分部积分就是乘积法则的积分形式
分部积分不是灵机一动的猜测——它是乘积法则经过基本定理的产物。设 u、v 在 [a,b] 上可导且 u′、v′ 可积。乘积法则给出 (uv)′ = u′v + uv′。把两边在 [a,b] 上积分，并对左边用基本定理第二部分。
u, v differentiable on [a,b]; u', v' integrable. Product rule: (uv)'(x) = u'(x) v(x) + u(x) v'(x). Integrate from a to b. The left side, by FTC Part 2 (uv is an antiderivative of (uv)'): integral from a to b of (uv)' = u(b)v(b) - u(a)v(a) = [uv] from a to b. So [uv] from a to b = integral of u'v + integral of u v'. Rearrange: integral from a to b of u v' = [uv] from a to b - integral from a to b of u' v. Worked example: integral from 0 to 1 of x e^x dx. Let u = x (u' = 1), v' = e^x (v = e^x). Then = [x e^x] from 0 to 1 - integral from 0 to 1 of 1 * e^x dx = (1*e^1 - 0) - [e^x] from 0 to 1 = e - (e - 1) = 1.
换元法就是链式法则的积分形式
换元法以同样方式来自链式法则。设 g 在 [a,b] 上连续可导，f 在 g 的值域上连续。则 f(g(x))·g′(x) 有原函数 F(g(x))，其中 F′ = f，基本定理第二部分把它化成一次干净的换限。
g in C^1 on [a,b], f continuous on g([a,b]); let F be an antiderivative of f.
Chain rule: d/dx [ F(g(x)) ] = F'(g(x)) g'(x) = f(g(x)) g'(x).
So F(g(x)) is an antiderivative of f(g(x)) g'(x). By FTC Part 2:
integral from a to b of f(g(x)) g'(x) dx = F(g(b)) - F(g(a)).
But also, with u = g(x):
integral from g(a) to g(b) of f(u) du = F(g(b)) - F(g(a)).
The two right sides agree, hence
integral_{a}^{b} f(g(x)) g'(x) dx = integral_{g(a)}^{g(b)} f(u) du.
Worked example: integral from 0 to 1 of 2x cos(x^2) dx, with u = x^2, du = 2x dx.
= integral from u=0 to u=1 of cos(u) du = [sin u] from 0 to 1 = sin(1) - sin(0) = sin 1.反常积分：越过边界去积分
黎曼积分按其构造，需要有界区间上的有界函数。反常积分则越过这道边界：对正经的黎曼积分取极限——让一个端点跑向 ∞，或趋近某个使 f 爆破的点。当这个极限作为有限数存在时，反常积分才存在（我们说它收敛）。
Definition (infinite interval):
integral from 1 to infinity of f := lim_{R -> infinity} integral from 1 to R of f,
when that limit exists and is finite.
Example (converges): integral from 1 to infinity of 1/x^2 dx.
integral from 1 to R of x^{-2} dx = [ -1/x ] from 1 to R = 1 - 1/R.
As R -> infinity, 1 - 1/R -> 1. So the integral CONVERGES to 1.
Example (diverges): integral from 1 to infinity of 1/x dx.
integral from 1 to R of 1/x dx = ln R - ln 1 = ln R.
As R -> infinity, ln R -> infinity. So the integral DIVERGES.
Definition (unbounded near 0):
integral from 0 to 1 of 1/sqrt(x) dx := lim_{t -> 0+} integral from t to 1 of x^{-1/2} dx
= lim_{t->0+} [ 2 sqrt(x) ] from t to 1 = lim_{t->0+} (2 - 2 sqrt(t)) = 2. CONVERGES.