## 浙大14年数分题：Dirichlet引理的证明

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

$f(x)$在$[0,1]$单增,证明:

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

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

\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*}

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

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

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

\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*}

## 级数中的一些反例

\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*}

## 多重积分计算的一些题

（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$\iint\limits_D\left(\frac{x}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x}+ \frac{y}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y.$

\begin{align*}&\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 \\&=\int_0^1\mathrm{d}r\int\limits_{x^2+y^2 = r^2}\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}s\\ &=\int_0^1\mathrm{d}r\int\limits_{x^2+y^2 = r^2}\frac{\partial f}{\partial n}\mathrm{d}s=\int_0^1\mathrm{d}r\iint\limits_{x^2+y^2 \leqslant r^2}\Delta f\mathrm{d}x\mathrm{d}y\\&=\int_0^1\mathrm{d}r\iint\limits_{x^2+y^2 \leqslant r^2}(x^2+y^2)\mathrm{d}x\mathrm{d}y=\int_0^1\frac{\pi r^4}{2}\mathrm{d}r =\frac{\pi}{10} .\end{align*}

（2）设$f$在$D:x^2+y^2\leq1$上二阶连续可微,且$\Delta f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=\exp{(-x^2-y^2)},$求

$\iint\limits_D\left(x\frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y.$

\begin{align*}&\iint\limits_D\left(x\frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y\\ &=\int_0^1\mathrm{d}r\int\limits_{x^2+y^2 = r^2}\left(x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}\right)\mathrm{d}s\\ &=\int_0^1\mathrm{d}r\int\limits_{x^2+y^2 =r^2}r\frac{\partial f}{\partial n}\mathrm{d}s=\int_0^1\mathrm{d}r\iint\limits_{x^2+y^2 \leqslant r^2}r\Delta f\mathrm{d}x\mathrm{d}y\\ &=\int_0^1\mathrm{d}r\iint\limits_{x^2+y^2 \leqslant r^2}r\exp{(-x^2-y^2)}\mathrm{d}x\mathrm{d}y=\int_0^1\pi r(1-e^{-r^2})\mathrm{d}r =\frac{\pi }{2e} .\end{align*}

## 北京大学数学科学学院2015年直博生摸底考试试题解答

1.(90分) 设$y=f(x)$是$\mathbb{R}$上的$C^\infty$函数,对任意整数$k\geq0$,记$M_k=\sup_{x\in\mathbb{R}}|f^{(k)}(x)|$.设$m$和$n$为两整数, $0\leq m<n$,试分别就下列情况,给出你的结论和证明.
（1）如果$M_m$和$M_n$均有界,那么对哪些整数$k$, $M_k$有界？对哪些整数$k$, $M_k$可以无界？
（2）如果$\lim_{x\to+\infty}|f^{(m)}(x)|$存在有限极限,而$M_n$有界,则对哪些自然数$k$,极限$\lim_{x\to+\infty}|f^{(k)}(x)|$也存在极限？
（3）如果$\lim_{x\to+\infty}|f^{(m)}(x)|$和$\lim_{x\to+\infty}|f^{(n)}(x)|$都存在有限极限,则对哪些自然数$k$,极限$\lim_{x\to+\infty}|f^{(k)}(x)|$也存在极限？

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.

\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*}
\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*}

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

\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*}

\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*}
\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*}

\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*}

\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*}

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.

${\left( {{A^ * }} \right)^*} = \left| {{A^ * }} \right|{\left( {{A^ * }} \right)^{ - 1}} = {\left| A \right|^{n - 1}} \cdot \frac{1}{{\left| A \right|}}A = {\left| A \right|^{n - 2}}A.$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.

5.(25分) 设$A$是一个3阶实方阵,考虑$A$所定义的线性变换$\mathbb{R}^3\to\mathbb{R}^3,\alpha\to A\alpha$ ( $\alpha$是列向量).试证明:若$AA'=A'A$ (其中$A'$是指$A$的转置矩阵),则上述线性变换必有一个2维不变子空间.

6.(25分) 设$A$和$B$是复数域$\mathbb{C}$上的两个$n$阶方阵,并且$A$有$n$个特征值$1,2,\cdots,n$, $B$也有$n$个特征值$\sqrt{p_1},\cdots,\sqrt{p_n}$,其中$p_1,\cdots,p_n$是前$n$个素数(比如$p_1=2,p_2=3$等).试证明: $M_n(\mathbb{C})$上的线性变换$X\to AXB$是可以对角化的.

7.(25分) 设$A = \left( {\begin{array}{*{20}{c}}{ - 2}&1&3\\{ - 2}&1&2\\{ - 1}&1&2\end{array}} \right),$

1). $f(A)$可对角化.    2). $\varphi (A)$是幂零矩阵.    3). $A=f(A)+\varphi (A)$.

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

8.几何部分共5道小题,每小题10分。

（1）三维欧氏空间中取定直角坐标系。有一直线$l$过点$(1,0,0)$且方向向量为$(0,1,1)$。$l$绕$z$轴旋转生成一个二次曲面$S$。试写出此二次曲面的代数方程(形如$f(x,y,z)=0$).
（2）设有一固定平面$\Sigma$,具有以下性质:上述直线$l$在绕$z$轴旋转过程中总是与$\Sigma$相交。考虑与$\Sigma$平行的平面族$\Sigma_t,t\in\mathbb{R},\Sigma_0=\Sigma$。试证明$\Sigma_t\cap S$总是椭圆.
（3）试证明$t$值变化过程中,上述各椭圆的中心总落在一条过原点的空间定直线$L$上.
（4）固定$L$上任一点$p$,试证明:由$p$向曲面$S$作的各条切线的切点都落在一条椭圆$\Gamma_p$上,且椭圆$\Gamma_p$所在平面是$\Sigma_t$之一.
（5）$S$把它在空间的补集分成内外两个连通分支,其中外部区域不包含原点。取上一小题所述椭圆$\Gamma$所在平面落在$S$外部的一点$\hat p$。试证明:从$\hat p$向$S$所作的各条切线之切点落在一条双曲线$\hat \Gamma$上,且$\hat \Gamma$所在平面过$p$点.

（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

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

• 若$A^2+B^2\neq0$时,作坐标系旋转

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

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

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

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

（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

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

（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$\left( {{x_{1,1}} - {x_{1,0}}} \right) \cdot \left( {2{x_{1,0}}} \right) + \left( {{y_{1,1}} - {y_{1,0}}} \right) \cdot \left( {2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{y_{1,0}}} \right) + \left( {{z_{1,1}} - {z_{1,0}}} \right) \cdot \left( { - 2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{z_{1,0}}} \right) = 0,$

$2{x_{1,1}}{x_{1,0}} + 2\frac{{{C^2} - {A^2} - {B^2}}}{{{A^2} + {B^2} + {C^2}}}{y_{1,1}}{y_{1,0}} - \frac{{8C\sqrt {{A^2} + {B^2}} }}{{{A^2} + {B^2} + {C^2}}} - 2 + \frac{{2\left( {{A^2} + {B^2} + {C^2}} \right) + 8C\sqrt {{A^2} + {B^2}} }}{{m\left( {{A^2} + {B^2} - {C^2}} \right)}}{z_{1,0}} = 0.$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.

PS：这份试卷是考完后的第一天根据好友同学提供的资料进行整理的，感谢他们的辛劳，同时也祝贺他们在昨天下午清华的初试中获得成功。

## 一个很好的积分题

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

\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*}

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

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

\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*}

\begin{align*}&\int_0^\pi  {{e^{p\cos x}}\cos \left( {p\sin x} \right)\cos qxdx}  = \int_0^\pi  {{e^{p\cos x}}\cos qx{\mathop{\rm Re}\nolimits} \left( {{e^{ip\sin x}}} \right)dx} \\= &\int_0^\pi  {\cos qx{\mathop{\rm Re}\nolimits} \left( {{e^{ip\sin x + p\cos x}}} \right)dx}  = \int_0^\pi  {\cos qx{\mathop{\rm Re}\nolimits} \left( {{e^{p{e^{ix}}}}} \right)dx} \\= &\int_0^\pi  {\cos qx{\mathop{\rm Re}\nolimits} \left( {\sum\limits_{k = 0}^\infty  {\frac{{{p^k}{e^{ikx}}}}{{k!}}} } \right)dx}  = \sum\limits_{k = 0}^\infty  {\frac{{{p^k}}}{{k!}}\int_0^\pi  {\cos qx\cos kxdx} } \\= &\frac{1}{2}\sum\limits_{k = 0}^\infty  {\frac{{{p^k}}}{{k!}}\left[ {\frac{{\sin \left( {k - q} \right)x}}{{k - q}} - \frac{{\sin \left( {k + q} \right)x}}{{k + q}}} \right]} _0^\pi  = \frac{\pi }{2}\frac{{{p^q}}}{{q!}}.\end{align*}

## 几个逼格稍高的积分级数题

\begin{align}\int_0^{ + \infty } {\frac{{\sin nx}}{{x + \frac{1}{{x + \frac{2}{{x + \frac{3}{{x +  \cdots }}}}}}}}dx}  &= \frac{{\sqrt {\frac{\pi }{2}} }}{{n + \frac{1}{{n + \frac{2}{{n + \frac{3}{{n +  \cdots }}}}}}}}\\\int_0^{ + \infty } {\frac{{\sin \frac{{n\pi x}}{2}}}{{x + \frac{{{1^2}}}{{x + \frac{{{2^2}}}{{x + \frac{{{3^2}}}{{x +  \cdots }}}}}}}}dx}  &= \frac{1}{{n + \frac{{{1^2}}}{{n + \frac{{{2^2}}}{{n + \frac{{{3^2}}}{{n +  \cdots }}}}}}}}.\end{align}

\begin{align}\sum\limits_{n = 0}^{ + \infty } {\left[ {\left( {1 + \frac{1}{3} + \frac{1}{5} +  \cdots  + \frac{1}{{2n + 1}}} \right) \cdot \frac{1}{{{5^n}\left( {2n + 1} \right)}}} \right]}  &= \frac{{{\pi ^2}}}{{4\sqrt 5 }} - \frac{{\sqrt 5 }}{{24}}{\left( {\ln \left( {2 + \sqrt 5 } \right)} \right)^2}\\\sum\limits_{n = 0}^{ + \infty } {\left[ {\left( {1 + \frac{1}{3} + \frac{1}{5} +  \cdots  + \frac{1}{{2n + 1}}} \right) \cdot \frac{1}{{{9^n}\left( {2n + 1} \right)}}} \right]}  &= \frac{{{\pi ^2}}}{8} - \frac{3}{8}{\left( {\ln 2} \right)^2}\end{align}

## 北大本科06数分期中试题(李伟固命题)

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: $f^{(k)}(0)>0,k=1,2,3,\cdots$.

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

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

${f^{\left( k \right)}}\left( 0 \right) = \sum\limits_{\substack{{k_1} + {k_2} + \cdots + {k_n} = k\\ \left( {{k_1},{k_2}, \cdots ,{k_n}} \right)}} {\lambda _1^{{k_1}}\lambda _2^{{k_2}} \cdots \lambda _n^{{k_n}}} \left( \text{其中}{{k_i} \in N} \right).$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.

$1 - \left\| {\nabla f\left( u \right)} \right\| = \left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {\nabla f\left( u \right)} \right\| \le \left\| {\nabla f\left( u \right) - \nabla f\left( {0,0} \right)} \right\| \le \left\| u \right\|,$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.

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

\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*}

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.

(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.

\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*}

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

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

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

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

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$f(c)f'(c)+f''(c)=0$.

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

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$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = c$.

## 碰到的数学分析中的反例

• 在$\mathbb{R}$上连续的有界函数一定一致连续吗?

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

• 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?

• 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?

## 这几天碰到的一些好题

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*}

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

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

\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

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

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

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

因此

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