定积分的概念和可积函数

单变量函数的积分学

张瑞
中国科学技术大学数学科学学院

定积分的概念和可积函数

求曲边图形的面积
\begin{tikzpicture}[x=2cm, y=2cm, global scale=1] % 上面,用 x=2cm, y=2cm 来设置x,y方向的单位长度,缺省是1cm \draw[->] (-0.1,0) -- (1.2,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,1.20) node[right] {$y$}; \fill[fill=yellow!80!black] (0.8,0)--(0.8,0.81)--(0.9,0.81)--(0.9,0); \draw[domain=0:1, color=blue, thick] plot (\x,{(\x)^2}) node[right] {$x^2$}; \foreach \x in {0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0 } { %\path[name path=main] (\x,0) -- (\x,2.0); %\path[name intersections={of=fx and main}] (intersection-1) coordinate (P); %\coordinate (a) at (intersection-1); \draw[red] (\x,0)-- (\x, {(\x)^2})-- ({(\x)-0.1}, {(\x)^2}) -- ({(\x)-0.1}, {(\x-0.1)^2}); } \draw[thick] (1,0) node[below] {$1$} --(1,1); \end{tikzpicture}

$[0,1]$区间分成$n$等分,每个曲边图形用一个矩形近似,面积为

\[\frac1n\left(\frac{i}n\right)^2, i=1,2,\cdots,n \]

因此,总的面积和为 $\displaystyle~\sum_{i=1}^n\frac1n\left(\frac{i}n\right)^2$

$n$趋于$+\infty$时,这个极限是多少?

\begin{tikzpicture}[x=2cm, y=2cm, global scale=1] % 上面,用 x=2cm, y=2cm 来设置x,y方向的单位长度,缺省是1cm \draw[->] (-0.1,0) -- (1.2,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,1.20) node[right] {$y$}; \fill[fill=yellow!80!black] (0.8,0)--(0.8,0.64)--(0.9,0.64)--(0.9,0); \draw[domain=0:1, color=blue, thick] plot (\x,{(\x)^2}) node[right] {$x^2$}; \foreach \x in {0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9} { %\path[name path=main] (\x,0) -- (\x,2.0); %\path[name intersections={of=fx and main}] (intersection-1) coordinate (P); %\coordinate (a) at (intersection-1); \draw[red] (\x,0)-- (\x, {(\x)^2})-- ({(\x)+0.1}, {(\x)^2}); } \draw[thick] (1,0) node[below] {$1$}--(1,1); \end{tikzpicture}

$[0,1]$区间分成$n$等分,每个曲边图形用一个矩形近似,面积为

\[\frac1n\left(\frac{i-1}n\right)^2, i=1,2,\cdots,n \]

因此,总的面积和为 $\displaystyle~\sum_{i=1}^n\frac1n\left(\frac{i-1}n\right)^2$

$n$趋于$+\infty$时,这个极限是多少?

\begin{tikzpicture}[x=2cm, y=2cm, global scale=1] % 上面,用 x=2cm, y=2cm 来设置x,y方向的单位长度,缺省是1cm \draw[->] (-0.1,0) -- (1.2,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,1.20) node[right] {$y$}; \fill[fill=yellow!80!black] (0.8,0)--(0.8,0.7225)--(0.9,0.7225)--(0.9,0); \draw[domain=0:1, color=blue, thick] plot (\x,{(\x)^2}) node[right] {$x^2$}; \foreach \x in {0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9} { %\path[name path=main] (\x,0) -- (\x,2.0); %\path[name intersections={of=fx and main}] (intersection-1) coordinate (P); %\coordinate (a) at (intersection-1); \draw[red] (\x,0)-- (\x, {(\x+0.05)^2})-- ({(\x)+0.1}, {(\x+0.05)^2}); } \draw[thick] (1,0) node[below] {$1$}--(1,1); \end{tikzpicture}

$[0,1]$区间分成$n$等分,每个曲边图形用一个矩形近似,面积为

\[\frac1n\left(\frac{2i-1}{2n}\right)^2, i=1,2,\cdots,n \]

因此,总的面积和为

\[\sum_{i=1}^n\frac1n\left(\frac{2i-1}{2n}\right)^2 \]

$n$趋于$+\infty$时,这个极限是多少?

求一个曲边梯形的面积
\usetikzlibrary{arrows} \usetikzlibrary{intersections, calc} \begin{tikzpicture}[scale=0.8,samples=200, >=latex,thick] \fill[fill=yellow!80!black] (2.0,0)--(2.0,1.3)--(2.5,1.3)--(2.5,0); % \clip (0,0) rectangle (5,5);% 切り抜き \draw[thick,->] (-0.5,0) -- (5,0) node[right] {$x$};% x軸 \draw[thick,->] (0,-0.1) -- (0,3) node[below right] {$y$};% y軸 \draw (0,0) node[below left] {O};% 原点 %\draw (0,0) to [out=60,in=120] (3,1); \draw[thin] (0.5,0)--(0.5,1.5); \draw[thin] (4,0) -- (4,2); \draw[name path=fx, thick] (0.5,1.5) .. controls (2,0) and (2,2) .. (4,2); \draw (2,2) node[above] {$y=f(x)$}; \draw (1.0, 0) node[below] {$x_1$}; \draw (2.5, 0) node[below] {$x_i$}; %\draw (3.5, 0) node[below] {$x_{n-1}$}; \draw (0.5, 0) node[below] {$a$}; \draw (4, 0) node[below] {$b$}; \foreach \x in {1.0, 1.7, 2.0, 2.5, 2.8, 3.5 } { \path[name path=main] (\x,0) -- (\x,2.0); \path[name intersections={of=fx and main}] (intersection-1) coordinate (P); %\coordinate (a) at (intersection-1); \draw[dotted] (\x,0)-- (P); } \end{tikzpicture}
  1. 先在$a$,$b$之间插入分割点$a=x_0<x_1<\cdots$ $<x_{n-1}<x_n=b$,将曲边梯形分成$n$个小“长条”
  2. 将每个“长条”近似为小矩形,则面积为
    \[f(\xi_i)(x_i-x_{i-1}) \]
    其中$\xi_i$$[x_{i-1},x_i]$中的任意一点。
\usetikzlibrary{arrows} \usetikzlibrary{intersections, calc} \begin{tikzpicture}[scale=0.8,samples=200, >=latex,thick] \fill[fill=yellow!80!black] (2.0,0)--(2.0,1.3)--(2.5,1.3)--(2.5,0); % \clip (0,0) rectangle (5,5);% 切り抜き \draw[thick,->] (-0.5,0) -- (5,0) node[right] {$x$};% x軸 \draw[thick,->] (0,-0.1) -- (0,3) node[below right] {$y$};% y軸 \draw (0,0) node[below left] {O};% 原点 %\draw (0,0) to [out=60,in=120] (3,1); \draw[thin] (0.5,0)--(0.5,1.5); \draw[thin] (4,0) -- (4,2); \draw[name path=fx, thick] (0.5,1.5) .. controls (2,0) and (2,2) .. (4,2); \draw (2,2) node[above] {$y=f(x)$}; \draw (1.0, 0) node[below] {$x_1$}; \draw (2.5, 0) node[below] {$x_i$}; %\draw (3.5, 0) node[below] {$x_{n-1}$}; \draw (0.5, 0) node[below] {$a$}; \draw (4, 0) node[below] {$b$}; \foreach \x in {1.0, 1.7, 2.0, 2.5, 2.8, 3.5} { \path[name path=main] (\x,0) -- (\x,2.0); \path[name intersections={of=fx and main}] (intersection-1) coordinate (P); %\coordinate (a) at (intersection-1); \draw[dotted] (\x,0)-- (P); } \end{tikzpicture}
  1. 把这些矩形面积加起来
    \[S_n=\sum_{i=1}^nf(\xi)(x_i-x_{i-1}) \]
    做为曲边梯形面积的“近似”
  2. 当插入点越来越多时,这个面积和的极限,就是曲边梯形的面积

类似地,可以计算质点作变速直线运动的路程。

把时间段$[\alpha,\beta]$分成多个小段,每个小段上的运动近似为匀速运动。

定积分

定义 1. (定积分)
$f(x)$$[a,b]$上有定义,在$[a,b]$的任一分割

\[T:a=x_0<x_1<x_2<\cdots<x_{n-1}<x_n=b \]

$\Delta x_i=x_i-x_{i-1}, i=1,2,\cdots,n$, $\lambda(T)=\max\{\Delta x_i\}_{i=1}^n$, 和式

\[\sigma(a,b,f;T,\xi_1,\cdots,\xi_n)=\sum_{i=1}^n f(\xi_i)\Delta x_i \]

称为积分和Riemann和,简记为$\sigma(T,\xi_i)$

$\displaystyle\lim_{\lambda(T)\to0+}\sigma(T,\xi_i)$存在,且与$T, \xi_i$无关,则

  • 称函数$f(x)$$[a,b]$可积Riemann可积

  • 并称这个极限为$f(x)$$[a,b]$上的定积分Riemann积分,记为

    \[\int_a^bf(x)dx=\lim_{\lambda(T)\to0+}\sum_{i=1}^nf(\xi_i)\Delta x_i \]
  • $f(x)$称为被积函数$a$积分下限$b$积分上限$f(x)dx$被积表达式$x$积分变量

  • 积分与变量名称无关,如
    \[\int_a^bf(x)dx=\int_a^bf(u)du=\int_a^bf(t)dt \]
  • $\epsilon-\delta$描述: $\forall \epsilon>0, \exists \delta>0$,使得
    \[|\sum_{i=1}^nf(\xi_i)\Delta x_i-I|<\epsilon, \forall \lambda(T)<\delta, \xi\in[x_{i-1},x_i] \]
    $\displaystyle\int_a^bf(x)dx=I$

例 1. 常值函数$f(x)\equiv c$是可积的。

. 对任意分割$T$,有

\[\sigma(T;\xi_i)=\sum_{i=1}^n c \Delta x_i=c(b-a) \]

例 2. Dirichlet函数 $D(x)=\begin{cases}1, x=\frac{p}{q} \\0, x\notin Q\end{cases}$ 是不可积的。

. 每个小区间上,取$\xi_i$为有理点和无理点得到的极限不一样。

定理 1. (可积则有界)
$f(x)$$[a,b]$上可积,则$f(x)$$[a,b]$上有界

推论 1.
函数无界则不可积

有界,也未必可积。如Dirichlet函数 $D(x)=\begin{cases}1, x=\frac{p}{q} \\0, x\notin Q\end{cases}$

问题. 函数$f(x)=x$$[0,1]$上可积吗?

证: (反证) 设$f(x)$$[a,b]$有无界。

$f(x)$可积,则有实数$I$,取$\epsilon=1$,则$\exists \delta>0$

\[|\sum_{i=1}^nf(\xi_i)\Delta x_i-I|<1, \forall \lambda(T)<\delta , \xi_i\in[x_{i-1},x_i] \]

$f(x)$无界,则一定在某个小区间上无界,不防设为$[x_0,x_1]$。 由

\[|\sum_{i=1}^nf(\xi_i)\Delta x_i|<|I|+1 \]

可得

\[|f(\xi_1)|\Delta x_1<I+1+|\sum_{i=2}^nf(\xi_i)\Delta x_i| \]

固定$\xi_i, i=2,3,\cdots,n$,则右边为一数,对$\forall \xi_1$都成立,即 $f(x)$$[x_0,x_1]$上有界。

矛盾

. 对Dirichlet函数,有

\[\sum_{i=1}^nD(\xi_i)\Delta x_i=\sum_{i=1}^n 1\cdot\Delta x_i=1 , \mbox{ when } \xi_i\in\mathbb{Q} \]
\[\sum_{i=1}^nD(\xi_i)\Delta x_i=\sum_{i=1}^n 0\cdot\Delta x_i=0 , \mbox{ when } \xi_i\notin\mathbb{Q} \]

可积的函数

定义 2. (达布和)
$f(x)$$[a,b]$上有界,分割$T:a=x_0<x_1<\cdots<x_n=b$

  • $f(x)$在小区间$[x_{i-1},x_i]$中的上确界和下确界为$M_i$$m_i$
  • $f(x)$在区间$[a,b]$中的上确界和下确界为$M$$m$
  • $\omega_i=M_i-m_i$,为$f(x)$在区间$[x_{i-1},x_i]$上的振幅
  • $\omega=M-m$$f(x)$在区间$[a,b]$上的振幅
  • $S(T)=\displaystyle\sum_{i=1}^nM_i\Delta x_i$达布上和
  • $s(T)=\displaystyle\sum_{i=1}^nm_i\Delta x_i$达布下和

定理 2.
$f(x)$$[a,b]$上有界,则$f(x)$可积当且仅当

\[\displaystyle\lim_{\lambda(T)\to0+}\sum_{i=1}^n\omega_i\Delta x_i=0 \]

定理 3.
$f(x)$$[a,b]$上有界,$f(x)$可积当且仅当, $\forall \epsilon>0$,存在分割$T$满足

\[|S(T)-s(T)|<\epsilon \]

定理 4.
$f(x)$$[a,b]$上连续,则$f(x)$可积

定理 5.
$f(x)$$[a,b]$上有界,且只有有限个间断点,则$f(x)$可积

定理 6.
$f(x)$$[a,b]$上单调有限,则$f(x)$可积

1.证明: $f(x)$$[a,b]$上连续,则一致连续。

$\forall \epsilon>0$$\exists \delta>0$,有

\[|f(x_1)-f(x_2)|<\epsilon, \forall |x_1-x_2|<\delta \]

取分割$T$$\lambda(T)<\delta$,则$\Delta x_i=x_i-x_{i-1}<\delta$,则有

\[\omega_i=M_i-m_i<\epsilon \]

所以有

\[\sum_{i=1}^n\omega_i\Delta x_i=\sum_{i=1}^n(M_i-m_i)\Delta x_i<\epsilon\sum_{i=1}^n\Delta x_i=\epsilon(b-a) \]

$ \displaystyle\lim_{\lambda(T)\to0+}\sum_{i=1}^n\omega_i\Delta x_i=0 $

2.证: 每个区间上都连续

3.证: 不防设$f(x)$单调增,则

\[\begin{aligned} \sum_{i=1}^n\omega_i\Delta x_i =&\sum_{i=1}^n(f(x_i)-f(x_{i-1}))\Delta x_i \\ <&\sum_{i=1}^n(f(x_i)-f(x_{i-1}))\lambda(T) \\ =&(f(b)-f(a))\lambda(T) \end{aligned} \]

$\forall \epsilon>0$,取$\delta<\frac{\epsilon}{f(b)-f(a)}$,则对$\lambda(T)<\delta$,有

\[\sum_{i=1}^n\omega_i\Delta x_i<\epsilon \]

即有$ \displaystyle\lim_{\lambda(T)\to0+}\sum_{i=1}^n\omega_i\Delta x_i=0 $

$f(x)$有无穷个间断点,是否就不可积?

Riemann函数

\[\begin{aligned} R(x)=\begin{cases} \dfrac1n ,& x=\dfrac{m}{n} \\ 0 ,& x\notin\mathbb{Q} \end{cases} \end{aligned} \]

为可积函数

证: (基本思路: 只有有限个点的值$>\epsilon$ )

$\forall \epsilon>0$,取$N>\dfrac{2}{\epsilon}$,则集合 $S=\{\frac{m}n: n\leq N\}$只有有限个,记个数为$k_N$。 取$\delta=\dfrac{\epsilon}{4k_N}$,则对 区间$[0,1]$上的任意分割$T$,当$\lambda(T)<\delta$时,

  • 把包含分母$n\leq N$的有理数$\dfrac{m}{n}$的区间归为第一类: 这些有理数只有有限($k_N$)。 则第一类区间个数不大于$2k$,而它们长度总和不超出$2k_N\lambda$
  • 其它区间列入第二类;这些区间上的振幅$\omega_i$小于$\dfrac1N$

\[\sum\omega_i\Delta x_i<2k_N\lambda(T)+\dfrac{1}{N} <2k_N\frac{\epsilon}{4k_N}+\frac{\epsilon}2 \]

算例

例 3. $\displaystyle\int_0^a x^k dx$$k\in\mathbb{Z}^+, a\in\mathbb{R}$

. $[0,a]$分为$n$等份,$x_i=\dfrac{a}{n}i, i=0,1,\cdots,n$, 则$\Delta x_i=\frac{a}{n}$。取$\xi_i=x_i$,(为什么?)

\[\sigma(T;\xi)=\sum_{i=1}^n(\dfrac{i}{n}a)^k\dfrac{a}n =a^{k+1}\dfrac{1^k+2^k+\cdots+n^k}{n^{k+1}} \]
\[\begin{aligned} \lim_{n\to+\infty}\dfrac{1^k+2^k+\cdots+n^k}{n^{k+1}} =\lim_{n\to+\infty}\dfrac{n^k}{n^{k+1}-(n-1)^{k+1}} \\ =\lim_{n\to+\infty}\dfrac{n^k}{(k+1)n^k+\cdots} =\frac1{k+1} \end{aligned} \]

函数可积吗?

2. $\displaystyle\int_a^b x^\mu dx$$\mu\in\mathbb{R}, b>a>0$

分割$[a,b]$$x_i=a\cdot q^i, i=0,1,\cdots,n$,其中$q=\sqrt[n]{\dfrac{b}{a}}$$\Delta x_i=a\cdot q^{i-1}(q-1)$,则$n\to+\infty, \lambda(T)\to0+$,取$\xi_i=x_{i-1}$

\[\sigma_n=\sum_{i=1}^n(a q^{i-1})^\mu (a q^{i-1}(q-1)) =a^{\mu+1}(q-1)\sum_{i=1}^n(q^{\mu+1})^{i-1} \]
  • $\mu\neq-1$,则
    \[\begin{aligned} \sigma_n=a^{\mu+1}(q-1)\dfrac{(q^{\mu+1})^n-1}{q^{\mu+1}-1} =\dfrac{q-1}{q^{\mu+1}-1}a^{\mu+1}((\dfrac{b}{a})^{\mu+1}-1) \\ =\dfrac{q-1}{q^{\mu+1}-1}(b^{\mu+1}-a^{\mu+1}) \end{aligned} \]
    \[\begin{aligned} \lim_{n\to+\infty}\dfrac{q-1}{q^{\mu+1}-1} =\lim_{n\to+\infty}\dfrac{q-1}{q^{\mu+1}-1} (set \alpha=q-1) \\ =\lim_{\alpha\to0}\dfrac{\alpha}{(\alpha+1)^{\mu+1}-1} =\dfrac1{\mu+1} \end{aligned} \]
  • $\mu=-1$,则
    \[\sigma_n=n(q-1)=n(\sqrt[n]{\dfrac{b}{a}}-1) \]
    \[\lim_{n\to+\infty}n(\sqrt[n]{\dfrac{b}{a}}-1)=\ln(\dfrac{b}{a}) \]

3. $\displaystyle\int_a^b \sin xdx$

$n$等分区间,$x_i=a+\dfrac{b-a}{n}i, i=0,1,\cdots,n$,取右端点$\xi_i=x_i, h=\dfrac{b-a}n$

\[\begin{aligned} \sigma_n=\sum_{i=1}^n\sin(a+ih)h=h\sum_{i=1}^n\sin(a+ih) \\ =h\dfrac1{2\sin(\frac{h}2)}\sum_{i=1}^n\sin(a+ih)2\sin(\frac{h}2) \\ \end{aligned} \]
\[\begin{aligned} \sigma_n=h\dfrac1{2\sin(\frac{h}2)}\sum_{i=1}^n(\cos(a+(i-\dfrac12)h)-\cos(a+(i+\dfrac12)h)) \\ =h\dfrac{\cos(a+\frac{h}2)-\cos(a+(n+\frac12)h)}{2\sin(\frac{h}2)} \\ =h\dfrac{\cos(a+\frac{h}2)-\cos(b+\frac{h}2)}{2\sin(\frac{h}2)} \\ \end{aligned} \]

则有

\[\lim_{n\to+\infty}\sigma_n=\cos(a)-\cos(b) \]

4. $\displaystyle\int_1^2 \ln x dx$

$n$等分区间,$x_i=1+\dfrac{i}{n}, i=0,1,\cdots,n$,取右端点$\xi_i=x_{i+1}, h=\dfrac{b-a}n$

\[\begin{aligned} \sigma_n=\sum_{i=1}^n\ln(1+\dfrac{i}n)\dfrac1n=\dfrac1n\sum_{i=1}^n\ln(1+\dfrac{i}n) \\ =\dfrac1n\ln(1+\dfrac1n)(1+\dfrac2n)\cdots(1+\dfrac{n}{n}) \\ =\dfrac1n\ln\dfrac{(n+1)\cdots(n+n)}{n^n} \end{aligned} \]
\[=\dfrac1n\ln\dfrac{(2n)!}{n^n n!} =\ln\dfrac{\sqrt[n]{(2n)!}}{n\sqrt[n]{n!}} \]

$\displaystyle\lim_{n\to+\infty} \dfrac{n}{\sqrt[n]{n!}}=e$

\[\begin{aligned} \lim_{n\to+\infty}\sigma_n=\ln(\lim_{n\to+\infty}(\dfrac{\sqrt[2n]{(2n)!}}{2n})^2\dfrac{n}{\sqrt[n]{n!}}4) \\ =\ln(\dfrac{4e}{e^2})=\ln4-1 \end{aligned} \]

或,取$x_i=(\sqrt[n]{2})^i, i=0,1,\cdots,n$$\xi_i=x_i$

例 4. $\displaystyle\lim_{n\to+\infty}((\dfrac1{\sqrt{4n^2-1}}+\cdots+\dfrac1{\sqrt{4n^2-n^2}}))$

例 5. $\displaystyle\lim_{n\to+\infty}(\dfrac1{\sqrt{n^2}}+\dfrac1{\sqrt{n^2-1}}+\cdots+\dfrac1{\sqrt{n^2-(n-1)^2}})$

\[\begin{aligned} &\displaystyle\lim_{n\to+\infty}((\dfrac1{\sqrt{4n^2-1}}+\cdots+\dfrac1{\sqrt{4n^2-n}}))\\ =&\lim_{n\to+\infty}(\dfrac1n(\dfrac1{\frac{\sqrt{4n^2-1}}{n}}+\cdots+\dfrac1{\frac{\sqrt{4n^2-n^2}}{n}})) \\ =&\lim_{n\to+\infty}(\dfrac1n(\dfrac1{\sqrt{4-(\frac1n)^2}}+ \cdots+\dfrac1{\sqrt{4-(\frac{n}{n})^2}})) \\ =&\lim_{n\to+\infty}\dfrac1n\sum_{i=1}^n\dfrac1{\sqrt{4-(\frac{i}{n})^2}} \\ =&\int_0^1\dfrac1{\sqrt{4-x^2}}dx \end{aligned} \]
\[\begin{aligned} &\displaystyle\lim_{n\to+\infty}(\dfrac1{\sqrt{n^2}}+\dfrac1{n^2-1}+\cdots+\dfrac1{\sqrt{n^2-(n-1)^2}}) \\ =&\lim_{n\to+\infty}\dfrac1n(\dfrac1{\frac{\sqrt{n^2}}{n}}+\dfrac1{\frac{\sqrt{n^2-1}}{n}}+\cdots+\dfrac1{\frac{\sqrt{n^2-(n-1)^2}}{n}}) \\ =&\lim_{n\to+\infty}\dfrac1n(\dfrac1{\sqrt1}+\dfrac1{\sqrt{1-\frac1{n^2}}}+\cdots+\dfrac1{\sqrt{1-(\frac{n-1}{n})^2}}) \\ =&\lim_{n\to+\infty}\dfrac1n\sum_{i=1}^n\dfrac1{\sqrt{1-(\frac{i-1}{n})^2}} \\ =&\int_0^1\dfrac1{\sqrt{1-x^2}}dx , (\xi_i=x_{i-1}=\dfrac{i-1}n) \end{aligned} \]

目录

本节读完

例 6. Thanks

6.

达布和

定义 3. (达布和)
$f(x)$$[a,b]$上有界,分割$T:a=x_0<x_1<\cdots<x_n=b$

  • $f(x)$在小区间$[x_{i-1},x_i]$中的上确界和下确界为$M_i$$m_i$
  • $f(x)$在区间$[a,b]$中的上确界和下确界为$M$$m$
  • $\omega_i=M_i-m_i$, $\omega=M-m$
  • $S(T)=\displaystyle\sum_{i=1}^nM_i\Delta x_i$达布上和
  • $s(T)=\displaystyle\sum_{i=1}^nm_i\Delta x_i$达布下和
  1. 对于给定的分割$T$,Riemann和$\sigma(T,\xi)$,有
    \[m(b-a)\leq s(T)\leq \sigma(T, \xi) \leq S(T) \leq M(b-a) \]
  2. 对于给定的分割$T$
    \[S(T)=\sup_{\xi_i}\{\sigma(T,\xi_i)\} , s(T)=\inf_{\xi_i}\{\sigma(T,\xi_i)\} \]

定理 7.
若在给定分割中加入一些新的节点,则达布上和减小,达布下和增加;

若在给定分割$T$中加入$l$个新的节点得到$T'$,则

\[\begin{aligned} &S(T)\geq S(T')\geq S(T)-l\omega\|T\| \\ &s(T)\leq s(T')\leq s(T)+l\omega\|T\| \\ &S(T)-s(T)\leq S(T')-s(T')+2l\omega\|T\| \end{aligned} \]

只需要讨论增加一个节点的情形。

$[x_k,x_{k+1}]$中增加了节点$x'$,记新的分割为$T'$。则$S(T')$$S(T)$仅在区间$[x_k,x_{k+1}]$涉及的项有有所不同。

$S(T)$中对应的项是

\[M_k(x_{k+1}-x_k) \]

其中,$M_k$是函数$f(x)$在区间$[x_k,x_{k+1}]$上的上确界。而$S(T')$中对应的项是

\[M'_k(x'-x_k)+M''_k(x_{k+1}-x') \]

其中$M'_k$$M''_{k+1}$是函数$f(x)$在区间$[x',x_{k+1}]$$[x_{k},x']$上的上确界。显然有

\[M'_k\leq M_k , M''_k\leq M_k \]

于是有

\[M'_k(x'-x_k)+M''_k(x_{k+1}-x')\leq M_k(x_{k+1}-x_k) \]

这样$S(T)-S(T')\geq0$

另一方面,记$f(x)$在区间$[a,b]$上的上确界为$M$, 下确界为$m$,则

\[\begin{aligned} S(T)-S(T')=&M_k(x_{k+1}-x_k)-[M'_k(x'-x_k)+M''_k(x_{k+1}-x')] \\ \leq &M(x_{k+1}-x_k)-[m(x'-x_k)+m(x_{k+1}-x')] \\ =&\omega(x_{k+1}-x_k) \\ \leq &\omega\|T\| \end{aligned} \]

定理 8.
对任意两个分割$T_1$$T_2$,有

\[s(T_1)\leq S(T_2) \]

进而易知(上和的下确界不小于下和的上确界),即

\[l=\sup_{T}(s(T))\leq\inf_{T}(S(T))=L \]

其中$l$称为函数的 下积分$L$称为函数的 上积分

定理 9.

\[\lim_{\|T\|\to 0} S(T)=L, \lim_{\lambda(T)\to 0} s(T)=l \]

$T_1$$T_2$为2个分割,$T_3$为这两个分割的合并。则有

\[S(T_3)\leq S(T_1), s(T_3)\geq s(t_1), S(T_3)\leq S(T_2), s(T_3)\geq s(T_2) \]

所以有

\[S(T_1)\geq S(T_3)\geq s(T_3)\geq s(T_2) \]

即,对任意不同的分割,下和不超过上和

可积性判断

定理 10. (定积分存在的判别准则)
$f(x)$$[a,b]$有界,则$f(x)$可积,当且仅当,

\[\lim_{\lambda(T)\to0+}(S(T)-s(T))=0 \]

\[\displaystyle\lim_{\lambda(T)\to0+}\sum_{i=1}^n\omega_i\Delta x_i=0 \]

证:

($\Rightarrow$) 当$f(x)$可积,则有

\[\lim_{\lambda(T)\to 0+}\sigma(T,\xi_i)=I \]

所以,$\forall \epsilon>0, \exists \delta$,有

\[|\sigma(T,\xi_i)-I|<\epsilon , \forall \lambda(T)<\delta, \forall \xi_i\in[x_{i-1},x_i] \]

即有

\[I-\epsilon<\sigma(T,\xi_i)<I+\epsilon \]

由性质2,有

\[I-\epsilon\leq s(T)\leq S(T)\leq I+\epsilon \]

则有

\[0\leq S(T)-s(T)\leq 2\epsilon \]

\[\lim_{\lambda(T)\to 0+}(S(T)-s(T))=0 \]

($\Leftarrow$) 若 $\lim_{\lambda(T)\to 0+}(S(T)-s(T))=0$,则

$\forall \epsilon>0, \exists \delta>0$,有

\[S(T)-s(T)<\epsilon, \forall \lambda(T)<\delta \]

由性质3,有

\[s(T)\leq l\leq L\leq S(T) \]

所以

\[0\leq L-l\leq S(T)-s(T)\leq\epsilon \]

$\lambda(T)\to0+$,则$l=L$,记为$l=L=I$,则有

\[s(T)\leq I \leq S(T) \]

由性质1,有

\[s(T)\leq\sigma(T,\xi_i)\leq S(T) \]
\[|\sigma(T,\xi_i)-I|\leq S(T)-s(T) , \lambda(T)<\delta \]

\[\lim_{\lambda(T)\to0+}\sigma(T,\xi_i)=I \]

定理 11.
$f(x)$$[a,b]$有界,则 $f(x)$可积,当且仅当, $l=L$

定理 12.
$f(x)$$[a,b]$可积的充要条件是: $\forall \epsilon>0$,存在分割$T$满足

\[|S(T)-s(T)|<\epsilon \]