阿贝尔分布求和法的应用(四)

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.

52. (Dirichlet-Jordan)设$f(x)$为以$2\pi$为周期的的可积函数,那么采用命题29的证明中的记法时,$f(x)$的傅里叶级数的部分和可表达成
$$S_{n}(x)=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x+\theta)D_{n}(\theta)d\theta$$
现设$f(x)$在$[-\pi,\pi]$上为连续的有界变差函数,试证$S_{n}(x)$一致收敛于$f(x)$.
证明:先论证$f(x)$的傅里叶级数的部分和可表达成
$$S_{n}(x)=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x+\theta)D_{n}(\theta)d\theta$$
其中
$$D_{n}(x)=\frac{\sin (n+\frac{1}{2})\theta}{2\sin\frac{\theta}{2}}$$
考察
$$\sum_{-N}^{N}e^{in\theta}=\sum_{-N}^{N}(\cos n\theta+\sin n\theta)=2D_{N}(\theta)$$
利用周期性,奇偶性,函数$f(x)$的傅里叶级数的部分和
\begin{align*}
S_{N}(f)(x)&=\sum_{-N}^{N}\hat{f}(n)e^{inx}\\
&=\sum_{-N}^{N}\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta)e^{-in\theta}d\theta\right)e^{inx}\\
&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)D_{N}(x-\theta)d\theta\\
&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x-\theta)D_{N}(\theta)d\theta\\
&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x+\theta)D_{N}(\theta)d\theta
\end{align*}
经计算
$$\frac{1}{\pi}\int_{-\pi}^{\pi}D_{N}(\theta)d\theta=1$$
拟合法
\begin{align*}
S_{N}(f)(x)-f(x)&=\frac{1}{\pi}\int_{-\pi}^{\pi}[f(x+\theta)-f(x)]D_{N}(\theta)\\
&=\frac{1}{\pi}\int_{0}^{\pi}[f(x+\theta)+f(x-\theta)-2f(x)]D_{N}(\theta)d\theta\\
&=\frac{1}{\pi}\int_{0}^{\pi}\Delta_{\theta}f(x)\frac{\sin n\theta}{\theta}d\theta+\frac{1}{\pi}\int_{0}^{\pi}\Delta_{\theta}f(x)\left(\frac{\cot \frac{\theta}{2}}{2}-\frac{1}{\theta}\right)d\theta\\
&+\frac{1}{2\pi}\int_{0}^{\pi}\Delta_{\theta}f(x)\cos n\theta d\theta\\
&=I_{1}+I_{2}+I_{3}
\end{align*}
由黎曼-勒贝格定理知$I_{2}\to 0, I_{3}\to 0$且一致收敛到0. $I_{1}$主要是在零点附近会出现问题
$$I_{1}=\int_{0}^{\frac{\pi}{n}}+\int_{\frac{\pi}{n}}^{\eta}+\int_{\eta}^{\pi}=I_{11}+I_{12}+I_{13}$$
分别估计
$$|I_{11}|\leq\max_{0\leq \theta\leq \frac{\pi}{n}}|\Delta_{\theta}f(x)|\int_{0}^{\pi}\frac{\sin\theta}{\theta}d\theta\leq \max_{0\leq \theta\leq \frac{\pi}{n}}|\Delta_{\theta}f(x)|\cdot \pi $$
$$|I_{12}|=\frac{n}{\pi}\int_{\frac{\pi}{n}}^{\xi}\Delta_{\theta}f(x)\sin n\theta d\theta\leq \frac{2}{\pi}\max_{0\leq \theta\leq \eta}|\Delta_{\theta}f(x)|+\frac{1}{\pi}\bigvee_{0}^{\eta}(\Delta_{\theta}f)$$
于是$I_{11},I_{12}$一致收敛到0. $I_{13}$可使用黎曼—勒贝格定理.
53. 设$f(x)$是$[a,b]$上的可积函数,
$$F(x)=\int_{a}^{x}f(t)dt,\,|F(x)|\leq M(x-a) (a\leq x \leq b)$$
又设$g(x)$为是$[a,b]$上的非负并且非增的可积函数,则
$$\left|\int_{a+}^{b}f(x)g(x)dx\right|\leq M\int_{a}^{b}g(x)dx$$
证明: 设$a<\alpha<\beta\leq b$. 则由分部积分法
\begin{align*}
&\int_{\alpha}^{\beta}f(x)g(x)dx-g(\beta)\int_{\alpha}^{\beta}f(x)dx\\
=&\int_{\alpha}^{\beta}f(x)[g(x)-g(\beta)]dx\\
=&-F(\alpha)[g(\alpha)-g(\beta)]-\int_{\alpha}^{\beta}F(x)dg(x)
\end{align*}
把关于$|F(x)|$的条件用到上式的末端可知其绝对值不超过
\begin{align*}
&M(\alpha-a)[g(\alpha)-g(\beta)]-\int_{\alpha}^{\beta}M(x-a)dg(x)\\
=&M(\alpha-a)[g(\alpha)-g(\beta)]-[M(x-a)g(x)]_{\alpha}^{\beta}+M\int_{\alpha}^{\beta}g(x)dx\\
=&M\int_{\alpha}^{\beta}g(x)dx+2M(\alpha-a)g(\alpha)-M(\alpha+\beta-2a)g(\beta)
\end{align*}
当$\alpha,\beta\to a+$时,右端两项是无穷小量,所以$\int_{a+}^{b}f(x)g(x)dx$收敛. 置$\beta=b,\alpha\to a+$, 由上式得
$$\left|\int_{a+}^{b}f(x)g(x)dx\right|\leq M\int_{a}^{b}g(x)dx-M(b-a)g(b)+g(b)\left|\int_{a}^{b}f(x)dx\right|\leq M\int_{a}^{b}g(x)dx$$
54. （Steffensen） 设$f(x)$和$g(x)$是$[a,b]$上的两个可积函数,$f(x)$不增(注:不增的意思就是递减),$g(x)$满足$0\leq g(x) \leq 1$,则
$$\int_{b-\lambda}^{b}f(t)dt\leq \int_{a}^{b}f(t)g(t)dt\leq \int_{a}^{a+\lambda}f(t)dt$$
证明: 非常有意思的不等式. 不妨设$f(t)$非负否则置$f(x)-m$代入原不等式.
\begin{align*}
&\int_{a}^{a+\lambda}f(x)dx-\int_{a}^{b}f(x)g(x)dx\\
=&\int_{a}^{a+\lambda}f(x)[1-g(x)]dx-\int_{a+\lambda}^{b}f(x)g(x)dx\\
\geq&f(a+\lambda)\int_{a}^{a+\lambda}[1-g(x)]dx-f(a+\lambda)\int_{a}^{\xi}g(x)dx\\
=&f(a+\lambda)\int_{\xi}^{b}g(x)dx\\
\geq&0
\end{align*}
另一方面
\begin{align*}
&\int_{a}^{b}f(x)g(x)dx-\int_{b-\lambda}^{b}f(x)dx\\
=&\int_{a}^{b-\lambda}f(x)g(x)dx+\int_{b-\lambda}^{b}f(x)[g(x)-1]dx\\
\geq&f(b-\lambda)\int_{a}^{b-\lambda}g(x)dx+f(b-\lambda)\int_{b-\lambda}^{b}[g(x)-1]dx\\
=&f(b-\lambda)\int_{a}^{b}g(x)dx-f(b-\lambda)\int_{b-\lambda}^{b}dx\\
=&0
\end{align*}
另证: 只需要注意到恒等变形
\begin{align*}
&\int_{a}^{a+\lambda}f(x)dx-\int_{a}^{b}f(x)g(x)dx\\
=&\int_{a}^{a+\lambda}[f(x)-f(a+\lambda)]\cdot [1-g(x)]dx
+\int_{a+\lambda}^{b}[f(a+\lambda)-f(x)]\cdot g(x)dx
\end{align*}
55. (Stefensen) 设$g_{1}(x)$与$g_{2}(x)$满足
$$\int_{a}^{x}g_{1}(x)\leq \int_{a}^{x}g_{2}(x)\,\,(a\leq x\leq b)$$
$$\int_{a}^{b}g_{1}(x)=\int_{a}^{b}g_{2}(x)$$
又设$f(x)$非增,则
$$\int_{a}^{x}g_{1}(x)\leq \int_{a}^{x}f(x)g_{2}(x)$$
证明: 使用第二积分中值定理,$f(x)-f(b)\geq 0$且非增.
$$\int_{a}^{b}f(x)\cdot[g_{2}(x)-g_{1}(x)]dx=\int_{a}^{b}[f(x)-f(b)]\cdot[g_{2}(x)-g_{1}(x)]dx
=[f(a)-f(b)]\int_{a}^{\xi}[g_{2}(x)-g_{1}(x)]dx\geq 0$$
56. 试由命题55推导出命题44.
证明:取截断函数即可
\begin{equation*}
g_{1}(x)=\left\{\begin{array}{ll}
1&\text{$b-\lambda\leq x\leq b$}\\
0&\text{其他}
\end{array}\right.
\end{equation*}
\begin{equation*}
g_{3}(x)=\left\{\begin{array}{ll}
1&\text{$a\leq x\leq a+\lambda$}\\
0&\text{其他}
\end{array}\right.
\end{equation*}
则当$b-\lambda \leq x\leq $且$0\leq g(x)\leq 1$时有
$$\int_{a}^{x}g_{1}(x)dx=\int_{a}^{b}g(x)dx+x-b\leq \int_{a}^{x}g(x)dx$$
且
$$\int_{a}^{b}g_{1}(x)dx=\int_{a}^{b}g(x)dx=\int_{a}^{b}g_{3}(x)$$
利用把$g_{1}(x),g_{2}(x),g_{3}(x)$代入命题55即得命题54.
57. (Hayashi)以$0\leq g(t)\leq A$代替命题54中的$0\leq g(x)\leq 1$,则有
$$A\int_{b-\lambda}^{b}f(t)dt\leq \int_{a}^{b}f(t)g(t)dt\leq A\int_{a}^{a+\lambda}f(t)dt$$
这里$A$是正的常数,$\lambda=\frac{1}{A}\int_{a}^{b}g(t)dt$.
证明:归一化处理
$$\tilde{g}(x)=\frac{g(x)}{A}\,\,(a\leq x\leq b)$$
则$0\leq g(x)\leq 1$,
$$\lambda=\int_{a}^{b}\tilde{g}(x)dx=\frac{1}{A}\int_{a}^{b}g(x)dx$$
由命题54得
$$\int_{b-\lambda}^{b}f(t)dt\leq \int_{a}^{b}f(t)\tilde{g}(t)dt\leq \int_{a}^{a+\lambda}f(t)dt$$
即
$$A\int_{b-\lambda}^{b}f(t)dt\leq \int_{a}^{b}f(t)g(t)dt\leq A\int_{a}^{a+\lambda}f(t)dt$$
58. 不等式
$$\int_{b-\lambda}^{b}f(t)dt\leq \int_{a}^{b}f(t)g(t)dt\leq \int_{a}^{a+\lambda}f(t)dt$$
式中$\lambda=\int_{a}^{b}g(t)dt$对\textbf{每个}不增的函数$f(t)$都成立的充分且必要条件是函数$g(t)$对所有的$x\in [a,b]$满足
$$0\leq\int_{x}^{b}g(t)dt\leq b-x\text{且}0\leq\int_{a}^{x}g(t)dt\leq x-a$$
证明: 先证明必要性，取
\begin{equation*}
f(t)=\left\{\begin{array}{ll}
1&\text{$a\leq t\leq x$}\\
0&\text{$x< t$}
\end{array}\right.
\end{equation*}
则$f(x)$为不增函数
$$\int_{a}^{b}f(t)g(t)dt=\int_{a}^{x}g(t)dt\leq \int_{a}^{x}dx=x-a$$
取
\begin{equation*}
f(t)=\left\{\begin{array}{ll}
1&\text{$x\leq t\leq b$}\\
0&\text{$t< x$}
\end{array}\right.
\end{equation*}
类似可得
$$0\leq\int_{x}^{b}g(t)dt\leq b-x$$
再证充分性
\begin{align*}
&\int_{a}^{a+\lambda}f(t)dt-\int_{a}^{b}f(t)g(t)dt\\
=&\int_{a}^{a+\lambda}f(t)\cdot[1-g(t)]dt-\int_{a+\lambda}^{b}f(t)g(t)dt\\
=&\int_{a}^{a+\lambda}[f(t)-f(a+\lambda)]\cdot [1-g(t)]dt+\int_{a+\lambda}^{b}[f(a+\lambda)-f(t)]g(t)dt\\
=&I_{1}+I_{2}
\end{align*}
利用分部积分
\begin{align*}
I_{1}&=\int_{a}^{a+\lambda}[f(t)-f(a+\lambda)]d\left[t-\int_{a}^{t}g(x)\right]\\
&=[f(t)-f(a+\lambda)]\left[t-\int_{a}^{t}g(x)\right]\Bigg|_{a}^{a+\lambda}+\int_{a}^{a+\lambda}\left[t-\int_{a}^{t}g(x)\right]d(-f)\\
&=[f(a)-f(a+\lambda)]a+\int_{a}^{a+\lambda}\left[t-\int_{a}^{t}g(x)\right]d(-f)\\
&\geq [f(a)-f(a+\lambda)]a+\int_{a}^{a+\lambda}a d(-f)\\
&\geq 0
\end{align*}
对$I_{2}$估计
\begin{align*}
I_{2}&=\int_{a+\lambda}^{b}[f(a+\lambda)-f(t)]g(t)dt\\
&=\int_{a+\lambda}^{b}[f(a+\lambda)-f(t)]d\int_{a}^{t}g(x)dx\\
&=\lim_{\|\pi\|\to 0}\sum_{k=1}^{n}[f(a+\lambda)-f(\xi_{k})]\cdot [\alpha(t_{k})-\alpha(t_{k-1})]\\
&\geq 0
\end{align*}
其中$\alpha(t)=\int_{a}^{t}g(x)dx$.
所以
$$\int_{a}^{b}f(t)g(t)dt\leq \int_{a}^{a+\lambda}f(t)dt$$
同理可证
$$\int_{b-\lambda}^{b}f(t)dt\leq \int_{a}^{b}f(t)g(t)dt$$

59. (Abel)设无穷积分$\int_{a}^{\infty}\varphi(x)dx$为收敛. 又设$\psi(x)$为一单调的有界函数.则积分$\int_{a}^{\infty}\varphi(x)\psi(x)dx$必收敛.
证明:由收敛的Cauchy准则知任意$\varepsilon>0$,存在$X$,$A>B\geq X$时
$$\left|\int_{A}^{B}\varphi(x)\right|\leq \varepsilon$$
利用积分第二中值定理
$$\left|\int_{A}^{B}\varphi(x)\psi(x)\right|=\left|\psi(A)\int_{A}^{\xi}\varphi(x)dx+\psi(B)\int_{\xi}^{B}\varphi(x)dx\right|\leq 2M\cdot \varepsilon$$
其中$a\leq \xi\leq b,M=\sup\limits_{a\leq x\leq \infty}{\psi(x)}$. 由Cauchy收敛准则积分$\int_{a}^{\infty}\varphi(x)\psi(x)dx$必收敛.
60. (Dirichlet)设函数$\alpha(x)=\int_{a}^{x}$为有界$(a\leq x\leq \infty)$. 又设当$x\to\infty$时$\psi(x)\downarrow 0$.则积分$\int_{a}^{\infty}\varphi(x)\psi(x)dx$必收敛.
证明: 由$\psi(x)\downarrow 0,x\to \infty$知任意$\varepsilon>0$存在$X$,当$A>B>X$时,$|\psi(x)|<\varepsilon$.
由积分第二中值定理
$$\left|\int_{A}^{B}\varphi(x)\psi(x)dx\right|=\left|\psi(B)\int_{A}^{\xi}\varphi(x)dx\right|\leq 2M\cdot \varepsilon$$
其中$M=\sup_{a\leq \infty}\left|\int_{a}^{x}\varphi(x)\right|$.由Cauchy收敛准则积分$\int_{a}^{\infty}\varphi(x)\psi(x)dx$必收敛.
61. 设$\varphi(x)\downarrow 0 (x\to\infty)$试证下列二积分必收敛
$$\int_{a}^{\infty}\varphi(x)\sin xdx,\,\int_{a}^{\infty}\varphi(x)\cos xdx$$
证明:
$$\left|\int_{a}^{x}\sin x\right|\leq 2;\,\left|\int_{a}^{x}\cos x\right|\leq 2$$
又$\varphi(x)\downarrow 0 (x\to\infty)$,由狄利克雷判别法即命题60知$\int_{a}^{\infty}\varphi(x)\sin xdx,\,\int_{a}^{\infty}\varphi(x)\cos xdx$收敛.
62. 设$a$是正的常数,试证不等式
$$\left|\int_{a}^{\infty}\cos (x^{2})dx\right|\leq \frac{1}{a}$$
证明:
$$\left|\int_{a}^{A}\cos (x^{2})dx\right|=\left|\int_{a^{2}}^{A^{2}}\frac{\cos t}{2\sqrt{t}}dt\right|=\frac{1}{2a}\left|\int_{a^{2}}^{\xi}\cos t\right|\leq \frac{2}{a}$$
令$A\to\infty$.由狄利克雷判别法知左边级数收敛.