方企勤 第七章 典型综合题分析 第10题
📝 题目
例 10 求证: $\displaystyle{\int }_{0}^{1}\frac{\ln \left( {1 + x}\right) }{1 + {x}^{2}}\mathrm{\;d}x = \frac{\pi }{8}\ln 2$ .
💡 答案解析
证法 1 令 $x = \tan t,\mathrm{\;d}x = {\sec }^{2}t\mathrm{\;d}t$ ,则
$$ {\int }_{0}^{1}\frac{\ln \left( {1 + x}\right) }{1 + {x}^{2}}\mathrm{\;d}x = {\int }_{0}^{\frac{\pi }{4}}\ln \left( {1 + \tan t}\right) \mathrm{d}t $$
$$ = {\int }_{0}^{\frac{\pi }{4}}\ln \left( {\sin t + \cos t}\right) \mathrm{d}t - {\int }_{0}^{\frac{\pi }{4}}\ln \left( {\cos t}\right) \mathrm{d}t $$
$$ = {\int }_{0}^{\frac{\pi }{4}}\ln \sqrt{2}\sin \left( {t + \frac{\pi }{4}}\right) \mathrm{d}t - {\int }_{0}^{\frac{\pi }{4}}\ln \left( {\cos t}\right) \mathrm{d}t $$
$$ = \frac{\pi }{8}\ln 2 + {\int }_{0}^{\frac{\pi }{4}}\ln \left( {\sin \left( {t + \frac{\pi }{4}}\right) }\right) \mathrm{d}t - {\int }_{0}^{\frac{\pi }{4}}\ln \left( {\cos t}\right) \mathrm{d}t. $$
(7.27)
又令 $t = \frac{\pi }{4} - u$ ,则
$$ {\int }_{0}^{\frac{\pi }{4}}\ln \left( {\sin \left( {t + \frac{\pi }{4}}\right) }\right) \mathrm{d}t = {\int }_{0}^{\frac{\pi }{4}}\ln \left( {\cos u}\right) \mathrm{d}u. \tag{7.28} $$
(7.28)式代入 (7.27) 式即得结论.
证法 2 设 $J\left( y\right) = {\int }_{0}^{1}\frac{\ln \left( {1 + {xy}}\right) }{1 + {x}^{2}}\mathrm{\;d}x$ ,则 $J\left( 0\right) = 0$ ,
$$ \frac{\mathrm{d}J}{\mathrm{\;d}y} = {\int }_{0}^{1}\frac{x}{\left( {1 + {x}^{2}}\right) \left( {1 + {xy}}\right) }\mathrm{d}x $$
$$ = {\int }_{0}^{1}\frac{1}{1 + {y}^{2}}\left\lbrack {\frac{x}{1 + {x}^{2}} + \frac{y}{1 + {x}^{2}} - \frac{y}{1 + {xy}}}\right\rbrack \mathrm{d}x $$
$$ = \frac{1}{1 + {y}^{2}}\left\lbrack {\frac{1}{2}\ln 2 + \frac{\pi }{4}y - \ln \left( {1 + y}\right) }\right\rbrack . $$
对上式两边从 0 到 1 积分,得 $J\left( 1\right) - J\left( 0\right) = \frac{\pi }{4}\ln 2 - J\left( 1\right)$ ,由此推出
$$ J\left( 1\right) = {\int }_{0}^{1}\frac{\ln \left( {1 + x}\right) }{1 + {x}^{2}}\mathrm{\;d}x = \frac{\pi }{8}\ln 2. $$
证法 3 设 $I\left( y\right) = {\int }_{0}^{y}\frac{\ln \left( {1 + {xy}}\right) }{1 + {x}^{2}}\mathrm{\;d}x$ ,则 $I\left( 0\right) = 0$ ,
$$ \frac{\mathrm{d}I}{\mathrm{\;d}y} = \frac{\ln \left( {1 + {y}^{2}}\right) }{1 + {y}^{2}} + {\int }_{0}^{y}\frac{x}{\left( {1 + {x}^{2}}\right) \left( {1 + {xy}}\right) }\mathrm{d}x $$
$$ = \frac{\ln \left( {1 + {y}^{2}}\right) }{2\left( {1 + {y}^{2}}\right) } + \frac{y\arctan y}{1 + {y}^{2}}. $$
上式两边从 0 到 $y$ 积分,得
$$ I\left( y\right) = \frac{1}{2}\left\{ {{\int }_{0}^{y}\ln \left( {1 + {t}^{2}}\right) \operatorname{darctan}t + {\int }_{0}^{y}\arctan t\operatorname{dln}\left( {1 + {t}^{2}}\right) }\right\} $$
$$ = \frac{1}{2}\ln \left( {1 + {y}^{2}}\right) \arctan y, $$
于是
$$ {\int }_{0}^{1}\frac{\ln \left( {1 + x}\right) }{1 + {x}^{2}}\mathrm{\;d}x = I\left( 1\right) = \frac{\pi }{8}\ln 2. $$
证法 4 设 $I = {\int }_{0}^{1}\frac{\ln \left( {1 + x}\right) }{1 + {x}^{2}}\mathrm{\;d}x$ . 因为 $\ln \left( {1 + x}\right) = {\int }_{0}^{1}\frac{x}{1 + {xy}}\mathrm{\;d}y$ ,所以
$$ I = {\int }_{0}^{1}\frac{1}{1 + {x}^{2}}\mathrm{\;d}x{\int }_{0}^{1}\frac{x}{1 + {xy}}\mathrm{\;d}y = {\int }_{0}^{1}\mathrm{\;d}y{\int }_{0}^{1}\frac{x\mathrm{\;d}x}{\left( {1 + {x}^{2}}\right) \left( {1 + {xy}}\right) } $$
$$ = {\int }_{0}^{1}\frac{1}{1 + {y}^{2}}\left\{ {\frac{1}{2}\ln 2 + \frac{\pi }{4}y - \ln \left( {1 + y}\right) }\right\} \mathrm{d}y = \frac{\pi }{4}\ln 2 - I, $$
从而
$$ I = \frac{\pi }{8}\ln 2. $$