9.3 习题课讲解

9.3.1 函数的可积性

解 (1) 有界闭区间上的连续函数是有界函数，由Lebesgue准则知结论成立。

(2) 有界闭区间上的单调函数是有界函数，单调函数至多只有可数无穷多个间断点，可数集是零测集。

(3) $f$是单调不减的有界函数，从而可积。

(4) $f+g,fg$都是有界函数，它们的间断点集是$f$的间断点集和$g$的间断点集的并集的子集，两个（甚至可数多个）零测集的并集是零测集。

(5) 如果$|f|$有正下界，则$1/f$有界，$1/f$的间断点是$f$的间断点，从而$1/f$可积。但如果$|f|$无下界，则$1/f$无界，从而$f$不可积。

(6) $g\circ f$有界。若$g$连续，则$g\circ f$的间断点集是$f$的间断点集的子集，从而是零测集，于是$g\circ f$可积。若$g$不连续则存在反例，设$g(x)=(\mathrm{sgn}x{)}^2$，$f$为Riemann函数，则$g\circ f$为Dirichlet函数，不可积。

(7) Dirichlet函数在所有点间断，不可积。

(8) Riemann函数有界，仅在有理数处间断，有理数集是零测集。所以Riemann函数可积。

(9) $f\in \mathcal{R}[\alpha ,\beta ]$。 $◻$

9.3.2 定积分的近似计算

解 设$f(x)=\{\begin{array}{ll}{x}^x,& x>0\\ 0,& x=0\end{array}$，容易验证$f$可积。注意到$$\begin{array}{}\text{(2)}& \frac{1}{n}[{\left(\frac{1}{n}\right)}^{\frac{1}{n}}+{\left(\frac{2}{n}\right)}^{\frac{2}{n}}+\cdots +{\left(\frac{n-1}{n}\right)}^{\frac{n-1}{n}}+1]=\sum _{i=1}^n\frac{1}{n}f\left(\frac{i}{n}\right)\end{array}$$ 因此$$\begin{array}{}\text{(3)}& \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}[{\left(\frac{1}{n}\right)}^{\frac{1}{n}}+{\left(\frac{2}{n}\right)}^{\frac{2}{n}}+\cdots +{\left(\frac{n-1}{n}\right)}^{\frac{n-1}{n}}+1]={\int }_0^1{x}^x\mathrm{d}x\end{array}$$ 由于$f(x)$是下凸函数且有极小值点$x={\mathrm{e}}^{-1}$，$f(x)=x^x={\mathrm{e}}^{x\ln x}\ge 1+x\ln x$，因此$$\begin{array}{}\text{(4)}& \begin{array}{rl}{\int }_0^1{x}^x\mathrm{d}x& >{\int }_0^1(1+x\ln x)\mathrm{d}x={[x+\frac{1}{2}{x}^2\ln x-\frac{1}{4}{x}^2]}_0^1=\frac{3}{4}\\ {\int }_0^1{x}^x\mathrm{d}x& <\frac{f(0)+f({\mathrm{e}}^{-1})}{2}({\mathrm{e}}^{-1}-0)+\frac{f({\mathrm{e}}^{-1})+f(1)}{2}(1-{\mathrm{e}}^{-1})\\ & =\frac{1+{\mathrm{e}}^{-{\mathrm{e}}^{-1}}}{2}<0.85\end{array}\end{array}$$ 因此对充分大的正整数$n$，有$$\begin{array}{}\text{(5)}& 0.75n<{\left(\frac{1}{n}\right)}^{\frac{1}{n}}+{\left(\frac{2}{n}\right)}^{\frac{2}{n}}+\cdots +{\left(\frac{n-1}{n}\right)}^{\frac{n-1}{n}}+1<0.85n\end{array}$$ $◻$

注 凸函数的定义：$\mathrm{\forall }{x}_1,x_2\in [a,b]$，$\mathrm{\forall }\lambda \in [0,1]$，有$$\begin{array}{}\text{(17)}& f(\lambda x_1+(1-\lambda )x_2)\le \lambda f(x_1)+(1-\lambda )f(x_2)\end{array}$$

解 记$R_k(h)=F_k(h)-F(h)$，对$R_k^{\prime }$进行Taylor展开，只保留$h$的偶次方项$$\begin{array}{}\text{(19)}& \begin{array}{rl}{R}_1^{\prime }(h)& =2f(0)-F^{\prime }(h)=2f(0)-f(h)-f(-h)\\ & =2f(0)-[f(0)+\frac{1}{2}{f}^{″}(0)h^2+\frac{1}{24}{f}^{(4)}({\xi }_1)h^4]-[f(0)+\frac{1}{2}{f}^{″}(0)h^2+\frac{1}{24}{f}^{(4)}({\eta }_1)h^4]\\ & =-f^{″}(0)h^2-\frac{1}{24}[f^{(4)}({\xi }_1)+f^{(4)}({\eta }_1)]h^4\end{array}\end{array}$$ $$\begin{array}{}\text{(20)}& \begin{array}{rl}{R}_2^{\prime }(h)& =f(-h)+f(h)+h[f^{\prime }(h)-f^{\prime }(-h)]-F^{\prime }(h)=h[f^{\prime }(h)-f^{\prime }(-h)]\\ & =h[2f^{″}(0)h+\frac{1}{6}{f}^{(4)}({\xi }_2)h^3+\frac{1}{6}{f}^{(4)}({\eta }_2)h^3]\\ & =2f^{″}(0)h^2+\frac{1}{6}[f^{(4)}({\xi }_2)+f^{(4)}({\eta }_2)]h^4\end{array}\end{array}$$ 故$R_1^{\prime },R_2^{\prime }$均为二阶小量，因此$R_1,R_2$均为三阶小量。注意到$$\begin{array}{}\text{(21)}& [2R_1(h)+R_2(h){]}^{\prime }=\mathcal{O}(h^4)⟹2R_1(h)+R_2(h)=\mathcal{O}(h^5),h\to 0\end{array}$$ 因此$2R_1+R_2$具有更高的阶，此时$$\begin{array}{}\text{(22)}& \mathcal{O}(h^5)=2R_1(h)+R_2(h)=4f(0)h-[f(h)+f(-h)]h-3{\int }_{-h}^{h}f(x)\mathrm{d}x\end{array}$$ 由此得到$$\begin{array}{}\text{(23)}& {\int }_{-h}^{h}f(x)\mathrm{d}x=\frac{f(-h)+4f(0)+f(h)}{3}h+\mathcal{O}(h^5)\end{array}$$ 此即Simpson公式，其更一般的形式为$$\begin{array}{}\text{(24)}& {\int }_a^{b}f(x)\mathrm{d}x=\frac{b-a}{6}[f(a)+4f\left(\frac{a+b}{2}\right)+f(b)]+\mathcal{O}((b-a{)}^5)\end{array}$$ $◻$

9.3.3 定积分计算

解 画图。 $◻$

解 (1) 记$G(m,n):={\int }_0^1\frac{x^{m-1}}{(m-1)!}\frac{(1-x{)}^{n-1}}{(n-1)!}\mathrm{d}x$，由分部积分可得$$\begin{array}{}\text{(25)}& \begin{array}{rl}G(m,n)& ={\int }_0^1\frac{(1-x{)}^{n-1}}{(n-1)!}\mathrm{d}\frac{x^m}{m!}={\frac{(1-x{)}^{n-1}}{(n-1)!}\frac{x^m}{m!}|}_0^1-{\int }_0^1\frac{x^m}{m!}\mathrm{d}\frac{(1-x{)}^{n-1}}{(n-1)!}\\ & ={\int }_0^1\frac{(1-x{)}^{n-2}}{(n-2)!}\frac{x^m}{m!}\mathrm{d}x=G(m+1,n-1)=\cdots =G(m+n-1,1)\\ & ={\int }_0^1\frac{x^{m+n-2}}{(m+n-2)!}\mathrm{d}x=\frac{1}{(m+n-1)!}\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(26)}& {\int }_0^1{x}^{m-1}(1-x{)}^{n-1}\mathrm{d}x=\frac{(m-1)!(n-1)!}{(m+n-1)!}\end{array}$$

(2) 记$I_n={\int }_0^{\pi /2}{\sin }^{n}x\mathrm{d}x$，由分部积分可得$$\begin{array}{}\text{(27)}& \begin{array}{rl}{I}_n& =-{\int }_0^{\pi /2}{\sin }^{n-1}x\mathrm{d}\cos x={-{\sin }^{n-1}x\cos x|}_0^{\pi /2}+(n-1){\int }_0^{\pi /2}{\sin }^{n-2}x{\cos }^{2}x\mathrm{d}x\\ & =(n-1)(I_{n-2}-I_n)=\frac{n-1}{n}{I}_{n-2}=\frac{(n-1)(n-3)}{n(n-2)}{I}_{n-4}=\cdots =\frac{(n-1)!!}{n!!}{I}_{n\mathrm{mod}2}\end{array}\end{array}$$ 其中$I_0=\frac{\pi }{2}$、$I_1=1$。

(3) $$\begin{array}{}\text{(28)}& {\int }_0^{\pi /4}\frac{\mathrm{d}x}{\cos x}={\ln (\sec x+\tan x)|}_0^{\pi /4}=\ln (1+\sqrt{2})\end{array}$$

(4) $$\begin{array}{}\text{(29)}& {\int }_0^1\frac{\sqrt{x}}{4-\sqrt{x}}\mathrm{d}x={(-8\sqrt{x}-x-32\ln (4-\sqrt{x}))}_0^1=-9+32\ln \frac{4}{3}\end{array}$$

(5) $$\begin{array}{}\text{(30)}& {\int }_0^1\sqrt{2x+x^2}\mathrm{d}x={[\frac{(x+1)\sqrt{x(x+2)}}{2}+\ln (\sqrt{x+2}-\sqrt{x})]}_0^1=\sqrt{3}+\ln (\sqrt{3}-1)-\frac{\ln 2}{2}\end{array}$$

(6) $$\begin{array}{}\text{(31)}& {\int }_0^{\ln 2}\sqrt{1+{\mathrm{e}}^x}\mathrm{d}x={[2\sqrt{1+{\mathrm{e}}^x}+x-2\ln (\sqrt{1+{\mathrm{e}}^x}+1)]}_0^{\ln 2}=2(\sqrt{3}-\sqrt{2})+\ln 2-2\ln \frac{\sqrt{3}+1}{\sqrt{2}+1}\end{array}$$

(7) $$\begin{array}{}\text{(32)}& {\int }_0^{\pi }\frac{\mathrm{d}x}{2+\cos x}={\frac{2}{\sqrt{3}}\arctan \frac{\tan \frac{x}{2}}{\sqrt{3}}|}_0^{\pi }=\frac{\pi }{\sqrt{3}}\end{array}$$

(8) $$\begin{array}{}\text{(33)}& {\int }_0^{3\pi /4}\frac{\sin x}{1+{\cos }^{2}x}\mathrm{d}x={-\arctan \cos x|}_0^{3\pi /4}=\arctan \frac{1}{\sqrt{2}}+\frac{\pi }{4}\end{array}$$ $◻$

解 (1) 令$t=\frac{\pi }{2}-x$，则有$$\begin{array}{}\text{(37)}& {\int }_0^{\pi /2}f(\sin x)\mathrm{d}x={\int }_{\pi /2}^{0}f(\cos t)\mathrm{d}(-t)={\int }_0^{\pi /2}f(\cos x)\mathrm{d}x\end{array}$$

令$t=\pi -x$，则有$$\begin{array}{}\text{(38)}& \begin{array}{rl}{\int }_0^{\pi }xf(\sin x)\mathrm{d}x& ={\int }_{\pi }^0(\pi -t)f(\sin t)\mathrm{d}(-t)=\pi {\int }_0^{\pi }f(\sin x)\mathrm{d}x-{\int }_0^{\pi }xf(\sin x)\mathrm{d}x\\ ⟹{\int }_0^{\pi }xf(\sin x)\mathrm{d}x& =\frac{\pi }{2}{\int }_0^{\pi }f(\sin x)\mathrm{d}x\end{array}\end{array}$$

(2) 由(1)可得$$\begin{array}{}\text{(39)}& {\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}\mathrm{d}x=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}\mathrm{d}x=\frac{\pi }{2}{(-\arctan \cos x)}_0^{\pi }=\frac{{\pi }^2}{4}\end{array}$$

(3) 令$t=\frac{\pi }{2}-x$，则有$$\begin{array}{}\text{(40)}& {\int }_0^{\pi /2}\frac{\mathrm{d}x}{1+{\tan }^{\alpha }x}={\int }_{\pi /2}^0\frac{\mathrm{d}(-t)}{1+{\cot }^{\alpha }t}={\int }_0^{\pi /2}\frac{\mathrm{d}x}{1+{\cot }^{\alpha }x}\end{array}$$ $◻$

解 (1) 因为$f$连续，故$F(t):={\int }_0^{t}f(s)\mathrm{d}s$可微，由分部积分可得$$\begin{array}{}\text{(43)}& {\int }_0^x{\int }_0^{t}f(s)\mathrm{d}s\mathrm{d}t={\int }_0^{x}F(t)\mathrm{d}t={tF(t)|}_0^x-{\int }_0^{x}t{F}^{\prime }(t)\mathrm{d}t={\int }_0^x(x-t)f(t)\mathrm{d}t\end{array}$$

(2) 由分部积分可得$$\begin{array}{}\text{(44)}& {\int }_0^{1}x{\int }_1^{x^2}{\mathrm{e}}^{-t^2}\mathrm{d}t\mathrm{d}x={\frac{x^2}{2}{\int }_1^{x^2}{\mathrm{e}}^{-t^2}\mathrm{d}t|}_{x=0}^1-{\int }_0^1\frac{x^2}{2}{\mathrm{e}}^{-x^4}\mathrm{d}(x^2)=\frac{1}{4}({\mathrm{e}}^{-1}-1)\end{array}$$ $◻$

9.3.4 有关定积分的证明题

另证 由定积分（极限）的保号性，很容易证明$$\begin{array}{}\text{(66)}& {\int }_a^{b}f(x)\mathrm{d}x\ge {\int }_a^{b}0\mathrm{d}x=0\end{array}$$

为了排除这个等式，我们考虑使用反证法。假设${\int }_a^{b}f(x)\mathrm{d}x=0$，任给$[a,b]$的划分$P$和标志点集$\{x_k{\}}_{k=1}^n$，当$\parallel P\parallel \to 0$时，Darboux上和$\bar{S}(f,P,\{x_k{\}}_{k=1}^n)$收敛于$0$。于是$\mathrm{\forall }{\epsilon }_1>0$，$\mathrm{\exists }[a_1,b_1]\subset [a,b]$使得$x\in [a_1,b_1]⟹f(x)<{\epsilon }_1$，否则$\mathrm{\exists }{\xi }_i\in [x_{i-1},x_i]$使得$f({\xi }_i)\ge {\epsilon }_1$，此时有$$\begin{array}{}\text{(67)}& \bar{S}(f,P,\{x_k{\}}_{k=1}^n)\ge \sum _{i=1}^n{\epsilon }_1(x_i-x_{i-1})={\epsilon }_1(b-a)\end{array}$$ 与其收敛于$0$矛盾。