Riemann-Lebesgue引理大家都很熟悉,那Dirichlet引理呢?

$f(x)$在$[0,1]$单增,证明:

\[\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {f\left( x \right)\frac{{\sin xy}}{x}dx} = \frac{\pi }{2}f\left( {{0_ + }} \right).\]

这是Dirichlet引理,菲赫金哥尔茨的《微积分教程》第三卷P358有详细的证明.另外,汪林的《数学分析问题研究与评注》P147上有他的推广及其证明.

对任意给出的$\varepsilon>0$, $\exists 0<\delta<1$,使得对于$0<t\leq \delta$,

\[0 \le g\left( t \right) - g\left( {{0_ + }} \right) < M_1\varepsilon ,\]

其中$M_1$是任意给定的常数.

考察积分

\begin{align*}\int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} &= \left( {\int_0^\delta {} + \int_\delta ^1 {} } \right)\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx\\&= {I_1} + {I_2}.\end{align*}

对于$I_1$,运用积分第二中值定理,我们有

\[{I_1} = \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\int_\eta ^\delta {\frac{{\sin xy}}{x}dx} = \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} ,\]

其中第二个因子对于一切值$y$一致有界.事实上,由反常积分$\displaystyle \int_0^\infty {\frac{{\sin z}}{z}dz}$的收敛性,可见当$z\to \infty$时, $z(z\geq 0)$的连续函数$\displaystyle \int_0^z {\frac{{\sin z}}{z}dz} $有有限的极限,并且对于一切值$z$有界

\[\left| {\int_0^z {\frac{{\sin z}}{z}dz} } \right| \le L\left( L \text{为常数}\right),\]从而

\[\left| {\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} } \right| = \left| {\int_0^{y\delta } {} + \int_0^{y\eta } {} } \right| \le 2L.\]

对于第一个因子,取$M_1=\frac{1 }{{4L}}$,则有$f\left( \delta \right) - f\left( {{0_ + }} \right) < \frac{\varepsilon }{{4L}}$.

因此\[\left| {{I_1}} \right| \le \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\left| {\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} } \right| < \frac{\varepsilon }{{4L}} \cdot 2L = \frac{\varepsilon }{2}.\]

至于$I_2$,由于$\displaystyle \int_\delta ^1 {\frac{{f\left( x \right) - f\left( {{0_ + }} \right)}}{x}dx} $存在,由Riemann-Lebesgue引理可知$\mathop {\lim }\limits_{y \to \infty } {I_2} = 0$,即对$\varepsilon >0,\exists M_2>0$,使得$y>M_2$时,有$\left| {{I_2}} \right| < \frac{\varepsilon }{2}$.

因此\[\left| {\int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} } \right| \le \left| {{I_1}} \right| + \left| {{I_2}} \right| < \varepsilon .\]

即\[\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} = 0.\]

从而

\begin{align*}&\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {f\left( x \right)\frac{{\sin xy}}{x}dx} = \frac{\pi }{2}f\left( {{0_ + }} \right)\\=& \mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} + f\left( {{0_ + }} \right)\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\frac{{\sin xy}}{x}dx} \\= &0 + f\left( {{0_ + }} \right)\int_0^{ + \infty } {\frac{{\sin z}}{z}dz} = \frac{\pi }{2}f\left( {{0_ + }} \right).\end{align*}

计算\[\mathop {\lim }\limits_{n \to  + \infty } \frac{{1 + n + \frac{{{n^2}}}{{2!}} +  \cdots  + \frac{{{n^n}}}{{n!}}}}{{{e^n}}} = \frac{1}{2}.\]

解.

法二.(小米)利用概率, 就是$n$个独立的参数为1的指数分布(泊松分布)和小于等于$n$的概率根据中心极限定理概率收敛到$1/2$.

判断题：若级数$\displaystyle \sum_{n=1}^{+\infty}a_n^3$ 收敛, 则$\displaystyle \sum_{n=1}^{+\infty}\frac{a_n}{n}$亦收敛.

反例：(老骥伏枥)令$a_{n}=\frac{1}{\log n}\sqrt[3]{\cos\frac{2n\pi}{3}}$,所以$a_{n}^3=\frac{1}{\log^3n}\cos\frac{2n\pi}{3}$.用狄利克雷判别法级数$\sum_{n=1}^{\infty}a_{n}^3$收敛.(因为$\frac{1}{\log^3n}$递减趋于零, $\cos\frac{2n\pi}{3}$部分和有界).而

\begin{align*}\sum_{n=1}^{\infty}\frac{a_{n}}{n}&=\sum_{k=0}^{\infty}\left[\frac{1}{(3k+3)\log(3k+3)}-\frac{1}{\sqrt[3]{2}}\left(\frac{1}{(3k+1)\log(3k+1)}+\frac{1}{(3k+2)\log(3k+2)}\right)\right]\\&\sim\sum_{k=0}^{\infty}\frac{1}{k\log k}\end{align*}

发散。（柯西积分判别法）

如果是正项级数,根据Holder不等式能够说明$\sum_{n=1}^{\infty}\frac{a_{n}}{n}$收敛.

（1）设$f$在$D:x^2+y^2\leq1$上二阶连续可微,且\[\Delta f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=x^2+y^2,\]求\[\iint\limits_D\left(\frac{x}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x}+ \frac{y}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y.\]

解：(Hansschwarzkopf)根据Gauss公式

这份试题本来已经写好答案了，但因为电脑的事，里面文件都没了。下面重新给出解答：

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2.(30分) 判断级数$\sum\nolimits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}}}{{\sqrt n  + {{\left( { - 1} \right)}^{\left[ {\sqrt n } \right]}}}}}$的敛散性,其中$[x]$表示$x$的取整.

证：\begin{align*}\sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}}}{{\sqrt n  + {{\left( { - 1} \right)}^{\left[ {\sqrt n } \right]}}}}}  &= \sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}\left( {\sqrt n  - {{\left( { - 1} \right)}^{\left[ {\sqrt n } \right]}}} \right)}}{{n - 1}}} \\&= \sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}\sqrt n }}{{n - 1}}}  - \sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^{n + \left[ {\sqrt n } \right]}}}}{{n - 1}}}.\end{align*}

由Leibniz判别法知,级数\[\sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}\sqrt n }}{{n - 1}}}  = \sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}}}{{\sqrt n  - \frac{1}{{\sqrt n }}}}} \]收敛.

当$k \le \sqrt n  < k + 1$,即${k^2} \le n < {\left( {k + 1} \right)^2}$时, ${\left[ {\sqrt n } \right]}=k$,则

\begin{align*}&\sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^{n + \left[ {\sqrt n } \right]}}}}{{n - 1}}}  =  - \frac{1}{2} + \sum\limits_{k = 2}^{ + \infty } {\sum\limits_{n = {k^2}}^{{k^2} + 2k} {\frac{{{{\left( { - 1} \right)}^{n + k}}}}{{n - 1}}} }  =  - \frac{1}{2} + \sum\limits_{k = 2}^{ + \infty } {{{\left( { - 1} \right)}^k}\sum\limits_{n = {k^2}}^{{k^2} + 2k} {\frac{{{{\left( { - 1} \right)}^n}}}{{n - 1}}} } \\&=  - \frac{1}{2} + \sum\limits_{k = 2}^{ + \infty } {{{\left( { - 1} \right)}^k}\sum\limits_{n = {k^2}}^{{k^2} + 2k} {{{\left( { - 1} \right)}^{{k^2}}}\left[ {\left( {\frac{1}{{{k^2} - 1}} + \frac{1}{{{k^2} + 1}} +  \cdots  + \frac{1}{{{k^2} + 2k - 1}}} \right)} \right.} } \\&\left. { - \left( {\frac{1}{{{k^2}}} + \frac{1}{{{k^2} + 2}} +  \cdots  + \frac{1}{{{k^2} + 2k - 2}}} \right)} \right] =  - \frac{1}{2} + \sum\limits_{k = 2}^{ + \infty } {\left[ {\left( {\frac{1}{{{k^2} - 1}} + \frac{1}{{{k^2} + 1}} +  \cdots  + \frac{1}{{{k^2} + 2k - 1}}} \right)} \right.} \\&\left. { - \left( {\frac{1}{{{k^2}}} + \frac{1}{{{k^2} + 2}} +  \cdots  + \frac{1}{{{k^2} + 2k - 2}}} \right)} \right] \le  - \frac{1}{2} + \sum\limits_{k = 2}^{ + \infty } {\left[ {\left( {\frac{{k + 1}}{{{k^2} - 1}} - \frac{k}{{{k^2} + 2k - 2}}} \right)} \right.} \\&\le  - \frac{1}{2} + \sum\limits_{k = 2}^{ + \infty } {\left[ {\left( {\frac{1}{{k - 1}} - \frac{1}{{k + 2}}} \right)} \right.}  =  - \frac{1}{2} + 1 + \frac{1}{2} + \frac{1}{3} = \frac{4}{3}\end{align*}

且

\begin{align*}{a_n} &= \left( {\frac{1}{{{k^2} - 1}} + \frac{1}{{{k^2} + 1}} +  \cdots  + \frac{1}{{{k^2} + 2k - 1}}} \right) - \left( {\frac{1}{{{k^2}}} + \frac{1}{{{k^2} + 2}} +  \cdots  + \frac{1}{{{k^2} + 2k - 2}}} \right)\\&= \left( {\frac{1}{{{k^2} - 1}} - \frac{1}{{{k^2}}}} \right) + \left( {\frac{1}{{{k^2} + 1}} - \frac{1}{{{k^2} + 2}}} \right) +  \cdots  + \left( {\frac{1}{{{k^2} + 2k - 3}} - \frac{1}{{{k^2} + 2k - 2}}} \right) + \frac{1}{{{k^2} + 2k - 1}}\\&\ge \frac{1}{{{k^2} + 2k - 1}} > 0.\end{align*}

故\[\sum\limits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^{n + \left[ {\sqrt n } \right]}}}}{{n - 1}}} \]收敛,从而数列$\sum\nolimits_{n = 2}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^n}}}{{\sqrt n  + {{\left( { - 1} \right)}^{\left[ {\sqrt n } \right]}}}}}$亦收敛.

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3.(30分) 证明\[\int_0^1 {\int_0^1 {\frac{1}{{{{\left( {xy} \right)}^{xy}}}}dxdy} }  = \int_0^1 {\frac{1}{{{x^x}}}dx}  = \sum\limits_{n = 1}^{ + \infty } {\frac{1}{{{n^n}}}} .\]

证：令$u = xy,v = x$,则$x=v,y=\frac uv$.由$0\leq x,y\leq 1$可知$0\leq u\leq v,0\leq v\leq 1$,

\begin{align*}\left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right| = \left| {\begin{array}{*{20}{c}}0&1\\{\frac{1}{v}}&{ - \frac{u}{{{v^2}}}}\end{array}} \right| = - \frac{1}{v}\,,\end{align*}

那么有

\begin{align*}&\int_0^1 {\int_0^1 {\frac{1}{{{{\left( {xy} \right)}^{xy}}}}dxdy} } = \int_0^1 {dv} \int_0^v {\frac{1}{{{u^u}v}}du} \\&= \int_0^1 {du} \int_u^1 {\frac{1}{{{u^u}v}}dv} = \int_0^1 {\frac{{ - \ln u}}{{{u^u}}}du} \\&= \int_0^1 {\frac{{ - \ln u - 1}}{{{u^u}}}du} + \int_0^1 {\frac{1}{{{u^u}}}du} \\&= \left[ {\frac{1}{{{u^u}}}} \right]_0^1 + \int_0^1 {\frac{1}{{{u^u}}}du} = \int_0^1 {\frac{1}{{{x^x}}}dx}.\end{align*}

而

\begin{align*}&\int_0^1 {\frac{1}{{{x^x}}}dx} = \int_0^1 {{e^{ - x\ln x}}dx} = \int_0^1 {{e^{ - x\ln x}}dx} \\&= \int_0^1 {\sum\limits_{n = 0}^{+\infty} {\frac{{{{\left( { - x\ln x} \right)}^n}}}{{n!}}} dx} = \sum\limits_{n = 0}^{+\infty} {\frac{1}{{n!}}\int_0^1 {{{\left( { - x\ln x} \right)}^n}dx} }.\end{align*}

令$t=-(n+1)\ln x$,有

\begin{align*}\int_0^1 {{{\left( { - x\ln x} \right)}^n}dx} &= \frac{1}{{{{\left( {n + 1} \right)}^{n + 1}}}}\int_0^{ + +\infty } {{t^n}{e^{ - t}}dt} \\&= \frac{{\Gamma \left( {n + 1} \right)}}{{{{\left( {n + 1} \right)}^{n + 1}}}} = \frac{{n!}}{{{{\left( {n + 1} \right)}^{n + 1}}}}.\end{align*}

因此有

\begin{align*}\int_0^1 {\frac{1}{{{x^x}}}dx} = \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{n!}}\int_0^1 {{{\left( { - x\ln x} \right)}^n}dx} } = \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{{{\left( {n + 1} \right)}^{n + 1}}}}} =\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{{n^n}}}} .\end{align*}

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4.(25分) 设$A$是一个$n$阶方阵,且$n\geq3$.$A^\ast$是$A$的伴随矩阵(即$A$的代数余子式所组成的矩阵).试证明,若${(A^\ast)}^\ast\neq O$ (零矩阵),则$A$可逆,且此时${(A^\ast)}^\ast$是$A$的一个纯量倍.

证：由于$A$可逆时,$A^\ast$必可逆,从而${(A^\ast)}^\ast$亦可逆;当$A$不可逆时,$A^\ast$的秩不大于$1$,从而${(A^\ast)}^\ast$必为零矩阵.

由此可知,当${(A^\ast)}^\ast\neq O$ 时,$A$可逆.再由\[A{A^ * } = \left| A \right|{I_n} \Rightarrow \left| {{A^ * }} \right| = {\left| A \right|^{n - 1}},{\left( {{A^ * }} \right)^{ - 1}} = \frac{1}{{\left| A \right|}}A\]及\[{A^ * }{\left( {{A^ * }} \right)^*} = \left| {{A^ * }} \right|{I_n}\]可知

\[{\left( {{A^ * }} \right)^*} = \left| {{A^ * }} \right|{\left( {{A^ * }} \right)^{ - 1}} = {\left| A \right|^{n - 1}} \cdot \frac{1}{{\left| A \right|}}A = {\left| A \right|^{n - 2}}A.\]由于${\left| A \right|^{n - 2}}$是个定值,我们得知${(A^\ast)}^\ast$是$A$的一个纯量倍.

解：\[f\left( \lambda \right) = \left| {\lambda {I_n} - A} \right| = \left( {\lambda + 1} \right){\left( {\lambda - 1} \right)^2}.\]

由Cayley---Hamilton定理可知,\[f\left( A \right) = \left( {A + 1} \right){\left( {A - 1} \right)^2} = {A^3} - {A^2} - A + {I_n} = 0.\]

我们有

\[A = {A^3} + \left( {{A^2} - {A^4}} \right).\]取\[f\left( x \right) = {x^3},\varphi \left( x \right) = {x^2} - {x^4}\]即可.

证：（1）记$A(1,0,0)$,设直线上有一点$B(x,y,z)$,则$\overrightarrow {AB} = \left( {x - 1,y,z} \right)=t(0,1,1)$,则$B$为$(1,t,t)$.对于给定$z=t$,其绕$z$轴旋转形成的图形为\[{x^2} + {y^2} = {t^2} + 1 = {z^2} + 1,\]故该二次曲面方程为\[{x^2} + {y^2} - {z^2} - 1 = 0.\]

（2）设固定平面$\Sigma$的方程为$Ax+By+Cz-a_t=0$

• 若$A^2+B^2=0$即$A=B=0$时,方程退化成

$z = \frac{{{a_t}}}{C}$,\[\left\{ \begin{array}{l}{x^2} + {y^2} - {z^2} = 1\\z = \frac{{{a_t}}}{C}\end{array} \right. \Rightarrow \frac{{{x^2}}}{{\frac{{a_t^2}}{{{C^2}}} + 1}} + \frac{{{y^2}}}{{\frac{{a_t^2}}{{{C^2}}} + 1}} = 1,\]

可知此时$\Sigma_t\cap S$为圆,当然可以看成是椭圆.

• 若$A^2+B^2\neq0$时,作坐标系旋转

\[\left\{ \begin{array}{l}{x_1} = \frac{B}{{\sqrt {{A^2} + {B^2}} }}x - \frac{A}{{\sqrt {{A^2} + {B^2}} }}y\\{y_1} = - \frac{{AC}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}x - \frac{{BC}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}y + \frac{{{A^2} + {B^2}}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}z\\{z_1} = \frac{A}{{\sqrt {{A^2} + {B^2} + {C^2}} }}x + \frac{B}{{\sqrt {{A^2} + {B^2} + {C^2}} }}y + \frac{C}{{\sqrt {{A^2} + {B^2} + {C^2}} }}z\end{array} \right.\]

即

\[\left\{ \begin{array}{l}x = \frac{B}{{\sqrt {{A^2} + {B^2}} }}{x_1} - \frac{{AC}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}{y_1} + \frac{A}{{\sqrt {{A^2} + {B^2} + {C^2}} }}{z_1}\\y = - \frac{A}{{\sqrt {{A^2} + {B^2}} }}{x_1} - \frac{{BC}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}{y_1} + \frac{B}{{\sqrt {{A^2} + {B^2} + {C^2}} }}{z_1}\\z = \frac{{\sqrt {{A^2} + {B^2}} }}{{\sqrt {{A^2} + {B^2} + {C^2}} }}{y_1} + \frac{C}{{\sqrt {{A^2} + {B^2} + {C^2}} }}{z_1}\end{array} \right.,\]

该平面方程化为${z_1} = \frac{{{a_t}}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}$, $S$化为

\[x_1^2 + \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}y_1^2 - \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}z_1^2 - \frac{{4C\sqrt {{A^2} + {B^2}} }}{{{A^2} + {B^2} + {C^2}}} - 1 = 0,\]

则截面方程为

\[x_1^2 + \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}y_1^2 - \frac{{{C^2} - {A^2} - {B^2}}}{{{{\left( {{A^2} + {B^2} + {C^2}} \right)}^2}}}a_t^2 - \frac{{4C\sqrt {{A^2} + {B^2}} }}{{{A^2} + {B^2} + {C^2}}} - 1 = 0.\]显然$\Sigma_t\cap S$为椭圆.

综上可知, $\Sigma_t\cap S$总是椭圆.

（3）由(2)可知,在$x_1y_1z_1$坐标系中,椭圆的中心为

\[\left( {0,0,\frac{{{a_t}}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}} \right),\]即中心在$z_1$轴上,\[\left\{ \begin{array}{l}{x_1} = \frac{B}{{\sqrt {{A^2} + {B^2}} }}x - \frac{A}{{\sqrt {{A^2} + {B^2}} }}y = 0\\{y_1} = - \frac{{AC}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}x - \frac{{BC}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}y + \frac{{{A^2} + {B^2}}}{{\sqrt {\left( {{A^2} + {B^2}} \right)\left( {{A^2} + {B^2} + {C^2}} \right)} }}z = 0\end{array} \right.,\]从而在$xyz$坐标系中,椭圆中心落在过原点的定直线

\[L: \left\{ \begin{array}{l}Bx - Ay = 0\\- ACx - BCy + \left( {{A^2} + {B^2}} \right)z = 0\end{array} \right.\]

上.

（4）设\[S: x_1^2 + \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}y_1^2 - \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}z_1^2 - \frac{{4C\sqrt {{A^2} + {B^2}} }}{{{A^2} + {B^2} + {C^2}}} - 1 = 0,\]上一点$P\left( {{x_{1,0}},{y_{1,0}},{z_{1,0}}} \right)$,则曲面$S$在$P$点处的切面为

\[\left( {{x_1} - {x_{1,0}}} \right) \cdot \left( {2{x_{1,0}}} \right) + \left( {{y_1} - {y_{1,0}}} \right) \cdot \left( {2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{y_{1,0}}} \right) + \left( {{z_1} - {z_{1,0}}} \right) \cdot \left( { - 2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{z_{1,0}}} \right) = 0,\]记坐标系$xyz$中,直线$L$上任一点$p$在$x_1,y_1,z_1$中的坐标为$p_1(0,0,m)$,则$p_1$满足切面方程,即\[\left( {0 - {x_{1,0}}} \right) \cdot \left( {2{x_{1,0}}} \right) + \left( {0 - {y_{1,0}}} \right) \cdot \left( {2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{y_{1,0}}} \right) + \left( {m - {z_{1,0}}} \right) \cdot \left( { - 2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{z_{1,0}}} \right) = 0,\]\[ - 2m\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{z_{1,0}} - \frac{{8C\sqrt {{A^2} + {B^2}} }}{{{A^2} + {B^2} + {C^2}}} - 2 = 0 \Rightarrow {z_{1,0}} = \frac{{\left( {{A^2} + {B^2} + {C^2}} \right) + 4C\sqrt {{A^2} + {B^2}} }}{{m\left( {{A^2} + {B^2} - {C^2}} \right)}},\]故此时的$P$的$z_1$坐标为定值, 在$x_1,y_1,z_1$中$P$形成的轨迹为椭圆,且对应在$x,y,z$中$\Sigma_t\cap S$的一个椭圆.

（5）设$x_1y_1z_1$坐标系中, $\Gamma_p: \left\{ \begin{array}{l}{z_1} = \frac{{\left( {{A^2} + {B^2} + {C^2}} \right) + 4C\sqrt {{A^2} + {B^2}} }}{{m\left( {{A^2} + {B^2} - {C^2}} \right)}}\\x_1^2 + \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}y_1^2 - \frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}z_1^2 - \frac{{4C\sqrt {{A^2} + {B^2}} }}{{{A^2} + {B^2} + {C^2}}} - 1 = 0\end{array} \right.$所在平面落在$S$外部的一点$\hat p_1$的坐标为$\left( {{x_{1,1}},{y_{1,1}},{z_{1,1}}} \right)$, $\hat p_1$满足(4)中$P$点处的切面方程,即\[\left( {{x_{1,1}} - {x_{1,0}}} \right) \cdot \left( {2{x_{1,0}}} \right) + \left( {{y_{1,1}} - {y_{1,0}}} \right) \cdot \left( {2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{y_{1,0}}} \right) + \left( {{z_{1,1}} - {z_{1,0}}} \right) \cdot \left( { - 2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{z_{1,0}}} \right) = 0,\]

则有

\[2{x_{1,1}}{x_{1,0}} + 2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{y_{1,1}}{y_{1,0}} - \frac{{8C\sqrt {{A^2} + {B^2}} }}{{{A^2} + {B^2} + {C^2}}} - 2 + \frac{{2\left( {{A^2} + {B^2} + {C^2}} \right) + 8C\sqrt {{A^2} + {B^2}} }}{{m\left( {{A^2} + {B^2} - {C^2}} \right)}}{z_{1,0}} = 0.\]显然其经过$p_1$.进一步地,经过与之前类似的坐标轴旋转,我们知道$P$的轨迹落在一条双曲线上.

背景是这个：

然后郝XX跟我说了下他的求法，还是挺有意思的！

求

\[\int_0^{2\pi } {{e^{\cos x}}\cos \left( {\sin x} \right)\cos nxdx} .\]

由Euler公式,我们有

\begin{align*}&{e^{y\cos x}}\cos \left( {y\sin x} \right) + i{e^{y\cos x}}\sin \left( {y\sin x} \right) = {e^{y\cos x}} \cdot {e^{iy\sin x}}\\=& {e^{y\cos x + iy\sin x}} = {e^{y\left( {\cos x + i\sin x} \right)}} = \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}}}{{n!}}{{\left( {\cos x + i\sin x} \right)}^n}} \\=& \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}}}{{n!}}{{\left( {\cos x + i\sin x} \right)}^n}} = \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}}}{{n!}}\left( {\cos nx + i\sin nx} \right)} \\= &\sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}\cos nx}}{{n!}}} + i\sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}\sin nx}}{{n!}}} .\end{align*}

因此有

\[{e^{y\cos x}}\cos \left( {y\sin x} \right) = \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}\cos nx}}{{n!}}} ,{e^{y\cos x}}\sin \left( {y\sin x} \right) = \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}\sin nx}}{{n!}}} .\]

在第一个式子中令$y=1$,我们有

\[{e^{\cos x}}\cos \left( {\sin x} \right) = \sum\limits_{n = 0}^{ + \infty } {\frac{{\cos nx}}{{n!}}} .\]

因此我们有

\begin{align*}&\int_0^{2\pi } {{e^{\cos x}}\cos \left( {\sin x} \right)\cos nxdx} = \int_0^{2\pi } {\left( {\cos nx \cdot \sum\limits_{n = 0}^{ + \infty } {\frac{{\cos nx}}{{n!}}} } \right)dx} \\=& \int_0^{2\pi } {\left( {\cos nx \cdot \frac{{\cos nx}}{{n!}}} \right)dx} = \left\{ \begin{array}{l}\frac{\pi }{{n!}},n \ge 1\\2\pi ,n = 0\end{array} \right..\end{align*}

而对于无穷乘积\[\prod\limits_{n = 3}^\infty  {\cos \frac{\pi }{{n!}}}  \approx 0.858314.\]也就是管理员一分钟都不能禁他！！！

事实上,我们有

先是证明两个积分成立：

前些年在百度文库(http://wenku.baidu.com/link?url=Y_HDeeYMcyEGuUJWt3fNqC7N08AEqMVfleNVGcv7hC2t9EVO0-MFFHWuqnLYyiDJ4H7ATgg1fOQGN1Lta2RW4Z4pOrm6aZ468WkrTqNqZzG)找到一份李伟固命制的06年北大期中考试题，当时感觉难度稍大，正好结识了Veer大神，便把这份题发给他。事实证明，V神秒得还是很顺利！

1.给定实数$\lambda_i(1\leq i\leq n)$,满足$\sum\limits_{i = 1}^n {\lambda _i^k} > 0\left( {k = 1,2,3, \cdots } \right)$.令$f\left( x \right) = \prod\limits_{i = 1}^n {\frac{1}{{1 - {\lambda _i}x}}}$.证明: $f^{(k)}(0)>0,k=1,2,3,\cdots$.

证明：令$$g\left( x \right) = \ln f\left( x \right) = - \ln \left( {1 - {\lambda _1}x} \right)-\ln \left( {1 - {\lambda _2}x} \right) -\cdots -\ln \left( {1 - {\lambda _n}x} \right)$$$\left( {x \in U\left( {0;\delta } \right)\text{使得对}\forall {\lambda _i},\text{有}1 - {\lambda _i}x > 0} \right)$,则

\[g\left( x \right) = \left( {\sum\limits_{i = 1}^n {{\lambda _i}} } \right)x + \left( {\frac{1}{2}\sum\limits_{i = 1}^n {\lambda _i^2} } \right){x^2} + \cdots + \left( {\frac{1}{k}\sum\limits_{i = 1}^n {\lambda _i^k} } \right){x^k} + \cdots \]由函数幂级数展开的唯一性可知${g^{\left( k \right)}}\left( 0 \right) = \frac{1}{k}\sum\limits_{i = 1}^n {\lambda _i^k} > 0\left( {x \in U\left( {0;\delta } \right)} \right)$.

另一方面$f\left( x \right) = {e^{g\left( x \right)}}\left( {x \in U\left( {0;\delta } \right)} \right)$.首先注意到对任意可导函数$F(x)$,有${\left( {{e^{F\left( x \right)}}} \right)^\prime } = F'\left( x \right){e^{F\left( x \right)}}$.其次注意到对可导函数组$F_1,F_2,\cdots,F_s$,有${\left( {{F_1}{F_2}{F_3} \cdots {F_s}} \right)^\prime } = {{F'}_1}{F_2}{F_3} \cdots {F_s} + {F_1}{F_2}^\prime {F_3} \cdots {F_s} + \cdots + {F_1}{F_2}{F_3} \cdots {F_s}^\prime$,从而归纳可证

\[{f^{\left( k \right)}}\left( x \right) = {\left( {{e^{g\left( x \right)}}} \right)^{\left( k \right)}} = \left( {\sum\limits_{j \in {N_ + },{k_i} \in {N_ + }} {{g^{\left( {{k_1}} \right)}}\left( x \right){g^{\left( {{k_2}} \right)}}\left( x \right) \cdots {g^{\left( {{k_j}} \right)}}\left( x \right)} } \right){e^{g\left( x \right)}}.\]由${g^{\left( k \right)}}\left( 0 \right) > 0,k=1,2,3,\cdots$且$g(0)=0$,所以${f^{\left( k \right)}}\left( 0 \right) > 0,k=1,2,3,\cdots$.

补充：可知\[f\left( x \right) = \prod\limits_{i = 1}^n {\frac{1}{{1 - {\lambda _i}x}}} = \left( {1 + {\lambda _1}x + \lambda _1^2{x^2} + \cdots } \right)\left( {1 + {\lambda _2}x + \lambda _2^2{x^2} + \cdots } \right) \cdots \left( {1 + {\lambda _n}x + \lambda _n^2{x^2} + \cdots } \right).\]由幂级数的乘积公式归纳可得

\[{f^{\left( k \right)}}\left( 0 \right) = \sum\limits_{\substack{{k_1} + {k_2} + \cdots + {k_n} = k\\ \left( {{k_1},{k_2}, \cdots ,{k_n}} \right)}} {\lambda _1^{{k_1}}\lambda _2^{{k_2}} \cdots \lambda _n^{{k_n}}} \left( \text{其中}{{k_i} \in N} \right).\]若能通过$\sum\limits_{i = 1}^n {\lambda _i^k} > 0\left( {k = 1,2,3, \cdots } \right)$得出$\sum\limits_{\substack{{k_1} + {k_2} + \cdots + {k_n} = k\\ \left( {{k_1},{k_2}, \cdots ,{k_n}} \right)}} {\lambda _1^{{k_1}}\lambda _2^{{k_2}} \cdots \lambda _n^{{k_n}}}>0$即可得到证明.

2.令$D = \left\{ {u = \left( {x,y} \right) \in {\mathbb{R}^2}\left| {\left\| u \right\| = \sqrt {{x^2} + {y^2}} \le \frac{1}{2}} \right.} \right\}$. $f(u)=f(x,y)$是全平面上的连续可微函数满足$\left\| {\nabla f\left( {0,0} \right)} \right\| = 1,\left\| {\nabla f\left( u \right) - \nabla f\left( v \right)} \right\| \le \left\| {u - v} \right\|$.那么对于任意的$u,v\in D$,证明函数$f|_D$在$D$中唯一点处达到其最大值.

证明：对$u\in D$,有$\left\| {\nabla f\left( u \right) - \nabla f\left( {0,0} \right)} \right\| \le \left\| {u - \left( {0,0} \right)} \right\|$,即

\[1 - \left\| {\nabla f\left( u \right)} \right\| = \left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {\nabla f\left( u \right)} \right\| \le \left\| {\nabla f\left( u \right) - \nabla f\left( {0,0} \right)} \right\| \le \left\| u \right\|,\]亦即$1 - \left\| u \right\| \le \left\| {\nabla f\left( u \right)} \right\|$,则$\nabla f\left( u \right) \ne \left( {0,0} \right)$,所以$f$的最大值只可能在边界上取得.

易知$f$在其边界上的函数可设为\[g\left( t \right) = f\left( {\frac{1}{2}\cos \theta ,\frac{1}{2}\sin \theta } \right),g'\left( \theta \right) = \frac{1}{2}\left( { - \sin \theta {f_x} + \cos \theta {f_y}} \right),\theta \in \left[ {0,2\pi } \right).\]现假设$f$在其边界上有两点取得最大值,不妨设为$\theta=\theta_1$和$\theta=\theta_2$,记${u_1} = \left( {\frac{1}{2}\cos {\theta _1},\frac{1}{2}\sin {\theta _1}} \right),{u_2} = \left( {\frac{1}{2}\cos {\theta _2},\frac{1}{2}\sin {\theta _2}} \right)$,则由$g'\left( {{\theta _1}} \right) = g'\left( {{\theta _2}} \right) = 0$可得$- \sin {\theta _1}{f_x}\left( {{u_1}} \right) + \cos {\theta _1}{f_y}\left( {{u_1}} \right) = 0$,即$\nabla f\left( {{u_1}} \right)$与$\left( { - \sin {\theta _1},\cos {\theta _1}} \right)$垂直即可.设$\nabla f\left( {{u_1}} \right) = {a_1}\left( {\cos {\theta _1},\sin {\theta _1}} \right)$,由于$f$在$u_1$处取得最大值,则$f$在$u_1$点沿方向$(\cos\theta_1,\sin \theta_1)$的方向导数需大于等于$0$,所以$a_1\geq 0$.

另一方面由$\left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {\nabla f\left( {{u_1}} \right)} \right\| \le \left\| {\nabla f\left( {0,0} \right) - \nabla f\left( {{u_1}} \right)} \right\|$当且仅当${\nabla f\left( {{0,0}} \right)}$与${\nabla f\left( {{u_1}} \right)}$异向取等可知

\[\left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {{u_1}} \right\| \le \left\| {\nabla f\left( {{u_1}} \right)} \right\| = \left| {{a_1}} \right|,\]

即$a_1\geq \frac12$.同理$\nabla f\left( {{u_2}} \right) = {a_2}\left( {\cos {\theta _2},\sin {\theta _2}} \right),a_2\geq \frac12$.由于不等式取等需要与${\nabla f\left( {0,0} \right)}$异向,故$a_1\geq\frac12,a_2\geq\frac12$中有一个是严格的.不妨设为$a_1>\frac12$,再设$\overrightarrow {{r_1}} = \left( {\cos {\theta _1},\sin {\theta _1}} \right),\overrightarrow {{r_2}} = \left( {\cos {\theta _2},\sin {\theta _2}} \right)$,则

\begin{align*}&{\left\| {\nabla f\left( {{u_1}} \right) - \nabla f\left( {{u_2}} \right)} \right\|^2} - {\left\| {{u_1} - {u_2}} \right\|^2}\\= &{\left\| {{a_1}\overrightarrow {{r_1}} - {a_2}\overrightarrow {{r_2}} } \right\|^2} - {\left\| {\frac{1}{2}\overrightarrow {{r_1}} - \frac{1}{2}\overrightarrow {{r_2}} } \right\|^2} = a_1^2 + a_2^2 - 2{a_1}{a_2}\overrightarrow {{r_1}} \overrightarrow {{r_2}} - \left( {\frac{1}{4} - \frac{1}{2}\overrightarrow {{r_1}} \overrightarrow {{r_2}} + \frac{1}{4}} \right)\\= &a_1^2 + a_2^2 - \frac{1}{2} - \left( {2{a_1}{a_2} - \frac{1}{2}} \right)\overrightarrow {{r_1}} \overrightarrow {{r_2}} \left( \text{由于}{\overrightarrow {{r_1}}\text{与} \overrightarrow {{r_2}} \text{不同向},\text{所以}\overrightarrow {{r_1}} \overrightarrow {{r_2}} < 1,\text{且}2{a_1}{a_2} - \frac{1}{2} > 0} \right)\\> &a_1^2 + a_2^2 - \frac{1}{2} - \left( {2{a_1}{a_2} - \frac{1}{2}} \right) = {\left( {{a_1} - {a_2}} \right)^2} \ge 0,\end{align*}

因此${\left\| {\nabla f\left( {{u_1}} \right) - \nabla f\left( {{u_2}} \right)} \right\|^2} - {\left\| {{u_1} - {u_2}} \right\|^2} > 0$,即$\left\| {\nabla f\left( {{u_1}} \right) - \nabla f\left( {{u_2}} \right)} \right\| > \left\| {{u_1} - {u_2}} \right\|$,矛盾,从而$f$在边界上只有一点取得最大值,即函数$f|_D$在$D$中唯一点处达到其最大值.

3.讨论级数$\sum\limits_{n = 1}^{ + \infty } {{{\left( { - 1} \right)}^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}}},\alpha\in \mathbb{R}$的收敛性.

解：(1)当$\alpha\leq0$时,我们可知$\mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}}$发散,从而级数$\sum\limits_{n = 1}^{ + \infty } {{{\left( { - 1} \right)}^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}}}$发散.

(2)当$\alpha>0$时,由于$\mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}} = 0$,我们讨论其前$2n$项和数列的收敛性即可,也就是级数$\sum\limits_{n = 1}^{ + \infty } {\left[ {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right]}$的收敛性.

而

\begin{align*}&\left| {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\= &\left| {\sin \left( {\ln \left( {2k - 1} \right)} \right)\left( {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right) + \frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right) - \sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\\le &\left| {\sin \left( {\ln \left( {2k - 1} \right)} \right)} \right|\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right] + \left| {\frac{{2\cos \left[ {\frac{{\ln \left( {2k - 1} \right) + \ln \left( {2k} \right)}}{2}} \right]\sin \left[ {\frac{{\ln \left( {2k - 1} \right) - \ln \left( {2k} \right)}}{2}} \right]}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\\le &\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right] + \left| {\ln \left( {1 - \frac{1}{{2k}}} \right)} \right|\frac{1}{{{{\left( {2k} \right)}^\alpha}}}.\end{align*}

已知$\left| {\ln \left( {1 - \frac{1}{{2k}}} \right)} \right|\frac{1}{{{{\left( {2k} \right)}^\alpha }}} \sim \frac{1}{{{{\left( {2k} \right)}^{\alpha + 1}}}}$,从而$\sum\limits_{k = 1}^{ + \infty } {\left| {\ln \left( {1 - \frac{1}{{2k}}} \right)} \right|\frac{1}{{{{\left( {2k} \right)}^\alpha }}}}$收敛.又$\sum\limits_{k = 1}^{ + \infty } {\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right]}$是收敛的,故$\sum\limits_{n = 1}^{ + \infty } {\left| {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|}$收敛,因此$\sum\limits_{n = 1}^{ + \infty } {\left[ {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right]}$亦收敛.

综上所述, 当$\alpha\leq 0$时,级数发散;当$\alpha>0$时级数收敛.

3.求\[\mathop {\inf }\limits_{n \ge 1} \left\{ {\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} } \right\}.\]

解：首先, $\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^1 {\frac{{\cos kx}}{k}} = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \cos x = 0,\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^2 {\frac{{\cos kx}}{k}} = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \left( {\cos x + \frac{{\cos 2x}}{2}} \right) = - \frac{1}{2}$.

现记${f_n}\left( x \right) = \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}}$,则$\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_1}\left( x \right) = 0 \ge - \frac{1}{2},\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_1}\left( x \right) = - \frac{1}{2} \ge - \frac{1}{2}$.

现假设$n=k-1(k\geq2,k\in\mathbb{N}_+)$时,有$\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_n}\left( x \right) = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_{k - 1}}\left( x \right) \ge - \frac{1}{2}$成立;

当$n=k$时,注意到\[{f_n}^\prime \left( x \right) = \sum\limits_{k = 1}^n {\left( { - \sin kx} \right)} = \sum\limits_{i = 1}^k {\left( { - \sin ix} \right)} = \sum\limits_{i = 1}^k {\frac{{\cos \left( {i + \frac{1}{2}} \right)x - \cos \left( {i - \frac{1}{2}} \right)x}}{{2\sin \frac{x}{2}}}} ,\]

则\[{f_k}^\prime \left( x \right) = \frac{{\cos \left( {k + \frac{1}{2}} \right)x - \cos \frac{x}{2}}}{{2\sin \frac{x}{2}}} = - \frac{{\sin \frac{{k + 1}}{2}x\sin \frac{k}{2}x}}{{\sin \frac{x}{2}}},x \in \left[ {0,\frac{\pi }{2}} \right].\]

由$f_k(x)$的连续性可知$\min f_k(x)$的点只可能在端点或稳定点,从而令$f'_k(x)=0$,则$x = \frac{{2j\pi }}{{k + 1}},\frac{{2j\pi }}{k},j \in \mathbb{Z}$且$\frac{{2j\pi }}{{k + 1}},\frac{{2j\pi }}{k} \in \left( {0,\frac{\pi }{2}} \right)$,而

\[{f_k}\left( {\frac{{2j\pi }}{{k + 1}}} \right) = \sum\limits_{i = 1}^n {\frac{{\cos i\left( {\frac{{2j\pi }}{{k + 1}}} \right)}}{i}} = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2kj\pi }}{{k + 1}}}}{k} = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2j\pi }}{{k + 1}}}}{k}.\]

由于$\frac{{2j\pi }}{{k + 1}} \in \left[ {0,\frac{\pi }{2}} \right]$,所以$\frac{{\cos \frac{{2j\pi }}{{k + 1}}}}{k} > 0$,则\[{f_k}\left( {\frac{{2j\pi }}{{k + 1}}} \right) = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2j\pi }}{{k + 1}}}}{k} \ge - \frac{1}{2}.\]

而${f_k}\left( {\frac{{2j\pi }}{k}} \right) = \sum\limits_{i = 1}^n {\frac{{\cos i\left( {\frac{{2j\pi }}{k}} \right)}}{i}} = {f_{k - 1}}\left( {\frac{{2j\pi }}{k}} \right) + \frac{1}{k} \ge - \frac{1}{2}$,又

${f_k}\left( 0 \right) = \sum\limits_{i = 1}^k {\frac{1}{i}} > - \frac{1}{2},{f_k}\left( {\frac{\pi }{2}} \right) = \sum\limits_{i = 1}^k {\frac{{\cos i\frac{\pi }{2}}}{i}} = \left\{ \begin{array}{l}\frac{1}{2}\left[ { - 1 + \left({\frac{1}{2} - \frac{1}{3}} \right) + \cdots + \left( {\frac{1}{{s - 1}} - \frac{1}{s}} \right)} \right],k = 2s + 1\\\frac{1}{2}\left[ { - 1 + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \cdots + \left( {\frac{1}{{s - 1}} - \frac{1}{s}} \right)} \right],k = 2s\end{array} \right.$,从而${f_k}\left( {\frac{\pi }{2}} \right) \ge - \frac{1}{2}$.

综上归纳可知$\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_n}\left( x \right) = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} \ge - \frac{1}{2}$,又$\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^2 {\frac{{\cos kx}}{k}} = - \frac{1}{2}$,因此

\[\mathop {\inf }\limits_{n \ge 1} \left\{ {\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} } \right\} = - \frac{1}{2}.\]

4.函数$f(x)$在$[0,1]$上二次可导, $f(0)=2,f'(0)=-2,f(1)=1$.证明存在$c\in (0,1)$,使得$f(c)f'(c)+f''(c)=0$.

证明：令$F\left( x \right) = \frac{1}{2}{f^2}\left( x \right) + f'\left( x \right)$,则$F(0)=\frac12\times 2^2-2=0$.现假设不存在$\xi\in (0,1]$使得$F(\xi)=0$,则$F$在$(0,1]$上不变号.倘若$\forall x\in (0,1]$都有$F(x)<0$,即$\frac{1}{2}{f^2}\left( x \right) + f'\left( x \right) < 0$,则$f'\left( x \right) < - \frac{1}{2}{f^2}\left( x \right) \le 0$,则$f$在$[0,1]$上单调递减,由$f(0)=2,f(1)=1$可知$1\leq f(x)\leq 2,f(x)\neq 0$,从而$\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}} < - \frac{1}{2}$.由$f,f'$的连续性可知$\int_0^1 {\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}}dx} < \int_0^1 {\left( { - \frac{1}{2}} \right)dx}$,即$\left. { - \frac{1}{{f\left( x \right)}}} \right|_0^1 < - \frac{1}{2}$,得到$-\frac12<-frac12$,矛盾.

倘若$\forall x\in (0,1]$都有$F(x)>0$,即$\frac{1}{2}{f^2}\left( x \right) + f'\left( x \right) > 0$,由于$f(1)=1\neq 0$,则$\forall x\in U^-(1,\delta)$,有$\int_x^1 {\frac{{f'\left( t \right)}}{{{f^2}\left( t \right)1}}dt} > \int_x^1 {\left( { - \frac{1}{2}} \right)dt}$,即$\left. { - \frac{1}{{f\left( t \right)}}} \right|_x^1 > \frac{{x - 1}}{2}$,从而$\frac{1}{{f\left( x \right)}} > \frac{{x + 1}}{2}$.

当$x\in U^-(1,\delta)$时,有$\frac{1}{{f\left( x \right)}} > \frac{{x + 1}}{2} > 0\left( {\delta \le 1} \right)$,故$f(x)$在$[0,1]$上不为$0$,则$\frac{1}{{f\left( 0 \right)}} > \frac{{0 + 1}}{2}$,则$\frac12>\frac12$,矛盾.

因此存在$\xi\in (0,1]$使得$F(\xi)=0$,从而$F(0)=F(\xi)=0$,由罗尔定理可知$\exists c\in (0,1)$,使得

\[F'\left( c \right) = f\left( c \right)f'\left( c \right) + f''\left( c \right) = 0.\]

5.$A$和$B$是自然数$\mathbb{N}$的两个无穷子集,满足$A\cap B=\text{空集},A\cup B=\mathbb{N}$,对于任意的自然数$c>0$,是否存在两个递增的数列$\{a_n\},\{b_n\},\{a_n\}\in A,\{b_n\}\in B$,使得$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = c$.

证明：对$\forall l\geq s\in \mathbb{N}_+$.记$s\sim l=\{s,s+1,\cdots,l\}$,则若对$\forall N\in \mathbb{N}_+$, $\exists n_2\geq n_1\geq N\geq c^2$有$x_{n_1}\sim x_{n_2}\subset B$.令$x_{n_1}=v_1,x_{n_2}=u_1$,则$v_1\sim u_1\subset B$.若$cv_1\sim cu_1+c[\sqrt{u_1}+1]\not\subset B$,则$\exists a\in A,b\in B$,使得$|a-cb|\leq c$或$|a-cb|\leq \sqrt{b}+1$.

若$cv_1\sim cu_1+c[\sqrt{u_1}+1]\subset B$,令$u_2=u_1+[\sqrt{u_1}+1]$,则$cv_1\sim cu_2\subset B$.以此类推,若始终成立$c^kv_1 \sim c^k u_{k+1}\subset B$,易知$u_{k+1}=u_k+[\sqrt{u_k}+1]\geq u_k+\sqrt{u_1}$,即$u_{k+1}-u_k\geq \sqrt{u_1}$,所以$u_k$有趋于无穷大的趋势,所以存在某个$k_0\in \mathbb{N}_+$,使得$\frac{u_{k_0}}{v_1}>c$.而$c^{k_0}v_1\sim c^{k_0} u_{k_0+1}\subset B$.令$x_{n_{k_0}}\in B$满足$c^{k_0}v_1\sim c^{k_0}u_{k_0+1}\subset c^{k_0}v_1\sim x_{n_{k_0}}\subset B,x_{n_{k_0}}+1\in A$,令$a_1=x_{n_{k_0}}+1$,则$c^{k_0}v_1<\frac{a_1}{c}\leq x_{n_{k_0}+1}$,则$\exists b_1\in B$,使得$\left| {\frac{{{a_1}}}{c} - {b_1}} \right| \le 1$,即$\left| {{a_1} - c{b_1}} \right| \le c$.

若不始终成立$c^kv_1\sim c^ku_{k+1}$,则$\exists b_1\in B,a_1\in A$有$|a_1-cb_1|\leq c$或$|a_1-cb_1|\leq \sqrt{b_1}+1$,即$|a_1-cb_1|\leq \sqrt{a_1}+1$.接着再取$n_1\geq n_2\geq N$,有$x_{n_2}\geq x_{n_1}>a_1,b_1$且$x_{n_1}\sim x_{n_2}\subset B$,则$\exists a_2\in A,b_2\in B$有$|a_2-cb_2|\leq \sqrt{b_2}+1,\cdots$,故存在递增数列$\{a_n\}\subset A,\{b_n\}\subset B$,有$|a_n-cb_n|\leq \sqrt{b_n}+1$,则$\left| {\frac{{{a_n}}}{{{b_n}}} - c} \right| \le \frac{1}{{\sqrt {{b_n}} }} + \frac{1}{{{b_n}}}$,所以$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = c$.

• 在$\mathbb{R}$上连续的有界函数一定一致连续吗?

考察函数$$f(x)=\sin x^2.$$

• 设函数$f(x)$在区间$[a,b]$上可微,则$f'(x)$在$[a,b]$上有界?

考察函数\[f(x)=x^{\frac32}\sin\frac1x.\]

• 存不存在这样函数, $f(a)=0$, $f(x)$在区间$[a,b]$上连续,在区间$(a,b)$上可导且$f(x)>0$,而对任意$\varepsilon>0$,函数在区间$(a,a+\varepsilon)$是不单调递增的?

考察函数\[x\sin \frac1x+10x.\]

• 是否存在仅在一点可导而在该点之外的每一点都不可导的函数?

考察函数\[f\left( x \right) = \left\{ \begin{array}{l}{x^2},x\text{为无理数}\\0,x\text{为有理数}\end{array} \right.\]在点$x=0$即可.

1. 求\[\iint\limits_D {{e^x}\cos ydxdy} ,\]其中$D: x^2+y^2\leq 1$.

    

   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:

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

   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*}

   而

   \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

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

   

   \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?

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

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

    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

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

   又

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

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

   因此

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