6.3 习题课讲解

6.3.1 计算含参积分的导数

解 分别对积分下限、积分上限和被积函数求导可得$$\begin{array}{}\text{(2)}& \frac{\mathrm{d}}{\mathrm{d}x}{\int }_{a(x)}^{b(x)}f(x,t)\mathrm{d}t=f(x,b(x))b^{\prime }(x)-f(x,a(x))a^{\prime }(x)+{\int }_{a(x)}^{b(x)}\frac{\partial f}{\partial x}(x,t)\mathrm{d}t\end{array}$$ $◻$

解 代入上题的公式可得$$\begin{array}{}\text{(4)}& I^{\prime }(x)={\mathrm{e}}^{{\cos }^{2}x+x\cos x}(-\sin x)-{\mathrm{e}}^{{\sin }^{2}x+x\sin x}\cos x+{\int }_{\sin x}^{\cos x}t{\mathrm{e}}^{t^2+xt}\mathrm{d}t\end{array}$$ 因此$$\begin{array}{}\text{(5)}& I^{\prime }(0)=-1+{\int }_0^{1}t{\mathrm{e}}^{t^2}\mathrm{d}t=-1+{\frac{1}{2}{\mathrm{e}}^{t^2}|}_0^1=\frac{1}{2}(\mathrm{e}-3)\end{array}$$ $◻$

解 由导数公式（先对积分限求导、再对被积函数求导）可得$$\begin{array}{}\text{(9)}& \begin{array}{rl}{F}^{\prime }(x)& ={\int }_{x^2}^{x}f(x,s)\mathrm{d}s+{\int }_0^x\mathrm{d}t\cdot \frac{\partial }{\partial x}{\int }_{t^2}^{x}f(t,s)\mathrm{d}s={\int }_{x^2}^{x}f(x,s)\mathrm{d}s+{\int }_0^{x}f(t,x)\mathrm{d}t\\ G^{\prime }(t)& ={\int }_0^{t}f(x,t)\mathrm{d}x+{\int }_0^t\mathrm{d}y\cdot \frac{\partial }{\partial t}{\int }_0^{t}f(x,y)\mathrm{d}x={\int }_0^{t}f(x,t)\mathrm{d}x+{\int }_0^{t}f(t,y)\mathrm{d}y\end{array}\end{array}$$ $◻$

6.3.2 计算含参积分

本小节内容以计算含参积分的方法为主，故略去对一致收敛性的讨论。

解 (1) 令$$\begin{array}{}\text{(10)}& F(b):={\int }_0^{\pi /2}\ln (a^2{\sin }^{2}x+b^2{\cos }^{2}x)\mathrm{d}x\end{array}$$ 显然$F(a)=0$，计算可得$$\begin{array}{}\text{(11)}& F^{\prime }(b)={\int }_0^{\pi /2}\frac{2b{\cos }^{2}x}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}\mathrm{d}x=2b{\int }_0^{\pi /2}\frac{\mathrm{d}x}{a^2{\tan }^{2}x+b^2}=\frac{\pi }{a+b}\end{array}$$ 故有$$\begin{array}{}\text{(12)}& F(b)=F(a)+{\int }_a^b{F}^{\prime }(t)\mathrm{d}t=\pi \ln \frac{a+b}{2}\end{array}$$

(2) 令$$\begin{array}{}\text{(13)}& G(\lambda ):={\int }_0^{\pi /2}\frac{\arctan (\lambda \tan x)}{\tan x}\mathrm{d}x\end{array}$$ 注意到$$\begin{array}{}\text{(14)}& \begin{array}{rl}\underset{x\to 0^{+}}{lim}\frac{\arctan (\lambda \tan x)}{\tan x}& =\underset{x\to 0^{+}}{lim}\frac{\lambda \tan x+o(\tan x)}{\tan x}=\lambda \\ \underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{\arctan (\lambda \tan x)}{\tan x}& =\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{\frac{\pi }{2}+o(1)}{\tan x}=0\end{array}\end{array}$$ 故此积分不为瑕积分。显然$G(0)=0$，计算可得$$\begin{array}{}\text{(15)}& G^{\prime }(\lambda )={\int }_0^{\pi /2}\frac{1}{1+{\lambda }^2{\tan }^{2}x}\mathrm{d}x=\frac{\pi }{2(1+\lambda )}\end{array}$$ 故有$$\begin{array}{}\text{(16)}& G(\lambda )=G(0)+{\int }_0^{\lambda }{G}^{\prime }(t)\mathrm{d}t=\frac{\pi }{2}\ln (1+\lambda )\end{array}$$ $◻$

注 若不限定$a,b,\lambda >0$，需要特别注意参数的取值范围，此时本题的答案为$$\begin{array}{}\text{(17)}& (1)=\pi \ln \frac{|a|+|b|}{2},(2)=\frac{\pi }{2}\mathrm{sgn}\lambda \ln (1+|\lambda |)\end{array}$$

解 记$$\begin{array}{}\text{(19)}& F(\alpha )={\int }_0^{\pi /2}\ln \frac{1+\alpha \cos x}{1-\alpha \cos x}\cdot \frac{1}{\cos x}\mathrm{d}x,-1<\alpha <1\end{array}$$ 容易发现$F(0)=0$，对参数求导可得$$\begin{array}{}\text{(20)}& \begin{array}{rl}{F}^{\prime }(\alpha )& ={\int }_0^{\pi /2}\frac{1}{1+\alpha \cos x}+\frac{1}{1-\alpha \cos x}\mathrm{d}x={\int }_0^{\pi /2}\frac{2}{1-{\alpha }^2{\cos }^{2}x}\mathrm{d}x\\ & ={\int }_0^{\pi /2}\frac{2\mathrm{d}\tan x}{1+{\tan }^{2}x-{\alpha }^2}\stackrel{u=\frac{\tan x}{\sqrt{1-{\alpha }^2}}}{=}\frac{2}{\sqrt{1-{\alpha }^2}}{\int }_0^{+\mathrm{\infty }}\frac{\mathrm{d}u}{1+u^2}=\frac{\pi }{\sqrt{1-{\alpha }^2}}\end{array}\end{array}$$ 积分可得$$\begin{array}{}\text{(21)}& F(\alpha )=F(0)+{\int }_0^{\alpha }{F}^{\prime }(t)\mathrm{d}t={\int }_0^{\alpha }\frac{\pi }{\sqrt{1-t^2}}\mathrm{d}t=\pi \arcsin \alpha \end{array}$$ 题目所求即为$F\left(\frac{1}{2}\right)=\frac{{\pi }^2}{6}$。 $◻$

解 不妨设$x\ge 0$，记$$\begin{array}{}\text{(23)}& F(x)={\int }_0^{\pi /2}\ln ({\sin }^2\theta +x^2{\cos }^2\theta )\mathrm{d}\theta \end{array}$$ 容易发现$F(1)=0$，对参数求导可得$$\begin{array}{}\text{(24)}& \begin{array}{rl}{F}^{\prime }(x)& ={\int }_0^{\pi /2}\frac{2x{\cos }^2\theta }{{\sin }^2\theta +x^2{\cos }^2\theta }\mathrm{d}\theta \stackrel{u=\tan \theta }{=}2x{\int }_0^{+\mathrm{\infty }}\frac{1}{u^2+x^2}\frac{1}{1+u^2}\mathrm{d}u\\ & =\frac{2x}{x^2-1}{\int }_0^{+\mathrm{\infty }}\frac{1}{1+u^2}-\frac{1}{u^2+x^2}\mathrm{d}u=\frac{2x}{x^2-1}\frac{\pi }{2}(1-\frac{1}{x})=\frac{\pi }{x+1}\end{array}\end{array}$$ 求导后的被积函数在$(0,+\mathrm{\infty })$上内闭一致收敛，故积分可得$$\begin{array}{}\text{(25)}& F(x)={\int }_1^x\frac{\pi }{t+1}\mathrm{d}t=\pi \ln \frac{x+1}{2},x>0\end{array}$$ 此后需要证明$$\begin{array}{}\text{(26)}& F(0)=2{\int }_0^{\pi /2}\ln \sin \theta \mathrm{d}\theta =-\frac{\pi }{2}\ln 2=\pi \ln \frac{1+0}{2}\end{array}$$ $◻$

解 任意给定$t>0$，作变量替换$y=tx$，则$$\begin{array}{}\text{(34)}& {\int }_0^{+\mathrm{\infty }}\frac{\sin tx}{x}\mathrm{d}x={\int }_0^{+\mathrm{\infty }}\frac{\sin y}{y}\mathrm{d}y=\frac{\pi }{2},\mathrm{\forall }t>0\end{array}$$ 注意到$$\begin{array}{}\text{(35)}& {\int }_0^{+\mathrm{\infty }}\frac{\cos ax-\cos bx}{x^2}\mathrm{d}x={\int }_0^{+\mathrm{\infty }}\mathrm{d}x{\int }_a^b\frac{\sin tx}{x}\mathrm{d}t\end{array}$$ 函数$\frac{\sin tx}{x}$可以连续延拓到${\mathbb{R}}^2$上，利用Dirichlet判别法可知${\int }_0^{+\mathrm{\infty }}\frac{\sin tx}{x}\mathrm{d}x$关于参变量$t\in [a,b]$一致收敛，因此上式右端可以交换积分次序，于是得到$$\begin{array}{}\text{(36)}& {\int }_0^{+\mathrm{\infty }}\frac{\cos ax-\cos bx}{x^2}\mathrm{d}x={\int }_a^b\mathrm{d}t{\int }_0^{+\mathrm{\infty }}\frac{\sin tx}{x}\mathrm{d}x={\int }_a^b\frac{\pi }{2}\mathrm{d}t=\frac{\pi }{2}(b-a)\end{array}$$ $◻$

解 构造含参积分$$\begin{array}{}\text{(38)}& I(\alpha ):={\int }_0^1\frac{\ln (1+\alpha x)}{1+x^2}\mathrm{d}x,\alpha >0\end{array}$$ 显然$I(0)=0$，对参数求导可得$$\begin{array}{}\text{(39)}& I^{\prime }(\alpha )={\int }_0^1\frac{x}{(1+\alpha x)(1+x^2)}\mathrm{d}x\end{array}$$ 考虑分解$$\begin{array}{}\text{(40)}& \frac{x}{(1+\alpha x)(1+x^2)}=\frac{Ax+B}{1+x^2}+\frac{C}{1+\alpha x}⟹x=(Ax+B)(1+\alpha x)+C(1+x^2)\end{array}$$ 令$x=-\frac{1}{\alpha }$可得$C=-\frac{\alpha }{1+{\alpha }^2}$；令$x=\mathrm{i}$可得$$\begin{array}{}\text{(41)}& \mathrm{i}=(A\mathrm{i}+B)(1+\alpha \mathrm{i})=(B-A\alpha )+(A+B\alpha )\mathrm{i}⟹A=\frac{1}{1+{\alpha }^2},B=\frac{\alpha }{1+{\alpha }^2}\end{array}$$ 因此$$\begin{array}{}\text{(42)}& \begin{array}{rl}{I}^{\prime }(\alpha )& =\frac{1}{1+{\alpha }^2}{\int }_0^1[\frac{x+\alpha }{1+x^2}-\frac{\alpha }{1+\alpha x}]\mathrm{d}x\\ & =\frac{1}{1+{\alpha }^2}{[\frac{1}{2}\ln (1+x^2)+\alpha \arctan x-\ln (1+\alpha x)]}_0^1\\ & =\frac{1}{1+{\alpha }^2}[\frac{1}{2}\ln 2+\frac{\pi }{4}\alpha -\ln (1+\alpha )]\end{array}\end{array}$$ 积分可得$$\begin{array}{}\text{(43)}& I(1)=I(0)+{\int }_0^1{I}^{\prime }(t)\mathrm{d}t=\frac{\ln 2}{2}{\int }_0^1\frac{\mathrm{d}t}{1+t^2}+\frac{\pi }{8}{\int }_0^1\frac{2t\mathrm{d}t}{1+t^2}-\underset{I(1)}{\underbrace{{\int }_0^1\frac{\ln (1+t)}{1+t^2}\mathrm{d}t}}\end{array}$$ 代入$\alpha =1$可解得$$\begin{array}{}\text{(44)}& I(1)=\frac{\ln 2}{4}\cdot \frac{\pi }{4}+\frac{\pi }{16}\cdot \ln 2=\frac{\pi }{8}\ln 2\end{array}$$ $◻$

解 显然$I(0)=0$。设$m>0$，对参数求导可得$$\begin{array}{}\text{(46)}& I^{\prime }(m)={\int }_0^{+\mathrm{\infty }}({\mathrm{e}}^{-\alpha x}-{\mathrm{e}}^{-\beta x})\cos mx\mathrm{d}x\end{array}$$ 利用待定系数法可得$$\begin{array}{}\text{(47)}& {\int }_0^{+\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\cos mx\mathrm{d}x={\frac{{\mathrm{e}}^{-\alpha x}}{m^2+{\alpha }^2}(-\alpha \cos mx+m\sin mx)|}_{x=0}^{+\mathrm{\infty }}=\frac{\alpha }{m^2+{\alpha }^2}\end{array}$$ 因此$$\begin{array}{}\text{(48)}& \begin{array}{rl}I(m)& =I(0)+{\int }_0^m{I}^{\prime }(t)\mathrm{d}t={\int }_0^m(\frac{\alpha }{t^2+{\alpha }^2}-\frac{\beta }{t^2+{\beta }^2})\mathrm{d}t=\arctan \frac{m}{\alpha }-\arctan \frac{m}{\beta }\\ & =\arctan \frac{\frac{m}{\alpha }-\frac{m}{\beta }}{1+\frac{m^2}{\alpha \beta }}=\arctan \frac{m(\beta -\alpha )}{\alpha \beta +m^2}\end{array}\end{array}$$ $◻$

解 构造含参积分$$\begin{array}{}\text{(50)}& I(\alpha )={\int }_0^{+\mathrm{\infty }}\frac{\cos \alpha x}{1+x^2}\mathrm{d}x,\alpha >0\end{array}$$ 显然$I(0)=\frac{\pi }{2}$，对参数求导可得$$\begin{array}{}\text{(51)}& \begin{array}{rl}{I}^{\prime }(\alpha )& =-{\int }_0^{+\mathrm{\infty }}\frac{x\sin \alpha x}{1+x^2}\mathrm{d}x=-{\int }_0^{+\mathrm{\infty }}(1-\frac{1}{1+x^2})\frac{\sin \alpha x}{x}\mathrm{d}x\\ & =-{\int }_0^{+\mathrm{\infty }}\frac{\sin \alpha x}{x}\mathrm{d}x+{\int }_0^{+\mathrm{\infty }}\frac{\sin \alpha x}{x(1+x^2)}\mathrm{d}x=-\frac{\pi }{2}+{\int }_0^{+\mathrm{\infty }}\frac{\sin \alpha x}{x(1+x^2)}\mathrm{d}x\end{array}\end{array}$$ 显然$I^{\prime }(0)=-\frac{\pi }{2}$，继续对参数求导可得$$\begin{array}{}\text{(52)}& I^{″}(\alpha )={\int }_0^{+\mathrm{\infty }}\frac{\cos \alpha x}{1+x^2}\mathrm{d}x=I(\alpha )\end{array}$$ 由此可得微分方程$$\begin{array}{}\text{(53)}& I^{″}(\alpha )-I(\alpha )=0,I(0)=\frac{\pi }{2},I^{\prime }(0)=-\frac{\pi }{2}\end{array}$$ 解得$$\begin{array}{}\text{(54)}& I(\alpha )=\frac{\pi }{2}{\mathrm{e}}^{-\alpha }⟹I(1)=\frac{\pi }{2\mathrm{e}}\end{array}$$ $◻$

另解 令$R>0$，注意到$$\begin{array}{}\text{(55)}& {\int }_0^R\frac{\cos \alpha x}{1+x^2}\mathrm{d}x={\int }_0^R\frac{{\mathrm{e}}^{\mathrm{i}\alpha x}+{\mathrm{e}}^{-\mathrm{i}\alpha x}}{2(1+x^2)}\mathrm{d}x=\frac{1}{2}{\int }_{-R}^R\frac{{\mathrm{e}}^{\mathrm{i}\alpha x}}{1+x^2}\mathrm{d}x,\alpha >0\end{array}$$ 函数$f(z):=\frac{{\mathrm{e}}^{\mathrm{i}\alpha z}}{1+z^2}$在上半平面除去一级极点$z=\mathrm{i}$外处处解析，取$C_R$为从$(R,0)$到$(-R,0)$的逆时针半圆弧、围道${\mathrm{\Gamma }}_R$为从$(-R,0)$到$(R,0)$的直线与$C_R$的拼接，由留数定理可得$$\begin{array}{}\text{(56)}& {\int }_{-R}^{R}f(z)\mathrm{d}z+{\int }_{C_R}f(z)\mathrm{d}z=2\pi \mathrm{i}\cdot \mathrm{Res}[f,\mathrm{i}]=2\pi \mathrm{i}\cdot \underset{z\to \mathrm{i}}{lim}(z-\mathrm{i})f(z)=2\pi \mathrm{i}\cdot \frac{{\mathrm{e}}^{-\alpha }}{2\mathrm{i}}=\pi {\mathrm{e}}^{-\alpha }\end{array}$$ 令$R\to +\mathrm{\infty }$，注意到$$\begin{array}{}\text{(57)}& |{\int }_{C_R}f(z)\mathrm{d}z|\le {\int }_{C_R}\left|\frac{{\mathrm{e}}^{\mathrm{i}\alpha z}}{1+z^2}\right|\cdot \left|\mathrm{d}z\right|\le {\int }_{C_R}\frac{\mathrm{d}\ell }{|z-\mathrm{i}||z+\mathrm{i}|}\le \frac{\pi R}{(R-1{)}^2}\to 0,R\to +\mathrm{\infty }\end{array}$$ 故有$$\begin{array}{}\text{(58)}& {\int }_0^{+\mathrm{\infty }}\frac{\cos \alpha x}{1+x^2}\mathrm{d}x=\frac{1}{2}\underset{R\to +\mathrm{\infty }}{lim}{\int }_{-R}^{R}f(z)\mathrm{d}z=\frac{1}{2}\cdot \pi {\mathrm{e}}^{-\alpha }=\frac{\pi }{2}{\mathrm{e}}^{-\alpha }\end{array}$$ $◻$

6.3.3 与含参积分有关的证明题

解 因为$F$是奇函数，故我们只需要考虑$t\ge 0$的情况。当$t_0>0$时，易知$\frac{t}{x^2+t^2}f(x)$在$[0,1]\times [\frac{t_0}{2},2t_0]$上连续，从而$F$在$t_0$处连续。

当$t_0=0$时，易知$F(0)=0$。由$f$的连续性可得$\mathrm{\forall }\epsilon >0$，$\mathrm{\exists }\delta (\epsilon )\in (0,1)$使得$|x|<\delta ⟹|f(x)-f(0)|<\epsilon$，于是$$\begin{array}{}\text{(59)}& \begin{array}{rl}F(t)& ={\int }_0^1\frac{t}{x^2+t^2}f(x)\mathrm{d}x\stackrel{x=ty}{=}{\int }_0^{1/t}\frac{1}{1+y^2}f(ty)\mathrm{d}y\\ & =\underset{I_1}{\underbrace{{\int }_0^{\delta /t}\frac{1}{y^2+1}f(0)\mathrm{d}y}}+\underset{I_2}{\underbrace{{\int }_0^{\delta /t}\frac{1}{y^2+1}[f(ty)-f(0)]\mathrm{d}y}}+\underset{I_3}{\underbrace{{\int }_{\delta /t}^{1/t}\frac{1}{y^2+1}f(ty)\mathrm{d}y}}\end{array}\end{array}$$ 其中当$t\to 0^{+}$时，记$M:=\underset{x\in [0,1]}{max}|f(x)|$，则有$$\begin{array}{}\text{(60)}& \begin{array}{rl}{I}_1& ={\int }_0^{\delta /t}\frac{1}{y^2+1}f(0)\mathrm{d}y=f(0)\arctan \frac{\delta }{t}\to \frac{\pi }{2}f(0)\\ |I_2|& =|{\int }_0^{\delta /t}\frac{1}{y^2+1}[f(ty)-f(0)]\mathrm{d}y|\le {\int }_0^{\delta /t}\frac{1}{y^2+1}\epsilon \mathrm{d}y=\epsilon \arctan \frac{\delta }{t}\to \frac{\pi }{2}\epsilon \\ |I_3|& =|{\int }_{\delta /t}^{1/t}\frac{1}{y^2+1}f(ty)\mathrm{d}y|\le {\int }_{\delta /t}^{1/t}\frac{1}{y^2+1}M\mathrm{d}y=M(\arctan \frac{1}{t}-\arctan \frac{\delta }{t})\\ & =M\frac{\frac{1}{t}-\frac{\delta }{t}}{1+\frac{{\xi }^2}{t^2}}\le \frac{M}{t\cdot \frac{{\delta }^2}{t^2}}=\frac{Mt}{{\delta }^2}\to 0,\xi \in (\delta ,1)\end{array}\end{array}$$ 再令$\epsilon \to 0^{+}$可得$\underset{t\to 0^{+}}{lim}F(t)=\frac{\pi }{2}f(0)$，即$F$在$t=0$处连续当且仅当$f(0)=0$。 $◻$

另解 也可以这么处理$\underset{t\to 0^{+}}{lim}F(t)$：$$\begin{array}{}\text{(61)}& \underset{t\to 0^{+}}{lim}F(t)=\underset{I_1}{\underbrace{\underset{t\to 0^{+}}{lim}{\int }_0^{t^{1/3}}\frac{t}{x^2+t^2}f(x)\mathrm{d}x}}+\underset{I_2}{\underbrace{\underset{t\to 0^{+}}{lim}{\int }_{t^{1/3}}^1\frac{t}{x^2+t^2}f(x)\mathrm{d}x}}\end{array}$$ 其中$$\begin{array}{}\text{(62)}& \begin{array}{rl}{I}_1& =\underset{t\to 0^{+}}{lim}f({\xi }_t)\underset{t\to 0^{+}}{lim}{\int }_0^{t^{1/3}}\frac{t}{x^2+t^2}\mathrm{d}x=f\left(\underset{t\to 0^{+}}{lim}{\xi }_t\right)\underset{t\to 0^{+}}{lim}\arctan \frac{t^{1/3}}{t}=\frac{\pi }{2}f(0)\\ |I_2|& \le \underset{t\to 0^{+}}{lim}\frac{t(1-t^{1/3})}{t^{2/3}+t^2}\underset{x\in [0,1]}{max}|f(x)|=0⟹I_2=0\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(63)}& \underset{t\to 0^{+}}{lim}F(t)=\frac{\pi }{2}f(0)\end{array}$$ $◻$

解 原积分可化为$$\begin{array}{}\text{(65)}& I\stackrel{t=\frac{x}{\alpha }}{=}\alpha {\int }_0^{+\mathrm{\infty }}{\mathrm{e}}^{-{(t-\frac{1}{{\alpha }^2})}^2}\mathrm{d}t\le \alpha {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\mathrm{e}}^{-{(t-\frac{1}{{\alpha }^2})}^2}\mathrm{d}t\stackrel{u=t-\frac{1}{{\alpha }^2}}{=}\alpha {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\mathrm{e}}^{-u^2}\mathrm{d}u=\alpha \sqrt{\pi }\le \sqrt{\pi }\end{array}$$ 由Weierstrass强函数判别法知原积分在$0<\alpha <1$上一致收敛。 $◻$

解 注意到$$\begin{array}{}\text{(67)}& {\int }_0^{+\mathrm{\infty }}\frac{x\sin \alpha x}{\alpha (1+x^2)}\mathrm{d}x={\int }_0^{+\mathrm{\infty }}(1-\frac{1}{1+x^2})\frac{\sin \alpha x}{\alpha x}\mathrm{d}x={\int }_0^{+\mathrm{\infty }}\frac{\sin \alpha x}{\alpha x}\mathrm{d}x-{\int }_0^{+\mathrm{\infty }}\frac{\sin \alpha x}{\alpha x(1+x^2)}\mathrm{d}x\end{array}$$ 后项一致收敛，因为由Weierstrass判别法可知$$\begin{array}{}\text{(68)}& {\int }_0^{+\mathrm{\infty }}\left|\frac{\sin \alpha x}{\alpha x}\frac{1}{1+x^2}\right|\mathrm{d}x\le {\int }_0^{+\mathrm{\infty }}\frac{1}{1+x^2}\mathrm{d}x=\frac{\pi }{2}\end{array}$$ 前项不一致收敛，亦即$\mathrm{\exists }{\epsilon }_0>0$使得$\mathrm{\forall }N>0$，$\mathrm{\exists }\alpha >0$、$\mathrm{\exists }A>B>N$使得$$\begin{array}{}\text{(69)}& \left|{\int }_B^A\frac{\sin \alpha x}{\alpha x}\mathrm{d}x\right|\stackrel{t=\alpha x}{=}\frac{1}{\alpha }\left|{\int }_{\alpha B}^{\alpha A}\frac{\sin t}{t}\mathrm{d}t\right|\ge {\epsilon }_0\end{array}$$ 选择合适的$\alpha ,A,B$满足以上条件，如$\alpha =\frac{{\epsilon }_0}{3n}$、$A=\frac{n\pi }{\alpha }$、$B=\frac{(n-1)\pi }{\alpha }$、$n=\lfloor \frac{N}{\pi }\rfloor +2$，则$$\begin{array}{}\text{(70)}& \left|{\int }_{\alpha B}^{\alpha A}\frac{\sin t}{t}\mathrm{d}t\right|={\int }_{(n-1)\pi }^{n\pi }\frac{|\sin t|}{t}\mathrm{d}t\ge {\int }_{n\pi -\frac{5\pi }{6}}^{n\pi -\frac{\pi }{6}}\frac{|\sin t|}{t}\mathrm{d}t\ge \frac{1}{2}\cdot \frac{1}{n\pi }\cdot \frac{4\pi }{6}=\frac{1}{3n}\end{array}$$ 综上所述，原含参积分在$\alpha >0$上非一致收敛。 $◻$