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分部求和法与积分中值定理
42. (分部积分法)设黎曼-斯提捷积分(Riemman-Stieltjes)积分$\int_{a}^{b}\alpha(x)df(x)$存在,则$\int_{a}^{b}f(x)d\alpha(x)$也存在并且有分部积分公式
$$\int_{a}^{b}f(x)d\alpha(x)=[f(x)\alpha(x)]\Big|_{a}^{b}-\int_{a}^{b}\alpha(x)df(x)$$
证明:一般的Riemman积分中的分部积分公式
$$\int_{a}^{b}f(x)d\alpha(x)=[f(x)\alpha(x)]\Big|_{a}^{b}-\int_{a}^{b}\alpha(x)df(x)$$
是与乘积的微分
$$d(f\alpha)=fd\alpha+\alpha df$$
对应的. 体现了微分与积分这一对矛盾. 但是要求$f,\alpha$为可微函数.在$R-S$积分中,这一条件过强.$\alpha(x)$实际上不必是可微函数甚至不必是连续函数.利用
$R-S$积分的定义,用离散与连续这对矛盾的眼光来看待分部积分.
作划分
$$\pi: a=x_{0}<x_{1}<x_{2}<\cdots<x_{n}=b$$
并且$\|\pi\|=\max ||\Delta x_{i}|,\Delta x_{i}=x_{i}-x_{i-1},x_{i-1}\leq \xi_{i}\leq x_{i}$.
应用分部求和法
\begin{align*}
\sigma(f,\pi,\xi)&=\sum_{k=1}^{n}f(\xi_{k})[\alpha(x_{k})-\alpha(x_{k-1})]\\
&=\sum_{k=1}^{n}f(\xi_{k})\alpha(x_{k})-\sum_{k=1}^{n}f(\xi_{k})\alpha(x_{k-1})\\
&=\sum_{k=1}^{n}f(\xi_{k})\alpha(x_{k})-\sum_{k=0}^{n-1}f(\xi_{k+1})\alpha(x_{k})\\
&=f(b)\alpha(b)-f(a)\alpha(a)+[f(\xi_{n})-f(x_{n})]\alpha(x_{n})+[f(x_{0})-f(\xi_{1})]\alpha(x_{0})\\
&-\sum_{1}^{n-1}[f(\xi_{k+1})-f(\xi_{k})]\alpha(x_{k})\\
&=f(b)\alpha(b)-f(a)\alpha(a)+[f(\xi_{n})-f(x_{n})]\alpha(x_{n})+[f(x_{0})-f(\xi_{1})]\alpha(x_{0})-\sigma(\alpha,\pi‘,x)
\end{align*}
序列
$$\pi‘: a=\xi_{1}<\xi_{2}<\cdots<\xi_{n+1}=b,\xi_{k}\leq x_{k}\leq\xi_{k+1}$$
并且$\|\pi‘\|\leq 2\|\pi\|$,是积分$\int_{a}^{b}\alpha df$的一个划分.从而
$$\lim_{\|\pi\|\to 0}\sigma(f,\pi,\xi)=f(x)\alpha(x)\Big|_{a}^{b}-\lim_{\|\pi‘\|\to 0}\sigma(\alpha,\pi‘,x)$$
即
$$\int_{a}^{b}f(x)d\alpha(x)=[f(x)\alpha(x)]\Big|_{a}^{b}-\int_{a}^{b}\alpha(x)df(x)$$
注意: 若$f$为连续函数而$\alpha$为有界变差函数,则$\int_{a}^{b}fd\alpha$与$\int_{a}^{b}\alpha df$都存在.
43. (第一积分中值定理) 设$\alpha(x)$为一单调函数而$f(x)$为实值连续函数则有中值公式
$$\int_{a}^{b}f(x)d\alpha(x)=f(\xi)[\alpha(b)-\alpha(a)],\,\,(a\leq \xi\leq b)$$
证明: 设
$$m(f)=\min_{x\in[a,b]} f(x);M(f)=\max_{x\in[a,b]}f(x)$$
则有
$$m(f)\leq \frac{1}{\alpha(b)-\alpha(a)}\int_{a}^{b}f(x)d\alpha(x)\leq M(f)$$
由实连续函数的介值定理知存在$\xi\in [a,b]$,使得
$$f(\xi)=\frac{1}{\alpha(b)-\alpha(a)}\int_{a}^{b}f(x)d\alpha(x)$$
44. 设$f(x)$连续而$\psi(x)$为$[a,b]$上的勒贝格(H.Lebesge1875-1941)可积函数(简写作$\psi \in L$),并设$\psi(x)\geq 0$,则必有$\xi,a\leq \xi\leq b$.使得
$$\int_{a}^{b}f(x)\psi(x)dx=f(\xi)\int_{a}^{b}\psi(x)dx$$
证明: 设
$$m(f)=\min_{x\in[a,b]} f(x);M(f)=\max_{x\in[a,b]}f(x)$$
则有
$$m(f)\leq \frac{1}{\int_{a}^{b}\psi(x) dx}\int_{a}^{b}f(x)\psi(x)dx\leq M(f)$$
由实连续函数的介值定理知存在$\xi\in [a,b]$,使得
$$\int_{a}^{b}f(x)\psi(x)dx=f(\xi)\int_{a}^{b}\psi(x)dx$$
45. (长大不等式) 设在$[a,b]$上$f(x)$为连续函数而$\alpha(x)$为有界变差函数,则
$$\left|\int_{a}^{b}f(x)dx\right|\leq M(f)\cdot \bigvee_{a}^{b}(\alpha)$$
这里$M(f)=\max_{a\leq x\leq b}f(x)$,而 $\bigvee\limits_{a}^{b}(\alpha)$为$\alpha$在$[a,b]$上的全变差.
证明:
\begin{align*}
\left|\int_{a}^{b}f(x)dx\right|&\leq \lim_{\|\pi\|\to 0}\sum_{k=1}^{n}\left|f(\xi_{k})\right|\cdot |\alpha(x_{k})-\alpha(x_{k-1})|\\
&\leq M(f)\cdot \lim_{\|\pi\|\to 0}\sum_{k=1}^{n}|\alpha(x_{k})-\alpha(x_{k-1})|\\
&\leq M(f)\cdot \bigvee_{a}^{b}(\alpha)
\end{align*}
46. (第二积分中值定理) 设在$[a,b]$上$\alpha(x)$为一实值连续函数而$f(x)$为一单调函数则必有$\xi,a\leq \xi\leq b$使得
$$\int_{a}^{b}f(x)d\alpha(x)=f(a)\int_{a}^{\xi}d\alpha(x)+f(b)\int_{\xi}^{b}d\alpha(x)$$
证明: 利用分部积分和第一中值定理(涉及到端点值和中间值)
\begin{align*}
\int_{a}^{b}f(x)d\alpha(x)&=f(x)\alpha(x)\Big|_{a}^{b}-\int_{a}^{b}\alpha(x)df(x)\\
&=f(b)\alpha(b)-f(a)\alpha(a)-\alpha(\xi)\int_{a}^{b}df(x)\\
&=f(b)[\alpha(b)-\alpha(\xi)]+f(a)[\alpha(\xi)-\alpha(a)]\\
&=f(a)\int_{a}^{\xi}d\alpha(x)+f(b)\int_{\xi}^{b}d\alpha(x)
\end{align*}
47. (Bonnet) 设$\varphi(x)\in L$,又设$f(x)$单调. 则必存在$\xi,a\leq \xi\leq b$
$$\int_{a}^{b}f(x)\varphi(x)dx=f(a)\int_{a}^{\xi}\varphi(x)dx+f(b)\int_{\xi}^{b}\varphi(x)dx$$
证明: 此命题是命题46的推论.
设$\alpha(x)=\int_{a}^{x}\varphi(t)dt$.
48. (Bonnet)设在$[a,b]$上$\alpha(x)$为实值连续函数而$f(x)\geq 0$且$f(x)\uparrow$,则必有$\xi,a\leq \xi\leq b$,使得
$$\int_{a}^{b}f(x)d\alpha(x)=f(b)\int_{\xi}^{b}d\alpha(x)$$
又若$f(x)\leq 0$且$f(x)\downarrow$,则必有$\xi,a\leq \xi\leq b$,使得
$$\int_{a}^{b}f(x)d\alpha(x)=f(a)\int_{a}^{\xi}d\alpha(x)$$
证明:假定$f(x)\leq 0$且$f(x)\downarrow$则
$$\sigma(f,\pi,\xi)=\sum_{k=1}^{n}f(\xi_{k})[\alpha(x_{k})-\alpha(x_{k-1})]$$
关于$f(\xi_{k})$
$$f(a)=f(\xi_{1})\geq f(\xi_{2})\geq \cdots \geq f(\xi_{n})\geq 0$$
而
$$\inf_{a\leq x\leq b}\int_{a}^{x}d\alpha(t)\leq \sum_{k=1}^{n}[\alpha(x_{k})-\alpha(x_{k-1})]\leq \sup_{a\leq x\leq b}\int_{a}^{x}d\alpha(t)$$
由Abel引理即命题 4 有
$$f(a)\inf_{a\leq x\leq b}\int_{a}^{x}d\alpha(t)\leq\sum_{k=1}^{n}f(\xi_{k})[\alpha(x_{k})-\alpha(x_{k-1})]\leq f(a)\sup_{a\leq x\leq b}\int_{a}^{x}d\alpha(t)$$
取极限$\|\pi\|\to 0$
$$f(a)\inf_{a\leq x\leq b}\int_{a}^{x}d\alpha(t)\leq\int_{a}^{b}f(x)d\alpha(x)\leq f(a)\sup_{a\leq x\leq b}\int_{a}^{x}d\alpha(t)$$
由于$\int_{a}^{x}d\alpha(t)=\alpha(x)-\alpha(a)$为连续函数,由连续函数的介质定理知,存在$\xi,a\leq \xi\leq b$
$$\int_{a}^{b}f(x)d\alpha(x)=f(a)\int_{a}^{\xi}d\alpha(t)$$
若$\alpha(x)=\int_{a}^{x}\varphi(t)dt$,结论为
$$\int_{a}^{b}f(x)\varphi(x)dx=f(a)\int_{a}^{\xi}\varphi(x)dx$$
类似证明另一等式.
49. 试由命题48导出命题46.
证明: 设$f(x)\downarrow$,命题 48要求$f(x)$非负,命题46无此要求只对单调性有要求.因此$f(x)-f(b)\geq 0,\,(a\leq x\leq b)$.
$$\int_{a}^{b}[f(x)-f(b)]d\alpha(x)=[f(a)-f(b)]\int_{a}^{\xi}d\alpha(x)=f(a)\int_{a}^{\xi}d\alpha+f(b)\int_{\xi}^{b}d\alpha-f(b)\int_{a}^{b}d\alpha$$
两边消去$f(b)\int_{a}^{b}d\alpha$即得命题46.
50. 设在$[a,b]$上$\alpha(x)$为一有界变差函数而$f(x)$为非负连续函数.若$f(x)\uparrow$,则
$$\int_{a}^{b}f(x)d\alpha(x)=Af(b)$$
此处
$$\inf_{a\leq x\leq b}\int_{x}^{b}d\alpha(x)\leq A\leq \sup_{a\leq x\leq b}\int_{x}^{b}d\alpha(x)$$
若$f(x)\downarrow$,则
$$\int_{a}^{b}f(x)d\alpha(x)=Bf(a)$$
此处
$$\inf_{a\leq x\leq b}\int_{a}^{x}d\alpha(x)\leq B\leq \sup_{a\leq x\leq b}\int_{a}^{x}d\alpha(x)$$
证明:若$f(x)\geq 0$且$f(x)\uparrow$则
$$f(b)\inf_{a\leq x\leq b}\int_{x}^{b}d\alpha(t)\leq \sum_{k=1}^{n}f(\xi_{k})[\alpha(x_{k})-\alpha(x_{k-1})]\leq f(b)\sup_{a\leq x\leq b}\int_{x}^{b}d\alpha(t) $$
取极限$\|\pi\|\to 0$
$$f(b)\inf_{a\leq x\leq b}\int_{x}^{b}d\alpha(t)\leq\int_{a}^{b}f(x)d\alpha(x)\leq f(b)\sup_{a\leq x\leq b}\int_{x}^{b}d\alpha(t)$$
所以
$$\inf_{a\leq x\leq b}\int_{x}^{b}d\alpha(t)\leq A=\frac{1}{f(b)}\int_{a}^{b}fd\alpha\leq \sup_{a\leq x\leq b}\int_{x}^{b}d\alpha(t)$$
同理可证$f(x)\downarrow$时
$$\inf_{a\leq x\leq b}\int_{a}^{x}d\alpha(t)\leq B=\frac{1}{f(a)}\int_{a}^{b}fd\alpha\leq \sup_{a\leq x\leq b}\int_{a}^{x}d\alpha(t)$$
51. (陈建功) 设$\alpha>0,A>0,0\leq a<b$.试证
$$\left|\int_{a}^{b}\cos\left(nt-\frac{A}{t^{\alpha}}\right)dt\right|<\frac{2}{n}$$
$$\left|\int_{a}^{b}\sin\left(nt-\frac{A}{t^{\alpha}}\right)dt\right|<\frac{2}{n}$$
证明: 作变量替换
$$\omega=t-\frac{A}{nt^{\alpha}}$$
则
$$\frac{d\omega}{dt}=1+\frac{A\alpha}{nt^{\alpha+1}}>0,\, t\in [a,b]$$
$$\frac{d^{2}(t)}{d\omega^{2}}=\left(1+\frac{A\alpha}{nt^{\alpha+1}}\right)^{-2}\frac{\alpha(\alpha+1)A}{nt^{\alpha+2}}\frac{dt}{d\omega}>0$$
则由积分第二中值定理
\begin{align*}
\left|\int_{a}^{b}\cos\left(nt-\frac{A}{t^{\alpha}}\right)dt\right|&=\left|\int_{\omega(a)}^{\omega(b)}\cos (n\omega)\frac{dt}{d\omega}d\omega\right|\\
&=\left(1+\frac{A\alpha}{nb^{\alpha+1}}\right)^{-1}\left|\int_{\xi}^{b}\cos(n\omega)d\omega\right|\\
&\leq \left|\frac{\sin (n\xi)-\sin(n\omega(b))}{n}\right|\\
&\leq \frac{2}{n}
\end{align*}
第二个不等式的证明完全类似.
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原文地址：http://www.cnblogs.com/zhangwenbiao/p/5731242.html
