6.5 定积分的计算

分部积分公式

由分部积分公式结合微积分基本定理, 有 $u v |_{a}^{b} = \int_{a}^{b} u d v + \int_{a}^{b} v d u,$ 也即 $\int_{a}^{b} u v^{'} d x = u v |_{a}^{b} - \int_{a}^{b} u^{'} v d x .$

例子: 计算

I_{n} = \int_{0}^{\frac{π}{2}} \sin^{n} x d x = \int_{0}^{\frac{π}{2}} \cos^{n} x d x

不妨设 $n \geq 2$ . 注意到 $\begin{aligned} \int_{0}^{\frac{π}{2}} \sin^{n} x d x = - \int_{0}^{\frac{π}{2}} \sin^{n - 1} x d \cos x \\ = & - \sin^{n - 1} x \cos x |_{0}^{\frac{π}{2}} + (n - 1) \int_{0}^{\frac{π}{2}} \cos^{2} x \sin^{n - 2} x d x \\ = & (n - 1) \int_{0}^{\frac{π}{2}} (1 - \sin^{2} x) \sin^{n - 2} x d x \\ = & (n - 1) I_{n - 2} - (n - 1) I_{n}, \end{aligned}$ 从而得到递推式 $I_{n} = \frac{n - 1}{n} I_{n - 2} .$ 而 $I_{0} = \frac{π}{2}$ , $I_{1} = 1$ , 从而 $n \geq 2$ 时 $I_{2 k} = \frac{(2 k - 1)!!}{(2 k)!!} \cdot \frac{π}{2}, I_{2 k + 1} = \frac{(2 k)!!}{(2 k + 1)!!} \cdot 1.$

带 Cauchy 余项的 Taylor 公式

设 $f (x)$ 在 $o (x_{0})$ 内有 $n + 1$ 阶导数, 则 $\forall x \in o (x_{0})$ , $f (x) = \sum_{i = 1}^{n} \frac{f^{(i)} (x_{0})}{i!} (x - x_{0})^{i} + \frac{1}{n!} \int_{x_{0}}^{x} f^{(n + 1)} (t) (x - t)^{n} d t .$

证明

计算可得 $\begin{aligned} \frac{1}{n!} \int_{x_{0}}^{x} f^{(n + 1)} (t) (x - t)^{n} d t = \frac{1}{n!} \int_{x_{0}}^{x} (x - t)^{n} d f^{(n)} (t) \\ = & {\frac{(x - t)^{n}}{n!} f^{(n)} (t) |}_{t = x_{0}}^{t = x} + \frac{1}{(n - 1)!} \int_{x_{0}}^{x} f^{(n)} (t) (x - t)^{n - 1} d t \\ = & - \frac{(x - x_{0})^{n}}{n!} f^{(n)} (t) + \frac{1}{(n - 1)!} \int_{x_{0}}^{x} f^{(n)} (t) (x - t)^{n - 1} d t \\ = & \dots = f (x) - \sum_{i = 0}^{n} \frac{f^{(i)} (x_{0})}{i!} (x - x_{0})^{i} . \end{aligned}$

换元法

这里我们统一假设 $f (x) \in C [a, b]$ , $φ (t) \in D [α, β]$ , 且 $a = φ (α), b = φ (β)$ , $φ ([α, β]) = [a, b]$ . 从而根据换元公式, $\int_{a}^{b} f (x) d x = \int_{α}^{β} f (φ (t)) φ^{'} (t) d t .$

证明

由于 $f (x) \in C [a, b]$ , 原函数 $F (x)$ 存在 (因为闭区间上的连续函数可积), 从而 $\int_{a}^{b} f (x) d x = F (x) |_{a}^{b}$ . 又因为 $(F (φ (t)))^{'} = F^{'} (φ (t)) φ^{'} (t) = f (φ (t)) φ^{'} (t),$ 从而 $\begin{aligned} \int_{α}^{β} f (φ (t)) φ^{'} (t) d t = F (φ (t)) |_{α}^{β} \\ = & F (φ (β)) - F (φ (α)) = F (b) - F (a) = F (x) |_{a}^{b} . \end{aligned}$

例子

$\int_{- 2}^{- \sqrt{2}} \frac{d x}{\sqrt{x^{2} - 1}}$ . 令 $x = \sec t$ , 则利用第6条, $\begin{aligned} I & = \int_{\frac{2 π}{3}}^{\frac{3 π}{4}} \frac{\sec t \tan t}{\sqrt{\sec^{2} t - 1}} d t = \int_{\frac{2 π}{3}}^{\frac{3 π}{4}} \frac{\sec t \tan t}{\tan t} d t \\ = - \ln | \sec t - \tan t | |_{\frac{2 π}{3}}^{\frac{3 π}{4}} . \end{aligned}$
$\int_{0}^{1} f (x) d x$ , 其中 $f (x) = [\frac{α}{x}] - α [\frac{1}{x}]$ , $α \in (0, 1]$ . 注意到 $f (x) = - (\frac{α}{x} - [\frac{α}{x}]) + α (\frac{1}{x} - [\frac{1}{x}]) = - f_{1} (x) + f_{2} (x) .$ 令 $t = x / α$ , 则 $\begin{aligned} \int_{0}^{1} f_{1} (x) d x & = \int_{0}^{1} \frac{α}{x} - [\frac{α}{x}] d x = \int_{0}^{α} (\frac{α}{x} - [\frac{α}{x}]) d x + \int_{0}^{1} (\frac{α}{x} - [\frac{α}{x}]) d x \\ = \int_{0}^{1} \frac{1}{t} - [\frac{1}{t}] α d t + α \ln x |_{α}^{1} \\ = \int_{0}^{1} f_{2} (t) d t + α \ln α, \end{aligned}$ 从而 $\int_{0}^{1} f (x) d x = - \int_{0}^{1} f_{1} (x) d x + \int_{0}^{1} f_{2} (x) d x = α \ln α .$

关于周期性的一些结论

$f (x)$ 是 $[- a, a]$ 上的奇函数, 则 $\int_{- a}^{a} f (x) d x = 0$ .
$f (x)$ 是 $[- a, a]$ 上的偶函数, 则 $\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ .
$f (x)$ 在 $R$ 上以 $T$ 为周期, 则 $\int_{α}^{α + T} f (x) d x = \int_{0}^{T} f (x) d x, \int_{α}^{α + n T} f (x) d x = n \int_{0}^{T} f (x) d x .$

例子

$\int_{0}^{\frac{π}{4}} \ln (1 + \tan x) d x$ . 令 $u = \frac{π}{4} - x$ , 则 $\begin{aligned} I & = \int_{\frac{π}{4}}^{0} \ln (1 + \tan (\frac{π}{4} - u)) (- d u) \\ = \int_{0}^{\frac{π}{4}} \ln \frac{2}{\tan u + 1} d u = \int_{0}^{\frac{π}{4}} \ln 2 d u - \int_{0}^{\frac{π}{4}} \ln (\tan u + 1) d u, \end{aligned}$ 解得 $I = \frac{π}{8} \ln 2$ .
$\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\sin^{2} x}{1 + e^{x}} d x$ . 令 $x = - u$ , 结合这个结论, $\begin{aligned} I & = \int_{0}^{\frac{π}{2}} \sin^{2} x (\frac{1}{1 + e^{- x}} + \frac{1}{1 + e^{x}}) d x \\ = \int_{0}^{\frac{π}{2}} \sin^{2} x d x = \frac{1!!}{2!!} \cdot \frac{π}{2} = \frac{π}{4} . \end{aligned}$

定积分下的 Jensen不等式

设 $φ (t) \in C [0, a]$ , $f (x) \in D^{2} (- \infty, + \infty)$ , $f^{″} (x) \geq 0$ . 证明 $f (\frac{1}{a} \int_{0}^{a} φ (t) d t) \leq \frac{1}{a} \int_{0}^{a} f (φ (t)) d t .$

证明

由条件, $f (x)$ 在 $(- \infty, + \infty)$ 上下凸, 从而回顾定义, $\begin{aligned} \frac{1}{a} \int_{0}^{a} φ (t) d t = \frac{1}{a} lim_{| | π | | \to 0} \sum_{i = 1}^{n} φ (ξ_{n}) Δ t_{i} . \end{aligned}$ 结合 $\sum_{i = 1}^{n} \frac{Δ t_{i}}{a} = 1$ , 有 $\begin{aligned} f (\frac{1}{a} \int_{0}^{a} φ (t) d t) = f (lim_{| | π | | \to 0} \sum_{i = 1}^{n} φ (ξ_{i}) \frac{Δ t_{i}}{a}) \\ = & lim_{| | π | | \to 0} f (\sum_{i = 1}^{n} φ (ξ_{i}) \frac{Δ t_{i}}{a}) \leq lim_{| | π | | \to 0} \sum_{i = 1}^{n} f (φ (ξ_{i})) \frac{Δ t_{i}}{a} = \frac{1}{a} \int_{0}^{a} f (φ (t)) d t . \end{aligned}$