📝 题目
解 (1) 令 ${x}^{1/4} = t$ ,有
$$ {\int }_{0}^{1}\frac{\mathrm{d}x}{\sqrt{1 - {x}^{1/4}}} = {\int }_{0}^{1}4{t}^{3}{\left( 1 - t\right) }^{-1/2}\mathrm{\;d}t = 4\mathrm{\;B}\left( {4,1/2}\right) $$
$$ = 4\frac{\Gamma \left( 4\right) \Gamma \left( {1/2}\right) }{\Gamma \left( {4 + 1/2}\right) } $$
$$ = 4\frac{3!\sqrt{\pi }}{\left( {3 + 1/2}\right) \left( {2 + 1/2}\right) \left( {1 + 1/2}\right) \sqrt{\pi }} $$
$$ = \frac{64}{35}\text{ . } $$
(2) $\displaystyle{\int }_{0}^{\frac{\pi }{2}}\sqrt{\tan x}\mathrm{\;d}x = {\int }_{0}^{\frac{\pi }{2}}{\sin }^{\frac{1}{2}}x{\cos }^{-\frac{1}{2}}x\mathrm{\;d}x = \frac{1}{2}\mathrm{\;B}\left( {\frac{3}{4},\frac{1}{4}}\right)$
$$ = \frac{1}{2}\frac{\Gamma \left( {3/4}\right) \Gamma \left( {1/4}\right) }{\Gamma \left( 1\right) } = \frac{1}{2} \cdot \frac{\pi }{\sin \pi /4} = \frac{\pi }{\sqrt{2}}. $$
(3)解法 1 令 ${x}^{2} = t$ ,我们有
$$ {\int }_{0}^{+\infty }{x}^{2n}{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x = {\int }_{0}^{+\infty }{t}^{n}{\mathrm{e}}^{-t}\frac{\mathrm{d}t}{2\sqrt{t}} = \frac{1}{2}{\int }_{0}^{+\infty }{t}^{n - \frac{1}{2}}{\mathrm{e}}^{-t}\mathrm{\;d}t $$
$$ = \frac{1}{2}\Gamma \left( {n + \frac{1}{2}}\right) = \frac{\left( {{2n} - 1}\right) !!}{{2}^{n + 1}}\sqrt{\pi }. $$
解法 2 因
$$ {\int }_{0}^{+\infty }{\mathrm{e}}^{-a{x}^{2}}\mathrm{\;d}x = \frac{\sqrt{\pi }}{2}{a}^{-\frac{1}{2}}, $$
上式对 $a$ 求导得
$$ {\int }_{0}^{+\infty }{x}^{2}{\mathrm{e}}^{-a{x}^{2}}\mathrm{\;d}x = \frac{\sqrt{\pi }}{2} \cdot \frac{1}{2}{a}^{-\frac{3}{2}}. $$
再对 $a$ 微一次得
$$ {\int }_{0}^{+\infty }{x}^{4}{\mathrm{e}}^{-a{x}^{3}}\mathrm{\;d}x = \frac{\sqrt{\pi }}{2} \cdot \frac{3!!}{{2}^{2}}{a}^{-\frac{5}{2}}. $$
依次微下去,微到第 $n$ 次得
$$ {\int }_{0}^{+\infty }{x}^{2n}{\mathrm{e}}^{-a{x}^{2}}\mathrm{\;d}x = \frac{\sqrt{\pi }}{2} \cdot \frac{\left( {{2n} - 1}\right) !!}{{2}^{n}}{a}^{-\frac{{2n} + 1}{2}}. $$
令 $a = 1$ ,即得
$$ {\int }_{0}^{+\infty }{x}^{2n}{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x = \frac{\left( {{2n} - 1}\right) !!}{{2}^{n + 1}}\sqrt{\pi }. $$
💡 答案与解析
解 (1) 令 ${x}^{1/4} = t$ ,有
$$ {\int }_{0}^{1}\frac{\mathrm{d}x}{\sqrt{1 - {x}^{1/4}}} = {\int }_{0}^{1}4{t}^{3}{\left( 1 - t\right) }^{-1/2}\mathrm{\;d}t = 4\mathrm{\;B}\left( {4,1/2}\right) $$
$$ = 4\frac{\Gamma \left( 4\right) \Gamma \left( {1/2}\right) }{\Gamma \left( {4 + 1/2}\right) } $$
$$ = 4\frac{3!\sqrt{\pi }}{\left( {3 + 1/2}\right) \left( {2 + 1/2}\right) \left( {1 + 1/2}\right) \sqrt{\pi }} $$
$$ = \frac{64}{35}\text{ . } $$
(2) $\displaystyle{\int }_{0}^{\frac{\pi }{2}}\sqrt{\tan x}\mathrm{\;d}x = {\int }_{0}^{\frac{\pi }{2}}{\sin }^{\frac{1}{2}}x{\cos }^{-\frac{1}{2}}x\mathrm{\;d}x = \frac{1}{2}\mathrm{\;B}\left( {\frac{3}{4},\frac{1}{4}}\right)$
$$ = \frac{1}{2}\frac{\Gamma \left( {3/4}\right) \Gamma \left( {1/4}\right) }{\Gamma \left( 1\right) } = \frac{1}{2} \cdot \frac{\pi }{\sin \pi /4} = \frac{\pi }{\sqrt{2}}. $$
(3)解法 1 令 ${x}^{2} = t$ ,我们有
$$ {\int }_{0}^{+\infty }{x}^{2n}{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x = {\int }_{0}^{+\infty }{t}^{n}{\mathrm{e}}^{-t}\frac{\mathrm{d}t}{2\sqrt{t}} = \frac{1}{2}{\int }_{0}^{+\infty }{t}^{n - \frac{1}{2}}{\mathrm{e}}^{-t}\mathrm{\;d}t $$
$$ = \frac{1}{2}\Gamma \left( {n + \frac{1}{2}}\right) = \frac{\left( {{2n} - 1}\right) !!}{{2}^{n + 1}}\sqrt{\pi }. $$
解法 2 因
$$ {\int }_{0}^{+\infty }{\mathrm{e}}^{-a{x}^{2}}\mathrm{\;d}x = \frac{\sqrt{\pi }}{2}{a}^{-\frac{1}{2}}, $$
上式对 $a$ 求导得
$$ {\int }_{0}^{+\infty }{x}^{2}{\mathrm{e}}^{-a{x}^{2}}\mathrm{\;d}x = \frac{\sqrt{\pi }}{2} \cdot \frac{1}{2}{a}^{-\frac{3}{2}}. $$
再对 $a$ 微一次得
$$ {\int }_{0}^{+\infty }{x}^{4}{\mathrm{e}}^{-a{x}^{3}}\mathrm{\;d}x = \frac{\sqrt{\pi }}{2} \cdot \frac{3!!}{{2}^{2}}{a}^{-\frac{5}{2}}. $$
依次微下去,微到第 $n$ 次得
$$ {\int }_{0}^{+\infty }{x}^{2n}{\mathrm{e}}^{-a{x}^{2}}\mathrm{\;d}x = \frac{\sqrt{\pi }}{2} \cdot \frac{\left( {{2n} - 1}\right) !!}{{2}^{n}}{a}^{-\frac{{2n} + 1}{2}}. $$
令 $a = 1$ ,即得
$$ {\int }_{0}^{+\infty }{x}^{2n}{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x = \frac{\left( {{2n} - 1}\right) !!}{{2}^{n + 1}}\sqrt{\pi }. $$