# 物竞速成[3] 微积分基础-3

### 正文

f(x)ab的定积分定义为\int_a^bf(x)\mathrm{d}x=\lim\limits_{n\rightarrow\infty}\sum\limits_{i=1}^n(\frac{1}{n}f(\frac{n-i}{n}a+\frac{i}{n}b)).^{[1]}

F(x)称为f(x)的原函数, 记为\int f(x)\mathrm{d}x=F(x)+C, 其中C为常数.

\int x^a\mathrm{d}x=\frac{x^{a+1}}{a+1}+C, 其中a\neq-1;
\int\frac{1}{x}\mathrm{d}x=\ln|x|+C;
\int\frac{1}{x^2+1}\mathrm{d}x=\arctan x+C;
\int\frac{1}{1-x^2}\mathrm{d}x=\tanh^{-1} x+C;
\int\frac{1}{\sqrt{1-x^2}}\mathrm{d}x=\arcsin x+C;
\int\frac{1}{\sqrt{x^2+1}}\mathrm{d}x=\sinh^{-1} x+C;
\int\frac{1}{\sqrt{x^2-1}}\mathrm{d}x=\cosh^{-1} x+C;
\int e^x\mathrm{d}x=e^x+C;
\int \sin x\mathrm{d}x=-\cos x+C;
\int \cos x\mathrm{d}x=\sin x+C;
\int\sec^2 x\mathrm{d}x=\tan x+C;
\int\tan x\sec x\mathrm{d}x=\sec x+C;
\int \sinh x\mathrm{d}x=\cosh x+C;
\int \cosh x\mathrm{d}x=\sinh x+C;

(1) \int\frac{x}{x^2+1}\mathrm{d}x=\frac{1}{2}\int\frac{\mathrm{d}(x^2+1)}{x^2+1}=\frac{\ln(x^2+1)}{2}+C

(2)
\begin{aligned}\int\frac{2-x}{\sqrt{9-4x^2}}\mathrm{d}x&=\int\frac{2}{\sqrt{9-4x^2}}\mathrm{d}x+\int\frac{-x}{\sqrt{9-4x^2}}\mathrm{d}x\\
&=\int\frac{2}{3\sqrt{1-(\frac{2x}{3})^2}}\mathrm{d}x+\frac{1}{8}\int\frac{\mathrm{d}(9-4x^2)}{\sqrt{9-4x^2}}\\
&=\int\frac{\mathrm{d}(\frac{2x}{3})}{\sqrt{1-(\frac{2x}{3})^2}}+\frac{\sqrt{9-4x^2}}{4}+C\\
&=\arcsin \frac{2x}{3}+\frac{\sqrt{9-4x^2}}{4}+C
\end{aligned}

(3)
\begin{aligned}\int\frac{\cos^2x-\sin x}{\cos x(1+e^{\sin x}\cos x)}\mathrm{d}x&=\int\frac{e^{\sin x}(\cos^2x-\sin x)}{e^{\sin x}\cos x(1+e^{\sin x}\cos x)}\mathrm{d}x\\
&=\int\frac{\mathrm{d}(e^{\sin x}\cos x)}{e^{\sin x}\cos x(1+e^{\sin x}\cos x)}\\
&=\int\frac{\mathrm{d}(e^{\sin x}\cos x)}{e^{\sin x}\cos x}-\int\frac{\mathrm{d}(1+e^{\sin x}\cos x)}{1+e^{\sin x}\cos x}\\
&=\ln|e^{\sin x}\cos x|-\ln|1+e^{\sin x}\cos x|+C
\end{aligned}

\begin{aligned}\int\sqrt{1-x^2}\mathrm{d}x&\mathop{=}\limits^{x=\sin t}\int\cos t\mathrm{d}\sin t\\
&=\int\cos^2t\mathrm{d}t\\
&=\int\frac{\cos 2t+1}{2}\mathrm{d}t\\
&=\frac{\sin 2t}{4}+\frac{t}{2}+C\\
&=\frac{x\sqrt{1-x^2}+\arcsin x}{2}+C
\end{aligned}

\begin{aligned}\int\sqrt{x^2+1}\mathrm{d}x&\mathop{=}\limits^{x=\sinh t}\int\cosh t\mathrm{d}\sinh t\\
&=\int\cosh^2t\mathrm{d}t\\
&=\int\frac{\cosh 2t+1}{2}\mathrm{d}t\\
&=\frac{\sinh 2t}{4}+\frac{t}{2}+C\\
&=\frac{x\sqrt{x^2+1}+\sinh^{-1}x}{2}+C
\end{aligned}

1. \deg(P(x))\geq\deg(Q(x)), 先用大除法将多项式部分剥离, 即将被积式化为T(x)+\frac{R(x)}{Q(x)}, 其中T(x), R(x)均为多项式, 且\deg(R(x))<\deg(Q(x)). 现在T(x)已经容易积分, 我们只关注后半部分.

2. Q(x)因式分解. 不妨设Q(x)首一, 根据虚根成对定理, Q(x)=(x-a_1)(x-a_2)\cdots(x-a_m)(x^2-b_1x+c_1)\cdots(x^2-b_nx+c_n).

3. 若无重根, 可将被积式后半部分写成\frac{w_1}{x-a_1}+\cdots+\frac{w_m}{x-a_m}+\frac{u_1x-v_1}{x^2-b_1x+c_1}+\cdots+\frac{u_nx-v_n}{x^2-b_nx+c_n}. 这些式子是好积的.

4. 若有重实根, 假设其重复k次, 则分母包含它的部分可写作\frac{w_{m1}}{x-a_m}+\frac{w_{m2}}{(x-a_m)^2}+\cdots+\frac{w_{mk}}{(x-a_m)^k}. 重虚根同理, 不过分子上应变为一次式.

\begin{aligned}\int\frac{x^3+x^2-3x+4}{x^3-3x+2}\mathrm{d}x&=\int\left(1+\frac{x^2+2}{(x-1)^2(x+2)}\right)\mathrm{d}x\\
&=x+\int\left(\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}\right)\mathrm{d}x
\end{aligned}

\begin{aligned}C&=\lim\limits_{x\rightarrow-2}\left(\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}\right)(x+2)\\
&=\lim\limits_{x\rightarrow-2}\frac{x^2+2}{(x-1)^2(x+2)}(x+2)\\
&=\frac{2}{3}
\end{aligned}

\begin{aligned}\int\frac{\sin x}{\sin x+\cos x}\mathrm{d}x&=\int\frac{2\tan\frac{x}{2}}{-\tan^2\frac{x}{2}+2\tan\frac{x}{2}+1}\mathrm{d}x\\
&\mathop{=}\limits^{t=\tan\frac{x}{2}}\int\frac{4t}{(-t^2+2t+1)(t^2+1)}\mathrm{d}t
\end{aligned}

1^{\circ}a>0, 令\sqrt{ax^2+bx+c}=\sqrt{a}x+t, 两边平方有bx+c=2\sqrt{a}t+t^2, 即有x=\frac{t^2-c}{b-2\sqrt{a}}, \sqrt{ax^2+bx+c}=t+\sqrt{a}\frac{t^2-c}{b-2\sqrt{a}}, 这样既有\frac{\mathrm{d}x}{\mathrm{d}t}t的有理分式, 又有原先被积函数为t的有理分式, 乘起来还是有理分式.

2^{\circ}c>0, 令\sqrt{ax^2+bx+c}=xt+\sqrt{c}, 则有x=\frac{2\sqrt{c}-b}{a-t^2}, \sqrt{ax^2+bx+c}=\frac{2\sqrt{c}-b}{a-t^2}t+\sqrt{c}.

3^{\circ}\sqrt{ax^2+bx+c}=\sqrt{a(x-p)(x-q)}, 其中p, q为实数且p\neq q. 令t=\sqrt{\frac{a(x-p)}{x-q}}, 则有x=\frac{qt^2-ap}{t^2-a}, \sqrt{ax^2+bx+c}=t(\frac{qt^2-ap}{t^2-a}-q)=\frac{a(q-p)t}{t^2-a}.

\begin{aligned}\int\frac{\mathrm{d}x}{x+\sqrt{x^2+x+1}}&\mathop{=}\limits^{\sqrt{x^2+x+1}=x+t}\int\frac{\mathrm{d}\frac{t^2-1}{1-2t}}{2\frac{t^2-1}{1-2t}+t}\\
&=\int\frac{2t^2-2t+2}{(2t-1)(t-2)}\mathrm{d}x\\
&=\int\left(1+\frac{3t}{(2t-1)(t-2)}\right)\mathrm{d}t\\
&=t+\int\left(\frac{2}{t-2}-\frac{1}{2t-1}\right)\mathrm{d}t\\
&=t+2\ln|t-2|-\frac{\ln|2t-1|}{2}+C\\
&=\sqrt{x^2+x+1}-x+2\ln|\sqrt{x^2+x+1}-x-2|\\
&-\frac{\ln|2\sqrt{x^2+x+1}-2x-1|}{2}+C
\end{aligned}

\begin{aligned}\int\ln x\mathrm{d}x&=x\ln x-\int x\mathrm{d}\ln x\\
&=x\ln x-\int x\frac{1}{x}\mathrm{d}x\\
&=x\ln x-x+C
\end{aligned}

\begin{aligned}\int e^x\sin x\mathrm{d}x&=-\int e^x\mathrm{d}\cos x\\
&=-e^x\cos x+\int\cos x\mathrm{d}e^x\\
&=-e^x\cos x+\int e^x\cos x\mathrm{d}x\\
&=-e^x\cos x+\int e^x\mathrm{d}\sin x\\
&=-e^x\cos x+e^x\sin x-\int e^x\sin x\mathrm{d}x\\
\end{aligned}

### 练习

3-1 求\int\frac{\mathrm{d}x}{(1+x)\sqrt{x}}.

3-2 求\int\frac{\mathrm{d}x}{(x^2+1)^{\frac{3}{2}}}.

3-3 求\int\frac{\sin x\cos x}{\sqrt{a^2\sin^2x+b^2\cos^2x}}\mathrm{d}x.

3-4 完成\int\frac{\sin x}{\sin x+\cos x}\mathrm{d}x.
(提示: 本题不一定要用万能公式)

3-5 求\int\frac{\mathrm{d}x}{\sqrt{1+\frac{1}{x}}}.

3-6 求\int\left(\frac{\ln x}{x}\right)^2\mathrm{d}x.

3-7 求\int\ln(x+\sqrt{x^2-1})\mathrm{d}x.

3-8 本题中, 我们将看见实变复值函数在不定积分中的应用. 设实变复值函数A(x)=f(x)+ig(x), 其中x, f(x), g(x)均取实值.

\begin{aligned}\int e^x\sin x\mathrm{d}x&=\int e^x(\frac{e^{ix}-e^{-ix}}{2i})\mathrm{d}x\\
&=\frac{1}{2i}\left(\int e^{(1+i)x}\mathrm{d}x-\int e^{(1-i)x}\mathrm{d}x\right)\\
&=\frac{1}{2i}\left(\frac{e^{(1+i)x}}{1+i}-\frac{e^{(1-i)x}}{1-i}\right)+C\\
&=\frac{1}{2i}\left(\frac{e^x(2i\sin x-2i\cos x)}{2}\right)+C\\
&=\frac{e^x(\sin x-\cos x)}{2}+C
\end{aligned}

### 上期答案

2-1 设\boldsymbol{E_0}=E_{01}\boldsymbol{i}+E_{02}\boldsymbol{j}+E_{03}\boldsymbol{k}, \boldsymbol{K}=K_1\boldsymbol{i}+K_2\boldsymbol{j}+K_3\boldsymbol{k}.

\nabla\times\boldsymbol{E}=ie^{-i\omega t}e^{i(K_1x+K_2y+K_3z)}(\boldsymbol{i}(K_2E_{03}-K_3E_{02})+\boldsymbol{j}(K_3E_{01}-K_1E_{03})+\boldsymbol{k}(K_1E_{02}-K_2E_{01}))
=i\boldsymbol{K}\times\boldsymbol{E}

2-2 (1) \nabla\times(\nabla A)=\nabla\times(A_x\boldsymbol{i}+A_y\boldsymbol{j}+A_z\boldsymbol{k})=\boldsymbol{i}(A_{zy}-A_{yz})+\boldsymbol{j}(A_{xz}-A_{zx})+\boldsymbol{k}(A_{yx}-A_{xy})=\boldsymbol{0}

(2) 设\boldsymbol{A}=P\boldsymbol{i}+Q\boldsymbol{j}+R\boldsymbol{k};

\nabla\cdot(\nabla\times\boldsymbol{A})=\nabla\cdot(\boldsymbol{i}(R_y-Q_z)+\boldsymbol{j}(P_z-R_x)+\boldsymbol{k}(Q_x-P_y))=R_{yx}-Q_{zx}+P_{zy}-R_{xy}+Q_{xz}-P_{yz}=0

2-3 (1) \boldsymbol{e}_\rho=\frac{(\cos\varphi,\sin\varphi,0)}{\sqrt{\cos^2\varphi+\sin^2\varphi}}=(\cos\varphi,\sin\varphi,0)

\boldsymbol{e}_\varphi=\frac{(-\rho\sin\varphi,\rho\cos\varphi,0)}{\sqrt{\rho^2\sin^2\varphi+\rho^2\cos^2\varphi}}=(-\sin\varphi,\cos\varphi,0)

\boldsymbol{e}_z=\frac{(0,0,1)}{\sqrt{1^2}}=(0,0,1)

(2) 经过观察我们可以发现, 这是一组正交基.

\begin{aligned}\nabla A&=(\nabla A\cdot\boldsymbol{e}_\rho)\boldsymbol{e}_\rho+(\nabla A\cdot\boldsymbol{e}_\varphi)\boldsymbol{e}_\varphi+(\nabla A\cdot\boldsymbol{e}_z)\boldsymbol{e}_z\\
&=(A_xx_\rho+A_yy_\rho+A_zz_\rho)\boldsymbol{e}_\rho+\frac{1}{\rho}(A_xx_\varphi+A_yy_\varphi+A_zz_\varphi)\boldsymbol{e}_\varphi\\
&+(A_xx_z+A_yy_z+A_zz_z)\boldsymbol{e}_z\\
&=A_\rho\boldsymbol{e}_\rho+\frac{1}{\rho}A_\varphi\boldsymbol{e}_\varphi+A_z\boldsymbol{e}_z\end{aligned}

### 注释

[1] 这是黎曼对积分的古典定义. 事实上, 这甚至是满足黎曼可积条件时的计算方法, 黎曼的严格定义是在[a,b]上取一组点列a=x_0<x_1<\cdots<x_n=b, 并在每个区间[x_{i-1},x_i]中任取一值\zeta_i, 定义黎曼和为S=\sum\limits_{i=1}^nf(\zeta_i)(x_i-x_{i-1}). 若\max(x_i-x_{i-1})\rightarrow0S收敛, 则定义\int_a^bf(x)\mathrm{d}x=S. 当今普遍采用的是基于勒贝格测度的定义.

THE END