大学数学公式集-微积分

“微积分是近代数学中最伟大的成就,对它的重要性无论做怎样的估计都不会。”— 冯·诺伊曼

“在一切理论成就中,未必再有什么像17世纪下半叶微积分的发明那样被看作人类精神的最高胜利了!” — 恩格斯


前言

十七世纪下半叶,微积分的创立,极大地推动了数学的发展。过去很多用初等数学无法解决的问题,运用微积分,这些问题往往迎刃而解,显示出微积分学的非凡威力。微积分是继欧几里得几何之后,数学界的一个最大的创造。
微积分是大学数学的必修课,但还不能代表大学数学的全部内容,后面还会介绍 线性代数、数论、群论、图论、概率论等内容。
中学数学的知识可以查看链接:中学数学公式集

极限与微分

基本定义

名称表达式
导数 f ′ ( x ) = d f ( x ) d x = lim ⁡ ε → 0 f ( x + ε ) − f ( x ) ε f'(x)=\frac{df(x)}{dx}=\lim_{\varepsilon \rightarrow 0}\frac {f(x+\varepsilon)-f(x)}{\varepsilon} f(x)=dxdf(x)=ε0limεf(x+ε)f(x)
链式法则 f ( g ( x ) ) = f ′ ( g ( x ) ) ⋅ g ′ ( x ) f(g(x))=f'(g(x)) \cdot g'(x) f(g(x))=f(g(x))g(x)
洛必达法则 lim ⁡ x → a f ( x ) g ( x ) = lim ⁡ x → a f ′ ( x ) g ′ ( x ) , \lim_{x \rightarrow a} \frac{f(x)}{g(x)}=\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}, xalimg(x)f(x)=xalimg(x)f(x), f ( a ) = g ( a ) = 0 f(a)=g(a)=0 f(a)=g(a)=0 ∣ f ( a ) ∣ = ∣ g ( a ) ∣ = + ∞ |f(a)|=|g(a)|=+\infty f(a)=g(a)=+
微积分基本定理 f ( b ) − f ( a ) = ∫ a b f ′ ( x ) d x f(b)-f(a)=\int_a^b f'(x)dx f(b)f(a)=abf(x)dx
分部积分法 ∫ f ( x ) g ′ ( x ) d x = ∫ f ( x ) d g ( x ) = f ( x ) g ( x ) − ∫ g ( x ) d f ( x ) \int f(x)g'(x) dx=\int f(x)dg(x)=f(x)g(x)-\int g(x)df(x) f(x)g(x)dx=f(x)dg(x)=f(x)g(x)g(x)df(x)

极限

极限方法
lim ⁡ n → + ∞ ( 1 + x n ) n = e x \lim_{n \rightarrow +\infty} (1+\frac{x}{n})^n = e^x n+lim(1+nx)n=ex ( 1 + 1 n ) n = ∑ k = 0 n C n k 1 n k = > lim ⁡ n → + ∞ ∑ k = 0 n n k k ! 1 n k = ∑ k = 0 ∞ 1 n ! (1+\frac{1}{n})^n=\sum_{k=0}^nC_n^k\frac{1}{n^k}=>\lim_{n \rightarrow +\infty}\sum_{k=0}^n\frac{n^k}{k!}\frac{1}{n^k}=\sum_{k=0}^\infty\frac{1}{n!} (1+n1)n=k=0nCnknk1=>n+limk=0nk!nknk1=k=0n!1
lim ⁡ x → 0 sin ⁡ ( x ) x = 1 \lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1 x0limxsin(x)=1 sin ⁡ ( x ) ≤ x ≤ tan ⁡ ( x ) \sin(x) \le x \le \tan(x) sin(x)xtan(x)
lim ⁡ n → + ∞ ( ∑ k = 1 n 1 k − ln ⁡ ( n ) ) = γ \lim_{n \rightarrow +\infty} (\sum_{k=1}^n \frac{1}{k} - \ln(n)) = \gamma n+lim(k=1nk1ln(n))=γ欧拉-马斯克若尼常数
lim ⁡ n → 0 2 π n ( n / e ) n n ! = 1 \lim_{n \rightarrow 0}\frac{\sqrt{2\pi n} (n/e) ^n }{n!}=1 n0limn!2πn (n/e)n=1斯特林公式

导数

原函数导数
f ( x ) g ( x ) f(x)g(x) f(x)g(x) f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) f'(x)g(x) + f(x)g'(x) f(x)g(x)+f(x)g(x)
f ( x ) g ( x ) \frac {f(x)}{g(x)} g(x)f(x) f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g 2 ( x ) \frac {f'(x)g(x) - f(x)g'(x) }{g^2(x)} g2(x)f(x)g(x)f(x)g(x)
f ( x ) g ( x ) {f(x)}^{g(x)} f(x)g(x) ( g ( x ) f ′ ( x ) f ( x ) + g ′ ( x ) l n ( f ( x ) ) ) f ( x ) g ( x ) (\frac{g(x)f'(x)}{f(x)}+g'(x)ln(f(x))){f(x)}^{g(x)} (f(x)g(x)f(x)+g(x)ln(f(x)))f(x)g(x)
x a x^a xa a x a − 1 ax^{a-1} axa1
a x a^x ax a x ln ⁡ ( a ) a^x\ln(a) axln(a)
exp ⁡ ( x ) \exp(x) exp(x) exp ⁡ ( x ) \exp(x) exp(x)
ln ⁡ ( x ) \ln(x) ln(x) 1 x \frac{1}{x} x1
sin ⁡ ( x ) \sin(x) sin(x) cos ⁡ ( x ) \cos(x) cos(x)
cos ⁡ ( x ) \cos(x) cos(x) − sin ⁡ ( x ) -\sin(x) sin(x)
tan ⁡ ( x ) \tan(x) tan(x) 1 cos ⁡ 2 ( x ) \frac{1}{\cos^2(x)} cos2(x)1
arccos ⁡ ( x ) , arcsin ⁡ ( x ) \arccos(x) , \arcsin(x) arccos(x),arcsin(x) 1 1 − x 2 \frac{1}{\sqrt{1-x^2}} 1x2 1
arctan ⁡ ( x ) \arctan(x) arctan(x) 1 1 + x 2 \frac{1}{1+x^2} 1+x21

泰勒级数

f ( x + a ) = ∑ n = 0 ∞ f ( n ) ( a ) n ! x n = f ( a ) + f ′ ( a ) x + f ′ ′ ( a ) 2 ! x 2 + f ′ ′ ′ ( a ) 3 ! x 3 + . . . f(x+a)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}x^n=f(a)+f'(a)x+\frac{f''(a)}{2!}x^2+\frac{f'''(a)}{3!}x^3+... f(x+a)=n=0n!f(n)(a)xn=f(a)+f(a)x+2!f′′(a)x2+3!f′′′(a)x3+...

函数泰勒级数展开通式收敛半径
( 1 + x ) n (1+x)^n (1+x)n 1 + n x + n ( n − 1 ) 2 x 2 + n ( n − 1 ) ( n − 2 ) 3 ! x 3 + n ( n − 1 ) ( n − 2 ) ( n − 3 ) 4 ! x 4 + . . . 1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\frac{n(n-1)(n-2)(n-3)}{4!}x^4+... 1+nx+2n(n1)x2+3!n(n1)(n2)x3+4!n(n1)(n2)(n3)x4+... ∑ k = 0 ∞ C n k k ! x k \sum_{k=0}^\infty \frac{C_n^k}{k!}x^k k=0k!Cnkxk 1 1 1
e x e^x ex 1 + x + 1 2 x 2 + 1 3 ! x 3 + 1 4 ! x 4 + . . . 1+x+\frac{1}{2}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+... 1+x+21x2+3!1x3+4!1x4+... ∑ k = 0 ∞ 1 k ! x k \sum_{k=0}^\infty \frac{1}{k!}x^k k=0k!1xk + ∞ +\infty +
ln ⁡ ( 1 + x ) \ln (1+x) ln(1+x) x − 1 2 x 2 + 1 3 x 3 − 1 4 x 4 + . . . x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+... x21x2+31x341x4+... ∑ k = 0 ∞ ( − 1 ) k + 1 k x k \sum_{k=0}^\infty \frac{(-1)^{k+1}}{k}x^k k=0k(1)k+1xk1
sin ⁡ x \sin x sinx x − 1 3 ! x 3 + 1 5 ! x 5 − 1 7 ! x 7 + . . . x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+... x3!1x3+5!1x57!1x7+... ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! x 2 k + 1 \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k+1} k=0(2k+1)!(1)kx2k+1 + ∞ +\infty +
cos ⁡ x \cos x cosx 1 − 1 2 ! x 2 + 1 4 ! x 4 − 1 6 ! x 6 + . . . 1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+... 12!1x2+4!1x46!1x6+... ∑ k = 0 ∞ ( − 1 ) k ( 2 k ) ! x 2 k \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!}x^{2k} k=0(2k)!(1)kx2k + ∞ +\infty +
arcsin ⁡ x \arcsin x arcsinx x + 1 6 x 3 + 3 40 x 5 + 5 112 x 7 + 35 1152 x 9 + . . . x+\frac{1}{6}x^3+\frac{3}{40}x^5+\frac{5}{112}x^7+\frac{35}{1152}x^9+... x+61x3+403x5+1125x7+115235x9+... ∑ k = 0 ∞ ( 2 n ) ! 4 n ( n ! ) 2 ( 2 n + 1 ) x 2 n + 1 \sum_{k=0}^\infty\frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1} k=04n(n!)2(2n+1)(2n)!x2n+11
arctan ⁡ x \arctan x arctanx x − 1 3 x 3 + 1 5 x 5 − 1 7 x 7 + . . . x-\frac{1}{3}x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+... x31x3+51x571x7+... ∑ k = 0 ∞ ( − 1 ) k 2 k + 1 x 2 k + 1 \sum_{k=0}^\infty\frac{(-1)^k}{2k+1}x^{2k+1} k=02k+1(1)kx2k+11

欧拉-麦克劳林(Euler-Maclaurin)求和公式

∑ n = a b f ( n ) = ∫ a b f ( x ) d x + f ( a ) + f ( b ) 2 + ∑ k = 1 m B 2 k ( 2 k ) ! [ f ( 2 k − 1 ) ( b ) − f ( 2 k − 1 ) ( a ) ] + R m \sum^b_{n=a}f(n)=\int_a^bf(x)dx+\frac{f(a)+f(b)}{2}+\sum^m_{k=1} \frac{B_{2k}}{(2k)!}[f^{(2k-1)}(b)-f^{(2k-1)}(a)]+R_m n=abf(n)=abf(x)dx+2f(a)+f(b)+k=1m(2k)!B2k[f(2k1)(b)f(2k1)(a)]+Rm
其中 B n B_n Bn是伯努利数, f 2 k − 1 f^{2k-1} f2k1表示 f ( x ) f(x) f(x)的 2k-1 阶导数。

如果固定a=1,则可以得到以下式子:
∑ k = 1 n f ( k ) = ∫ f ( n ) d n + C + f ( n ) / 2 + f ′ ( n ) / 12 + O ( f ′ ′ ′ ( n ) ) \sum^n_{k=1}f(k)=\int f(n)dn+C+f(n)/2+f'(n)/12+O(f'''(n)) k=1nf(k)=f(n)dn+C+f(n)/2+f(n)/12+O(f′′′(n))

∑ k = 1 n f ( k ) \sum^n_{k=1} f(k) k=1nf(k) ∫ f ( n ) d n \int f(n)dn f(n)dn C C C f ( n ) / 2 f(n)/2 f(n)/2 f ′ ( n ) / 12 f'(n)/12 f(n)/12 O ( f ′ ′ ′ ( n ) ) O(f'''(n)) O(f′′′(n))备注
∑ k = 1 n k \sum^n_{k=1} k k=1nk n 2 / 2 n^2/2 n2/20 n / 2 n/2 n/200
∑ k = 1 n k 2 \sum^n_{k=1} k^2 k=1nk2 n 3 / 3 n^3/3 n3/30 n 2 / 2 n^2/2 n2/2 n / 6 n/6 n/60
∑ k = 1 n k 3 \sum^n_{k=1} k^3 k=1nk3 n 4 / 4 n^4/4 n4/40 n 3 / 2 n^3/2 n3/2 n 2 / 4 n^2/4 n2/40
∑ k = 1 n 1 / k \sum^n_{k=1} 1/k k=1n1/k ln ⁡ n \ln n lnn γ \gamma γ 1 / 2 n 1/2n 1/2n − 1 / 12 n 2 -1/12n^2 1/12n2 O ( 1 / n 4 ) O(1/n^4) O(1/n4) γ = 0.577215664... \gamma=0.577215664... γ=0.577215664...
∑ k = 1 n 1 / k 2 \sum^n_{k=1} 1/k^2 k=1n1/k2 − 1 / n -1/n 1/n ζ ( 2 ) \zeta(2) ζ(2) 1 / 2 n 2 1/2n^2 1/2n2 − 1 / 6 n 3 -1/6n^3 1/6n3 O ( 1 / n 5 ) O(1/n^5) O(1/n5) ζ ( 2 ) = π 2 / 6 \zeta(2)=\pi^2/6 ζ(2)=π2/6
∑ k = 1 n k \sum^n_{k=1} \sqrt{k} k=1nk 2 n n / 3 2n\sqrt{n}/3 2nn /3 ζ ( − 0.5 ) \zeta(-0.5) ζ(0.5) n / 2 \sqrt{n}/2 n /2 1 / 24 n 1/24\sqrt{n} 1/24n O ( 1 / n 2.5 ) O(1/n^{2.5}) O(1/n2.5) ζ ( − 0.5 ) = − 0.2078862249 \zeta(-0.5)=-0.2078862249 ζ(0.5)=0.2078862249
∑ k = 1 n 1 / k \sum^n_{k=1} 1/\sqrt{k} k=1n1/k 2 n 2\sqrt{n} 2n ζ ( 0.5 ) \zeta(0.5) ζ(0.5) 1 / 2 k 1/2\sqrt{k} 1/2k − 1 / 6 n 3 -1/6n^3 1/6n3 O ( 1 / n 3.5 ) O(1/n^{3.5}) O(1/n3.5) ζ ( 0.5 ) = − 1.4603545088... \zeta(0.5)=-1.4603545088... ζ(0.5)=1.4603545088...
∑ k = 1 n ln ⁡ k \sum^n_{k=1} \ln k k=1nlnk n ( ln ⁡ n − 1 ) n(\ln n-1) n(lnn1) − ζ ′ ( 0 ) -\zeta'(0) ζ(0) ln ⁡ n / 2 \ln n/2 lnn/2 1 / 12 n 1/12n 1/12n O ( 1 / n 3 ) O(1/n^3) O(1/n3) ζ ′ ( 0 ) = − ln ⁡ 2 π \zeta'(0)=-\ln \sqrt{2\pi} ζ(0)=ln2π
∑ k = 1 n k ln ⁡ k \sum^n_{k=1} k\ln k k=1nklnk n 2 ( 2 ln ⁡ n − 1 ) / 4 n^2(2\ln{n}-1)/4 n2(2lnn1)/4 − ζ ′ ( − 1 ) -\zeta'(-1) ζ(1) n ln ⁡ n / 2 n\ln n/2 nlnn/2 ( ln ⁡ n + 1 ) / 12 (\ln n+1)/12 (lnn+1)/12 O ( 1 / n 2 ) O(1/n^2) O(1/n2) ζ ′ ( − 1 ) = − 0.16542113... \zeta'(-1)=-0.16542113... ζ(1)=0.16542113...
∑ k = 1 n ln ⁡ k / k \sum^n_{k=1} \ln k/k k=1nlnk/k ( ln ⁡ n ) 2 / 2 (\ln{n})^2/2 (lnn)2/2 γ 1 \gamma_1 γ1 ln ⁡ n / 2 n \ln n/2n lnn/2n ( 1 − ln ⁡ n ) / 12 n 2 (1-\ln n)/12n^2 (1lnn)/12n2 O ( ln ⁡ n / n 4 ) O(\ln n/n^4) O(lnn/n4) γ 1 = − 0.0728158454... \gamma_1=-0.0728158454... γ1=0.0728158454...
∑ k = 1 n ln ⁡ k / k \sum^n_{k=1} \ln k/\sqrt{k} k=1nlnk/k 2 n ( ln ⁡ n − 2 ) 2\sqrt{n}(\ln{n}-2) 2n (lnn2) − ζ ′ ( 0.5 ) -\zeta'(0.5) ζ(0.5) ln ⁡ n / 2 n \ln n/2\sqrt{n} lnn/2n ( 2 − ln ⁡ n ) / 24 n n (2-\ln n)/24n\sqrt{n} (2lnn)/24nn O ( ln ⁡ n / n 3.5 ) O(\ln n/n^{3.5}) O(lnn/n3.5) ζ ′ ( 0.5 ) = − 3.92264613... \zeta'(0.5)=-3.92264613... ζ(0.5)=3.92264613...
∑ k = 1 n k ln ⁡ k \sum^n_{k=1} \sqrt{k}\ln k k=1nk lnk n n ( 6 ln ⁡ n − 4 ) / 9 n\sqrt{n}(6\ln{n}-4)/9 nn (6lnn4)/9 − ζ ′ ( − 0.5 ) -\zeta'(-0.5) ζ(0.5) n ln ⁡ n / 2 \sqrt{n}\ln n/2 n lnn/2 ( 2 + ln ⁡ n ) / 2 n (2+\ln n)/2\sqrt{n} (2+lnn)/2n O ( ln ⁡ n / n 2.5 ) O(\ln n/n^{2.5}) O(lnn/n2.5) ζ ′ ( − 0.5 ) = − 0.36085436... \zeta'(-0.5)=-0.36085436... ζ(0.5)=0.36085436...

无穷级数求和

级数和方法
ζ ( s ) = ∏ p r i m e   p 1 1 − p − s = ∑ n = 1 ∞ 1 n s \zeta(s)=\prod_{prime\ p}\frac{1}{1-p^{-s}}=\sum_{n=1}^\infty \frac{1}{n^s} ζ(s)=prime p1ps1=n=1ns1黎曼 ζ \zeta ζ函数
ζ ( 2 ) = ∑ n = 1 ∞ 1 n 2 = 1 + 1 4 + 1 9 + 1 16 + . . . = π 2 6 \zeta(2)=\sum_{n=1}^\infty \frac{1}{n^2}=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...=\frac {\pi^2}{6} ζ(2)=n=1n21=1+41+91+161+...=6π2巴塞尔问题, sin ⁡ x x = ∏ n = 1 ∞ ( 1 − x 2 n 2 π 2 ) \frac{\sin x}{x}=\prod_{n=1}^\infty (1-\frac{x^2}{n^2\pi^2}) xsinx=n=1(1n2π2x2)
∑ n = 1 ∞ ( − 1 ) n + 1 n = 1 − 1 2 + 1 3 − 1 4 + . . . = ln ⁡ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...=\ln 2 n=1n(1)n+1=121+3141+...=ln2 ln ⁡ ( 1 + x ) \ln(1+x) ln(1+x)的泰勒展开,在 x = 1 x=1 x=1的情况
∑ n = 0 ∞ ( − 1 ) n 2 n + 1 = 1 − 1 3 + 1 5 − 1 7 + . . . = π 4 \sum_{n=0}^\infty \frac{(-1)^{n}}{2n+1}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...=\frac{\pi}{4} n=02n+1(1)n=131+5171+...=4π arctan ⁡ ( x ) \arctan(x) arctan(x)的泰勒展开,在 x = 1 x=1 x=1的情况

积分

不定积分

公式方法
∫ x n d x = x n + 1 n + 1 + C n ≠ − 1 \int x^n dx=\frac{x^{n+1}}{n+1}+C\quad n\ne -1 xndx=n+1xn+1+Cn=1
∫ 1 x d x = ln ⁡ x + C \int \frac{1}{x} dx=\ln x+C x1dx=lnx+C
∫ 1 x 2 − a 2 d x = 1 2 a ln ⁡ ∣ x − a x + a ∣ + C \int \frac{1}{x^2-a^2} dx=\frac{1}{2a} \ln |\frac{x-a}{x+a}| +C x2a21dx=2a1lnx+axa+C 1 x 2 − a 2 = 1 2 a ( 1 x − a − 1 x + a ) \frac{1}{x^2-a^2}=\frac{1}{2a}(\frac{1}{x-a}-\frac{1}{x+a}) x2a21=2a1(xa1x+a1)
∫ 1 x 2 + a 2 d x = 1 a arctan ⁡ ( x a ) + C \int \frac{1}{x^2+a^2} dx=\frac{1}{a} \arctan(\frac{x}{a})+C x2+a21dx=a1arctan(ax)+C ( arctan ⁡ x ) ′ = 1 x 2 + 1 (\arctan x)'=\frac{1}{x^2+1} (arctanx)=x2+11
∫ 1 ( x 2 + a 2 ) 2 d x = 1 2 a 3 arctan ⁡ x a + x 2 a 2 ( a 2 + x 2 ) + C \int \frac{1}{(x^2+a^2)^2} dx=\frac{1}{2a^3}\arctan \frac{x}{a}+\frac{x}{2a^2(a^2+x^2)}+C (x2+a2)21dx=2a31arctanax+2a2(a2+x2)x+C ∫ ∂ ∂ a 1 x 2 + a 2 d x = ∂ ∂ a ( 1 a arctan ⁡ x a ) + C \int \frac{\partial}{\partial a} \frac{1}{x^2+a^2} dx=\frac{\partial}{\partial a}(\frac{1}{a} \arctan \frac{x}{a})+C ax2+a21dx=a(a1arctanax)+C
∫ a 2 − x 2 d x = x a 2 − x 2 + a 2 4 arcsin ⁡ ( x a ) + C \int \sqrt{a^2-x^2}dx=x\sqrt{a^2-x^2}+\frac{a^2}{4}\arcsin(\frac{x}{a})+C a2x2 dx=xa2x2 +4a2arcsin(ax)+C x = a sin ⁡ t x=a\sin t x=asint
∫ x 2 + a d x = 1 2 ( x x 2 + a + a ln ⁡ ∣ x 2 + a ∣ ) + C \int \sqrt{x^2+a}dx=\frac{1}{2}(x\sqrt{x^2+a}+a\ln|\sqrt{x^2+a}|)+C x2+a dx=21(xx2+a +alnx2+a )+C x = a sinh ⁡ t x=a \sinh t x=asinht
∫ 1 a 2 − x 2 d x = arcsin ⁡ x a + C \int \frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \frac{x}{a}+C a2x2 1dx=arcsinax+C arcsin ⁡ ( x ) ′ = 1 1 − x 2 \arcsin(x)'=\frac{1}{\sqrt{1-x^2}} arcsin(x)=1x2 1
∫ 1 x 2 + a d x = ln ⁡ ∣ x + x 2 + a ∣ + C \int \frac{1}{\sqrt{x^2+a}}dx=\ln|x+\sqrt{x^2+a}|+C x2+a 1dx=lnx+x2+a +C a r c s i n h ( x ) ′ = 1 1 + x 2 arcsinh(x)'=\frac{1}{\sqrt{1+x^2}} arcsinh(x)=1+x2 1
∫ x x 2 + a d x = x 2 + a + C \int \frac{x}{\sqrt{x^2+a}}dx=\sqrt{x^2+a}+C x2+a xdx=x2+a +C t = x 2 + a t=x^2+a t=x2+a
∫ sin ⁡ ( x ) d x = − cos ⁡ ( x ) + C \int \sin(x) dx=-\cos(x)+C sin(x)dx=cos(x)+C
∫ cos ⁡ ( x ) d x = sin ⁡ ( x ) + C \int \cos(x) dx=\sin(x)+C cos(x)dx=sin(x)+C
∫ tan ⁡ ( x ) d x = − ln ⁡ ∣ cos ⁡ ( x ) ∣ + C \int \tan(x)dx=-\ln|\cos(x)|+C tan(x)dx=lncos(x)+C ∫ s i n ( x ) c o s ( x ) d x = ∫ − d c o s ( x ) c o s ( x ) \int \frac{sin(x)}{cos(x)}dx=\int \frac{-d cos(x)}{cos(x)} cos(x)sin(x)dx=cos(x)dcos(x)
∫ sin ⁡ 2 ( x ) d x = 1 4 ( 2 x − sin ⁡ 2 x ) + C \int \sin^2(x) dx= \frac{1}{4}(2x-\sin 2x)+C sin2(x)dx=41(2xsin2x)+C
∫ sin ⁡ 3 ( x ) d x = 1 12 ( cos ⁡ 3 x − 9 cos ⁡ x ) + C \int \sin^3(x) dx= \frac{1}{12}(\cos 3x-9\cos x)+C sin3(x)dx=121(cos3x9cosx)+C
∫ sin ⁡ ( x ) cos ⁡ ( x ) d x = 1 2 sin ⁡ 2 x + C \int \sin (x)\cos (x) dx= \frac{1}{2}\sin^2 x+C sin(x)cos(x)dx=21sin2x+C
∫ 1 sin ⁡ ( x ) d x = l n ∣ 1 − c o s ( x ) s i n ( x ) ∣ + C = ln ⁡ ∣ tan ⁡ x 2 ∣ + C \int \frac{1}{\sin(x)}dx=ln|\frac{1-cos(x)}{sin(x)}|+C=\ln|\tan\frac{x}{2}|+C sin(x)1dx=lnsin(x)1cos(x)+C=lntan2x+C ∫ s i n ( x ) d x s i n 2 ( x ) = ∫ d c o s ( x ) 1 − c o s 2 ( x ) \int\frac{sin(x)dx}{sin^2(x)}=\int\frac{dcos(x)}{1-cos^2(x)} sin2(x)sin(x)dx=1cos2(x)dcos(x)
∫ 1 a + cos ⁡ x d x = 1 1 − a 2 ln ⁡ tan ⁡ ( x / 2 ) 1 − a 2 + a + 1 tan ⁡ ( x / 2 ) 1 − a 2 − a − 1 + C \int \frac{1}{a+\cos x}dx=\frac{1}{\sqrt{1-a^2}}\ln \frac{\tan(x/2)\sqrt{1-a^2}+a+1}{\tan(x/2)\sqrt{1-a^2}-a-1}+C a+cosx1dx=1a2 1lntan(x/2)1a2 a1tan(x/2)1a2 +a+1+C换元法 t = tan ⁡ x 2 t=\tan \frac{x}{2} t=tan2x
∫ 1 sin ⁡ 2 ( x ) d x = − 1 tan ⁡ ( x ) + C \int \frac{1}{\sin^2(x)}dx=-\frac{1}{\tan(x)}+C sin2(x)1dx=tan(x)1+C
∫ x sin ⁡ ( x ) d x = sin ⁡ ( x ) − x cos ⁡ ( x ) + C \int x\sin(x)dx=\sin(x)-x\cos(x)+C xsin(x)dx=sin(x)xcos(x)+C分部积分法
∫ x cos ⁡ ( x ) d x = cos ⁡ ( x ) + x sin ⁡ ( x ) + C \int x\cos(x)dx= \cos(x)+x\sin(x)+C xcos(x)dx=cos(x)+xsin(x)+C分部积分法
∫ x 2 sin ⁡ ( x ) d x = 2 x sin ⁡ ( x ) − ( x 2 − 2 ) cos ⁡ ( x ) + C \int x^2\sin(x)dx= 2x\sin(x)-(x^2-2)\cos(x)+C x2sin(x)dx=2xsin(x)(x22)cos(x)+C分部积分法
∫ x 2 cos ⁡ ( x ) d x = 2 x cos ⁡ ( x ) + ( x 2 − 2 ) sin ⁡ ( x ) + C \int x^2\cos(x)dx= 2x\cos(x)+(x^2-2)\sin(x)+C x2cos(x)dx=2xcos(x)+(x22)sin(x)+C分部积分法
∫ x 3 cos ⁡ ( x ) d x = ( 3 x 2 − 6 ) cos ⁡ ( x ) + ( x 3 − 6 x ) sin ⁡ ( x ) + C \int x^3\cos(x)dx= (3x^2-6 )\cos(x) + (x^3-6x)\sin(x)+C x3cos(x)dx=(3x26)cos(x)+(x36x)sin(x)+C
∫ x 3 sin ⁡ ( x ) d x = ( 3 x 2 − 6 ) sin ⁡ ( x ) − ( x 3 − 6 x ) cos ⁡ ( x ) + C \int x^3 \sin(x)dx= (3x^2 - 6)\sin(x) - (x^3 - 6x)\cos(x) + C x3sin(x)dx=(3x26)sin(x)(x36x)cos(x)+C
∫ a x d x = 1 ln ⁡ ( a ) a x + C \int a^xdx=\frac{1}{\ln(a)}a^x+C axdx=ln(a)1ax+C a x = e l n ( a ) x a^x=e^{ln(a)x} ax=eln(a)x
∫ x e x d x = ( x − 1 ) e x + C \int xe^xdx=(x-1)e^x+C xexdx=(x1)ex+C分部积分法
∫ x 2 e x d x = ( x 2 − 2 x + 2 ) e x + C \int x^2e^xdx=(x^2-2x+2)e^x+C x2exdx=(x22x+2)ex+C分部积分法
∫ x 3 e x d x = ( x 3 − 3 x 2 + 6 x − 6 ) e x + C \int x^3e^xdx=(x^3-3x^2+6x-6)e^x+C x3exdx=(x33x2+6x6)ex+C分部积分法
∫ e a x sin ⁡ ( b x ) d x = a sin ⁡ ( b x ) − b cos ⁡ ( b x ) a 2 + b 2 e a x \int e^{ax}\sin(bx)dx=\frac{a\sin(bx)-b\cos(bx)}{a^2+b^2}e^{ax} eaxsin(bx)dx=a2+b2asin(bx)bcos(bx)eax分部积分法
∫ e a x cos ⁡ ( b x ) d x = a cos ⁡ ( b x ) + b sin ⁡ ( b x ) a 2 + b 2 e a x \int e^{ax}\cos(bx)dx=\frac{a\cos(bx)+b\sin(bx)}{a^2+b^2}e^{ax} eaxcos(bx)dx=a2+b2acos(bx)+bsin(bx)eax分部积分法
∫ ln ⁡ ( x ) d x = x ( ln ⁡ x − 1 ) + C \int \ln(x)dx=x(\ln x-1)+C ln(x)dx=x(lnx1)+C分部积分法
∫ x ln ⁡ ( x ) d x = 1 4 x 2 ( 2 ln ⁡ x − 1 ) + C \int x \ln(x)dx=\frac{1}{4}x^2(2\ln x-1)+C xln(x)dx=41x2(2lnx1)+C分部积分法

定积分

公式方法
∫ 0 ∞ x n e x d x = n ! \int_0^\infty x^ne^xdx=n! 0xnexdx=n!
∫ 0 π 2 sin ⁡ 2 n + 1 ( x ) d x = ( 2 n ) ! ! ( 2 n + 1 ) ! ! ∫ 0 π 2 sin ⁡ 2 n ( x ) d x = π 2 ( 2 n ) ! ! ( 2 n + 1 ) ! ! n ≥ 0 \int_0^{\frac{\pi}{2}} \sin^{2n+1}(x)dx=\frac{(2n)!!}{(2n+1)!!}\\\int_0^{\frac{\pi}{2}} \sin^{2n}(x)dx=\frac{\pi}{2}\frac{(2n)!!}{(2n+1)!!}\\n\ge0 02πsin2n+1(x)dx=(2n+1)!!(2n)!!02πsin2n(x)dx=2π(2n+1)!!(2n)!!n0
∫ 0 2 π sin ⁡ ( k x ) sin ⁡ ( m x ) d x = π δ k m ∫ 0 2 π cos ⁡ ( k x ) cos ⁡ ( m x ) d x = π δ k m ∫ 0 2 π sin ⁡ ( k x ) cos ⁡ ( m x ) d x = 0 k , m > 0 \int_0^{2\pi}\sin(kx)\sin(mx)dx=\pi\delta_{km}\\\int_0^{2\pi}\cos(kx)\cos(mx)dx=\pi\delta_{km}\\\int_0^{2\pi}\sin(kx)\cos(mx)dx=0\\k,m > 0 02πsin(kx)sin(mx)dx=πδkm02πcos(kx)cos(mx)dx=πδkm02πsin(kx)cos(mx)dx=0k,m>0 δ k m \delta_{km} δkm是克罗内克函数,仅当 k = m k=m k=m时值为1,其他情况值为0
∫ − ∞ ∞ e − x 2 d x = π \int_{-\infty}^\infty e^{-x^2}dx= \sqrt{\pi} ex2dx=π 极坐标变换 ∬ R 2 e − x 2 − y 2 d x d y = ∫ 0 ∞ ∫ 0 2 π e − r 2 r d θ d r \iint_{R^2} e^{-x^2-y^2}dxdy=\int_0^\infty \int_0^{2\pi}e^{-r^2}rd\theta dr R2ex2y2dxdy=002πer2rdθdr
∫ 0 ∞ e − x 2 cos ⁡ ( x ) d x = π 2 e − 1 4 \int_0^\infty e^{-x^2}\cos(x)dx={\frac{\sqrt\pi}{2}}e^{\frac{-1}{4}} 0ex2cos(x)dx=2π e41费曼积分法 ∫ 0 ∞ e − a x 2 cos ⁡ ( b x ) d x = π 4 a e − b 2 4 a \int_0^\infty e^{-ax^2}\cos(bx)dx=\sqrt{\frac{\pi}{4a}}e^{\frac{-b^2}{4a}} 0eax2cos(bx)dx=4aπ e4ab2
∫ − ∞ ∞ sin ⁡ ( x ) x d x = π \int_{-\infty}^\infty \frac{\sin(x)}{x}dx=\pi xsin(x)dx=π费曼积分法 ∫ 0 ∞ s i n ( x ) x e − a x d x = − arctan ⁡ ( a ) + π 2 \int_0^\infty \frac{sin(x)}{x}e^{-ax}dx=-\arctan(a)+\frac{\pi}{2} 0xsin(x)eaxdx=arctan(a)+2π
∫ 0 π ln ⁡ ( 1 + cos ⁡ x ) d x = π ln ⁡ 2 \int_0^\pi\ln(1+\cos x)dx=\pi\ln 2 0πln(1+cosx)dx=πln2费曼积分法 ∫ 0 π ln ⁡ ( 1 + a cos ⁡ x ) d x = π ln ⁡ ∣ 1 + 1 − a 2 2 ∣ \int_0^\pi\ln(1+a\cos x)dx=\pi\ln|\frac{1+\sqrt{1-a^2}}{2}| 0πln(1+acosx)dx=πln21+1a2

常微分方程

名称方程通解
分离变量法 d y d x = f ( x ) g ( y ) \frac{dy}{dx}=\frac{f(x)}{g(y)} dxdy=g(y)f(x) ∫ g ( y ) d y = ∫ f ( x ) d x + C \int g(y)dy = \int f(x) dx +C g(y)dy=f(x)dx+C
齐次微分方程 d y d x = f ( y x ) \frac{dy}{dx}=f(\frac{y}{x}) dxdy=f(xy) ∫ d u f ( u ) − u = ∫ d x x + C \int \frac{du}{f(u)-u} = \int \frac{ dx}{x} +C f(u)udu=xdx+C其中 u = y / x u=y/x u=y/x
一阶线性微分方程 d y d x + P ( x ) y = Q ( x ) \frac{dy}{dx}+P(x)y=Q(x) dxdy+P(x)y=Q(x) y = e − ∫ P ( x ) d x ( ∫ Q ( x ) e ∫ P ( x ) d x d x + C ) y=e^{-\int P(x)dx}(\int Q(x)e^{\int P(x)dx}dx+C) y=eP(x)dx(Q(x)eP(x)dxdx+C)
伯努利方程 d y d x + P ( x ) y = Q ( x ) y n \frac{dy}{dx}+P(x)y=Q(x)y^n dxdy+P(x)y=Q(x)yn z = y 1 − n , n ≠ 1 z=y^{1-n}, n \ne 1 z=y1n,n=1,则 d z d x + ( 1 − n ) P ( x ) z = ( 1 − n ) Q ( x ) \frac{dz}{dx}+(1-n)P(x)z=(1-n)Q(x) dxdz+(1n)P(x)z=(1n)Q(x)
常系数线性微分方程 y ′ ′ + b y ′ + c y = 0 y''+by'+cy=0 y′′+by+cy=0 y = { ( C 1 + C 2 x ) e p x , p = q C 1 e p x + C 2 e q x , p ≠ q ( C 1 cos ⁡ β x + C 2 sin ⁡ β x ) e α x , p = q ‾ = α + i β y=\begin{cases} (C_1+C_2x)e^{px}, p = q \\C_1e^{px}+C_2e^{qx}, p \ne q\\(C_1 \cos \beta x+ C_2 \sin \beta x)e^{\alpha x}, p = \overline q = \alpha + i\beta \end{cases} y= (C1+C2x)epx,p=qC1epx+C2eqx,p=q(C1cosβx+C2sinβx)eαx,p=q=α+iβ 其中 p , q p,q p,q是方程 x 2 + b x + c = 0 x^2+bx+c=0 x2+bx+c=0的根
欧拉方程 x 2 y ′ ′ + b x y ′ + c y = f ( x ) x^2y''+bxy'+cy=f(x) x2y′′+bxy+cy=f(x) x = e t x=e^t x=et,则 d y d x = 1 x d d t y , d 2 y d x 2 = 1 x 2 d d t ( d d t − 1 ) y , d 3 y d x 3 = 1 x 3 d d t ( d d t − 1 ) ( d d t − 2 ) y \frac{dy}{dx}=\frac{1}{x}\frac{d}{dt}y,\\ \frac{d^2y}{dx^2}=\frac{1}{x^2}\frac{d}{dt}(\frac{d}{dt}-1)y,\\ \frac{d^3y}{dx^3}=\frac{1}{x^3}\frac{d}{dt}(\frac{d}{dt}-1)(\frac{d}{dt}-2)y dxdy=x1dtdy,dx2d2y=x21dtd(dtd1)y,dx3d3y=x31dtd(dtd1)(dtd2)y 简化为常系数微分方程

多重积分、曲线与曲面积分

名称定义
二重积分极坐标转换 ∬ S f ( x , y )   d x   d y = ∬ S f ( r cos ⁡ θ , r sin ⁡ θ ) r   d r   d θ \iint_S f(x,y)\ dx\ dy=\iint_S f(r\cos \theta,r \sin \theta)r\ dr\ d\theta Sf(x,y) dx dy=Sf(rcosθ,rsinθ)r dr dθ
三重积分球坐标转换 ∭ V f ( x , y , z )   d x   d y   d z = ∭ V f ( r cos ⁡ θ sin ⁡ φ , r sin ⁡ θ sin ⁡ φ , r cos ⁡ φ ) r 2 sin ⁡ φ   d r   d θ   d φ \iiint_V f(x,y,z)\ dx\ dy\ dz=\iiint_V f(r\cos \theta \sin\varphi,r \sin \theta \sin\varphi, r\cos \varphi)r^2 \sin \varphi \ dr\ d\theta \ d\varphi Vf(x,y,z) dx dy dz=Vf(rcosθsinφ,rsinθsinφ,rcosφ)r2sinφ dr dθ dφ
第一类曲线积分 ∫ L f ( x , y ) d s = ∫ a b f [ x ( t ) , y ( t ) ] x ′ 2 ( t ) + y ′ 2 ( t ) d t \int_L f(x,y)ds=\int_a^b f[x(t),y(t)]\sqrt{x'^2(t)+y'^2(t)}dt Lf(x,y)ds=abf[x(t),y(t)]x′2(t)+y′2(t) dt
第二类曲线积分 ∫ L ( P ( x , y ) , Q ( x , y ) ) ⋅ d L ⃗ = ∫ L P d x + Q d y = ∫ a b [ P x ′ ( t ) + Q y ′ ( t ) ] d t \int_L (P(x,y),Q(x,y))\cdot d\vec L=\int_L Pdx+Qdy=\int_a^b [Px'(t)+Qy'(t)]dt L(P(x,y),Q(x,y))dL =LPdx+Qdy=ab[Px(t)+Qy(t)]dt d L ⃗ = v ⃗ d L d \vec L = \vec v dL dL =v dL 表示带方向的线微元, v ⃗ \vec v v 是线的单位向量
第一类曲面积分 ∬ S f ( x , y ) d S = ∬ D f ( x , y ) 1 + z ′ 2 ( x ) + z ′ 2 ( y ) d x d y \iint_S f(x,y)dS=\iint_D f(x,y)\sqrt{1+z'^2(x)+z'^2(y)}dxdy Sf(x,y)dS=Df(x,y)1+z′2(x)+z′2(y) dxdy
第二类曲面积分 ∬ S ( P , Q , R ) ⋅ d S ⃗ = ∬ D y z P d y d z − ∬ D x z Q d x d z + ∬ D x y R d x d z \iint_S (P,Q,R)\cdot d \vec S=\iint_{Dyz}Pdydz-\iint_{Dxz}Qdxdz+\iint_{Dxy}Rdxdz S(P,Q,R)dS =DyzPdydzDxzQdxdz+DxyRdxdz d S ⃗ = n ⃗ s d S d \vec S = \vec n_s dS dS =n sdS 表示带方向的曲面微元, n ⃗ s \vec n_s n s是单位法向量
哈密顿算子 ∇ = i ∂ ∂ x + j ∂ ∂ y + k ∂ ∂ z \nabla=i \frac\partial{\partial x} + j \frac\partial{\partial y} +k \frac\partial{\partial z} =ix+jy+kz
梯度 g r a d ( f ) = ∇ f = i ∂ f ∂ x + j ∂ f ∂ y + k ∂ f ∂ z grad(f)=\nabla f=i \frac{\partial f}{\partial x} + j \frac{\partial f}{\partial y} +k \frac{\partial f}{\partial z} grad(f)=f=ixf+jyf+kzf
散度 d i v ( f ⃗ ) = ∇ ⋅ f ⃗ = ∂ f x ∂ x + ∂ f y ∂ y + ∂ f z ∂ z div(\vec f)=\nabla \cdot \vec f=\frac{\partial f_x}{\partial x} + \frac{\partial f_y}{\partial y} +\frac{\partial f_z}{\partial z} div(f )=f =xfx+yfy+zfz
旋度 c u r l ( f ⃗ ) = ∇ × f ⃗ = ∣ i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z f x f y f z ∣ curl(\vec f)=\nabla \times \vec f = \begin{vmatrix}i&j&k\\\frac\partial{\partial x}&\frac\partial{\partial y}&\frac\partial{\partial z}\\f_x&f_y&f_z\end{vmatrix} curl(f )=×f = ixfxjyfykzfz
拉普拉斯算子 ∇ 2 f = ∇ ⋅ ( ∇ f ) = ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2 + ∂ 2 f ∂ z 2 \nabla ^2 f=\nabla \cdot (\nabla f) = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} +\frac{\partial^2 f}{\partial z^2} 2f=(f)=x22f+y22f+z22f
高斯公式 ∯ S f ⃗ ⋅ d S ⃗ = ∭ v ∇ ⋅ f ⃗ d V \oiint_S \vec f \cdot d \vec S=\iiint_v \nabla \cdot \vec f dV Sf dS =vf dV
格林公式 ∮ Γ f x d x + f y d y = ∬ S ( ∂ f y ∂ x − ∂ f x ∂ y ) d x d y \oint_\Gamma f_x dx+f_y dy = \iint_S (\frac{\partial f_y}{\partial x} - \frac{\partial f_x}{\partial y}) dxdy Γfxdx+fydy=S(xfyyfx)dxdy
经典斯托克斯公式 ∮ L f ⃗ ⋅ d L ⃗ = ∬ S ∇ × f ⃗ ⋅ d S ⃗ \oint_L \vec f \cdot d \vec L = \iint_S \nabla \times \vec f \cdot d \vec S Lf dL =S×f dS
广义斯托克斯公式 ∮ ∂ S ω = ∫ S d ω \oint_{\partial S} \omega=\int_S d\omega Sω=Sdω S S S k k k维流形, ∂ S {\partial S} S是其边界, d ω d\omega dω ω \omega ω的外微分

广义斯托克斯公式 是 微积分基本定理、高斯公式,格林公式,经典斯托克斯公式 的推广。
相关介绍

傅里叶级数

f N ( x ) = a 0 2 + ∑ n = 1 N ( a 0 cos ⁡ 2 π n x L + b 0 sin ⁡ 2 π n x L ) = ∑ n = − N N c n e 2 i π n x L f_N(x)=\frac{a_0}{2}+\sum_{n=1}^N(a_0\cos \frac {2\pi n x}{L}+b_0\sin \frac {2\pi n x}{L})=\sum_{n=-N}^{N}c_ne^{\frac{2i\pi n x}{L}} fN(x)=2a0+n=1N(a0cosL2πnx+b0sinL2πnx)=n=NNcneL2iπnx
其中 L L L是周期函数 f ( x ) f(x) f(x)的周期, a n = 2 L ∫ 0 L f ( x ) cos ⁡ ( 2 π n L x ) d x a_n=\frac{2}{L}\int_0^L f(x)\cos (\frac{2\pi n}{L}x) dx an=L20Lf(x)cos(L2πnx)dx, b n = 2 L ∫ 0 L f ( x ) sin ⁡ ( 2 π n L x ) d x b_n=\frac{2}{L}\int_0^L f(x)\sin (\frac{2\pi n}{L}x) dx bn=L20Lf(x)sin(L2πnx)dx, c n = 1 L ∫ 0 L f ( x ) exp ⁡ ( − 2 π n L x ) d x c_n=\frac{1}{L}\int_0^L f(x)\exp (-\frac{2\pi n}{L}x) dx cn=L10Lf(x)exp(L2πnx)dx
帕塞瓦尔恒等式: 如果 f ( x ) f(x) f(x)的周期为 2 π 2\pi 2π, 则有 ∑ n = − ∞ + ∞ ∣ c n ∣ 2 = 1 2 π ∫ − π π ∣ f ( x ) ∣ 2 d x \sum_{n=-\infty}^{+\infty}|c_n|^2=\frac{1}{2\pi}\int_{-\pi}^\pi |f(x)|^2dx n=+cn2=2π1ππf(x)2dx

函数,定义域 − π < x ≤ π -\pi < x \le \pi π<xπ傅里叶级数
冲激函数 δ ( t ) \delta(t) δ(t) cos ⁡ ( x ) + cos ⁡ ( 2 x ) + cos ⁡ ( 3 x ) + cos ⁡ ( 4 x ) + . . . \cos(x)+\cos(2x)+\cos(3x)+\cos(4x)+... cos(x)+cos(2x)+cos(3x)+cos(4x)+...
符号函数 s g n ( x ) sgn(x) sgn(x) 4 π ( sin ⁡ x 1 + sin ⁡ 3 x 3 + sin ⁡ 5 x 5 + sin ⁡ 7 x 7 + . . . ) \frac{4}{\pi}(\frac{\sin x}{1}+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+\frac{\sin 7x}{7}+...) π4(1sinx+3sin3x+5sin5x+7sin7x+...)
x x x 2 ( sin ⁡ x 1 − sin ⁡ 2 x 2 + sin ⁡ 3 x 3 − sin ⁡ 4 x 4 + . . . ) 2(\frac{\sin x}{1} -\frac{ \sin 2x}{2} + \frac {\sin 3x}{3} - \frac {\sin 4x}{4} + ...) 2(1sinx2sin2x+3sin3x4sin4x+...)
∣ x ∣ |x| x π 2 − 4 π ( cos ⁡ x 1 + cos ⁡ 3 x 3 2 + sin ⁡ 5 x 5 2 + sin ⁡ 7 x 7 2 + . . . ) \frac{\pi}{2}-\frac{4}{\pi}(\frac{\cos x}{1}+\frac{\cos 3x}{3^2}+\frac{\sin 5x}{5^2}+\frac{\sin 7x}{7^2}+...) 2ππ4(1cosx+32cos3x+52sin5x+72sin7x+...)
x 2 x^2 x2 π 2 3 + 4 ( cos ⁡ x 1 2 − cos ⁡ 2 x 2 2 + cos ⁡ 3 x 3 2 − cos ⁡ 4 x 4 2 + . . . ) \frac{\pi^2}{3}+4(\frac{\cos x}{1^2}-\frac{\cos 2x}{2^2}+\frac{\cos 3x}{3^2}-\frac{\cos 4x}{4^2}+...) 3π2+4(12cosx22cos2x+32cos3x42cos4x+...)
∣ cos ⁡ x ∣ |\cos x| cosx 4 π ( 1 2 + cos ⁡ 2 x 2 2 − 1 − cos ⁡ 4 x 4 2 − 1 + sin ⁡ 6 x 6 2 − 1 − sin ⁡ 8 x 8 2 − 1 + . . . ) \frac{4}{\pi}(\frac{1}{2}+\frac{\cos 2x}{2^2-1}-\frac{\cos 4x}{4^2-1}+\frac{\sin 6x}{6^2-1}-\frac{\sin 8x}{8^2-1}+...) π4(21+221cos2x421cos4x+621sin6x821sin8x+...)

拉普拉斯变换

函数拉普拉斯变换
f ( t ) f(t) f(t) L [ f ( t ) ] = ∫ 0 + ∞ f ( t ) e s t d t L[f(t)]=\int_0^{+\infty}f(t)e^{st}dt L[f(t)]=0+f(t)estdt
f ′ ( t ) f'(t) f(t) s L [ f ( t ) ] − f ( 0 ) sL[f(t)]-f(0) sL[f(t)]f(0)
f ′ ′ ( t ) f''(t) f′′(t) s 2 L [ f ( t ) ] − s f ( 0 ) − f ′ ( 0 ) s^2L[f(t)]-sf(0)-f'(0) s2L[f(t)]sf(0)f(0)
∫ 0 t f ( τ ) d τ \int_0^t f(\tau)d\tau 0tf(τ)dτ 1 s L [ f ( t ) ] \frac{1}{s}L[f(t)] s1L[f(t)]
t n f ( t ) t^nf(t) tnf(t) ( − 1 ) n d n d s n L [ f ( t ) ] (-1)^n\frac{d^n}{ds^n}L[f(t)] (1)ndsndnL[f(t)]
1 t f ( t ) \frac{1}{t}f(t) t1f(t) ∫ s + ∞ L [ f ( t ) ] d s \int_s^{+\infty}L[f(t)]ds s+L[f(t)]ds
∫ 0 t f ( t − u ) g ( u ) d u \int_0^t f(t-u) g(u)du 0tf(tu)g(u)du L [ f ( t ) ] L ( g ( t ) ] L[f(t)]L(g(t)] L[f(t)]L(g(t)]
冲激函数 δ ( t ) \delta(t) δ(t) 1 1 1
1 1 1 1 s \frac{1}{s} s1
t n t^n tn n ! s n + 1 \frac{n!}{s^{n+1}} sn+1n!
e a t e^{at} eat 1 s − a \frac{1}{s-a} sa1
sin ⁡ a t \sin{at} sinat a s 2 + a 2 \frac{a}{s^2+a^2} s2+a2a
cos ⁡ a t \cos{at} cosat s s 2 + a 2 \frac{s}{s^2+a^2} s2+a2s
sinh ⁡ a t \sinh{at} sinhat a s 2 − a 2 \frac{a}{s^2-a^2} s2a2a
cosh ⁡ a t \cosh{at} coshat s s 2 − a 2 \frac{s}{s^2-a^2} s2a2s
e − b t sin ⁡ a t e^{-bt}\sin at ebtsinat a ( s + b ) 2 + a 2 \frac{a}{(s+b)^2+a^2} (s+b)2+a2a
e − b t cos ⁡ a t e^{-bt}\cos at ebtcosat s + b ( s + b ) 2 + a 2 \frac{s+b}{(s+b)^2+a^2} (s+b)2+a2s+b
ln ⁡ ( t ) \ln(t) ln(t) − 1 s ( ln ⁡ ( s ) + γ ) -\frac{1}{s}(\ln(s)+\gamma) s1(ln(s)+γ)
贝赛尔函数 J n ( a t ) J_n(at) Jn(at) ( s 2 + a 2 − s ) n a n s 2 + a 2 \frac{(\sqrt{s^2+a^2}-s)^n}{a^n\sqrt{s^2+a^2}} ans2+a2 (s2+a2 s)n
高斯误差函数 e r f ( t ) erf(t) erf(t) 1 s e s 2 4 ( 1 − e r f s 2 ) \frac{1}{s}e^{s^2}{4}(1-erf\frac {s}{2}) s1es24(1erf2s)

总结

本文列举了微积分入门级的知识,希望对大家有所帮助。
因篇幅问题,1.大多数公式都缺少详细说明或证明;2.除了上述内容,还有变分法、偏微分方程、微分几何、复变函数积分等内容未曾提及。这些内容也会单独进行介绍,敬请期待。
写这博客源自于我对数学的喜好,对学过的数学知识进行积累,温故而知新。如果有缺漏或错误的,可以在评论区指出。

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