1.讨论下列广义积分的一致收敛性:

（1） $\displaystyle \int_0^{ + \infty } {{e^{ - \left( {1 + {a^2}} \right)t}}\sin tdt} ,\quad a \in \left( { - \infty , + \infty } \right)$;

（2） $\displaystyle \int_0^{ + \infty } {\frac{{\cos xy}}{{\sqrt {x + y} }}dx} ,\quad y \in \left[ {{y_0}, + \infty } \right)$,其中$y_0>0$;

（3） $\displaystyle \int_0^{ + \infty } {{e^{ - t{x^2}}}dx} ,\quad t \in \left( {0, + \infty } \right)$;

（4） $\displaystyle \int_1^{ + \infty } {{e^{ - \alpha x}}\frac{{\cos x}}{{\sqrt x }}dx} ,\quad \alpha \in \left[ {0, + \infty } \right)$;

（5） $\displaystyle \int_0^{ + \infty } {{e^{ - {{\left( {x - y} \right)}^2}}}dx} ,\quad y \in \left( { - \infty , + \infty } \right)$;

（6） $\displaystyle \int_0^{ + \infty } {x\ln x{e^{ - t\sqrt x }}dx}$,$\quad$ (1) $t\in [t_0,+\infty)$,其中$t_0>0$,$\quad$ (2) $t\in (0,+\infty)$;

（7） $\displaystyle \int_1^{ + \infty } {\frac{{1 - {e^{ - ut}}}}{t}\cos tdt} ,\quad u \in \left[ {0,1} \right]$;

（8） $\displaystyle \int_0^{ + \infty } {\frac{{\alpha t}}{{1 + {\alpha ^2} + {t^2}}} \cdot {e^{ - {\alpha ^2}{t^2}}}\cos {\alpha ^2}{t^2}dt} ,\quad \alpha \in \left( {0, + \infty } \right)$;

（9） $\displaystyle \int_0^{ + \infty } {{e^{ - {x^2}\left( {1 + {y^2}} \right)}}\sin ydy} ,\quad x \in \left( {0, + \infty } \right)$;

（10） $\displaystyle \int_0^{ + \infty } {\frac{{\alpha dx}}{{1 + {\alpha ^2}{x^2}}}} ,\quad \alpha \in \left( {0,1} \right)$;

（11） $\displaystyle \int_0^2 {\frac{{{x^t}}}{{\sqrt[3]{{\left( {x - 1} \right)\left( {x - 2} \right)}}}}dx} ,\quad \left| t \right| < \frac{1}{2}$;

（12） $\displaystyle \int_0^1 {{{\left( {1 - x} \right)}^{u - 1}}dx}$,$\quad$ (1) $u\in [a,+\infty)$,其中$a>0$,$\quad$ (2) $u\in (0,+\infty)$.

解.

（1） 一致收敛.由于

$$\left| {{e^{ - \left( {1 + {a^2}} \right)t}}\sin t} \right| \le {e^{ - t}},\quad 0 \le t < + \infty , - \infty < a < + \infty ,$$

而$\int_0^{ + \infty } {{e^{ - t}}dt} = 1$收敛,由Weierstrass判别法知, $\int_0^{ + \infty } {{e^{ - \left( {1 + {a^2}} \right)t}}\sin tdt}$在$\left( { - \infty , + \infty } \right)$上一致收敛.

（2） 一致收敛.由于

$$\left| {\int_0^A {\cos xydx} } \right| = \left| {\frac{{\sin Ay}}{y}} \right| \le \frac{1}{y} \le \frac{1}{{{y_0}}}, \quad A \ge 0,y \ge {y_0},$$

因此它在$[y_0,+\infty)$一致有界.而$1/\sqrt{x+y}$是$x$的单调减少函数且$0<1/\sqrt{x+y}\leq 1/\sqrt{x+y_0}$,而$\lim_{x\to +\infty} \frac{1}{\sqrt{x+y_0}}=0$, 故这个极限关于$y$在$[y_0,+\infty)$上是一致的.于是由Dirichlet判别法知$\int_0^{ + \infty } {\frac{{\cos xy}}{{\sqrt {x + y} }}dx}$在$\left[ {{y_0}, + \infty } \right)$上一致收敛.

（3） 非一致收敛.对于正整数$n$,取$t_n=\frac1{n^2}$,这时

$$\begin{align*}\left| {\int_n^{2n} {{e^{ - {t_n}{x^2}}}dx} } \right| &= \int_n^{2n} {{e^{ - \frac{1}{{{n^2}}}{x^2}}}dx} > \int_n^{2n} {{e^{ - \frac{1}{{{n^2}}}{{\left( {2n} \right)}^2}}}dx} \\&= \int_n^{2n} {{e^{ - 4}}dx} = {e^{ - 4}}n \ge {e^{ - 4}}.\end{align*}$$

因此,只要取$\varepsilon_0=e^{-4}$,则对于任意大的正数$A_0$,总存在正整数$n$满足$n>A_0$,及$t_n=1/n^2\in (0,+\infty)$,使得$\left| {\int_n^{2n} {{e^{ - {t_n}{x^2}}}dx} } \right| > {e^{ - 4}} = {\varepsilon _0}$.由Cauchy收敛原理的推论可知$\int_0^{ + \infty } {{e^{ - t{x^2}}}dx}$关于$t$在$\left( {0, + \infty } \right)$上非一致收敛.

（4） $\int_1^A {\cos xdx}$显然有界, $1/\sqrt{x}$在$[1,+\infty)$上单调且$\lim_{x\to +\infty}\frac{1}{\sqrt{x}}=0$,由Dirichlet判别法, $\int_1^{+\infty}\frac{\cos x}{\sqrt{x}}$收敛,它当然关于$\alpha$一致收敛.显然$e^{-\alpha x}$关于$x$单调,且 $$0\leq e^{-\alpha x}\leq 1,\quad 0\leq \alpha<+\infty,1\leq x<+\infty,$$ 即$e^{-\alpha x}$一致有界.由Abel判别法, $\int_1^{ + \infty } {{e^{ - \alpha x}}\frac{{\cos x}}{{\sqrt x }}dx}$在$\left[ {0, + \infty } \right)$上一致收敛.

（5） 不一致收敛.注意到$\displaystyle J\left( A \right) = \int_A^{2A} {{e^{ - {{\left( {x - y} \right)}^2}}}dx} = \int_{A - y}^{2A - y} {{e^{ - {u^2}}}du}$,并让$y$取$A$值,则得$\displaystyle J\left( A \right) = \int_0^A {{e^{ - {u^2}}}du} \to \int_0^{ + \infty } {{e^{ - {u^2}}}du} \left( {A \to + \infty } \right)$,即$J(A)$在$A\to +\infty$时不趋于$0$.

（6） 先证明$\int_0^{ + \infty } {x\ln x{e^{ - t\sqrt x }}dx}$在$[t_0,+\infty)(t_0>0)$上一致收敛.由于

$$\left| {x\ln x{e^{ - t\sqrt x }}} \right| \le \left| {x\ln x} \right|{e^{ - {t_0}\sqrt x }},\quad 0\leq x<+\infty,t_0\leq t<+\infty,$$

而$\int_0^1 {x\ln \frac{1}{x}{e^{ - {t_0}\sqrt x }}dx}$与$\int_1^{ + \infty } {x\ln x{e^{ - {t_0}\sqrt x }}dx}$均收敛,由Weierstrass判别法知,

$\int_0^{ + \infty } {x\ln x{e^{ - t\sqrt x }}dx}$在$[t_0,+\infty)$上一致收敛.

再证明$\int_0^{ + \infty } {x\ln x{e^{ - t\sqrt x }}dx}$在$(0,+\infty)$上非一致收敛.对于正整数$n$,取$t_n=\frac{1}{\sqrt{n}}$,这时

$$\begin{align*}&\left| {\int_n^{2n} {x\ln x{e^{ - {t_n}\sqrt x }}dx} } \right| = \left| {\int_n^{2n} {x\ln x{e^{ - \frac{1}{{\sqrt n }}\sqrt x }}dx} } \right|\\>& n\ln n\int_n^{2n} {{e^{ - \frac{1}{{\sqrt n }}\sqrt x }}dx} > n\ln n\int_n^{2n} {{e^{ - \frac{1}{{\sqrt n }}\sqrt {2n} }}dx} \\=& {n^2}\ln n \cdot {e^{ - \sqrt 2 }} \ge 4\ln 2 \cdot {e^{ - \sqrt 2 }}.\end{align*}$$

因此,只要取${\varepsilon _0} = 4\ln 2 \cdot {e^{ - \sqrt 2 }}$,则对于任意大的正数$A_0$,总存在正整数$n$满足$n>A_0$,及$y_n=\frac1{\sqrt n}\in (0,+\infty)$,使得$\left| {\int_n^{2n} {x\ln x{e^{ - {t_n}\sqrt x }}dx} } \right| > 4\ln 2 \cdot {e^{ - \sqrt 2 }}=\varepsilon_0$.由Cauchy收敛原理的推论知$\int_0^{ + \infty } {x\ln x{e^{ - t\sqrt x }}dx}$关于$t$在$(0,+\infty)$上非一致收敛.

（7） 一致收敛.由于$\int_1^{ + \infty } {\frac{{\cos t}}{t}dt}$收敛,它当然关于$u$一致收敛.显然$1-e^{-ut}$关于$t$单调,且

$$0\leq 1-e^{-ut}\leq 1,\quad 0\leq u\leq1,1\leq t<+\infty,$$ 即$1-e^{-ut}$一致有界.由Abel判别法, $\int_1^{ + \infty } {\frac{{1 - {e^{ - ut}}}}{t}\cos tdt}$在$\left[ {0,1} \right]$上一致收敛.

（8） 一致收敛.由于 $$\left| {\int_0^A {\alpha t \cdot {e^{ - {\alpha ^2}{t^2}}}\cos {\alpha ^2}{t^2}dt} } \right| \le \int_0^A {\alpha t \cdot {e^{ - {\alpha ^2}{t^2}}}dt} = \frac{{1 - {e^{ - {\alpha ^2}{A^2}}}}}{{2\alpha }},$$ 且 $$\mathop {\lim }\limits_{\alpha \to 0} \frac{{1 - {e^{ - {\alpha ^2}{A^2}}}}}{{2\alpha }} = 0,\mathop {\lim }\limits_{\alpha \to + \infty } \frac{{1 - {e^{ - {\alpha ^2}{A^2}}}}}{{2\alpha }} = 0,$$

因此它在$(0,+\infty)$一致有界,而$\frac{1}{{1 + {\alpha ^2} + {t^2}}}$是$x$的单调减少函数且$\frac{1}{{1 + {\alpha ^2} + {t^2}}} \le \frac{1}{{1 + {t^2}}},\lim_{t\to \infty}\frac1{1+t^2}=0$,因此$\lim_{t\to +\infty}\frac{1}{{1 + {\alpha ^2} + {t^2}}}=0$关于$\alpha$在$(0,+\infty)$上是一致的,于是由Dirichlet判别法知$\int_0^{ + \infty } {\frac{{\alpha t}}{{1 + {\alpha ^2} + {t^2}}} \cdot {e^{ - {\alpha ^2}{t^2}}}\cos {\alpha ^2}{t^2}dt}$在$\left( {0, + \infty } \right)$上一致收敛.

（9） 非一致收敛.对于正整数$n$,取${x_n} = \frac{1}{{\sqrt {1 + {{\left( {2n + 1} \right)}^2}{\pi ^2}} }}$,这时

$$\begin{align*}\left| {\int_{2n\pi }^{\left( {2n + 1} \right)\pi } {{e^{ - x_n^2\left( {1 + {y^2}} \right)}}\sin ydy} } \right| &= \left| {\int_{2n\pi }^{\left( {2n + 1} \right)\pi } {{e^{ - \frac{1}{{1 + {{\left( {2n + 1} \right)}^2}{\pi ^2}}}\left( {1 + {y^2}} \right)}}\sin ydy} } \right|\\&> \frac{1}{e}\int_{2n\pi }^{\left( {2n + 1} \right)\pi } {\sin ydy} = \frac{2}{e}.\end{align*}$$

因此,只要取$\varepsilon_0=2/e$,则对于任意大的正数$A_0$,总存在正整数$n$满足$2n\pi>A_0$,及$y_n=\frac{1}{{\sqrt {1 + {{\left( {2n + 1} \right)}^2}{\pi ^2}} }} \in (0,+\infty)$,使得$\left| {\int_{2n\pi }^{\left( {2n + 1} \right)\pi } {{e^{ - x_n^2\left( {1 + {y^2}} \right)}}\sin ydy} } \right| > \frac{2}{e}=\varepsilon_0$.由Cauchy收敛原理的推论知$\int_0^{ + \infty } {{e^{ - {x^2}\left( {1 + {y^2}} \right)}}\sin ydy}$关于$x$在$\left( {0, + \infty } \right)$上非一致收敛.

（10） 非一致收敛.对于任意取定的正数$A$,由于 $$\int_A^{ + \infty } {\frac{{\alpha dx}}{{1 + {\alpha ^2}{x^2}}}} = \frac{\pi }{2} - \arctan \left( {\alpha A} \right),$$ 取$\alpha=1/A$,则有 $$\int_A^{ + \infty } {\frac{{\alpha dx}}{{1 + {\alpha ^2}{x^2}}}} = \int_A^{ + \infty } {\frac{{1/A}}{{1 + {x^2}/{A^2}}}dx} = \frac{\pi }{4} .$$ 因此$\int_0^{ + \infty } {\frac{{\alpha dx}}{{1 + {\alpha ^2}{x^2}}}}$在$\left( {0,1} \right)$上不一致收敛.

（11） 一致收敛.见周民强207页.利用

当$0<x<1$时,我们有

$$0 \le \left| {\frac{{{x^t}}}{{\sqrt[3]{{\left( {x - 1} \right)\left( {x - 2} \right)}}}}} \right| < \frac{1}{{\sqrt x \sqrt[3]{{\left( {x - 1} \right)\left( {x - 2} \right)}}}}.$$

当$1<x<2$时,有

$$0 \le \left| {\frac{{{x^t}}}{{\sqrt[3]{{\left( {x - 1} \right)\left( {x - 2} \right)}}}}} \right| < \frac{{\sqrt x }}{{\sqrt[3]{{\left( {x - 1} \right)\left( {2 - x} \right)}}}}.$$

因此有

$$\int_0^2 {\frac{{{x^t}}}{{\sqrt[3]{{\left( {x - 1} \right)\left( {x - 2} \right)}}}}dx} < \int_0^1 {\frac{1}{{\sqrt x \sqrt[3]{{\left( {1 - x} \right)\left( {2 - x} \right)}}}}dx} + \int_1^2 {\frac{{\sqrt x }}{{\sqrt[3]{{\left( {x - 1} \right)\left( {2 - x} \right)}}}}dx} .$$

注意到下列渐进估计

$$\begin{align*}{\frac{1}{{\sqrt x \sqrt[3]{{\left( {1 - x} \right)\left( {2 - x} \right)}}}}} &= O\left( {\frac{1}{{\sqrt x }}} \right),x \to {0^ + },\\{\frac{1}{{\sqrt x \sqrt[3]{{\left( {1 - x} \right)\left( {2 - x} \right)}}}}} &= O\left( {\frac{1}{{\sqrt[3]{{x - 1}}}}} \right),x \to 1,\\\frac{{\sqrt x }}{{\sqrt[3]{{\left( {x - 1} \right)\left( {2 - x} \right)}}}} &= O\left( {\frac{1}{{\sqrt[3]{{x - 1}}}}} \right),x \to 1,\\\frac{{\sqrt x }}{{\sqrt[3]{{\left( {x - 1} \right)\left( {2 - x} \right)}}}} &= O\left( {\frac{1}{{\sqrt[3]{{2 - x}}}}} \right),x \to 2,\end{align*}$$

可知右端积分均收敛.由Weierstrass判别法可知,原积分关于$|t|<1/2$一致收敛.

（12） 先证明$\int_0^1 {{{\left( {1 - x} \right)}^{u - 1}}dx}$在$[a,+\infty)(a>0)$上一致收敛.由于 $${\left( {1 - x} \right)^{u - 1}} \le {\left( {1 - x} \right)^{a - 1}},\quad 0 \le x \le 1,a \le u < + \infty,$$ 而$\int_0^1 {{{\left( {1 - x} \right)}^{a - 1}}dx} = \frac{1}{a}$收敛,由Weierstrass判别法知, $\int_0^1 {{{\left( {1 - x} \right)}^{u - 1}}dx}$在$[a,+\infty)$上一致收敛.

再证明$\int_0^1 {{{\left( {1 - x} \right)}^{u - 1}}dx}$在$(0,+\infty)$上非一致收敛.对于任意取定的正数$A$且$A\to 0$,由于 $$\int_A^1 {{{\left( {1 - x} \right)}^{u - 1}}dx} = \frac{1}{A},$$

取$u=A\in (0,+\infty )$,当$A$足够小时,我们有 $$\int_A^1 {{{\left( {1 - x} \right)}^{u - 1}}dx} = \frac{{{{\left( {1 - A} \right)}^u}}}{u} = \frac{{{{\left( {1 - A} \right)}^A}}}{A} > 1.$$ 因此$\int_0^1 {{{\left( {1 - x} \right)}^{u - 1}}dx}$在$(0,+\infty)$上非一致收敛.

2.设$\displaystyle \int_0^{ + \infty } {{x^\lambda }f\left( x \right)dx}$当$\lambda=a,\lambda=b$时收敛$(a<b)$.证明$\displaystyle \int_0^{ + \infty } {{x^\lambda }f\left( x \right)dx}$当$\lambda=a,\lambda=b$关于$\lambda\in [a,b]$一致收敛.

证.这题来自菲哥第二册P577.积分$\int_0^1 {{x^a}f\left( x \right)dx}$是收敛的,而$x^{\lambda-a}$对于$\lambda\geq a$的值是$x$的单调函数,并以$1$为界.因此积分

$$\int_0^1 {{x^\lambda }f\left( x \right)dx} = \int_0^1 {{x^{\lambda - a}} \cdot {x^a}f\left( x \right)dx}$$ 关于$\lambda$一致收敛.类似地可以看出以下积分

$$\int_1^{ + \infty } {{x^\lambda }f\left( x \right)dx} = \int_1^{ + \infty } {{x^{\lambda - b}} \cdot {x^b}f\left( x \right)dx} ,$$

关于$\lambda\leq b$一致收敛.因此原积分一致收敛.

3.证明积分$\int_0^{ + \infty } {x{e^{ - xy}}dy}$在$(0,+\infty)$上不一致收敛.

证.对于任意取定的正数$A$,由于

$$\int_A^{ + \infty } {x{e^{ - xy}}dy} = {e^{ - Ax}},$$

取$x=1/A\in (0,+\infty)$,则有

$$\int_A^{ + \infty } {x{e^{ - xy}}dy} = \frac{1}{e}.$$ 因此$\int_0^{ + \infty } {x{e^{ - xy}}dy}$在$(0,+\infty)$上不一致收敛.

谢惠民下册P238第21章的一个参考题:

证明与曲面$ax^2+by^2+cz^2=1(abc\neq0)$相切的三个互相垂直的平面的交点在球面$x^2+y^2+z^2=\frac1a+\frac1b+\frac1c$上.

证:(幸子)椭球面方程为$ax^2+by^2+cz^2=1(abc\neq0)$,则法向量${n_i} = \left( {a{x_i},b{y_i},c{z_i}} \right)$,切平面方程为$a{x_i}x + b{y_i}y + c{z_i}z = 1$.

设三个切点分别为${\alpha _i}\left( {{x_i},{y_i},{z_i}} \right)\left( {i = 1,2,3} \right)$,三平面交点为$(x,y,z)$.由三平面垂直可知

$$\overrightarrow {{n_i}} \cdot \overrightarrow {{n_j}} = {a^2}{x_i}{x_j} + {b^2}{y_i}{y_j} + {c^2}{z_i}{z_j} = 0\left( {i \ne j} \right).$$

原点到三切平面的距离分别为

$$\frac{1}{{\sqrt {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} }}\left( {i = 1,2,3} \right).$$

由几何关系(考虑长方体对角线)可知

$${x^2} + {y^2} + {z^2} = \sum\limits_{i = 1}^3 {\frac{1}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} .$$

设$\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right)$,则对任意的$i = 1,2,3$,有

$$1 = \left( {\begin{array}{*{20}{c}}{a{x_i}}&{b{y_i}}&{c{z_i}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{a{x_i}}&{b{y_i}}&{c{z_i}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right) = {t_i}\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right).$$

并且

$$\begin{align*}&{x^2} + {y^2} + {z^2} = \left( {\begin{array}{*{20}{c}}x&y&z\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{t_1}}&{{t_2}}&{{t_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{b{y_1}}&{c{z_1}}\\{a{x_2}}&{b{y_2}}&{c{z_2}}\\{a{x_3}}&{b{y_3}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right)\\=& \sum\limits_{i = 1}^3 {{t_i}^2\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)} = \sum\limits_{i = 1}^3 {\frac{1}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} = \sum\limits_{i = 1}^3 {\frac{{a{x_i}^2 + b{y_i}^2 + c{z_i}^2}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} \\=& a\left( {\frac{{{x_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{x_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{x_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right)\\+& b\left( {\frac{{{y_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{y_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{y_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right)\\+& c\left( {\frac{{{z_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{z_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{z_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right).\end{align*}$$

注意到对任意的$w$,记$w = \sum\limits_{i = 1}^3 {{s_i}{n_i}}$,则${s_i} = \frac{{w \cdot {n_i}}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}$,即 $$w = \sum\limits_{i = 1}^3 {\frac{{w \cdot {n_i}}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}{n_i}} .$$

分别令$w = \sum\limits_{i = 1}^3 {\frac{{w \cdot {n_i}}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}{n_i}}$,得

$$\begin{align*}a\left( {\frac{{{x_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{x_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{x_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{a},\\b\left( {\frac{{{y_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{y_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{y_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{b},\\c\left( {\frac{{{z_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{z_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{z_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{c}.\end{align*}$$

因此有 $${x^2} + {y^2} + {z^2} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$

解法二.记$d_i=\sqrt{a^2x_i^2+b^2y_i^2+c^2z_i^2},\ i=1,2,3$,则$\begin{pmatrix}\frac{ax_1}{d_1}&\frac{by_1}{d_1}&\frac{cz_1}{d_1}\\\frac{ax_2}{d_2}&\frac{by_2}{d_2}&\frac{cz_2}{d_2}\\\frac{ax_3}{d_3}&\frac{by_3}{d_3}&\frac{cz_3}{d_3}\end{pmatrix}$是正交矩阵,从而有

$$\frac{a^2x_1^2}{d_1^2}+\frac{a^2x_2^2}{d_2^2}+\frac{a^3x_3^2}{d_3^2}=1\Rightarrow \frac{ax_1^2}{d_1^2}+\frac{ax_2^2}{d_2^2}+\frac{ax_3^2}{d_3^2}=\frac{1}{a}.$$

类似得到另外两式, 相加便有

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ax_1^2+by_1^2+cz_1^2}{d_1^2}+\frac{ax_2^2+by_2^2+cz_2^2}{d_2^2}+\frac{ax_3^2+by_3^2+cz_3^2}{d_3^2}=\frac{1}{d_1^2}+\frac{1}{d_2^2}+\frac{1}{d_3^2}.$$

7.设$u_i\in L^{p_i}(\Omega),p_i>0,i=1,2,\cdots,m$,且${\sum\limits_{i = 1}^{m } {\frac{1}{{{p_i}}}} }=1$.证明 $$\iint\limits_\Omega {{u_1} \cdots {u_m}dxdy} \le {\left\| {{u_1}} \right\|_{{p_1}}} \cdots {\left\| {{u_m}} \right\|_{{p_m}}}.$$

解.利用P270的Holder不等式以及数学归纳法即可.假设$m-1$时不等式成立,则有

$$\begin{align*}&\iint\limits_\Omega {{u_1} \cdots {u_{m - 1}}{u_m}dxdy} \le \iint\limits_\Omega {\left| {{u_1} \cdots {u_{m - 1}}{u_m}} \right|dxdy} \\= &{\left\| {{u_1} \cdots {u_{m - 1}}{u_m}} \right\|_1} \le {\left\| {{u_1} \cdots {u_{m - 1}}} \right\|_{1/\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }}{\left\| {{u_m}} \right\|_{{p_m}}}\\\le& {\left( {{{\left\| {u_1^{1/\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }} \right\|}_{{p_1}\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }} \cdots {{\left\| {u_{m - 1}^{1/\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }} \right\|}_{{p_{m - 1}}\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }}} \right)^{\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }}{\left\| {{u_m}} \right\|_{{p_m}}}\\= &{\left\| {{u_1}} \right\|_{{p_1}}} \cdots {\left\| {{u_{m - 1}}} \right\|_{{p_{m - 1}}}}{\left\| {{u_m}} \right\|_{{p_m}}}.\end{align*}$$

另解:利用$n$元的Holder不等式

$$\prod\limits_{i = 1}^m {x_i^{{\theta _i}}} \le \sum\limits_{i = 1}^m {{\theta _i}{x_i}} ,\quad \text{其中}\sum\limits_{i = 1}^m {{\theta _i}} = 1,{\theta _i} \ge 0.$$

取 $${x_i} = \frac{{{{\left| {{u_i}} \right|}^{{p_i}}}}}{{\left\| {{u_i}} \right\|_{{p_i}}^{{p_i}}}},{\theta _i} = \frac{1}{{{p_i}}}$$ 再积分即可.

证明 $$1 < \iiint\limits_{{{\left[ {0,1} \right]}^3}} {\left( {\cos \left( {xyz} \right) + \sin \left( {xyz} \right)} \right)dxdydz} < \sqrt 2 .$$

证.注意到 $$1 < \cos \left( {xyz} \right) + \sin \left( {xyz} \right) = \sqrt 2 \sin \left( {xyz + \frac{\pi }{4}} \right) < \sqrt 2$$ 即可.

证明

$${\left\{ {{{\int_a^b {dx\left[ {\int_c^d {f\left( {x,y} \right)dy} } \right]} }^2}} \right\}^{1/2}} \le \int_c^d {dy{{\left[ {\int_a^b {{f^2}\left( {x,y} \right)dx} } \right]}^{1/2}}} ,$$

其中$f$是连续函数.

证.利用Cauchy-Schwarz不等式我们有

$$\begin{align*}&{\left\{ {\int_c^d {dy{{\left[ {\int_a^b {{f^2}\left( {x,y} \right)dx} } \right]}^{1/2}}} } \right\}^2}\\= &\int_c^d {dy{{\left[ {\int_a^b {{f^2}\left( {x,y} \right)dx} } \right]}^{1/2}}} \cdot \int_c^d {dz{{\left[ {\int_a^b {{f^2}\left( {x,z} \right)dx} } \right]}^{1/2}}} \\\ge& \int_c^d {dy} \int_c^d {dz} \int_a^b {f\left( {x,y} \right)f\left( {x,z} \right)dx} = {\int_a^b {dx\left[ {\int_c^d {f\left( {x,y} \right)dy} } \right]} ^2}.\end{align*}$$

前几天徐半仙问了我谢惠民下册P325页上一道难度稍大的全微分题目，利用今晚美好的独处时间（笑哭），做了下，解答如下：

对于以下一阶微分形式$\omega$,求函数$M(x,y)\neq0$, 使得在适当的区域内$M\omega$为全微分,并求其原函数:

（1） $\displaystyle \omega = \left[ { - y\sqrt {{x^2} + {y^2} + 1} - x\left( {{x^2} + {y^2}} \right)} \right]dx + \left[ {x\sqrt {{x^2} + {y^2} + 1} - y\left( {{x^2} + {y^2}} \right)} \right]dy$;

（2） $\displaystyle \omega = x\left[ {{{\left( {ay + bx} \right)}^3} + a{y^3}} \right]dx + y\left[ {{{\left( {ay + bx} \right)}^3} + b{x^3}} \right]dy$.

解：（1）取 $$M = \frac{1}{{\left( {{x^2} + {y^2}} \right)\sqrt {{x^2} + {y^2} + 1} }},$$ 我们有

$$M\omega = \left( { - \frac{y}{{{x^2} + {y^2}}} - \frac{x}{{\sqrt {{x^2} + {y^2} + 1} }}} \right)dx + \left( {\frac{x}{{{x^2} + {y^2}}} - \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }}} \right)dy.$$

则有$P = - \frac{y}{{{x^2} + {y^2}}} - \frac{x}{{\sqrt {{x^2} + {y^2} + 1} }},Q = \frac{x}{{{x^2} + {y^2}}} - \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }}$,且 $$\frac{{\partial P}}{{\partial y}} = \frac{{{y^2} - {x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} + \frac{{xy}}{{{{\left( {{x^2} + {y^2} + 1} \right)}^{3/2}}}} = \frac{{\partial Q}}{{\partial x}}.$$

此时原函数为

$$\begin{align*}\varphi \left( {x,y} \right) = &\int_{{x_0}}^x {P\left( {x,{y_0}} \right)dx} + \int_{{y_0}}^y {Q\left( {x,y} \right)dy} + C'\\= &\int_{{x_0}}^x {\left( { - \frac{{{y_0}}}{{{x^2} + y_0^2}} - \frac{x}{{\sqrt {{x^2} + y_0^2 + 1} }}} \right)dx} + \int_{{y_0}}^y {\left( {\frac{x}{{{x^2} + {y^2}}} - \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }}} \right)dy} + C'\\= &\left( { - \arctan \frac{x}{{{y_0}}} - \sqrt {{x^2} + y_0^2 + 1} + \arctan \frac{{{x_0}}}{{{y_0}}} + \sqrt {x_0^2 + y_0^2 + 1} } \right)\\&+ \left( {\arctan \frac{y}{x} - \sqrt {{x^2} + {y^2} + 1} - \arctan \frac{{{y_0}}}{x} + \sqrt {{x^2} + y_0^2 + 1} } \right) + C'\\=& \arctan \frac{y}{x} - \sqrt {{x^2} + {y^2} + 1} + C.\end{align*}$$

值得一提的是：本题的积分因子是通过Wolfram Alpha求解出ODE，然后分别对$x,y$求偏导得来的.

（2）丁同仁书上一定理：

齐次方程$P(x,y)dx+Q(x,y)dy=0$有积分因子$M=\frac{1}{xP+yQ}$.

定理的证明:作变换$y=ux$,则由$P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy = 0$是齐次方程,我们有 $$P\left( {x,ux} \right)dx + Q\left( {x,ux} \right)\left( {udx + xdu} \right) = \left[ {{x^m}P\left( {1,u} \right) + u{x^m}Q\left( {1,u} \right)} \right]dx + {x^{m + 1}}Q\left( {1,u} \right)du = 0.$$

方程两边同乘 $$\frac{1}{{xP + yQ}} = \frac{1}{{{x^{m + 1}}\left[ {P\left( {1,u} \right) + uQ\left( {1,u} \right)} \right]}},$$ 则有

$$\frac{1}{x}dx + \frac{{Q\left( {1,u} \right)}}{{P\left( {1,u} \right) + uQ\left( {1,u} \right)}}du = 0.$$ 显然此方程为全微分方程.证毕.

取 $$M = \frac{1}{{xP + yQ}} = \frac{1}{{\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)}}.$$

则有

$$P' = \frac{{x{{\left( {ay + bx} \right)}^3} + ax{y^3}}}{{\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)}},Q' = \frac{{y{{\left( {ay + bx} \right)}^3} + b{x^3}y}}{{\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)}}.$$

我猜此时一定成立 $$\frac{{\partial P'}}{{\partial y}} = \frac{{\partial Q'}}{{\partial x}}.$$

事实上

$$\begin{align*}\frac{{\partial P'}}{{\partial y}} = & - 2xy{\left( {ay + bx} \right)^6} - 2{x^3}y{\left( {ay + bx} \right)^4} + \left( {5a{x^3}{y^2} + ax{y^4}} \right){\left( {ay + bx} \right)^3}\\& - 3{a^2}x{y^3}\left( {{x^2} + {y^2}} \right){\left( {ay + bx} \right)^2} + a{x^3}{y^4}\left( {ay + bx} \right) - {a^2}{x^3}{y^5}\\\frac{{\partial Q'}}{{\partial x}} = &- 2xy{\left( {ay + bx} \right)^6} - 2x{y^3}{\left( {ay + bx} \right)^4} + \left( {5b{x^2}{y^3} + b{x^4}y} \right){\left( {ay + bx} \right)^3}\\& - 3{b^2}{x^3}y\left( {{x^2} + {y^2}} \right){\left( {ay + bx} \right)^2} + b{x^4}{y^3}\left( {ay + bx} \right) - {b^2}{x^5}{y^3}.\end{align*}$$

于是

$$\begin{align*}\varphi \left( {x,y} \right) &= \int_{{x_0}}^x {P'\left( {x,{y_0}} \right)dx} + \int_{{y_0}}^y {Q'\left( {x,y} \right)dy} + C'\\&= \frac{1}{2}\ln \left[ {\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)} \right] - \frac{3}{2}\ln \left( {ay + bx} \right) + C.\end{align*}$$

事实上,我们还可取 $$M = \frac{1}{{{{\left( {ay + bx} \right)}^3}}},$$ 由此得到

$$\varphi \left( {x,y} \right) = \frac{{{x^2} + {y^2}}}{2} + \frac{{{x^2}{y^2}}}{{2{{\left( {ay + bx} \right)}^2}}} + C.$$

往事如烟！

谢上P9的一道不等式题，以前写过，但文件丢失，先前的解答难以回忆起，现在重新给出解答。

用向前-向后数学归纳法证明:设$\displaystyle 0<x_i\leq \frac12,i=1,2,\cdots,n$,则

$$\frac{{\prod\limits_{i = 1}^n {{x_i}} }}{{{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^n}}} \le \frac{{\prod\limits_{i = 1}^n {\left( {1 - {x_i}} \right)} }}{{{{\left[ {\sum\limits_{i = 1}^n {\left( {1 - {x_i}} \right)} } \right]}^n}}}.$$

(这个不等式是由在美国数学界有重大影响的华裔数学家樊畿(Fan Ky)得到的,关于它的许多研究和推广见[30].)

首先,

$$\frac{{{{\left[ {\sum\limits_{i = 1}^n {\left( {1 - {x_i}} \right)} } \right]}^n}}}{{{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^n}}} \le \frac{{\prod\limits_{i = 1}^n {\left( {1 - {x_i}} \right)} }}{{\prod\limits_{i = 1}^n {{x_i}} }} \Leftrightarrow {\left[ {\frac{n}{{\sum\limits_{i = 1}^n {{x_i}} }} - 1} \right]^n} \le \prod\limits_{i = 1}^n {\left( {\frac{1}{{{x_i}}} - 1} \right)} .$$

当$n=2$时,即证

$$\left( {\frac{1}{{{x_1}}} - 1} \right)\left( {\frac{1}{{{x_2}}} - 1} \right) \ge {\left[ {\frac{2}{{{x_1} + {x_2}}} - 1} \right]^2}.$$

展开得

$$\frac{1}{{{x_1}{x_2}}} - \frac{1}{{{x_1}}} - \frac{1}{{{x_2}}} + 1 \ge \frac{4}{{{{\left( {{x_1} + {x_2}} \right)}^2}}} - \frac{4}{{{x_1} + {x_2}}} + 1.$$

等价于证明

$$\frac{1}{{{x_1}{x_2}}} - \frac{4}{{{{\left( {{x_1} + {x_2}} \right)}^2}}} \ge \frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} - \frac{4}{{{x_1} + {x_2}}} \Leftrightarrow \frac{{{{\left( {{x_1} - {x_2}} \right)}^2}}}{{{{\left( {{x_1} + {x_2}} \right)}^2}{x_1}{x_2}}} \ge \frac{{{{\left( {{x_1} - {x_2}} \right)}^2}}}{{\left( {{x_1} + {x_2}} \right){x_1}{x_2}}}.$$

注意到$x_1+x_2\leq 1$,上式显然成立.

从$n=2$的已知情况出发,可以得到如下$n=4$时的情形:

$$\begin{align*}&\prod\limits_{i = 1}^4 {\left( {\frac{1}{{{x_i}}} - 1} \right)} = \prod\limits_{i = 1}^2 {\left( {\frac{1}{{{x_i}}} - 1} \right)} \cdot \prod\limits_{i = 3}^4 {\left( {\frac{1}{{{x_i}}} - 1} \right)} \ge {\left[ {\frac{2}{{\sum\limits_{i = 1}^2 {{x_i}} }} - 1} \right]^2}{\left[ {\frac{2}{{\sum\limits_{i = 3}^4 {{x_i}} }} - 1} \right]^2}\\= &{\left[ {\left( {\frac{1}{{\frac{1}{2}\sum\limits_{i = 1}^2 {{x_i}} }} - 1} \right)\left( {\frac{1}{{\frac{1}{2}\sum\limits_{i = 3}^4 {{x_i}} }} - 1} \right)} \right]^2} \le {\left[ {{{\left( {\frac{2}{{\frac{1}{2}\sum\limits_{i = 1}^2 {{x_i}} + \frac{1}{2}\sum\limits_{i = 3}^4 {{x_i}} }} - 1} \right)}^2}} \right]^2} = {\left[ {\frac{4}{{\sum\limits_{i = 1}^4 {{x_i}} }} - 1} \right]^4}.\end{align*}$$

同样可知,若$n=2^k$时不等式已成立,则可得到$n=2^{k+1}$时的不等式

$$\begin{align*}&\prod\limits_{i = 1}^{{2^{k + 1}}} {\left( {\frac{1}{{{x_i}}} - 1} \right)} = \prod\limits_{i = 1}^{{2^k}} {\left( {\frac{1}{{{x_i}}} - 1} \right)} \cdot \prod\limits_{i = {2^k} + 1}^{{2^{k + 1}}} {\left( {\frac{1}{{{x_i}}} - 1} \right)} \ge {\left[ {\frac{{{2^k}}}{{\sum\limits_{i = 1}^{{2^k}} {{x_i}} }} - 1} \right]^{{2^k}}}{\left[ {\frac{{{2^k}}}{{\sum\limits_{i = {2^k} + 1}^{{2^{k + 1}}} {{x_i}} }} - 1} \right]^{{2^k}}}\\= & {\left[ {\left( {\frac{1}{{\frac{1}{{{2^k}}}\sum\limits_{i = 1}^{{2^k}} {{x_i}} }} - 1} \right)\left( {\frac{1}{{\frac{1}{{{2^k}}}\sum\limits_{i = {2^k} + 1}^{{2^{k + 1}}} {{x_i}} }} - 1} \right)} \right]^{{2^k}}} \le {\left[ {{{\left( {\frac{2}{{\frac{1}{{{2^k}}}\sum\limits_{i = 1}^{{2^k}} {{x_i}} + \frac{1}{{{2^k}}}\sum\limits_{i = {2^k} + 1}^{{2^{k + 1}}} {{x_i}} }} - 1} \right)}^2}} \right]^{{2^k}}} = {\left[ {\frac{{{2^{k + 1}}}}{{\sum\limits_{i = 1}^{{2^{k + 1}}} {{x_i}} }} - 1} \right]^{{2^{k + 1}}}}.\end{align*}$$

这样就证明了当$n$为$2$的所有方幂时平均值不等式已成立.这是“向前”部分.

第二步要证明,当平均值不等式对某个$n>2$成立时,则它对$n-1$也一定成立.这是证明中的“向后”部分.写出

$$\begin{align*}&{\left[ {\frac{{n - 1}}{{\sum\limits_{i = 1}^{n - 1} {{x_i}} }} - 1} \right]^{n - 1}} = {\left[ {\frac{n}{{\sum\limits_{i = 1}^{n - 1} {{x_i}} + \frac{1}{{n - 1}}\sum\limits_{i = 1}^{n - 1} {{x_i}} }} - 1} \right]^n} \cdot {\left[ {\frac{{n - 1}}{{\sum\limits_{i = 1}^{n - 1} {{x_i}} }} - 1} \right]^{ - 1}}\\\le &\prod\limits_{i = 1}^{n - 1} {\left( {\frac{1}{{{x_i}}} - 1} \right)} \cdot \left( {\frac{1}{{\frac{1}{{n - 1}}\sum\limits_{i = 1}^{n - 1} {{x_i}} }} - 1} \right) \cdot {\left[ {\frac{{n - 1}}{{\sum\limits_{i = 1}^{n - 1} {{x_i}} }} - 1} \right]^{ - 1}} = \prod\limits_{i = 1}^{n - 1} {\left( {\frac{1}{{{x_i}}} - 1} \right)} .\end{align*}$$

于是$n-1$时不等式也成立.合并以上向前和向后两部分,可见不等式对每个自然数$n$成立.

事实上,我们有

$${\left[ {\frac{n}{{\sum\limits_{i = 1}^n {{x_i}} }} - 1} \right]^n} \le \prod\limits_{i = 1}^n {\left( {\frac{1}{{{x_i}}} - 1} \right)}  \Leftrightarrow \frac{1}{n}\sum\limits_{i = 1}^n {\ln \left( {\frac{1}{{{x_i}}} - 1} \right)}  \ge \ln \left( {\frac{1}{{\frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} }} - 1} \right).$$ 结合函数$y = \ln \left( {\frac{1}{x} - 1} \right)$的凹凸性便可得证.

大风起兮云飞扬，一生挚爱美娇娘.

爱情，本来就是勇敢者的游戏！

1.设已知$\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}{a_n}} = A,\sum\limits_{n = 1}^\infty {{a_{2n - 1}}} = B$,证明: $\sum\limits_{n = 1}^\infty {{a_n}}$收敛并求其和.

解:显然有

$$\sum\limits_{n = 1}^\infty {{a_n}} = 2\sum\limits_{n = 1}^\infty {{a_{2n - 1}}} - \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}{a_n}} = 2B - A.$$

2.设$P(x)=a_0+a_1x+\cdots+a_mx^m$为$m$次多项式,求级数$\sum\limits_{n = 0}^\infty {\frac{{P\left( n \right)}}{{n!}}}$的和.

解:事实上,

$$\begin{align*}{b_k} &= \sum\limits_{n = 0}^\infty {\frac{{{n^k}}}{{n!}}} = \sum\limits_{n = 1}^\infty {\frac{{{n^{k - 1}}}}{{\left( {n - 1} \right)!}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( {n + 1} \right)}^{k - 1}}}}{{n!}}} \\&= {b_{k - 1}} + C_{k - 1}^1{b_{k - 2}} + \cdots + C_{k - 1}^{k - 2}{b_1} + {b_0},\end{align*}$$

其中$b_0=e$.

由此得到的数叫Bell数,记为$B_n$,并且

$$B\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{B\left( n \right)}}{{n!}}{x^n}} = {e^{{e^x} - 1}}.$$

回到原题,我们有 $$\sum\limits_{n = 0}^\infty {\frac{{P\left( n \right)}}{{n!}}} = e\sum\limits_{k = 0}^m {{a_k}{B_k}} .$$

3.求$1 - \frac{{{2^3}}}{{1!}} + \frac{{{3^3}}}{{2!}} - \frac{{{4^3}}}{{3!}} + \cdots$的和.

解:事实上,

$$\begin{align*}{b_k} &= \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}\frac{{{n^k}}}{{n!}}} = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{{n^{k - 1}}}}{{\left( {n - 1} \right)!}}} = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^{n+1}}\frac{{{{\left( {n + 1} \right)}^{k - 1}}}}{{n!}}} \\& =- {b_{k - 1}} - C_{k - 1}^1{b_{k - 2}} - \cdots - C_{k - 1}^{k - 2}{b_1} - {b_0},\end{align*}$$

其中$b_0=1/e$.因此$b_1=-1/e,b_2=0,b_3=1/e$.

因此

$$\begin{align*}& 1 - \frac{{{2^3}}}{{1!}} + \frac{{{3^3}}}{{2!}} - \frac{{{4^3}}}{{3!}} + \cdots = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}\frac{{{{\left( {n + 1} \right)}^3}}}{{n!}}} \\=& {b_3} + 3{b_2} + 3{b_1} + {b_0} = - \frac{1}{e}.\end{align*}$$

4.求下列级数的和:(1) $\sum\limits_{n = 1}^\infty {\arctan \frac{1}{{2{n^2}}}}$; (2) $\sum\limits_{n = 1}^\infty {\arctan \frac{2}{{{n^2}}}}$.

解:事实上

$$\sum\limits_{n = 1}^\infty {\arctan \frac{1}{{2{n^2}}}} = \sum\limits_{n = 1}^\infty {\left( {\arctan \frac{1}{{2n - 1}} - \arctan \frac{1}{{2n + 1}}} \right)} = \frac{\pi }{4}.$$

而

$$\sum\limits_{n = 1}^\infty {\arctan \frac{2}{{{n^2}}}} = \sum\limits_{n = 1}^\infty {\left( {\arctan \frac{1}{{n - 1}} - \arctan \frac{1}{{n + 1}}} \right)} = \frac{\pi }{2} + \frac{\pi }{4} = \frac{{3\pi }}{4}.$$

5.设$a>1$,求$\sum\limits_{n = 0}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}} + 1}}}$的和.

解:事实上

$$\begin{align*}\sum\limits_{n = 0}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}} + 1}}} &= \frac{1}{{a + 1}} + \sum\limits_{n = 1}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}} + 1}}} = \frac{1}{{a + 1}} - \frac{1}{{a - 1}} + \frac{1}{{a + 1}} + \sum\limits_{n = 1}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}} + 1}}} \\&= \frac{1}{{a + 1}} - \frac{2}{{{a^2} - 1}} + \sum\limits_{n = 1}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}} + 1}}} = \frac{1}{{a + 1}} - \frac{{{2^2}}}{{{a^{{2^2}}} - 1}} + \sum\limits_{n = 2}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}} + 1}}} \\&= \frac{1}{{a + 1}} - \mathop {\lim }\limits_{n \to \infty } \frac{{{2^{n + 1}}}}{{{a^{{2^{n + 1}}}} - 1}} = \frac{1}{{a + 1}}.\end{align*}$$

6.求$1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{{11}} - \cdots$的和.

解:

$$\begin{align*}&\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{8n - 7}} + \frac{1}{{8n - 5}} - \frac{1}{{8n - 3}} - \frac{1}{{8n - 1}}} \right)} = \sum\limits_{n = 1}^\infty {\int_0^1 {\left( {{x^{8n - 8}} + {x^{8n - 6}} - {x^{8n - 4}} - {x^{8n - 2}}} \right)} } \\=& \int_0^1 {\sum\limits_{n = 1}^\infty {\left( {{x^{8n - 8}} + {x^{8n - 6}} - {x^{8n - 4}} - {x^{8n - 2}}} \right)} dx} = \int_0^1 {\frac{{1 + {x^2} - {x^4} - {x^6}}}{{1 - {x^8}}}dx} \\= &\left. {\frac{{\arctan \left( {1 + \sqrt 2 x} \right) - \arctan \left( {1 - \sqrt 2 x} \right)}}{{\sqrt 2 }}} \right|_0^1 = \frac{\pi }{{2\sqrt 2 }}.\end{align*}$$

7.求$1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \cdots$的和.

解:

$$\begin{align*}&\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{8n - 7}} - \frac{1}{{8n - 1}}} \right)} = \sum\limits_{n = 1}^\infty {\int_0^1 {\left( {{x^{8n - 8}} - {x^{8n - 2}}} \right)} } \\=& \int_0^1 {\sum\limits_{n = 1}^\infty {\left( {{x^{8n - 8}} - {x^{8n - 2}}} \right)} dx} = \int_0^1 {\frac{{1 - {x^6}}}{{1 - {x^8}}}dx} \\= &\left. {\frac{{2\arctan x + \sqrt 2 \arctan \left( {1 + \sqrt 2 x} \right) - \arctan \left( {1 - \sqrt 2 x} \right)}}{4}} \right|_0^1 = \frac{{\sqrt 2 + 1}}{8}\pi .\end{align*}$$

8.求$1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{{10}} + \cdots$的和.

解:

$$\begin{align*}&\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{6n - 5}} - \frac{1}{{6n - 2}}} \right)} = \sum\limits_{n = 1}^\infty {\int_0^1 {\left( {{x^{6n - 6}} - {x^{6n - 3}}} \right)} } \\= &\int_0^1 {\sum\limits_{n = 1}^\infty {\left( {{x^{6n - 6}} - {x^{6n - 3}}} \right)} dx} = \int_0^1 {\frac{{1 - {x^3}}}{{1 - {x^6}}}dx} = \int_0^1 {\frac{1}{{1 + {x^3}}}dx} \\=& \left. {\left( { - \frac{1}{6}\ln \left( {{x^2} - x + 1} \right) + \frac{1}{3}\ln \left( {x + 1} \right) + \frac{{\arctan \frac{{2x - 1}}{{\sqrt 3 }}}}{{\sqrt 3 }}} \right)} \right|_0^1 = \frac{{\sqrt 3 \pi + 3\ln 2}}{9}.\end{align*}$$

9.设${a_n} = 1 + \frac{1}{2} + \cdots + \frac{1}{n},n = 1,2, \cdots$,求$\sum\limits_{n = 1}^\infty {\frac{{{a_n}}}{{n\left( {n + 1} \right)}}}$的和.

解:

$$\begin{align*}&\sum\limits_{n = 1}^\infty {\frac{{{a_n}}}{{n\left( {n + 1} \right)}}} = \sum\limits_{n = 1}^\infty {\frac{{1 + \frac{1}{2} + \cdots + \frac{1}{n}}}{{n\left( {n + 1} \right)}}} \\=&\sum\limits_{n = 1}^\infty {\left( {\frac{{1 + \frac{1}{2} + \cdots + \frac{1}{n}}}{n} - \frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1}}} \right)} + \sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {n + 1} \right)}^2}}}} \\= & 1 - \mathop {\lim }\limits_{n \to \infty } \frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1}} + \left( {\frac{{{\pi ^2}}}{6} - 1} \right) = \frac{{{\pi ^2}}}{6} - \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{{n + 2}}}}{1} = \frac{{{\pi ^2}}}{6}.\end{align*}$$

10.求$\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 1}} + \frac{1}{{4n + 3}} - \frac{1}{{2n + 2}}} \right)}$的和.

解:

$$\begin{align*}&\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 1}} + \frac{1}{{4n + 3}} - \frac{1}{{2n + 2}}} \right)} = \sum\limits_{n = 0}^\infty {\int_0^1 {\left( {{x^{4n}} + {x^{4n + 2}} - {x^{2n + 1}}} \right)} } \\= &\int_0^1 {\sum\limits_{n = 0}^\infty {\left( {{x^{4n}} + {x^{4n + 2}} - {x^{2n + 1}}} \right)} dx} = \int_0^1 {\left( {\frac{{1 + {x^2}}}{{1 - {x^4}}} - \frac{x}{{1 - {x^2}}}} \right)dx} \\=& \int_0^1 {\frac{1}{{1 + x}}dx} = \ln 2.\end{align*}$$

11.求$1 - \frac{1}{4} + \frac{1}{6} - \frac{1}{9} + \frac{1}{{11}} - \frac{1}{{14}} + \cdots$的和.

解:

$$\begin{align*}&\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{5n - 4}} - \frac{1}{{5n - 1}}} \right)} = \sum\limits_{n = 1}^\infty {\int_0^1 {\left( {{x^{5n - 5}} - {x^{5n - 2}}} \right)dx} } = \int_0^1 {\sum\limits_{n = 1}^\infty {\left( {{x^{5n - 5}} - {x^{5n - 2}}} \right)} dx} \\=& \int_0^1 {\frac{{1 - {x^3}}}{{1 - {x^5}}}dx} = \int_0^1 {\left( {\frac{{\left( {5 - \sqrt 5 } \right)/10}}{{{x^2} + \frac{{\sqrt 5 + 1}}{2}x + 1}} + \frac{{\left( {5 + \sqrt 5 } \right)/10}}{{{x^2} + \frac{{ - \sqrt 5 + 1}}{2}x + 1}}} \right)dx} \\=& \left. {\left[ {\frac{{5 - \sqrt 5 }}{{10}}\sqrt {\frac{{10 + 2\sqrt 5 }}{5}} \arctan \frac{{4x + \sqrt 5 + 1}}{{\sqrt {10 - 2\sqrt 5 } }} + \frac{{5 + \sqrt 5 }}{5}\sqrt {\frac{2}{{5 + \sqrt 5 }}} \arctan \frac{{4x - \sqrt 5 + 1}}{{\sqrt {10 + 2\sqrt 5 } }}} \right]} \right|_0^1 \\=& \frac{{\sqrt {25 + 10\sqrt 5 } }}{{25}}\pi .\end{align*}$$

12.求$\frac{{{x^3}}}{{3!}} + \frac{{{x^9}}}{{9!}} + \frac{{{x^{15}}}}{{15!}} + \cdots$的和函数.

解:事实上,方程$\omega^3=1$有三个根$1,{ - \frac{1}{2} + \frac{{\sqrt 3 i}}{2}},{ - \frac{1}{2} - \frac{{\sqrt 3 i}}{2}}$.利用$\sinh$便可得到所需函数

$$\begin{align*}&\frac{{\sinh x + \sinh \left( { - \frac{1}{2} + \frac{{\sqrt 3 i}}{2}} \right)x + \sinh \left( { - \frac{1}{2} - \frac{{\sqrt 3 i}}{2}} \right)x}}{3}\\= & - \frac{2}{3}\sinh \frac{x}{2}\cos \frac{{\sqrt 3 x}}{2} + \frac{{\sinh x}}{3} = \frac{{{x^3}}}{{3!}} + \frac{{{x^9}}}{{9!}} + \frac{{{x^{15}}}}{{15!}} + \cdots .\end{align*}$$

我们还有

$$\begin{align*}&{\frac{{\sin x +\sin \left( { - \frac{1}{2} + \frac{{\sqrt 3 i}}{2}} \right)x + \sin \left( { - \frac{1}{2} - \frac{{\sqrt 3 i}}{2}} \right)x}}{{ - 3}}}\\= &\frac{2}{3}\sin \frac{x}{2}\cosh \frac{{\sqrt 3 x}}{2} - \frac{{\sin x}}{3} = \frac{{{x^3}}}{{3!}} - \frac{{{x^9}}}{{9!}} + \frac{{{x^{15}}}}{{15!}} - \frac{{{x^{21}}}}{{21!}} + \cdots .\end{align*}$$

13.求$\sum\limits_{n = 1}^\infty {\frac{{{{\left[ {\left( {n - 1} \right)!} \right]}^2}}}{{\left( {2n} \right)!}}{{\left( {2x} \right)}^{2n}}}$的和函数.

解:在$|x|<1$上对$S(x)$逐项求导,知$S'\left( x \right) = 2\sum\limits_{n = 1}^\infty {\frac{{{{\left[ {\left( {n - 1} \right)!} \right]}^2}}}{{\left( {2n - 1} \right)!}}{{\left( {2x} \right)}^{2n - 1}}}$,且$S''\left( x \right) = 4\sum\limits_{n = 1}^\infty {\frac{{{{\left[ {\left( {n - 1} \right)!} \right]}^2}}}{{\left( {2n - 2} \right)!}}{{\left( {2x} \right)}^{2n - 2}}}$.由此可得$(1-x^2)S''(x)-xS'(x)=4$.在两端乘以${(1-x^2)}^{-1/2}$,我们有

$${\left( {\sqrt {1 - {x^2}} S'\left( x \right)} \right)^\prime } = \frac{4}{{\sqrt {1 - {x^2}} }},$$ 故

$$S\left( x \right) = \frac{{4\arcsin x}}{{\sqrt {1 - {x^2}} }} + \frac{1}{{\sqrt {1 - {x^2}} }},\quad \left| x \right| < 1.$$

14.求$\sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{\left( {1 - {x^n}} \right)\left( {1 - {x^{n + 1}}} \right)}}}$的和函数.

解:注意到

$$\begin{align*}&\left( {1 - \frac{1}{x}} \right)\sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{\left( {1 - {x^n}} \right)\left( {1 - {x^{n + 1}}} \right)}}} \\=& \sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{\left( {1 - {x^n}} \right)\left( {1 - {x^{n + 1}}} \right)}}} - \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{\left( {1 - {x^n}} \right)\left( {1 - {x^{n + 1}}} \right)}}} \\= &\sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}} - {x^n}}}{{\left( {1 - {x^n}} \right)\left( {1 - {x^{n + 1}}} \right)}}} = \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{1 - {x^{n + 1}}}} - \frac{1}{{1 - {x^n}}}} \right)} \\=& \mathop {\lim }\limits_{n \to \infty } \frac{1}{{1 - {x^{n + 1}}}} - \frac{1}{{1 - x}} = \begin{cases}\frac{1}{{x - 1}},&\left| x \right| > 1\\\frac{x}{{x - 1}},&\left| x \right| < 1\end{cases} .\end{align*}$$

因此

$$\sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{\left( {1 - {x^n}} \right)\left( {1 - {x^{n + 1}}} \right)}}} = \begin{cases}\frac{x}{{{{\left( {x - 1} \right)}^2}}}, &\left| x \right| > 1\\\frac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}}}, &\left| x \right| < 1\end{cases} .$$

15.设$\sum\limits_{n = 1}^\infty {\frac{1}{{{a_n}}}}$为发散的正项级数, $x>0$,求$\sum\limits_{n = 1}^\infty {\frac{{{a_1}{a_2} \cdots {a_n}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_{n + 1}} + x} \right)}}}$的和函数.

解:首先,

$$\begin{align*}&\sum\limits_{n = 1}^\infty {\frac{{{a_1}{a_2} \cdots {a_n}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_{n + 1}} + x} \right)}}} \\=& \frac{{{a_1}}}{{{a_2} + x}} + \frac{1}{x}\sum\limits_{n = 2}^\infty {\left[ {\frac{{{a_1}{a_2} \cdots {a_n}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_n} + x} \right)}} - \frac{{{a_1}{a_2} \cdots {a_{n + 1}}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_{n + 1}} + x} \right)}}} \right]} \\=& \frac{{{a_1}}}{{{a_2} + x}} + \frac{1}{x}\left[ {\frac{{{a_1}{a_2}}}{{{a_2} + x}} - \mathop {\lim }\limits_{n \to \infty } \frac{{{a_1}{a_2} \cdots {a_{n + 1}}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_{n + 1}} + x} \right)}}} \right].\end{align*}$$

当$n$足够大时, $$1 + \frac{x}{{{a_{n + 1}}}} \sim {e^{x/{a_{n + 1}}}}.$$

因此${\left( {1 + \frac{x}{{{a_2}}}} \right) \cdots \left( {1 + \frac{x}{{{a_{n + 1}}}}} \right)}$与$\exp \left\{ {x\sum\limits_{n = 1}^\infty {\frac{1}{{{a_n}}}} } \right\}$具有相同的收敛性,均发散,故

$$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_1}{a_2} \cdots {a_{n + 1}}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_{n + 1}} + x} \right)}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_1}}}{{\left( {1 + \frac{x}{{{a_2}}}} \right) \cdots \left( {1 + \frac{x}{{{a_{n + 1}}}}} \right)}} = 0.$$

从而

$$\sum\limits_{n = 1}^\infty {\frac{{{a_1}{a_2} \cdots {a_n}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_{n + 1}} + x} \right)}}} = \frac{{{a_1}}}{{{a_2} + x}} + \frac{{{a_1}{a_2}}}{{x\left( {{a_2} + x} \right)}} = \frac{{{a_1}}}{x}.$$

16.设$x>1$,求$\frac{x}{{x + 1}} + \frac{{{x^2}}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} + \frac{{{x^4}}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right)}} + \cdots$的和函数.

解: $$\begin{align*}I &= \left( {1 - \frac{1}{{x + 1}}} \right) + \frac{{{x^2}}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} + \frac{{{x^4}}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right)}} + \cdots \\&= 1 + \left( { - \frac{1}{{x + 1}} + \frac{{{x^2}}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}} \right) + \frac{{{x^4}}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right)}} + \cdots \\&= 1 - \frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} + \frac{{{x^4}}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right)}} + \cdots \\&= 1 - \frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right)}} + \cdots \\&= \cdots = 1 - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right) \cdots \left( {{x^{{2^{n - 1}}}} + 1} \right)}} = 1.\end{align*}$$