这几天碰到的一些好题 - Eufisky - The lost book
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这几天碰到的一些好题

Eufisky posted @ 2015年6月12日 02:45 in 数学分析 with tags 积分计算 , 1189 阅读
  1. 求\[\iint\limits_D {{e^x}\cos ydxdy} ,\]其中$D: x^2+y^2\leq 1$.

     

    enlightened先证明\[\int_0^{2\pi } {{e^{r\cos \theta }}\cos \left( {r\sin \theta } \right)d\theta } = 2\pi .\]

     

    事实上,在积分号下求导,得$F'\left( r \right) = \int_0^{2\pi } {{e^{r\cos \theta }}\cos \left( {\theta + r\sin \theta } \right)d\theta } $.由归纳法,可知

    \[{F^{\left( n \right)}}\left( r \right) = \int_0^{2\pi } {{e^{r\cos \theta }}\cos \left( {n\theta + r\sin \theta } \right)d\theta } .\tag{1}\]从而导出${F^{\left( n \right)}}\left( 0 \right) = 0\left( {n = 1,2, \cdots } \right)$.因此根据Taylor公式,我们有

    \[F\left( r \right) - F\left( 0 \right) = \sum\limits_{k = 1}^{n - 1} {\frac{{{F^{\left( k \right)}}\left( 0 \right)}}{{k!}}{r^k}} + \frac{{{F^{\left( n \right)}}\left( {{\theta _1}r} \right)}}{{n!}}{r^n} = \frac{{{F^{\left( n \right)}}\left( {{\theta _1}r} \right)}}{{n!}}{r^n}\left( {0 < {\theta _1} < 1} \right).\]

    注意到由(1)可得$\left| {{F^{\left( n \right)}}\left( {{\theta _1}r} \right)} \right| \le 2\pi {e^r}$,故知

    \[\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{F^{\left( n \right)}}\left( {{\theta _1}r} \right)}}{{n!}}{r^n}} \right| \le \mathop {\lim }\limits_{n \to \infty } \frac{{2\pi {e^r}}}{{n!}}{r^n} = 0,F\left( r \right) \equiv F\left( 0 \right) = 2\pi .\]

     

    \begin{align*}\iint\limits_D {{e^x}\cos ydxdy} &= \int_0^1 {rdr\int_0^{2\pi } {{e^{r\cos \theta }}\cos \left( {r\sin \theta } \right)d\theta } } \\&= 2\pi \int_0^1 {rdr} = \pi.\end{align*}

  2. 证明:

    \[\mathop {\lim }\limits_{a \to \frac{\pi }{2}} \int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\sin x - \sin a} }}dx} = \frac{\pi }{{\sqrt 2 }}.\]

    enlightened证法一: 利用椭圆函数的一些性质.

    \begin{align*}&\mathop {\lim }\limits_{a \to \frac{\pi }{2}} \int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\sin x - \sin a} }}dx} = \mathop {\lim }\limits_{a \to \frac{\pi }{2}} \int_0^{\frac{\pi }{2} - a} {\frac{1}{{\sqrt {\cos x - \cos \left( {\frac{\pi }{2} - a} \right)} }}dx} \\= &\mathop {\lim }\limits_{a \to \frac{\pi }{2}} \sqrt 2 K\left( {\sin \frac{1}{2}\left( {\frac{\pi }{2} - a} \right)} \right) = \sqrt 2 K\left( 0 \right) = \frac{\pi }{{\sqrt 2 }},\end{align*}

    其中$K(x)$为Complete Elliptic Integral of the First Kind,以及参考Elliptic Integral of the First Kind.

     

    证法二: 根据

    \begin{align*}&\int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\sin x - \sin a} }}dx} = \int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {2\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}} }}dx} \\= &\frac{1}{{\sqrt 2 }}\int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\frac{{\cos \frac{{x + a}}{2}}}{{\frac{\pi }{2} - \frac{{x + a}}{2}}}\frac{{\sin \frac{{x - a}}{2}}}{{\frac{{x - a}}{2}}}} \sqrt {\left( {\frac{\pi }{2} - \frac{{x + a}}{2}} \right)\frac{{x - a}}{2}} }}dx} \\= &\frac{1}{{\sqrt 2 }}\sqrt {\frac{{\frac{\pi }{2} - \frac{{\xi + a}}{2}}}{{\cos \frac{{\xi + a}}{2}}}} \cdot \sqrt {\frac{{\frac{{\xi - a}}{2}}}{{\sin \frac{{\xi - a}}{2}}}} \int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\left( {\frac{\pi }{2} - \frac{{x + a}}{2}} \right)\frac{{x - a}}{2}} }}dx} \left( {a < \xi < \frac{\pi }{2}} \right).\end{align*}

    \begin{align*}&\int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\left( {\frac{\pi }{2} - \frac{{x + a}}{2}} \right)\frac{{x - a}}{2}} }}dx} = 2\int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\pi \left( {x - a} \right) - {x^2} + {a^2}} }}dx} = 2\int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {{{\left( {a - \frac{\pi }{2}} \right)}^2} - {{\left( {x - \frac{\pi }{2}} \right)}^2}} }}dx} \\= &2\int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {1 - {{\left( {\frac{{2x - \pi }}{{\pi - 2a}}} \right)}^2}} }}d\left( {\frac{{2x - \pi }}{{\pi - 2a}}} \right)} = \left. {2\arcsin \left( {\frac{{2x - \pi }}{{\pi - 2a}}} \right)} \right|_a^{\frac{\pi }{2}} = 2 \times \frac{\pi }{2} = \pi .\end{align*}

     

     

    证法三: 由于

    \begin{align*}&\mathop {\lim }\limits_{a \to \frac{\pi }{2}} \int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\sin x - \sin a} }}dx} = \mathop {\lim }\limits_{a \to \frac{\pi }{2}} \int_0^1 {\frac{{\frac{\pi }{2} - a}}{{\sqrt {\sin \left( {a + \left( {\frac{\pi }{2} - a} \right)t} \right) - \sin a} }}dt} \\= &\mathop {\lim }\limits_{a \to \frac{\pi }{2}} \int_0^1 {\frac{{\frac{\pi }{2} - a}}{{\sqrt {2\sin \left[ {\frac{t}{2}\left( {\frac{\pi }{2} - a} \right)} \right]\cos \left[ {a + \frac{t}{2}\left( {\frac{\pi }{2} - a} \right)} \right]} }}dt} \\= &\int_0^1 {\mathop {\lim }\limits_{a \to \frac{\pi }{2}} \frac{{\frac{\pi }{2} - a}}{{\sqrt {2\sin \left[ {\frac{t}{2}\left( {\frac{\pi }{2} - a} \right)} \right]\sin \left[ {\left( {1 - \frac{t}{2}} \right)\left( {\frac{\pi }{2} - a} \right)} \right]} }}dt} = \int_0^1 {\frac{1}{{\sqrt 2 \cdot \sqrt {\frac{t}{2}\left( {1 - \frac{t}{2}} \right)} }}dt} .\end{align*}

    令$\sqrt {\frac{t}{2}} = \sin \theta $,我们有

     

    \[\int_0^1 {\frac{1}{{\sqrt 2 \cdot \sqrt {\frac{t}{2}\left( {1 - \frac{t}{2}} \right)} }}dt} = \int_0^{\frac{\pi }{4}} {2\sqrt 2 d\theta } = \frac{\pi }{{\sqrt 2 }}.\]

     

    证法四: 显然等价于求

    \[\mathop {\lim }\limits_{a \to {0^ + }} \int_0^a {\frac{1}{{\sqrt {\cos x - \cos a} }}dx} = \mathop {\lim }\limits_{a \to {0^ + }} \int_0^a {\frac{1}{{\sqrt {2\sin \frac{{a - x}}{2}\sin \frac{{a + x}}{2}} }}dx} .\]

    由$0\leq x<a$,则有

    \begin{align*}&{\left( {2\sin \frac{{a - x}}{2}\sin \frac{{a + x}}{2}} \right)^{ - \frac{1}{2}}} = {\left[ {2\left( {\frac{{a - x}}{2} + O\left( {{{\left( {a - x} \right)}^3}} \right)} \right)\left( {\frac{{a + x}}{2} + O\left( {{{\left( {a + x} \right)}^3}} \right)} \right)} \right]^{ - \frac{1}{2}}}\\= &{\left( {\frac{{{a^2} - {x^2}}}{2} + O\left( {\left( {{a^2} - {x^2}} \right){{\left( {a + x} \right)}^2}} \right)} \right)^{ - \frac{1}{2}}} = \frac{{\sqrt 2 }}{{\sqrt {{a^2} - {x^2}} }} + O\left( {\frac{{{{\left( {a + x} \right)}^2}}}{{\sqrt {{a^2} - {x^2}} }}} \right).\end{align*}

    即有

    \begin{align*}\int_0^a {\frac{1}{{\sqrt {\cos x - \cos a} }}dx} &= \int_0^a {\frac{{\sqrt 2 }}{{\sqrt {{a^2} - {x^2}} }}dx} + O\left( {\int_0^a {\frac{{{{\left( {a + x} \right)}^2}}}{{\sqrt {{a^2} - {x^2}} }}dx} } \right)\\&=\frac{\pi }{{\sqrt 2 }} + O\left( {{a^2}} \right) \to \frac{\pi }{{\sqrt 2 }}\left( {a \to 0} \right).\end{align*}

    其中在算阶的时候用到了

    \[\max \left\{ {\left( {{a^2} - {x^2}} \right){{\left( {a + x} \right)}^2},\left( {{a^2} - {x^2}} \right){{\left( {a - x} \right)}^2},{{\left( {{a^2} - {x^2}} \right)}^3}} \right\} = \left( {{a^2} - {x^2}} \right){\left( {a + x} \right)^2},\]

    这里$0\leq x<a$.

     

    推论:设$f(x)$在$[0,a]$具有二阶连续导数,且满足$f'(x)=0,f''(0)<0$,则有

    \[\mathop {\lim }\limits_{a \to {0^ + }} \int_0^a {\frac{{dx}}{{\sqrt {f\left( x \right) - f\left( a \right)} }}} = \frac{\pi }{{\sqrt {2\left| {f''\left( 0 \right)} \right|} }}.\]

    enlightened

    \begin{align*}&\mathop {\lim }\limits_{a \to {0^ + }} \int_0^a {\frac{{dx}}{{\sqrt {f\left( x \right) - f\left( a \right)} }}} = \mathop {\lim }\limits_{a \to {0^ + }} \int_0^1 {\frac{{adt}}{{\sqrt {f\left( {at} \right) - f\left( a \right)} }}} \\=& \mathop {\lim }\limits_{a \to {0^ + }} \int_0^1 {\frac{{\sqrt 2 dt}}{{\sqrt {f''\left( {a{\xi _1}} \right){t^2} - f''\left( {a{\xi _2}} \right)} }}} \left( {0 < {\xi _1} < 1,0 < {\xi _2} < 1} \right)\\= &\int_0^1 {\mathop {\lim }\limits_{a \to {0^ + }} \frac{{\sqrt 2 dt}}{{\sqrt {f''\left( {a{\xi _1}} \right){t^2} - f''\left( {a{\xi _2}} \right)} }}} = \int_0^1 {\mathop {\lim }\limits_{a \to {0^ + }} \frac{{\sqrt 2 dt}}{{\sqrt {\left| {f''\left( 0 \right)} \right|\left( {1 - {t^2}} \right)} }}} \\= &\frac{\pi }{{\sqrt {2\left| {f''\left( 0 \right)} \right|} }}.\end{align*}

  3. 证明:

    \[\mathop {\lim }\limits_{n \to \infty } \int_{n\pi }^{\left( {n + 1} \right)\pi } {\frac{x}{{1 + {x^2}{{\sin }^2}x}}dx} = \pi .\]

    更一般地,以下极限

    \[\mathop {\lim }\limits_{n \to \infty } \int_{n\pi }^{\left( {n + 1} \right)\pi } {\frac{x}{{1 + {x^\alpha }{{\sin }^2}x}}dx} \]

    与$n$的关系是什么?

    enlightened对$x\in [n\pi,(n+1)\pi]$,则有

    \[\frac{{n\pi }}{{1 + {{\left( {n + 1} \right)}^2}{\pi ^2}{{\sin }^2}x}} < \text{被积函数} < \frac{{\left( {n + 1} \right)\pi }}{{1 + {n^2}{\pi ^2}{{\sin }^2}x}}.\]

  4. 求\[\int_0^1 {\left( {1 + \ln x} \right)\ln \left( {1 + x} \right)\ln \ln \frac{1}{x}dx} .\]

    enlightened令$x=e^{-t}$,我们有

    $$ I = \int_{0}^{+\infty}e^{-t}(1-t)\log(1+e^{-t})\log t\,dt$$

    由于

    $$ \int_{0}^{+\infty}e^{-(n+1)t}(1-t)\log t\,dt = -\frac{1}{(n+1)^2}\left(1+n\gamma+n\log (n+1)\right),$$

    我们有

     

    $$ I = \sum_{n\geq 1}\frac{(-1)^n}{n(n+1)^2}\left(1+n\gamma+n\log (n+1)\right).$$

    因此

    $$ I = 2-2\log 2-\gamma-\zeta(2)\left(\frac{1}{2}+\log\sqrt{4\pi}\right)+\pi^2\log A,$$

    其中 $A$ 是Glaisher-Kinkelin constant.

  5. 证明:\[I = \int_0^1 {{x^{ - x}}\left( {{{\ln }^2}x - 2} \right)dx} < 0.\]

    enlightened 注意到

    \[{x^{ - x}} = {e^{ - x\ln x}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k} \cdot \frac{{{x^k}{{\ln }^k}x}}{{k!}}} .\]

    故有

    \[I = \sum\limits_{k = 0}^\infty {\left[ {\frac{{{{\left( { - 1} \right)}^k}}}{{k!}}\int_0^1 {\left( {{x^k}{{\ln }^{k + 2}}x - 2{x^k}{{\ln }^k}x} \right)dx} } \right]} .\]

    \[\int_0^1 {{x^k}{{\ln }^{k + 2}}xdx} = {\left( { - 1} \right)^k}\int_0^\infty {{e^{ - \left( {k + 1} \right)t}}{t^{k + 2}}dt} = \frac{{{{\left( { - 1} \right)}^k}\left( {k + 2} \right)!}}{{{{\left( {k + 1} \right)}^{k + 3}}}},x = {e^{ - t}}.\]

    和\[\int_0^1 {{x^k}{{\ln }^k}xdx} = \frac{{{{\left( { - 1} \right)}^k}k!}}{{{{\left( {k + 1} \right)}^{k + 1}}}}.\]

    因此

    \[I = \sum\limits_{k = 0}^\infty {\left[ {\frac{{{{\left( { - 1} \right)}^k}}}{{k!}}\int_0^1 {\left( {\frac{{{{\left( { - 1} \right)}^k}\left( {k + 2} \right)!}}{{{{\left( {k + 1} \right)}^{k + 3}}}} - 2\frac{{{{\left( { - 1} \right)}^k}k!}}{{{{\left( {k + 1} \right)}^{k + 1}}}}} \right)dx} } \right]} = - \sum\limits_{k = 0}^\infty {\frac{k}{{{{\left( {k + 1} \right)}^{k + 2}}}}} < 0.\]


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