方企勤 第六章 多元函数积分学 第5题
📝 题目
例 5 设 $p,q,s \geq 0$ ,证明
$$ I = {\iint }_{\begin{matrix} {x \geq 0,y \geq 0} \\ {x + y \leq 1} \end{matrix}}{x}^{p}{y}^{q}{\left( 1 - x - y\right) }^{s}\mathrm{\;d}x\mathrm{\;d}y $$
$$ = \frac{\Gamma \left( {p + 1}\right) \Gamma \left( {q + 1}\right) \Gamma \left( {s + 1}\right) }{\Gamma \left( {p + q + s + 3}\right) }. $$
💡 答案解析
证法 1 令 $x = r{\cos }^{2}\theta ,y = r{\sin }^{2}\theta$ ,变换的雅可比行列式为
$$ \frac{\partial \left( {x,y}\right) }{\partial \left( {r,\theta }\right) } = \left| \begin{matrix} {\cos }^{2}\theta & - {2r}\cos \theta \sin \theta \\ {\sin }^{2}\theta & {2r}\sin \theta \cos \theta \end{matrix}\right| $$
$$ = {2r}\sin \theta \cos \theta , $$
所以
$$ I = 2{\int }_{0}^{\frac{\pi }{2}}{\cos }^{{2p} + 1}\theta {\sin }^{{2q} + 1}\theta \mathrm{d}\theta {\int }_{0}^{1}{r}^{p + q + 1}{\left( 1 - r\right) }^{s}\mathrm{\;d}r $$
$$ = \mathrm{B}\left( {p + 1,q + 1}\right) \mathrm{B}\left( {p + q + 2,s + 1}\right) $$
$$ = \frac{\Gamma \left( {p + 1}\right) \Gamma \left( {q + 1}\right) }{\Gamma \left( {p + q + 2}\right) } \cdot \frac{\Gamma \left( {p + q + 2}\right) \Gamma \left( {s + 1}\right) }{\Gamma \left( {p + q + s + 3}\right) } $$
$$ = \frac{\Gamma \left( {p + 1}\right) \Gamma \left( {q + 1}\right) \Gamma \left( {s + 1}\right) }{\Gamma \left( {p + q + s + 3}\right) }. $$
证法 2 因 $I = {\int }_{0}^{1}\mathrm{\;d}x{\int }_{0}^{1 - x}{x}^{p}{y}^{q}{\left( 1 - x - y\right) }^{s}\mathrm{\;d}y$ ,对内层积分作定积分变换 $y = \left( {1 - x}\right) t$ ,则
$$ I = {\int }_{0}^{1}\mathrm{\;d}x{\int }_{0}^{1}{x}^{p}{\left( 1 - x\right) }^{q + s + 1}{t}^{q}{\left( 1 - t\right) }^{s}\mathrm{\;d}t $$
$$ = {\int }_{0}^{1}{x}^{p}{\left( 1 - x\right) }^{q + s + 1}\mathrm{\;d}x \cdot {\int }_{0}^{1}{t}^{q}{\left( 1 - t\right) }^{s}\mathrm{\;d}t $$
$$ = \mathrm{B}\left( {p + 1,q + s + 2}\right) \cdot \mathrm{B}\left( {q + 1,s + 1}\right) $$
$$ = \frac{\Gamma \left( {p + 1}\right) \Gamma \left( {q + 1}\right) \Gamma \left( {s + 1}\right) }{\Gamma \left( {p + q + s + 3}\right) }. $$