📝 题目
例 1 试证明
(1) $\mathrm{B}\left( {x,y}\right) = {\int }_{0}^{+\infty }\frac{{u}^{x - 1}}{{\left( 1 + u\right) }^{x + y}}\mathrm{\;d}u$ ,
(2) $\mathrm{B}\left( {x,y}\right) = {\int }_{0}^{1}\frac{{u}^{x - 1} + {u}^{y - 1}}{{\left( 1 + u\right) }^{x + y}}\mathrm{\;d}u$ .
💡 答案与解析
证明 在下面的积分中做变元替换
$$ t = \frac{u}{1 + u} = 1 - \frac{1}{1 + u} $$
就得到
$$ {\int }_{0}^{1}{t}^{x - 1}{\left( 1 - t\right) }^{y - 1}\mathrm{\;d}t $$
$$ = {\int }_{0}^{+\infty }\frac{{u}^{x - 1}}{{\left( 1 + u\right) }^{x + y}}\mathrm{\;d}u $$
$$ = {\int }_{0}^{1}\frac{{u}^{x - 1}}{{\left( 1 + u\right) }^{x + y}}\mathrm{\;d}u + {\int }_{1}^{+\infty }\frac{{u}^{x - 1}}{{\left( 1 + u\right) }^{x + y}}\mathrm{\;d}u. $$
在最后一个积分中做变元替换
$$ u = \frac{1}{v} $$
又得到
$$ {\int }_{1}^{+\infty }\frac{{u}^{x - 1}}{{\left( 1 + u\right) }^{x + y}}\mathrm{\;d}u = {\int }_{0}^{1}\frac{{v}^{y - 1}}{{\left( 1 + v\right) }^{x + y}}\mathrm{\;d}v. $$
这样, 我们证明了
$$ \mathrm{B}\left( {x,y}\right) = {\int }_{0}^{+\infty }\frac{{u}^{x - 1}}{{\left( 1 + u\right) }^{x + y}}\mathrm{\;d}u $$
$$ = {\int }_{0}^{1}\frac{{u}^{x - 1} + {u}^{y - 1}}{{\left( 1 + u\right) }^{x + y}}\mathrm{\;d}u. $$