{2\pi }\ln {r}^{2}\mathrm{\;d}x + I\left( {1/r}\right) = {4\pi }\ln r. $$
评注 重积分部分讨论单层位势时,需要用到这个积分.