6.3 习题课讲解

6.3.1 含参积分

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 (1) 令$$\begin{array}{}\text{(3)}& F(b):={\int }_0^{\pi /2}\ln (a^2{\sin }^{2}x+b^2{\cos }^{2}x)\mathrm{d}x\end{array}$$ 显然$F(a)=0$，计算可得$$\begin{array}{}\text{(4)}& F^{\prime }(b)={\int }_0^{\pi /2}\frac{2b{\cos }^{2}x}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}\mathrm{d}x=2b{\int }_0^{\pi /2}\frac{\mathrm{d}x}{a^2{\tan }^{2}x+b^2}=\frac{\pi }{a+b}\end{array}$$ 故有$$\begin{array}{}\text{(5)}& F(b)=F(a)+{\int }_a^b{F}^{\prime }(t)\mathrm{d}t=\pi \ln \frac{a+b}{2}\end{array}$$

(2) 令$$\begin{array}{}\text{(6)}& G(\lambda ):={\int }_0^{\pi /2}\frac{\arctan (\lambda \tan x)}{\tan x}\mathrm{d}x\end{array}$$ 注意到$$\begin{array}{}\text{(7)}& \begin{array}{rl}\underset{x\to 0^{+}}{lim}\frac{\arctan (\lambda \tan x)}{\tan x}& =\underset{x\to 0^{+}}{lim}\frac{\lambda \tan x+o(\tan x)}{\tan x}=\lambda \\ \underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{\arctan (\lambda \tan x)}{\tan x}& =\underset{x\to {\frac{\pi }{2}}^{-}}{lim}\frac{\frac{\pi }{2}+o(1)}{\tan x}=0\end{array}\end{array}$$ 故此积分不为瑕积分。显然$G(0)=0$，计算可得$$\begin{array}{}\text{(8)}& G^{\prime }(\lambda )={\int }_0^{\pi /2}\frac{1}{1+{\lambda }^2{\tan }^{2}x}\mathrm{d}x=\frac{\pi }{2(1+\lambda )}\end{array}$$ 故有$$\begin{array}{}\text{(9)}& G(\lambda )=G(0)+{\int }_0^{\lambda }{G}^{\prime }(t)\mathrm{d}t=\frac{\pi }{2}\ln (1+\lambda )\end{array}$$ $◻$

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$$\begin{array}{}\text{(10)}& (1)=\pi \ln \frac{|a|+|b|}{2},(2)=\frac{\pi }{2}\mathrm{sgn}\lambda \ln (1+|\lambda |)\end{array}$$

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$$\begin{array}{}\text{(11)}& \begin{array}{rl}F(t)& ={\int }_0^1\frac{t}{x^2+t^2}f(x)\mathrm{d}x\stackrel{x=ty}{=}{\int }_0^{1/t}\frac{1}{1+y^2}f(ty)\mathrm{d}y\\ & =\underset{I_1}{\underbrace{{\int }_0^{\delta /t}\frac{1}{y^2+1}f(0)\mathrm{d}y}}+\underset{I_2}{\underbrace{{\int }_0^{\delta /t}\frac{1}{y^2+1}[f(ty)-f(0)]\mathrm{d}y}}+\underset{I_3}{\underbrace{{\int }_{\delta /t}^{1/t}\frac{1}{y^2+1}f(ty)\mathrm{d}y}}\end{array}\end{array}$$ 其中当$t\to 0^{+}$时，记$M:=\underset{x\in [0,1]}{max}|f(x)|$，则有$$\begin{array}{}\text{(12)}& \begin{array}{rl}{I}_1& ={\int }_0^{\delta /t}\frac{1}{y^2+1}f(0)\mathrm{d}y=f(0)\arctan \frac{\delta }{t}\to \frac{\pi }{2}f(0)\\ |I_2|& =|{\int }_0^{\delta /t}\frac{1}{y^2+1}[f(ty)-f(0)]\mathrm{d}y|\le {\int }_0^{\delta /t}\frac{1}{y^2+1}\epsilon \mathrm{d}y=\epsilon \arctan \frac{\delta }{t}\to \frac{\pi }{2}\epsilon \\ |I_3|& =|{\int }_{\delta /t}^{1/t}\frac{1}{y^2+1}f(ty)\mathrm{d}y|\le {\int }_{\delta /t}^{1/t}\frac{1}{y^2+1}M\mathrm{d}y=M(\arctan \frac{1}{t}-\arctan \frac{\delta }{t})\\ & =M\frac{\frac{1}{t}-\frac{\delta }{t}}{1+\frac{{\xi }^2}{t^2}}\le \frac{M}{t\cdot \frac{{\delta }^2}{t^2}}=\frac{Mt}{{\delta }^2}\to 0,\xi \in (\delta ,1)\end{array}\end{array}$$ 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$\underset{t\to 0^{+}}{lim}F(t)$：$$\begin{array}{}\text{(13)}& \underset{t\to 0^{+}}{lim}F(t)=\underset{I_1}{\underbrace{\underset{t\to 0^{+}}{lim}{\int }_0^{t^{1/3}}\frac{t}{x^2+t^2}f(x)\mathrm{d}x}}+\underset{I_2}{\underbrace{\underset{t\to 0^{+}}{lim}{\int }_{t^{1/3}}^1\frac{t}{x^2+t^2}f(x)\mathrm{d}x}}\end{array}$$ 其中$$\begin{array}{}\text{(14)}& \begin{array}{rl}{I}_1& =\underset{t\to 0^{+}}{lim}f({\xi }_t)\underset{t\to 0^{+}}{lim}{\int }_0^{t^{1/3}}\frac{t}{x^2+t^2}\mathrm{d}x=f\left(\underset{t\to 0^{+}}{lim}{\xi }_t\right)\underset{t\to 0^{+}}{lim}\arctan \frac{t^{1/3}}{t}=\frac{\pi }{2}f(0)\\ |I_2|& \le \underset{t\to 0^{+}}{lim}\frac{t(1-t^{1/3})}{t^{2/3}+t^2}\underset{x\in [0,1]}{max}|f(x)|=0⟹I_2=0\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(15)}& \underset{t\to 0^{+}}{lim}F(t)=\frac{\pi }{2}f(0)\end{array}$$ $◻$

另证 如不借助复变函数，可以注意到$$\begin{array}{}\text{(22)}& \begin{array}{rl}{I}^{\prime }(t)& ={\int }_0^{2\pi }{\mathrm{e}}^{t\cos \theta }[\cos \theta \cos (t\sin \theta )-\sin \theta \sin (t\sin \theta )]\mathrm{d}\theta \\ & =\frac{1}{t}{\int }_0^{2\pi }{\mathrm{e}}^{t\cos \theta }\mathrm{d}\sin (t\sin \theta )-{\int }_0^{2\pi }{\mathrm{e}}^{t\cos \theta }\sin \theta \sin (t\sin \theta )\mathrm{d}\theta \\ & ={\frac{1}{t}{\mathrm{e}}^{t\cos \theta }\sin (t\sin \theta )|}_{\theta =0}^{2\pi }-\frac{1}{t}{\int }_0^{2\pi }{\mathrm{e}}^{t\cos \theta }(-t\sin \theta )\sin (t\sin \theta )\mathrm{d}\theta \\ & -{\int }_0^{2\pi }{\mathrm{e}}^{t\cos \theta }\sin \theta \sin (t\sin \theta )\mathrm{d}\theta =0+0=0\end{array}\end{array}$$ 也可以利用数学归纳法证明$$\begin{array}{}\text{(23)}& F^{(n)}(t)={\int }_0^{2\pi }{\mathrm{e}}^{t\cos \theta }\cos (n\theta +t\sin \theta )\mathrm{d}\theta ,n\in {\mathbb{N}}^{\ast }\end{array}$$ 且$F^{(n)}(t)=0$、$F(0)=2\pi$，由带Lagrange余项的Taylor公式即可得到$F(t)=2\pi$。 $◻$