微积分复习笔记（1）：单变量微积分

P.S. 本来不想分篇的，但居然字数超上限了。

$$\frac{2}{\pi }x<\sin x<x(0<x<\frac{\pi }{2}).$$

$$x<\tan x<\frac{4}{\pi }x(0<x<\frac{\pi }{4}).$$

$$\frac{1}{x+1}<\ln (1+\frac{1}{x})<\frac{1}{x}(x>0).$$

$$\frac{x}{1+x}<\ln (1+x)<x(x>0).$$

$$\arctan x\le x\le \arcsin x(0\le x\le 1).$$

$$\sin (\arccos x)=\sqrt{1-x^2};\tan (\arcsin x)=\frac{x}{\sqrt{1-x^2}};\sin (\arctan x)=\frac{x}{\sqrt{1+x^2}}.$$

$$\underset{u\to 1,v\to \infty }{\lim }{u}^v=\exp \underset{u\to 1,v\to \infty }{\lim }(u-1)v.$$

$$f(x)-f(x_0)=\sum _{i=1}^n\frac{f^{(i)}(x_0)}{i!}(x-x_0{)}^i+\frac{f^{(n+1)}(\xi )}{(n+1)!}(x-x_0{)}^{n+1},\xi \in (x,x_0)\vee (x_0,x).$$

$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$

$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$$

$$\tan x=x+\frac{1}{3}{x}^3+...$$

$$\arcsin x=x+\frac{1}{6}{x}^3+...$$

$$\arctan x=x-\frac{1}{3}{x}^3+...$$

$$\ln (1+x)=x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-\frac{1}{4}{x}^4+...$$

$$(1+x{)}^{\frac{1}{x}}=e-\frac{e}{2}x+...$$

$$\frac{1}{1+x^n}=1-x^n+x^{2n}-x^{3n}+...$$

$$(1+x{)}^a=1+ax+\frac{a(a-1)}{2!}{x}^2+\frac{a(a-1)(a-2)}{3!}{x}^3+...$$

$$[\ln (x+\sqrt{x^2\pm a^2}){]}^{\prime }=\frac{1}{\sqrt{x^2\pm a^2}}.$$

$$[\sin (ax+b){]}^{(n)}=a^n\sin (ax+b+\frac{n\pi }{2}).$$

$$[\cos (ax+b){]}^{(n)}=a^n\cos (ax+b+\frac{n\pi }{2}).$$

$$[\ln (ax+b){]}^{(n)}=(-1{)}^{n-1}{a}^n\frac{(n-1)!}{(ax+b{)}^n}.$$

$$(\frac{1}{ax+b}{)}^{(n)}=(-1{)}^n{a}^n\frac{n!}{(ax+b{)}^{n+1}}.$$

$$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{\frac{\mathrm{d}y}{\mathrm{d}t}/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}.$$

$$K=|\frac{\mathrm{d}\theta }{\mathrm{d}S}|=\frac{|\mathrm{d}\arctan \frac{\mathrm{d}y}{\mathrm{d}x}|}{\sqrt{(\mathrm{d}x{)}^2+(\mathrm{d}y{)}^2}}=\left|\frac{\mathrm{d}\arctan \frac{\mathrm{d}y}{\mathrm{d}x}}{\mathrm{d}x}\right|\cdot \frac{|\mathrm{d}x|}{\sqrt{(\mathrm{d}x{)}^2+(\mathrm{d}y{)}^2}}=\frac{|{\mathrm{d}}^{2}y\mathrm{d}x|}{[(\mathrm{d}x{)}^2+(\mathrm{d}y{)}^2{]}^{\frac{3}{2}}}=\frac{|y^{\prime \prime }|}{[1+(y^{\prime }{)}^2{]}^{\frac{3}{2}}}.$$

多项式函数 $f(x)=(x-a{)}^{n}g(x)$ 中 $g(a)\ne 0$, $n>1$: $n$ 为偶数时 $x=a$ 为极值点; 奇数时为拐点.

$f(x)={\prod }_{i=1}^s(x-a_i{)}^{n_i}$ 中 $n_i\in {\mathbb{N}}_{+}$, $a_i\ne a_j(i\ne j)$: $k_1$ 为 $n_i=1$ 个数, $k_2$ 为 $n_i>1$ 为偶数个数, $k_3$ 为 $n_i>1$ 为奇数个数 $⟹$ 极值点个数为 $k_1+2k_2+k_3-1$; 拐点个数为 $k_1+2k2+3k_3-2$.

$f^{(n)}(x)=0$ 有 $k$ 个根 $⟹f(x)=0$ 至多有 $n+k$ 个根.

可导点不可同时为极值点和拐点.

$$\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{i=1}^{n}f(\frac{i}{n})={\int }_0^{1}f(x)\mathrm{d}x.$$

$$\int \tan x\mathrm{d}x=-\ln |\cos x|+C.$$

$$\int \cot x\mathrm{d}x=\ln |\sin x|+C.$$

$$\int \sec x\mathrm{d}x=\ln |\sec x+\tan x|+C.$$

$$\int \csc x\mathrm{d}x=\ln |\csc x-\cot x|+C.$$

$$\int \frac{1}{a^2+x^2}\mathrm{d}x=\frac{1}{|a|}\arctan \frac{x}{|a|}+C.$$

$$\int \frac{1}{\sqrt{a^2-x^2}}\mathrm{d}x=\arcsin \frac{x}{|a|}+C.$$

$$\int \sqrt{a^2-x^2}\mathrm{d}x=\frac{a^2}{2}\arcsin \frac{x}{|a|}+\frac{x}{2}\sqrt{a^2-x^2}+C.$$

$$\int \frac{1}{\sqrt{x^2\pm a^2}}\mathrm{d}x=\ln (x+\sqrt{x^2\pm a^2})+C.$$

$$\int \sqrt{x^2\pm a^2}\mathrm{d}x=\frac{x}{2}\sqrt{x^2\pm a^2}\pm \frac{a^2}{2}\ln (x+\sqrt{x^2\pm a^2})+C.$$

$$\int \frac{1}{(x^2+a^2{)}^2}\mathrm{d}x=\frac{1}{2a^2}\int [\frac{a^2-x^2}{(x^2+a^2{)}^2}+\frac{1}{x^2+a^2}]\mathrm{d}x=\frac{1}{2a^2}(\frac{x}{x^2+a^2}+\frac{1}{|a|}\arctan \frac{x}{|a|}).$$

$$t=\tan \frac{x}{2};\tan x=\frac{2t}{1-t^2},\sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2};\mathrm{d}x=\frac{2}{1+t^2}\mathrm{d}t.$$

$$t=\tan x;{\sin }^{2}x=\frac{t}{1+t^2},{\cos }^{2}x=\frac{1}{1+t^2};\mathrm{d}x=\frac{1}{1+t^2}\mathrm{d}t.$$

$$\int f\cdot f^{\prime }=f^2.$$

$$\int (f^{\prime }{)}^2+f^{\prime \prime }\cdot f=f\cdot f^{\prime }.$$

$$\int (f^{\prime }+f\cdot {\phi }^{\prime })\cdot e^{\phi }=f\cdot e^{\phi }.$$

$$\int (f^{\prime }\cdot x-f)/x^2=\frac{f}{x}.$$

$$\int [f^{\prime \prime }\cdot f-(f^{\prime }{)}^2]/f^2=\frac{f^{\prime }}{f}.$$

$$\int \frac{f^{\prime }}{f}=\ln f.$$

$$\int u{v}^{(n+1)}=\sum _{i=0}^n(-1{)}^i{u}^{(i)}{v}^{(n-i)}+(-1{)}^{n+1}\int u^{(n+1)}v.$$

$$\int e^{ax}\sin bx\mathrm{d}x=\frac{1}{a^2}(e^{ax}{)}^{\prime }\sin bx-\frac{1}{a^2}(\sin bx{)}^{\prime }{e}^{ax}-\frac{b^2}{a^2}\int e^{ax}\sin bx\mathrm{d}x.$$

$${\int }_b^{a}f(x)\mathrm{d}x={\int }_b^{a}f(a+b-x)\mathrm{d}x.$$

$${\int }_{-a}^a\sqrt{a^2-x^2}\mathrm{d}x=\frac{\pi a^2}{2}(a>0).$$

$${\int }_0^1\arcsin \sqrt{1-x^2}\mathrm{d}x=1.$$

$${\int }_0^{\frac{\pi }{2}}{\sin }^{n}x\mathrm{d}x={\int }_0^{\frac{\pi }{2}}{\cos }^{n}x\mathrm{d}x=\{\begin{array}{ll}\frac{(n-1)!!}{n!!},& n=2k+1\\ \frac{(n-1)!!}{n!!}\cdot \frac{\pi }{2},& n=2k\end{array}(k\in {\mathbb{N}}^{\ast }).$$

$$\frac{\mathrm{d}}{\mathrm{d}x}{\int }_{{\phi }_1(x)}^{{\phi }_2(x)}f(t)\mathrm{d}t=f[{\phi }_1(x)]{\phi }_1^{\prime }(x)-f[{\phi }_2(x)]{\phi }_2^{\prime }(x).$$

$$\frac{\mathrm{d}}{\mathrm{d}x}{\int }_0^{x}tf(x-t)\mathrm{d}t=\frac{\mathrm{d}}{\mathrm{d}x}{\int }_0^x(x-t)f(t)\mathrm{d}t={\int }_0^{x}f(t)\mathrm{d}t$$

变限积分必连续; 被积函数跳跃间断点处不可导, 可去间断点处可导.

${\int }_0^1\frac{1}{x^p}$ 收敛时 $0<p<1$; 发散时 $p\ge 1$.
${\int }_1^{+\infty }\frac{1}{x^p}$ 收敛时 $p>1$; 发散时 $p\le 1$.
${\sum }_{n=1}^{\infty }\frac{1}{n^p}$ 收敛时 $p>1$; 发散时 $p\le 1$.
${\sum }_{n=1}^{\infty }(-1{)}^{n-1}\frac{1}{n^p}$ 绝对收敛时 $p>1$; 条件收敛时 $0<p\le 1$.

恒正函数 ${\lim }_{x\to \mathcal{B}}\frac{f(x)}{g(x)}=\lambda$: ($⟹$ 比较判敛法极限形式)
$\lambda \ne 0,\infty$ 时, $\exists ϵ>0$, $(\lambda -ϵ)g(x)\le f(x)\le (\lambda -ϵ)g(x)$.
$\lambda =0$ 时, $\exists ϵ>0$, $f(x)\le ϵg(x)$.

$$\begin{gathered} \Gamma (a)={\int }_0^{+\infty }{x}^{a-1}{e}^{-x}\mathrm{d}x=2{\int }_0^{+\infty }{x}^{2a-1}{e}^{-x^2}\mathrm{d}x; \\ \Gamma (a)=(a-1)\Gamma (a-1);\Gamma (n)=(n-1)!(n\in {\mathbb{N}}_{+});{\int }_{-\infty }^{+\infty }{e}^{-x^2}\mathrm{d}x\equiv \Gamma (\frac{1}{2})=\sqrt{\pi }. \end{gathered}$$

平面曲线 $y=f(x),x\in [a,b]$ 绕直线 $Ax+By+C=0$ 旋转得到的体积:
$$\begin{gathered} V=\pi {\int }_{L}r(s{)}^{2}h(s)\mathrm{d}l=\pi {\int }_a^b(\frac{|Ax+Bf(x)+C|}{\sqrt{A^2+B^2}}{)}^2|1-\frac{A}{B}{f}^{\prime }(x)|\frac{1}{\sqrt{1+\frac{A^2}{B^2}}}\mathrm{d}x \\ =\frac{\pi }{(A^2+B^2{)}^{\frac{3}{2}}}{\int }_a^b[Ax+Bf(x)+C{]}^2|A{f}^{\prime }(x)-B|\mathrm{d}x. \end{gathered}$$

微分方程通解含有的独立常数个数等于微分方程的阶数.

$$\begin{gathered} y^{\prime }+P(x)y=Q(x)⟹[e^{\int P(x)\mathrm{d}x}y{]}^{\prime }=Q(x)e^{\int P(x)\mathrm{d}x} \\ ⟹y=e^{-\int P(x)\mathrm{d}x}[\int Q(x)e^{\int P(x)\mathrm{d}x}\mathrm{d}x+C]. \end{gathered}$$

$$\begin{gathered} y^{\prime }+P(x)y=Q(x)y^n⟹y^{-n}{y}^{\prime }+P(x)y^{-n+1}=Q(x)⟹ \\ \frac{1}{1-n}{z}_x^{\prime }+P(x)z=Q(x),z=y^{-n+1}. \end{gathered}$$

$$y^{\prime \prime }=f(y,y^{\prime })⟹p_y^{\prime }p=f(y,p),p=y^{\prime }.$$

$$x^2{y}^{\prime \prime }+px{y}^{\prime }+qy=f(x)⟹y_t^{\prime \prime }+(p-1)y_t^{\prime }+qy=f(e^t),x=e^t>0.$$

$y^{\prime \prime }+p{y}^{\prime }+qy=0$, 特征方程 ${\lambda }^2+p\lambda +q=0$.
有不等实根时 $y=C_1{e}^{{\lambda }_{1}x}+C_2{e}^{{\lambda }_{2}x}$;
有相等实根时 $y=(C_1+C_{2}x)e^{\lambda x}$;
仅有复根 $\alpha \pm \beta i$ 时 $y=e^{\alpha x}(C_1\cos \beta x+C_2\sin \beta x)$.

$y^{(n)}+{\sum }_{i=1}^{n-1}{p}_i{y}^{(n-i)}+p_{n}y=0$, 特征方程 ${\lambda }^n+{\sum }_{i=1}^{n-1}{p}_i{\lambda }^{n-i}+p_n=0$.
有单实根时 $y=C{e}^{\lambda x}$;
有 $k$ 重实根时 $y=({\sum }_{i=1}^k{C}_i{x}^{i-1})e^{\lambda x}$;
有一对单复根 $\alpha \pm \beta i$ 时 $y=e^{\alpha x}(C_1\cos \beta x+C_2\sin \beta x)$;
有一对 $k$ 重复根 $\alpha \pm \beta i$ 时 $y=e^{\alpha x}(C_1\cos \beta x+C_2\sin \beta x+C_{3}x\cos \beta x+C_{4}x\sin \beta x)$.

$y^{\prime \prime }+p{y}^{\prime }+qy=f(x)$.
$f(x)=e^{\alpha x}{P}_n(x)$ 时 $y^{\ast }=e^{\alpha x}{Q}_n(x)x^k$: $\alpha$ 不为特征根时 $k=0$; 单特征根时 $k=1$; 二重特征根时 $k=2$.
$f(x)=e^{\alpha x}[P_m(x)\cos \beta x+P_n(x)\sin \beta x]$ 时 $y^{\ast }=e^{\alpha x}[Q_l(x)\cos \beta x+R_l(x)\sin \beta x]x^k$, l = max ⁡ { m , n } l=\max\{m,n\} l=max{m,n}: $\alpha \pm \beta i$ 不为特征根时 $k=0$; 特征根时 $k=1$.

{ u n } \{u_n\} {un​} 单调不增, ${\lim }_{n\to \infty }{u}_n=0⟹$ 交错级数 ${\sum }_{n=1}^{\infty }(-1{)}^{n-1}{u}_n$ 收敛 (莱布尼茨判敛法).

绝对收敛: ${\sum }_{n=1}^{\infty }|u_n|$ 收敛时 ($⟹{\sum }_{n=1}^{\infty }{u}_n$ 收敛).
条件收敛: ${\sum }_{n=1}^{\infty }{u}_n$ 收敛且 ${\sum }_{n=1}^{\infty }|u_n|$ 发散.

函数项级数 ${\sum }_{n=0}^{\infty }{u}_n(x)$ 收敛域 (阿贝尔定理): ${\lim }_{n\to \infty }\frac{|u_{n+1}(x)|}{|u_n(x)|}<1$ (达朗贝尔判敛法) 或 ${\lim }_{n\to \infty }\sqrt[n]{|u_n(x)|}<1$ (柯西判敛法) 给出收敛区间 $(a,b)$ 并单独判断 $a$ 和 $b$ 处敛散性.

幂级数 ${\sum }_{n=0}^{\infty }{a}_n{x}^n\cdot {\sum }_{n=0}^{\infty }{b}_n{x}^n={\sum }_{n=0}^{\infty }({\sum }_{i=1}^n{a}_i{b}_{n-i})x^n$, 收敛半径 R ≥ min ⁡ { R a , R b } R\geq\min\{R_a,R_b\} R≥min{Ra​,Rb​}.

幂级数恒等变换: ${\sum }_{n=k}^{\infty }{a}_n{x}^n={\sum }_{n=k+l}^{\infty }{a}_{n-l}{x}^{n-l}$; ${\sum }_{n=k}^{\infty }{a}_n{x}^n={\sum }_{n=k}^{k+l-1}{a}_n{x}^n+{\sum }_{n=k+l}^{\infty }{a}_n{x}^n$; ${\sum }_{n=k}^{\infty }{a}_n{x}^n=x^l{\sum }_{n=k}^{\infty }{a}_n{x}^{n-l}$.

幂级数的和函数 $S(x)={\sum }_{n=0}^{\infty }{a}_n{x}^n$ 在收敛域上连续.
幂级数及和函数在收敛域上可积(可导)时, 有逐项积分(求导): ${\int }_0^{x}S(x)\mathrm{d}x={\sum }_{n=0}^{\infty }\frac{a_n}{n+1}{x}^{n+1}$; $S^{\prime }(x)={\sum }_{n=1}^{\infty }n{a}_n{x}^{n-1}$.

$$\sum _{n=1}^{\infty }\frac{x^n}{n}={\int }_0^x(\sum _{n=0}^{\infty }{x}^n)\mathrm{d}x=-\ln (1-x),x\in [-1,1).$$

$$\sum _{n=1}^{\infty }n{x}^{n-1}=(\sum _{n=1}^{\infty }{x}^n{)}^{\prime }=\frac{1}{(1-x{)}^2},x\in (-1,1).$$

$$\frac{1}{1+x}=\sum _{n=0}^{\infty }(-1{)}^n{x}^n,x\in (-1,1).$$

$$e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!},x\in \mathbb{R}.$$

$$\sin x=\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{(2n+1)!},x\in \mathbb{R}.$$

$$\begin{gathered} f(x+2l)=f(x)⟹f(x)\sim (\frac{a_0}{2}+\sum _{n=1}^{\infty }{a}_n\cos \frac{n\pi }{l}x)+(\sum _{n=1}^{\infty }{b}_n\sin \frac{n\pi }{l}x); \\ {a}_n=\frac{1}{l}{\int }_{-l}^{l}f(x)\cos \frac{n\pi }{l}x\mathrm{d}x,b_n=\frac{1}{l}{\int }_{-l}^{l}f(x)\sin \frac{n\pi }{l}x\mathrm{d}x. \end{gathered}$$

$f(x)$ 以 $2l$ 为周期, 在 $[-l,l]$ 上可积, 至多有有限个第一类间断点及极值点 $⟹f(x)$ 傅里叶级数几乎处处收敛, 和函数 $S(x)=f(x)$, $x$ 处连续; $\frac{f(x-0)+f(x+0)}{2}$, $x$ 处间断; $\frac{f(-l+0)+f(l-0)}{2}$, $x=\pm l$.

$f(x)$ 在 $[0,l]$ 上可积:
奇延拓: $F(x)=f(x),kl<x\le (k+1)l;-f(-x),(k-1)l\le x<kl;0,x=kl;k\in \mathbb{Z}$.
偶延拓: $F(x)=f(x),kl\le x\le (k+1)l;f(-x),(k-1)l\le x<kl;k\in \mathbb{Z}$.