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基础公式
三次方和差公式
\begin{aligned}
(a \pm b)^3 &= a^3 \pm 3a^2b + 3ab^2 \pm b^3 \\
a^3 \pm b^3 &= (a \pm b)(a^2 \mp ab + b^2) \\
\end{aligned}
排列组合公式
A_n^m = \frac{n!}{(n-m)!}
C_n^m = \frac{A_n^m}{m!} = \frac{n!}{m!(n-m)!} = C_n^{n-m}
\sum_{i=0}^{k} C_m^iC_n^{k-i} = C_{m+n}^k
三角函数公式
基础公式
\sin \alpha \cdot \csc \alpha = 1
\cos \alpha \cdot \sec \alpha = 1
\tan \alpha \cdot \cot \alpha = 1
\tan \alpha = \frac{\sin \alpha}{\cos \alpha}
\sin^2 \alpha + \cos^2 \alpha = 1
\sec^2 \alpha - \tan^2 \alpha = 1
\csc^2 \alpha - \cot^2 \alpha = 1
倍角公式
\sin 2\alpha = 2\sin \alpha \cos \alpha
\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = 1 - 2\sin^2 \alpha = 2\cos^2 \alpha - 1
\sin 3\alpha = -4\sin^3 \alpha + 3\sin \alpha
\sin^2 \alpha = \frac{1}{2}(1-\cos 2\alpha)
\cos^2 \alpha = \frac{1}{2}(1 + \cos 2\alpha)
\tan 2\alpha = \frac{2\tan\alpha}{1-\tan^2\alpha}
\cot 2\alpha = \frac{\cot^2 \alpha - 1}{2\cot \alpha}
半角公式
\sin^2 \frac{\alpha}{2} = \frac{1}{2}(1 - \cos \alpha)
\cos^2 \frac{\alpha}{2} = \frac{1}{2}(1 + \cos \alpha)
\sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}}
\cos \frac{\alpha}{2} = \pm \sqrt{\frac{1 + \cos \alpha}{2}}
\tan \frac{\alpha}{2} = \frac{1 - \cos\alpha}{\sin \alpha} = \frac{\sin \alpha}{1 + \cos \alpha} = \pm \sqrt{\frac{1 - \cos\alpha}{1+\cos\alpha}}
\cot \frac{\alpha}{2} = \frac{\sin \alpha}{1 - \cos \alpha} = \frac{1 + \cos\alpha}{\sin \alpha} = \pm \sqrt{\frac{1+\cos\alpha} {1 - \cos\alpha}}
和差公式
\sin(\alpha \pm \beta) = \sin\alpha \cos\beta \pm \cos\alpha \sin\beta
\cos(\alpha \pm \beta) = \cos\alpha \cos\beta \mp \sin\alpha \sin\beta
\tan(\alpha\pm\beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta}
\cot(\alpha\pm\beta) = \frac{\cot\alpha\cot\beta \mp 1}{\cot\beta\pm\cot\alpha}
积化和差公式
\sin\alpha \cos\beta = \frac{1}{2}[
\sin(\alpha + \beta) + \sin(\alpha - \beta)
]
\cos\alpha \sin\beta = \frac{1}{2}[
\sin(\alpha + \beta) - \sin(\alpha - \beta)
]
\cos\alpha \cos\beta = \frac{1}{2}[
\cos(\alpha + \beta) + \cos(\alpha - \beta)
]
\sin\alpha \sin\beta = \frac{1}{2}[
\cos(\alpha - \beta) - \cos(\alpha + \beta)
]
和差化积公式
\sin \alpha + \sin \beta = 2 \sin\frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}
\sin \alpha - \sin \beta = 2 \sin\frac{\alpha - \beta}{2} \cos \frac{\alpha + \beta}{2}
\cos \alpha + \cos \beta = 2\cos\frac{\alpha + \beta}{2} \cos\frac{\alpha - \beta}{2}
\cos \alpha - \cos \beta = - 2\sin\frac{\alpha + \beta}{2} \sin\frac{\alpha - \beta}{2}
经典不等式公式
a b
2|ab| \leqslant a^2 + b^2
| a \pm b| \leqslant | a | + | b |
\big| |a| - |b| \big| \leqslant | a-b|
|a_1 \pm a_1 \pm \cdots \pm a_n| \leqslant | a_1 | + | a_1 | + \cdots + | a_n |
\left| \int_a^bf(x)dx \right| \leqslant \int_a^b| f(x) | dx\ (a<b)
a_1,a_2,\cdots,a_n>0
几何平均数大于调和平均数
\frac{n}{\displaystyle \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}} \leqslant \sqrt[n]{a_1a_2\cdots a_n}
算术平均数大于几何平均数
\sqrt[n]{a_1a_2\cdots a_n} \leqslant \frac{a_1 + a_2 + \cdots + a_n}{n}
平方平均数大于算术平均数
\frac{a_1 + a_2 + \cdots + a_n}{n} \leqslant \sqrt{\frac{a_1^2 + a_2^2 + \cdots + a_n^2}{n}}
次方平均数的大小
\begin{aligned}
\sqrt[p]{\frac{a_1^p + a_2^p + \cdots + a_n^p}{n}} \leqslant \sqrt[q]{\frac{a_1^q + a_2^q + \cdots + a_n^q}{n}} & (0 < p < q )
\end{aligned}
以上，当且仅当 a_1 = a_2 = \cdots = a_n
更多不等式，参见 不等式
基本放缩法
\begin{cases}
n\cdot u_{\min} \leqslant u_1 + u_2 + \cdots + u_n \leqslant n \cdot u_{\max} \\
u_{\max} \leqslant u_1 + u_2 + \cdots + u_n \leqslant n \cdot u_{\max} &(u_i \geqslant 0) \\
\end{cases}
两个重要极限
\lim_{x \to 0} \frac{\sin x}{ x } = 1
\lim_{x \to \infty} (1 + \frac{1}{x})^x = e
常用泰勒公式
\begin{aligned}
f(x) &= f(x_0) + f'(x_0)(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n + R_n(x)\\
R_n(x) &= \frac{f^{(n+1)}(\xi)}{(n+1)!}(x - x_0)^{(n+1)}\ \xi \in (x, x_0) \\
R_n(x) &= o[(x - x_0)^n]
\end{aligned}
sin\ x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots + (-1)^n\frac{x^{2n+1}}{(2n+1)!} + o(x^{2n+1})
cos\ x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots + (-1)^n\frac{x^{2n}}{(2n)!} + o(x^{2n})
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} + o(x^n)
ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + \cdots + (-1)^{n-1}\frac{x^n}{n} + o(x^n)
tan\ x = x + \frac{x^3}{3} + o(x^3)
arcsin\ x = x + \frac{x^3}{3!} + o(x^3)
arctan\ x = x - \frac{x^3}{3} + o(x^3)
(1+x)^a = 1 + ax + \frac{a(a-1)}{2!}x^2 + o(x^2)
重要幂级数展开式
e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} + \cdots, -\infty < x < +\infty
\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n = 1 - x + x^2 - \cdots + (-1)^n x^n + \cdots, -1 < x < 1
\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots + x^n + \cdots , -1 < x < 1
ln(1+x) = \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n} = x - \frac{x^2}{2} + \cdots + (-1)^{n-1}\frac{x^n}{n} + \cdots, -1 < x \leqslant 1
\sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n + 1)!} = x - \frac{x^3}{3!} + \cdots + (-1)^n \frac{x^{2n+1}}{(2n + 1)!} + \cdots, -\infty < x < + \infty
\cos x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \cdots + (-1)^n \frac{x^{2n}}{(2n)!} + \cdots, -\infty < x < + \infty
基本求导公式
(x^\mu)' = \mu x^{\mu - 1}
(\sec x)' = \sec x \tan x
(\csc x)' = - \csc x \cot x
(\arcsin x)' = \frac{1}{\sqrt{1 - x^2}}
(\arccos x)' = \frac{-1}{\sqrt{1 - x^2}}
(\arctan x)' = \frac{1}{1+ x^2}
(arccot\ x)' = \frac{-1}{1+ x^2}
(\log_ax)' = \frac{1}{x\ln a}
\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}
莱布尼茨求导公式
(u \cdot v)'' = u''\cdot v + 2 u' \cdot v' + u \cdot v''
(u \cdot v) ^{(n)} = \sum_{k=0}^{n} C_n^k u^{(n-k)} v^{(k)}
高阶导数
(a^x)^{(n)} = (\ln a)^n\cdot a^x
(e^x)^{(n)} = e^x
\displaystyle (\sin kx)^{(n)} = k^n\sin \left(\frac{n\pi}{2} + kx\right)
\displaystyle (\cos kx)^{(n)} = k^n\cos \left(\frac{n\pi}{2} + kx\right)
\displaystyle (\ln x)^{(n)} = (-1)^{n-1}\frac{(n-1)!}{x^n}
\displaystyle [\ln(1 + x)]^{(n)} = (-1)^{(n-1)}\frac{(n-1)!}{(1 + x)^n}
\displaystyle \left(\frac{1}{x}\right)^{(n)} = (-1)^{n}\frac{n!}{x^{n + 1}}
\displaystyle \left(\frac{1}{x + a}\right)^{(n)} = (-1)^{n}\frac{n!}{(x + a)^{n + 1}}
\displaystyle (x^m)^{(n)} = m(m-1)\cdots(m - n + 1) x^{m - n}
\displaystyle (u \pm v)^{(n)} = u^{(n)} \pm v^{(n)}
中值定理
基本定义及证明，参见 中值定理的证明
辅助函数类型：证明过程参见 中值定理辅助函数类型
中值等式 G(\xi) = 0
辅助函数 F(x)
\displaystyle\frac{f'(x)}{f(x)} = 0 \ln f(x)
f'(\xi) + A\xi^k + B = 0 \displaystyle f(x) + \frac{Ax^{k+1}}{k+1} + Bx
f(a)g'(\xi)-f'(\xi)g(a) - k = 0 f(a)g(x) - f(x)g(a) - kx
\displaystyle\sum_{i=0}^{n-1}a_i(n-i)\xi^{n-1-i} = 0 \displaystyle\sum_{i=0}^{n-1}a_i x^{n-i}
f'(\xi)g(\xi) + f(\xi)g'(\xi) = 0 f(x)g(x)
f(\xi)g''(\xi) - f''(\xi)g(\xi) = 0 f(x)g'(x) - f'(x)g(x)
\xi f'(\xi) + kf(\xi) = 0 x^kf(x)
(\xi - 1)f'(\xi) + kf(\xi) = 0 (x-1)^kf(x)
f'(\xi)g(1-\xi) - kf(\xi)g'(1-\xi) = 0 g^k(1-x)f(x)
f'(\xi) + \lambda f(\xi) = 0 e^{\lambda x}f(x)
f'(\xi) + g'(\xi)f(\xi) = 0 e^{g(x)}f(x)
\xi f'(\xi) - kf(\xi) = 0 \displaystyle\frac{f(x)}{x^k}
f'(\xi) - kf(\xi) = 0 \displaystyle\frac{f(x)}{e^{kx}}
\displaystyle f(\xi) + \frac{x - b}{a}f'(\xi) = 0 (x - b)^a f(x)
f'(\xi)g(\xi) - f(\xi)g'(\xi) = 0 \displaystyle\frac{f(x)}{g(x)}
\displaystyle\frac{1-\xi^2}{(1+ \xi^2)^2} = 0 \displaystyle\frac{x}{1+x^2}
f'(\xi) - f(\xi) + k\xi - k = 0 \displaystyle\frac{f(x) - kx}{e^{x}}
f''(\xi) + f'(\xi) - k = 0 e^x[f'(x) - k]
f'(\xi) + k[f(\xi)-\xi] -1 = 0 e^{kx}[f(x) - x]
f''(\xi) - f(\xi) = 0 e^x[f(x) - f'(x)]
定积分
\begin{cases}
\displaystyle\sum_{i=1}^{\infty} f\left(\frac{i}{n}\right) \frac{1}{n} = \int_0^1f(t)dt \\
\displaystyle\sum_{i=1}^{\infty} f\left(\frac{x\cdot i}{n}\right) \frac{x}{n} = \int_0^xf(t)dt \\
\displaystyle\sum_{i=1}^{\infty} f\left(a + \frac{(b - a)\cdot i}{n}\right) \frac{b - a}{n} = \int_a^bf(t)dt
\end{cases}
弧长公式（第一型曲线积分）
\begin{aligned}
S =& \int_a^b \sqrt{1 + y'^2} dx \\
\xlongequal{x=\varphi(t);y=\psi(t)}& \int_{a_t}^{b_t} \sqrt{[\varphi'(t)]^2 + [\psi'(t)]^2} dt \\
\xlongequal{x=r\cos\theta;y=r\sin\theta}& \int_{a_\theta}^{b_\theta} \sqrt{[r'\cos\theta - r\sin \theta]^2 + [r'\sin\theta + r\cos\theta]^2} d\theta \\
=& \int_{a_\theta}^{b_\theta} \sqrt{r^2 + (r')^2} d\theta \\
\end{aligned}
曲率
\begin{aligned}
\frac{\Delta \varphi}{\Delta s} &=
\frac{|\arctan f'(x + \Delta x) - \arctan f'(x)|}
{s(x + \Delta x) - s(x)} \\
&=\frac{|\frac{\arctan f'(x + \Delta x) - \arctan f'(x)}{\Delta x}|}
{\frac{s(x + \Delta x) - s(x)}{\Delta x}}
\end{aligned}
|\frac{d}{dx}(\arctan f'(x))| = |\frac{d}{dx}(\arctan y')| = \frac{|y''|}{1 + (y')^2}
\frac{ds(x)}{dx} = \sqrt{1 + (y')^2}
k = \lim_{\Delta s \to 0} \frac{\Delta \varphi}{\Delta s} = \frac{|y''|}{[1 + (y')^2]^{3/2}}
若 \begin{cases}\displaystyle x=x(t) \\ y=y(t) \\ \end{cases}
k = \frac{|y''x' - y'x''|}{(x'^2 + y'^2)^{\frac{3}{2}}}
基本积分公式
\int x^\mu dx = \frac{1}{\mu + 1}\cdot x^{\mu + 1} + C \ (\mu \neq -1)
\int \frac{dx}{x} = ln|x| + C
\int \cos x dx = \sin x + C
\int \sin x dx = -\cos x +C
\int \frac{dx}{\cos^2 x} = \int \sec^2 x dx = \tan x + C
\int \frac{dx}{\sin^2 x} = \int \csc^2 x dx = -\cot x + C
\int \sec x \tan x dx = \sec x + C
\int \csc x \cot x dx = - \csc x + C
\int \frac{dx}{1+x^2} = \arctan x + C
\int \frac{dx}{\sqrt{1 - x^2}} = \arcsin x + C
\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \arctan\frac{x}{a} + C
\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C
\int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln \left| \frac{x-a}{x+a}\right| + C
\int \frac{dx}{a^2 - x^2} = \frac{1}{2a}\ln \left| \frac{x+a}{x-a}\right| + C
\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln(x + \sqrt{x^2 + a^2}) + C
\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln|x + \sqrt{x^2 - a^2}| + C
\int \sqrt{x^2 + a^2} dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln(x + \sqrt{x^2 + a^2}) + C
\int \sqrt{a^2 - x^2} dx = - \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin \frac{x}{a} + C
\int \tan x dx = \int \frac{\sin x}{\cos x} dx = -\int \frac{d \cos x}{\cos x} = - \ln | \cos x| + C
\int \cot x dx = \ln | \sin x | + C
\int \csc x dx = \ln |\tan \frac{x}{2}| + C = \ln |\csc x - \cot x| + C
\int \sec x dx = \ln |\sec x + \tan x| + C
\int a^x dx = \frac{1}{\ln a} a^x + C
万能代换
\begin{aligned}
u =& \tan\frac{x}{2} \\
\sin x =& \frac{2u}{1+u^2} \\
\cos x =& \frac{1-u^2}{1+u^2} \\
dx =& \frac{2}{1+u^2}du
\end{aligned}
万能代换可以将三角函数换成有理分式的形式来进行积分
\left\{
\begin{aligned}
\sqrt{a^2 - x^2} \rightarrow x &= asin\ t\\
\sqrt{a^2 + x^2} \rightarrow x &= atan\ t\\
\sqrt{x^2 - a^2} \rightarrow x &= asec\ t\\
\end{aligned}
\right.
变限积分函数求导
\begin{aligned}
F(x) &= \int_{\varphi_{1}(x)}^{\varphi_{2}(x)} f(t) dt\\
F'(x)&= \frac{d}{dx}[\int_{\varphi_{1}(x)}^{\varphi_{2}(x)} f(t) dt] = f[\varphi_{2}(x)]\varphi'_{2}(x) - f[\varphi_{1}(x)]\varphi'_{1}(x)
\end{aligned}
旋转体体积
曲边梯形 0 \leqslant y \leqslant f(x) (a\leqslant x\leqslant b)
绕 x
V_x = \pi \int_a^b [f(x)]^2 dx
a \geqslant 0 y
V_y = 2\pi \int_a^b xf(x) dx
旋转体的侧面积
由 y = f(x) f(x) \geqslant 0 a < x < b x x 侧面积 为：
S = 2\pi \int_a^b f(x)\sqrt{1 + [f'(x)]^2} dx
若曲线由参数方程 x = x(t) y=y(t)\geqslant 0 t\in[\alpha, \beta] 侧面积 为：
S = 2\pi \int_\alpha^\beta y(t)\sqrt{[x'(t)]^2 + [y'(t)]^2} dt
二重积分旋转体体积
区域 D(x, y) \geqslant 0
绕 x \displaystyle V_x = 2\pi \iint\limits_D y d\sigma
绕 y \displaystyle V_y = 2\pi \iint\limits_D x d\sigma
若 D=\{(r, \theta)|0\leqslant \alpha \leqslant \theta \leqslant \beta \leqslant \pi, 0 \leqslant r \leqslant r(\theta)\}
绕 极轴 （x
\begin{aligned}
V_x =& 2\pi \iint\limits_D y d\sigma \\
=& 2\pi \int_\alpha^\beta d\theta \int_0^{r(\theta)} r\sin \theta r dr \\
=&\frac{2\pi}{3}\int_\alpha^\beta r^3(\theta) \sin \theta d\theta
\end{aligned}
古尔丁定理
上半平面内一个有界比区域 D 绕 x 旋转一周生成的旋转体的体积，等于 该区域的 形心纵坐标 为半径的圆的周长 与 D A
\begin{aligned}
V_x =& 2\pi \iint\limits_D y d\sigma \\
=& 2\pi A\cdot \frac{1}{A} \iint\limits_D y d\sigma \\
=& 2\pi A \overline{y} \\
\end{aligned}
区间再现公式
\begin{aligned}
& \int_a^b f(x) dx \\
\xlongequal{x = a+b-t}& \int_b^a f(a+b-t) d(-t) \\
=& \int_a^b f(a+b-t) dt \\
=& \int_a^b f(a+b-x) dx
\end{aligned}
无穷级数
关于 级数 和 和 是很微妙的，严格的说级数是一个形如 \displaystyle\sum_{i=m}^n a_i 1 + 2 + 3 + 4 + 5 15 15 1 + 2 + 3 + 4 + 5
当级数的项数 n\to \infty
正项级数 \displaystyle\sum_{n=1}^\infty u_n \Leftrightarrow
比较判别法
若 0 \leqslant u_n \leqslant v_n
\displaystyle\sum_{n=1}^\infty v_n \Rightarrow \displaystyle\sum_{n=1}^\infty u_n
\displaystyle\sum_{n=1}^\infty u_n \Rightarrow \displaystyle\sum_{n=1}^\infty v_n
比较判别法的极限形式
设 \displaystyle\lim_{n\to\infty} \frac{u_n}{v_n} = l (0 \leqslant l \leqslant +\infty)
若 0 < l < +\infty \displaystyle\sum_{n=1}^\infty u_n \displaystyle\sum_{n=1}^\infty v_n
若 l = 0
\displaystyle\sum_{n=1}^\infty v_n \Rightarrow \displaystyle\sum_{n=1}^\infty u_n \displaystyle\sum_{n=1}^\infty u_n \Rightarrow \displaystyle\sum_{n=1}^\infty v_n
若 l = +\infty
\displaystyle\sum_{n=1}^\infty u_n \Rightarrow \displaystyle\sum_{n=1}^\infty v_n \displaystyle\sum_{n=1}^\infty v_n \Rightarrow \displaystyle\sum_{n=1}^\infty u_n
比值判别法
设 \displaystyle\sum_{n=1}^\infty u_n \displaystyle \lim_{n\to \infty} \frac{u_{n + 1}}{u_n} = \rho \displaystyle\sum_{n=1}^\infty u_n
\rho < 1 \rho > 1 \rho = 1
根值判别法
设 \displaystyle\sum_{n=1}^\infty u_n \displaystyle \lim_{n\to \infty} \sqrt[n]{u_n} = \rho \displaystyle\sum_{n=1}^\infty u_n
\rho < 1 \rho > 1 \rho = 1
莱布尼茨判别法
若 u_n \geqslant u_{n+1} (n = 1,2,\cdots) \displaystyle \lim_{n\to\infty} = 0 \displaystyle\sum_{n=1}^\infty (-1)^n u_n
无穷级数的和函数
\begin{cases}
\displaystyle\sum_{n=1}^{\infty} n x^{n - 1} = \frac{1}{(1 - x)^2} & (-1 < x < 1) \\
\displaystyle\sum_{n=1}^{\infty} \frac{1}{n} x^{n} = -\ln(1 - x) & (-1 \leqslant x < 1) \\
\end{cases}
反常积分审敛法
反常积分 \displaystyle \int_a^{+\infty} \frac{dx}{x^p} (a>0, p>0)
\begin{aligned}
\int_a^{+\infty} \frac{dx}{x^p} = \begin{cases}
\frac{x^{1-p}}{1-p}\bigg|_a^{+\infty} &(p \neq 1) \\
\ln x\bigg|_a^{+\infty} &(p = 1)
\end{cases}
\end{aligned}
\begin{aligned}
\int_a^{+\infty} \frac{dx}{x^p} = \begin{cases}
\infty &(0 <p \leqslant 1) \\
\frac{a^{1-p}}{p-1} &(p > 1)
\end{cases}
\end{aligned}
反常积分 \displaystyle \int_a^b \frac{dx}{(x-a)^q} (q>0)
\begin{aligned}
\int_a^{+\infty} \frac{dx}{x^p} = \begin{cases}
\infty &(q \geqslant 1) \\
\frac{(b - a)^{1-q}}{1-q} &(0 < q < 1)
\end{cases}
\end{aligned}
总结上两题：大大小小
如果积分区间有无穷，则 p > 1
如果积分区间有瑕点，则 0 < p < 1
反常积分 \displaystyle \int_a^{+\infty} \frac{dx}{x^p\ln^q x} (a>0)
当 p > 1
当 p = 1
q > 1 q \leqslant 1
当 p < 1
定理一： 设函数 f(x) [a, +\infty) f(x) \geqslant 0 p > 1 \displaystyle \lim_{x \to +\infty} x^pf(x) = c \displaystyle \int_a^{+\infty} f(x) dx \displaystyle \lim_{x \to +\infty} xf(x) = d>0 \displaystyle \int_a^{+\infty} f(x) dx
通俗解释是：要想判断 \displaystyle \int_a^{+\infty} f(x) dx x \to \infty f(x) \displaystyle \frac{1}{x^p} \displaystyle \int_a^{+\infty} f(x) dx \displaystyle \int_a^{+\infty} \frac{1}{x^p} dx \displaystyle \int_a^{+\infty} f(x) dx
定理二： 设函数 f(x) (a, b] f(x) \geqslant 0 x = a f(x) 0 < q < 1 \displaystyle \lim_{x \to +\infty}(x - a)^qf(x) = c \displaystyle \int_a^b f(x) dx \displaystyle \lim_{x \to +\infty} (x - a)f(x) = d>0 \displaystyle \int_a^b f(x) dx
通俗解释是：要想判断 \displaystyle \int_a^b f(x) dx x \to \infty f(x)
常微分方程公式
变量可分离型
\begin{aligned}
\frac{dy}{dx} &= f(x)g(y) \\
\int \frac{dy}{g(y)} &=\int f(x)dx
\end{aligned}
可化为变量可分离型
\begin{aligned}
\frac{dy}{dx} &= f(ax + by + c) \\
u &= ax + by + c \\
\frac{du}{dx} &= a + b\frac{dy}{dx}\\
\frac{du}{dx} &= a + bf(u)
\end{aligned}
齐次微分方程
\begin{aligned}
\frac{dy}{dx} &= \varphi(\frac{y}{x}) \\
u &= \frac{y}{x} \\
y &= ux \\
\frac{dy}{dx} &= u + x\frac{du}{dx}\\
x\frac{du}{dx} + u &= \varphi(u) \\
\frac{du}{\varphi(x) - u} &= \frac{dx}{x}
\end{aligned}
一阶线性微分方程
\begin{aligned}
y' + p(x)y &= q(x) \\
v(x) &= e^{\int p(x)dx} \\
y &= \frac{1}{v(x)}\left[\int v(x) \cdot q(x) dx + C \right]
\end{aligned}
推导过程参见 一阶线性方程公式推导
伯努利微分方程
\begin{aligned}
y' + p(x)y &= q(x)y^n \\
y^{-n}\cdot y' + p(x) y^{1-n} &= q(x) \\
z &= y^{1-n} \\
\frac{dz}{dx} &= (1-n)y^{-n}\frac{dy}{dx}\\
\frac{1}{1-n}\frac{dz}{dx} + p(x)z &= q(x) \\
\frac{dz}{dx} + (1-n)p(x)z &= (1-n)q(x) \\
v(x) &= e^{(1-n)\int p(x)dx} \\
z &= \frac{1}{v(x)}\left[(1-n)\int v(x) \cdot q(x) dx + C \right] \\
y & = \sqrt[1-n]{z}
\end{aligned}
可降阶的高阶微分方程
可通过 n
y'' = f(x, y') y
作变换 y' = p(x) \displaystyle y'' = \frac{dp}{dx} \displaystyle \frac{dp}{dx} = f(x, p) x
y'' = f(y, y') x
作变换 y' = p(y) \displaystyle y'' = \frac{dp}{dx} = \frac{dp}{dy}\cdot\frac{dy}{dx} = p\frac{dp}{dy} \displaystyle \frac{dp}{dx} = f(x, p) p, y
二阶常系数齐次线性微分方程
\begin{aligned}
y'' + py' + qy &= 0 \\
\lambda^2 + p\lambda + q &= 0 \Rightarrow \lambda_1,\lambda_2\\
\Downarrow
\end{aligned}
\begin{aligned}
\lambda_1 \neq \lambda_2 &\Rightarrow y = C_1e^{\lambda_1x} + C_2e^{\lambda_2x} \\
\lambda_1 = \lambda_2 = \lambda &\Rightarrow y = (C_1 + C_2x)e^{\lambda x} \\
\lambda = \alpha \pm \beta i &\Rightarrow y = C_1e^{\lambda_1x} + C_2e^{\lambda_2x} \\ \xlongequal[e^{ix} = \cos x + i \sin x]{欧拉公式} e^{\alpha x}& (C_1 \cos \beta x + C_2\sin \beta x)
\end{aligned}
二阶常系数非齐次线性微分方程
当 f(x) = P_n(x)e^{\alpha x}
设 y^* = e^{\alpha x}Q_n(x)x^k
其中 e^{\alpha x}
Q_n(x) x n
k =
\begin{cases}
0, & \alpha \neq \lambda_1, \alpha \neq \lambda_2, \\
1, & \alpha = \lambda_1\ or\ \alpha = \lambda_2, \\
2, & \alpha = \lambda_1 = \lambda_2, \\
\end{cases}
当 f(x) = e^{\alpha x}[P_m(x)\cos\beta x + P_n(x) \sin \beta x]
设 y^* = e^{\alpha x}[Q^{(1)}_l(x)\cos\beta x + Q^{(2)}_l(x) \sin \beta x]x^k
其中 e^{\alpha x}
l=\max\{m,n\}
Q^{(1)}_l,Q^{(2)}_l x l
k =
\begin{cases}
0, &\alpha \pm \beta i 不是特征根,\\
1, &\alpha \pm \beta i 是特征根,\\
\end{cases}
二阶常系数线性微分方程求特解的微分算子法，参见文章末尾，可以加快计算速度。
欧拉方程：
形如 x^ny^{(n)} + p_1x^{n - 1}y^{(n-1)} \cdots + p_{n-1}xy'+ p_ny = f(x) 欧拉方程 。
可以通过欧拉变换 x = e^t t = \ln x \displaystyle \frac{dy}{dx} = \frac{1}{x}\frac{dy}{dt} \displaystyle \frac{d^2y}{dx^2} = \frac{1}{x^2}\frac{d^2y}{dt^2}
空间解析几何公式
向量表达式
a = (a_x, a_y, a_z) = a_x i + a_y j + a_z k
两点 (x_0, y_0, z_0) (x_1, y_1, z_1)
d = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2}
点 (x_0, y_0, z_0) Ax + By + Cz + D = 0
d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
点 M_0(x_0, y_0, z_0) \displaystyle\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}
\begin{aligned}
&M_1(x_1, y_1, z_1) \\
s &= (a, b, c) \\
d &= \frac{|\overrightarrow{M_0M_1} \times s|}{|s|}
\end{aligned}
上面这个公式不好记，有必要说明一下，两个向量的叉乘的模长是对应包围着的平行四边形的面积，除以底边 向量 |\overrightarrow{s}| s
空间曲面向量场
梯度场：曲面的法向量的方向，此方向的方向导数绝对值最大
环量：向量场在曲线或曲面切线方向上的分量的积分 ，是一个标量
通量：向量场在曲线或曲面法线方向上的分量的积分 ，是一个标量
旋度：环量的密度，就是向量场在曲线或曲面切线方向上的分量
散度：通量的密度，就是向量场在曲线或曲面法线方向上的分量
向量场 \boldsymbol{F} = P\boldsymbol{i} + Q\boldsymbol{j} + R\boldsymbol{k} (x_0, y_0, z_0)
梯度场
\nabla \boldsymbol{F} = \frac{\partial P}{\partial x}\boldsymbol{i} + \frac{\partial Q}{\partial y}\boldsymbol{j} + \frac{\partial R}{\partial z}\boldsymbol{k}
梯度
grad \boldsymbol{F} = \left(\frac{\partial P}{\partial x} , \frac{\partial Q}{\partial y} , \frac{\partial R}{\partial z}\right)\Big|_{(x_0, y_0, z_0)}
散度
div \boldsymbol{F} = \left(\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}\right)\Big|_{(x_0, y_0, z_0)}
旋度
curl \boldsymbol{F} = rot \boldsymbol{F} = \nabla \boldsymbol{F} = \left|
\begin{matrix}
\boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\
\frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\
P & Q & R
\end{matrix}
\right| =
(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}) \boldsymbol{i} +
(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}) \boldsymbol{j} +
(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \boldsymbol{k}
方向导数
\left(\frac{\partial f}{\partial s} \right)_{u, p_0} = (\nabla f)_{p_0} \cdot u
空间曲面方程
椭球面
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1
单叶双曲面
\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1
双叶双曲面
\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1
椭圆抛物面
\frac{x^2}{a^2} + \frac{y^2}{b^2} = z
椭圆锥面
\frac{x^2}{a^2} + \frac{y^2}{b^2} = z^2
双曲抛物面（马鞍面）
\frac{x^2}{a^2} - \frac{y^2}{b^2} = z
椭圆柱面
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
双曲柱面
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
抛物柱面
以上函数图像参见 考研常用函数图像
更多曲线曲面方程函数图像参见 一些重要的函数图像
多元函数微分学
全微分
若 \Delta z = f(x_0 + \Delta x, y_0 + \Delta y) - f(x_0, y_0) = A\Delta x + B\Delta y + o(\rho)
其中 \rho = \sqrt{(\Delta x)^2 + (\Delta y)^2} f(x, y) (x_0, y_0) A\Delta x + B\Delta y z=f(x, y) (x_0, y_0) dz dz = A\Delta x + B\Delta y f'_x(x_0, y_0) = A f'_y(x_0, y_0) = B
z = f(x, y) P(x_0, y_0) P z (A, B, -1)
该思想可以很容易的拓展到高维空间，可微即是 切超平面 存在。
可微的必要条件
若 z = f(x, y) (x_0, y_0) (x_0, y_0) f'_x(x_0, y_0) = A f'_y(x_0, y_0) = B
可微的充分条件
如果 f(x, y) f'_x(x, y) f'_y(x, y) (x_0, y_0) f(x, y) (x_0, y_0)
多元函数几个概念之间的关系
偏导数连续 \Rightarrow \Rightarrow \displaystyle\begin{cases}函数连续 \\偏导数存在 \\ \end{cases}
复合函数求导法则
设函数 z = f(u, v) u = u(x, y) v = v(x, y) z (x, y) D D D
\begin{aligned}
\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x} \\
\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y}
\end{aligned}
隐函数求导法则
设函数 F(x, y, z) P_0(x_0, y_0, z_0) F(x_0, y_0, z_0) = 0 F'_z(x_0, y_0, z_0) \neq 0 F(x_0, y_0, z_0) = 0 P_0 z=f(x, y)
z_0 = f(x_0, y_0) F(x, y, f(x, y)) \equiv 0 z = f(x, y)
则：
\begin{aligned}
\frac{\partial z}{\partial x} = -\frac{F_x'(x, y, z)}{F'_z(x, y, z)} \\
\frac{\partial z}{\partial y} = -\frac{F_y'(x, y, z)}{F'_z(x, y, z)}
\end{aligned}
二元函数求极值
\begin{cases}
f''_{xx}(x, y) = A \\
f''_{xy}(x, y) = B \\
f''_{yy}(x, y) = C \\
\end{cases}
\Delta = B^2 - AC \begin{cases}
< 0 \Rightarrow 极值
\begin{cases}
A < 0 \Rightarrow 极大值 \\
A > 0 \Rightarrow 极小值
\end{cases}\\
> 0 \Rightarrow 非极值\\
= 0 \Rightarrow 方法失效，另谋他法
\end{cases}
拉格朗日乘数法
求目标函数 u = f(x, y, z) \begin{cases} \varphi(x,y,z)=0 \\ \psi(x, y, z) = 0 \\\end{cases}
构造辅助函数 F(x, y, z, \lambda, \mu) = f(x, y, z) + \lambda\varphi(x, y, z) + \mu\psi(x, y, z)
\begin{cases}
F'_x = f'_x + \lambda\varphi'_x + \mu\psi'_x = 0, \\
F'_y = f'_y + \lambda\varphi'_y + \mu\psi'_y = 0, \\
F'_z = f'_z + \lambda\varphi'_z + \mu\psi'_z = 0, \\
F'_\lambda = \varphi(x, y, z) = 0, \\
F'_\mu = \psi(x, y, z) = 0, \\
\end{cases}
根据上述方程组求出备选点，然后代数函数求出最大值和最小值。
关于拉格朗日乘数法的理解：实际上上面的等式如果只有一个约束条件的化可以改写成梯度的形式：
\displaystyle\left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right) = \lambda\left(\frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial \varphi}{\partial z}\right)
于是可以很容易的理解，只有极值函数和条件函数的法向量线性相关（重合）时，取得极值；而梯度可以表示法向量，也就是说极值函数与条件函数相切时取得极值。
另外利用梯度的表示方式，很容易能消去 \lambda \lambda = 0
二重积分
二重积分可以化为累次积分的定积分，表示曲顶柱体的体积。
\begin{aligned}
& \iint\limits_{D} f(x, y) d\sigma \\
\xlongequal{笛卡尔坐标}& \int_{a_x}^{{b_x}} dx \int_{a_y}^{{b_y}} f(x, y) dy \\
\xlongequal{极坐标}& \int_{a_\theta}^{{b_\theta}} d\theta \int_{a_r}^{{b_r}} f(r\cos \theta, r\sin \theta) rdr \\
\end{aligned}
\sum_{i=1}^\infty\sum_{j=1}^\infty f\left(\frac{i}{n}, \frac{j}{n}\right) \frac{1}{n^2} = \int_0^1\int_0^1 f(x, y) dxdy
三重积分
三重积分可以化为累次积分的定积分，表示空间区域 以密度 \rho=f(x, y, z)
\begin{aligned}
& \iiint\limits_{\Omega} f(x, y, z) dv \\
\xlongequal{笛卡尔坐标}& \int_{a_x}^{{b_x}} dx \int_{a_y}^{{b_y}} dy \int_{a_z}^{{b_z}} f(x, y, z) dz\\
\xlongequal{笛卡尔坐标}& \int_{a_z}^{{b_z}} dz \iint\limits_{D} f(x, y， z) d\sigma \\
\xlongequal{柱坐标}& \iiint\limits_\Omega f(r\cos \theta, r\sin \theta, z) rdr d\theta \\
\xlongequal{球坐标}& \iiint\limits_\Omega f(r\sin\varphi \cos \theta, r \sin\varphi \sin \theta, r\cos\varphi) \cdot r^2\sin\varphi dr d \varphi d\theta \\
\end{aligned}
曲面面积公式（第一型曲面积分）
由方程 z=z(x, y) \Sigma
S = \iint\limits_D \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} dxdy
若曲面以 \rho = f(x, y, z(x, y))
M = \iint\limits_D f(x, y, z(x, y)) \cdot \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} dxdy
形心与重心
设平面薄片 D \rho(x, y) D
\begin{aligned}
\overline{x} = \frac{\displaystyle \iint\limits_D x \rho(x, y)d \sigma}{\displaystyle \iint\limits_D \rho(x, y)d \sigma} \\
\overline{y} = \frac{\displaystyle \iint\limits_D y \rho(x, y)d \sigma}{\displaystyle \iint\limits_D \rho(x, y)d \sigma} \\
\end{aligned}
设空间物体 \Omega p(x,y,z) \Omega
\begin{aligned}
\overline{x} = \frac{\displaystyle \iiint\limits_\Omega x \rho(x, y, z)dv}{\displaystyle \iiint\limits_\Omega \rho(x, y, z)d v} \\
\overline{y} = \frac{\displaystyle \iiint\limits_\Omega y \rho(x, y, z)d v}{\displaystyle \iiint\limits_\Omega \rho(x, y, z)d v} \\
\overline{z} = \frac{\displaystyle \iiint\limits_\Omega z \rho(x, y, z)dv}{\displaystyle \iiint\limits_\Omega \rho(x, y, z)dv} \\
\end{aligned}
以上两种重心坐标若 \rho \equiv 1
有必要解释一下重心：首先找一根细绳，在物体上找一点，用绳悬挂，待物体静止后，通过悬挂点连一条竖直线。在该竖直线外再找一点悬挂，两条竖直线的交点就是不规则物体的重心。
第二型曲线积分
平面第二型曲线积分
若平面有 力场 可以表示为 \vec{f}(x, y) = P(x, y) \cdot \vec{i} + Q(x, y) \cdot \vec{j} P Q x,y \vec{f} L
\begin{aligned}
W =& \int_L P(x, y)dx + \int_L Q(x, y)dy \\
=& \int_L P(x, y)dx + Q(x, y)dy \\
\end{aligned}
格林公式
若曲线 L D
\begin{aligned}
& \oint_L P(x, y)dx + Q(x, y)dy \\
\xlongequal{格林公式}& \iint\limits_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) d\sigma \\
\end{aligned}
若 \displaystyle \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 任一围成的区域包含所有不连续点的曲线 ，再进行积分。一般不连续点为 原点 ，若围成的区域都有定义，则积分为 0
空间第二型曲线积分
若空间有 力场 可以表示为 \vec{f}(x, y, z) = P(x, y, z) \cdot \vec{i} + Q(x, y, z) \cdot \vec{j} + R(x, y, z) \cdot \vec{k} P,Q,R x,y,z \vec{f} L
\begin{aligned}
W =& \int_L P(x, y, z)dx + \int_L Q(x, y, z)dy + \int_L R(x, y, z)dz \\
=& \int_L P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz \\
\end{aligned}
斯托克斯公式
若曲线 L \Sigma
\begin{aligned}
& \oint_L P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz \\
\xlongequal{1}& \iint\limits_\Sigma \begin{vmatrix}
dydz & dzdx & dxdy \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
P & Q & R \\
\end{vmatrix} \\
=& \iint\limits_\Sigma
\left(
\frac{\partial R}{\partial y} -
\frac{\partial Q}{\partial z}
\right) dydz +
\left(
\frac{\partial P}{\partial z} -
\frac{\partial R}{\partial x}
\right) dzdx +
\left(
\frac{\partial Q}{\partial x} -
\frac{\partial P}{\partial y}
\right) dxdy \\
\xlongequal{2}& \iint\limits_\Sigma \begin{vmatrix}
\cos\alpha & \cos\beta & \cos\gamma \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
P & Q & R \\
\end{vmatrix} dS \\
\end{aligned}
其中 \vec{n} = (\cos\alpha, \cos\beta, \cos\gamma) \Sigma (x, y, z) 单位法向量 ，有人叫 方向余弦 。
以上有比较解释一下：
斯托克斯公式将空间第二型曲线积分 化成 第二型曲面积分
斯托克斯公式将空间第二型曲线积分 化成 第一型曲面积分
第二型曲面积分 也可以化成 第一型曲面积分
第二型曲面积分
若曲面 \Sigma 散度 （流量场） 可以表示为 \vec{f}(x, y, z) = P(x, y, z) \cdot \vec{i} + Q(x, y, z) \cdot \vec{j} + R(x, y, z) \cdot \vec{k} P, Q, R x,y, z \vec{f} \Sigma 通量 （流量）：
\begin{aligned}
I =& \iint\limits_\Sigma P dydz + \iint\limits_\Sigma Q dzdx + \iint\limits_\Sigma R dxdy \\
=& \iint\limits_\Sigma P dydz + Q dzdx + R dxdy \\
\xlongequal{化第一型}& \iint\limits_\Sigma (\cos\alpha P + \cos\beta dzdx + \cos\gamma R) dS \\
\xlongequal{化二重积分}& \iint\limits_D (\cos\alpha P + \cos\beta dzdx + \cos\gamma R) \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} dxdy \\
\xlongequal{化二重积分}& \iint\limits_D (-\frac{\partial z}{\partial x} P - \frac{\partial z}{\partial y}Q+ R) d\sigma\\
\end{aligned}
其中：
\vec{n} = (\cos\alpha, \cos\beta, \cos\gamma) \Sigma (x, y, z) 单位法向量 。D \Sigma xOy
以上积分，均为曲面 \Sigma
高斯公式
若积分曲面 \Sigma \Omega
\begin{aligned}
I =& \iint\limits_\Sigma P dydz + Q dzdx + R dxdy \\
\xlongequal{高斯公式}& \iiint\limits_\Omega \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}dv \\
\end{aligned}
若 \displaystyle \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = 0 任一围成的区域包含所有不连续点的曲面 ，再进行积分。一般不连续点为 原点，若围成的区域都有定义，则积分为 0
雅可比行列式
\begin{aligned}
x =& r \cos \theta \\
y =& r \sin \theta \\
J =& (x, y)^T\left(\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}\right)
= \left[
\begin{array}{cc}
\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta}\\\\
\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\\
\end{array}
\right] = \left[
\begin{array}{cc}
\cos \theta & -r\sin \theta\\\\
\sin \theta & r\cos \theta \\
\end{array}
\right]\\
| J | =& r\cos^2\theta + r\sin^2\theta = r \\
dxdy =& |J|drd\theta = r drd\theta
\end{aligned}
Γ 函数
\begin{aligned}
\Gamma(\alpha) &= \int_0^{+\infty} t^{\alpha-1}e^{-t} dt & (t>0) \\
\Gamma(\alpha) &= 2\int_0^{+\infty} t^{2\alpha-1}e^{-t^2} dt & (t>0)
\end{aligned}
\begin{aligned}
\Gamma(x + 1) &= x\Gamma(x) \\
\Gamma(1) &= 1 \\
\Gamma(n + 1) &= n! \\
\Gamma\left(\frac{1}{2}\right) &= \sqrt{\pi} \\
\Gamma(1 - z)\Gamma(z) &= \frac{\pi}{\sin(\pi z)}
\end{aligned}
\int_0^{+\infty} e^{-t^2} dt = \frac{\sqrt{\pi}}{2}
华莱士公式
\begin{aligned}
I_n &= \int_{0}^{\frac{\pi}{2}} \sin^n(x)dx = \int_{0}^{\frac{\pi}{2}} \cos^n(x)dx \\
&=\begin{cases}
\frac{n-1}{n} \cdot\frac{n-3}{n-2} \cdot \cdots \cdot \frac{3}{4}\cdot\frac{1}{2} \cdot \frac{\pi}{2} & (n=even)\\
\frac{n-1}{n} \cdot\frac{n-3}{n-2} \cdot \cdots \cdot \frac{4}{5}\cdot\frac{2}{3} & (n=odd)\\
\end{cases}
\end{aligned}
B 函数 β
\begin{aligned}
B(x, y) =& \int_0^1t^{x - 1}(1 - t)^{y - 1} dt \\
B(x, y) =& \int_0^{+\infty} \frac{t^{x - 1}}{(1 + t)^{x + y}} dt \\
B(x, y) =& \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)}\\
B(x, y) =& \frac{x - 1}{x + y} B(x - 1, y) = \frac{y - 1}{x + y} B(x, y - 1)\\
\end{aligned}
傅里叶级数
\begin{aligned}
S(x) &= \frac{a_0}{2} + \sum_{n=1}^\infty(a_n\cos\frac{n\pi x}{l} + b_n\sin\frac{n\pi x}{l}) \\
a_0 &= \frac{1}{l}\int_{-l}^{l} f(x)dx \\
a_n &= \frac{1}{l}\int_{-l}^{l} f(x)\cos \frac{n\pi x}{l} dx \\
b_n &= \frac{1}{l}\int_{-l}^{l} f(x)\sin \frac{n\pi x}{l} dx \\
\end{aligned}
使用计算技巧
有理函数积分裂项
来源：https://www.youtube.com/watch?v=ljzWO23htVs
求谁 、挡谁 、代谁 、特解
\begin{aligned}
y &= \int \frac{1}{(x^2-x-2)(x-1)^2} dx
= \frac{1}{(x+1)(x-2)(x-1)^2}dx \\
f(x) &= \frac{1}{(x+1)(x-2)(x-1)^2}
= \frac{a}{(x-1)^2} + \frac{b}{x-1} + \frac{m}{x + 1} + \frac{n}{x - 2} \\
a &= \lim_{x \to 1} (x-1)^2f(x)
= \lim_{x \to 1} \frac{1}{(x+1)(x-2)} = -\frac{1}{2} \\
b &= \lim_{x \to 1}\frac{d[(x - 1)^2f(x)]}{dx}
= \lim_{x \to 1}\frac{1-2x}{[(x+1)(x-2)]^2} = - \frac{1}{4} \\
n &= \lim_{x\to2}(x - 2)f(x) = \lim_{x \to 2}\frac{1}{(x+1)(x-1)^2} = \frac{1}{3} \\
m &= \lim_{x \to -1} (x + 1) f(x)
= \lim_{x \to -1}\frac{1}{(x-2)(x-1)^2} = -\frac{1}{12} \\
y &= \int \left[ \frac{-\frac{1}{2}}{(x-1)^2} + \frac{-\frac{1}{4}}{x-1} + \frac{-\frac{1}{12}}{x + 1} + \frac{\frac{1}{3}}{x - 2}\right] dx \\
&= \frac{1}{2(x-1)} - \frac{\ln|x-1|}{4}
- \frac{\ln|x+1|}{12} + \frac{\ln|x-2|}{3} + C
\end{aligned}
非奇非偶函数的偶倍奇零
来源：https://www.youtube.com/watch?v=_XAzSUSnr9s
\int_{-a}^{a} f(x) dx = \int_0^{a} [f(x) + f(-x)] dx
微分算子法求 y^*
来源：https://www.youtube.com/watch?v=GN_zOiODG40
基础规定：
D 求导 \frac{1}{D} 积分
\begin{aligned}
y'' + py' + qy = f(x)\\
y^* = \frac{1}{D^2 + pD + q}f(x)
\end{aligned}
f(x) = ae^{kx}
见到 D k
如果分母为0，则提出一个 x
式子中不含 D
f(x) = \sin ax f(x) = \cos ax
见到 D^2 -a^2
如果没有 D^2 D D^2
出现 \frac{1}{D}
式子中不含 D
f(x) = P_n(x)
分子等比级数和展开
如果分母中没有1，则提出\frac{1}{D}
\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \ (-1 < x < 1)
f(x) = e^{kx}y(x)
将特解写成如下形式：
y^* = e^{kx}\frac{1}{(D+k)^2 + p(D+k) + q}y(x)
然后再利用上面的几种方法进行计算
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