# 微积分笔记（56）——含参变量积分（1）

Contents

## 含参变量积分

### 含参变量积分

#### 推广：含多参变量积分

$$\int_\Omega f(x, y) \, \mathrm{d} x$$

$$\varphi(y) = \int_\Omega f(x, y) \, \mathrm{d} x$$

#### 实例

Gamma 函数：
$$\Gamma(t) = \int_0^{+\infty} x^{t - 1} e^{-x} \, \mathrm{d} x$$
$t > 0$ 时积分收敛。

Beta 函数：
$$B(u, v) = \int_0^1 (1 - x)^{u - 1} x^{v - 1} \, \mathrm{d} x$$
$u > 0, v > 0$ 时积分收敛。

### 含参变量积分的性质

$$\varphi(y) = \int_a^b f(x, y) \, \mathrm{d} x$$

#### 连续性分析

$\forall y, y + \Delta y \in D$，考虑：
$$\varphi(y + \Delta y) - \varphi(y) = \int_a^b [f(x, y + \Delta y) - f(x, y)] \, \mathrm{d} x$$

$\forall \varepsilon > 0, \exists \delta > 0$，使得 $\forall |\Delta y| < \delta$： $$|f(x, y + \Delta y) - f(x, y)| < \frac{\varepsilon}{b - a}, \forall x \in [a, b]$$ 因此： $$|\varphi(y + \Delta y) - \varphi(y)| \le \int_a^b |f(x, y + \Delta y) - f(x, y)| \, \mathrm{d} x \\ < \frac{\varepsilon}{b - a} \int_a^b \, \mathrm{d} x = \varepsilon$$

#### 含参变量积分的连续性

$$\varphi(y) = \int_a^b f(x, y) \, \mathrm{d} x$$

$$\lim_{y \to y_0} \varphi(y) = \varphi(y_0)$$
：上式可写成：
$$\lim_{y \to y_0} \int_a^b f(x, y) \, \mathrm{d} x = \int_a^b f(x, y_0) \, \mathrm{d} x = \int_a^b \lim_{y \to y_0} f(x, y) \, \mathrm{d} x$$
——极限与积分换序，也就积分号下求极限。

$$\lim_{x \to x_0} \sum_{n = 1}^\infty u_n(x) = \sum_{n = 1}^\infty \lim_{x \to x_0} u_n(x)$$
（一致连续）

#### 可微性分析

$$\frac{\varphi(y + \Delta y) - \varphi(y)}{\Delta y} = \int_a^b \left[\frac{f(x, y + \Delta y) - f(x, y)}{\Delta y} \right] \, \mathrm{d} x$$

$$\therefore \frac{\varphi(y + \Delta y) - \varphi(y)}{\Delta y} = \int_a^b D_y f(x, y + \theta \Delta y) \, \mathrm{d} x$$

$$\lim_{\Delta y \to 0} \frac{\varphi(y + \Delta y) - \varphi(y)}{\Delta y} = \int_a^b D_y f(x, y) \, \mathrm{d} x$$

$$D_y \varphi(y) = \int_a^b D_y f(x, y) \, \mathrm{d} x$$

#### 含参变量积分的可微性

$$\varphi(y) = \int_a^b f(x, y) \, \mathrm{d} x$$

$$D_{y_i} \varphi(y) = \int_a^b D_{y_i} f(x, y) \, \mathrm{d} x$$
：上式也即：
$$D_{y_i} \left[\int_a^b f(x, y) \, \mathrm{d} x \right] = \int_a^b D_{y_i} f(x, y) \, \mathrm{d} x$$
——求导与积分换序，也称积分号下求导。

$$\left[\sum_{n = 1}^\infty u_n(x)\right]^\prime = \sum_{n = 1}^\infty u_n^\prime(x)$$
（一致连续）

#### 含参变量积分的积分（Fubini 定理）

$$\int_\alpha^\beta \, \mathrm{d} y \int_a^b f(x, y) \, \mathrm{d} x = \int_a^b \, \mathrm{d} x \int_\alpha^\beta f(x, y) \, \mathrm{d} y$$

$$F(u, y) = \int_a^u f(x, y) \, \mathrm{d} x$$

$$\frac{\mathrm{d}}{\mathrm{d} u} \int_\alpha^\beta F(u, y) \, \mathrm{d} y = \int_\alpha^\beta D_u F(u, y) \, \mathrm{d} y = \int_\alpha^\beta f(u, y) \, \mathrm{d} y$$

$$\int_\alpha^\beta F(b, y) \, \mathrm{d} y - \int_\alpha^\beta F(a, y) \, \mathrm{d} y = \int_a^b \, \mathrm{d} u \int_\alpha^\beta f(u, y) \, \mathrm{d} y$$

### 含参变量无穷积分的性质

#### 含参变量无穷积分

$$\int_a^{+\infty} f(x, y) \, \mathrm{d} x$$

$$\varphi(y) = \int_a^{+\infty} f(x, y) \, \mathrm{d} x, y \in D$$

#### 连续性初步分析

$\forall y_0, y_0 + \Delta y \in D$，仿照普通含参变量积分处理：
$$\varphi(y_0 + \Delta y) - \varphi(y_0) = \int_a^{+\infty} [f(x, y_0 + \Delta y) - f(x, y_0)] \, \mathrm{d} x$$

$$\varphi(y_0 + \Delta y) - \varphi(y_0) = \int_a^A [f(x, y_0 + \Delta y) - f(x, y_0)] \, \mathrm{d} x + \int_A^{+\infty} [f(x, y_0 + \Delta y) - f(x, y_0)] \, \mathrm{d} x$$

$$\int_A^{+\infty} [f(x, y_0 + \Delta y) - f(x, y_0)] \, \mathrm{d} x$$

#### 含参变量无穷积分的一致收敛

$\forall \varepsilon > 0, \exists A > a$，使得 $\forall A^\prime, A^{\prime \prime} \ge A, \forall y \in D$：
$$\left| \int_{A^\prime}^{A^{\prime\prime}} f(x, y) \, \mathrm{d} x \right| < \varepsilon$$ 这时称无穷积分： $$\int_a^{+\infty} f(x, y) \, \mathrm{d} x$$ 关于 $y \in D$ 一致收敛。

：等价于对于 $\forall y \in D$：
$$\left|\int_A^{+\infty} f(x, y) \, \mathrm{d} x\right| \le \varepsilon$$

$$\sup_{y \in D} \left|\int_A^{+\infty} f(x, y) \, \mathrm{d} x\right| \le \varepsilon \Rightarrow \lim_{A \to +\infty} \sup_{y \in D} \int_A^{+\infty} f(x, y) \, \mathrm{d} x = 0$$

$$\int_1^{+\infty} \frac{\mathrm{d} x}{x^y}$$

$$\int_A^{+\infty} \frac{\mathrm{d} x}{x^y} = \frac{1}{(y - 1) A^{y - 1}} \to + \infty$$
$y$ 接近 $1$，收敛变慢。

#### 一致收敛判别法（Weierstrass）

$$\int_a^{+\infty} F(x) \, \mathrm{d} x$$

$$\sup_{y \in D} |f(x, y)| \le F(x)$$

$$\int_a^{+\infty} f(x, y) \, \mathrm{d} x$$

$$\left|\int_A^{+\infty} f(x, y) \, \mathrm{d} x \right| \le \int_A^{+\infty} |f(x, y)| \, \mathrm{d} x \le \int_a^{+\infty} F(x) \, \mathrm{d} x \\ \therefore \sup_{y \in D} \left|\int_A^{+\infty} f(x, y) \, \mathrm{d} x\right| \le \int_A^{+\infty} F(x) \, \mathrm{d} x$$

$$\int_a^{+\infty} F(x) \, \mathrm{d} x$$

$$\lim_{A \to +\infty} \int_A^{+\infty} F(x) \, \mathrm{d} x = 0$$

：根据上面收敛过程，除了一般收敛之外还导出了：
$$\int_a^{+\infty} f(x, y) \, \mathrm{d} x$$

#### 连续性分析

$$\int_a^{+\infty} f(x, y) \, \mathrm{d} x$$

$$\varphi(y_0 + \Delta y) - \varphi(y_0) = \int_a^A [f(x, y_0 + \Delta y) - f(x, y_0)] \, \mathrm{d} x + \int_A^{+\infty} [f(x, y_0 + \Delta y) - f(x, y_0)] \, \mathrm{d} x$$

$$\left|\int_A^{+\infty} f(x, y) \, \mathrm{d} x\right| \le \frac{\varepsilon}{4}$$

#### 含参变量无穷积分的连续性

$$\varphi(y) = \int_a^{+\infty} f(x, y) \, \mathrm{d} x$$

1. 上面结论实际表示 $\forall y_0 \in D$：
$$\lim_{y \to y_0} \int_a^{+\infty} f(x, y) \, \mathrm{d} x = \int_a^{+\infty} \lim_{y \to y_0} f(x, y) \, \mathrm{d} x$$
由证明过程可见，无穷积分不必在整个 $D$ 上一致收敛——只须要求在 $y_0$ 附近无穷积分一致收敛，即可导出上式。

2. 一致收敛函数级数也有类似的极限求和换序结论：
$$\lim_{x \to x_0} \sum_{n = 1}^\infty u_n(x) = \sum_{n = 1}^\infty \lim_{x \to x_0} u_n(x)$$

#### 可微性分析

$$\frac{\varphi(y_0 + \Delta y) - \varphi(y_0)}{\Delta y} = \int_a^{+\infty} \left[\frac{f(x, y_0 + \Delta y) - f(x, y_0)}{\Delta y} \right] \, \mathrm{d} x$$

$$\therefore \frac{\varphi(y_0 + \Delta y) - \varphi(y_0)}{\Delta y} = \int_a^{+\infty} D_y f(x, y_0 + \theta \Delta y) \, \mathrm{d} x$$

$$\int_a^{+\infty} D_y f(x, y) \, \mathrm{d} x$$

$$D_y \varphi(y_0) = \lim_{\Delta y \to 0} \frac{\varphi(y_0 + \Delta y) - \varphi(y_0)}{\Delta y} = \int_a^{+\infty} D_y f(x, y_0) \, \mathrm{d} x$$

#### 含参变量无穷积分的可微性

1. $\displaystyle \varphi(y) = \int_a^{+\infty} f(x, y) \, \mathrm{d} x$ 对于每个 $y \in D$ 收敛。
2. $D_{y_i} f \in C(\tilde{D}), \displaystyle \int_a^{+\infty} D_{y_i} f(x, y) \, \mathrm{d} x$ 关于 $y \in D$ 一致收敛。

$$D_{y_i} \varphi(y) = \int_a^{+\infty} D_{y_i} f(x, y) \, \mathrm{d} x$$

：回忆函数级数有类似结论：
$$\left[\sum_{n = 1}^\infty u_n(x) \right]^\prime = \sum_{n = 1}^\infty u^\prime_n(x)$$

$$\sum_{n = 1}^\infty u_n(x)$$

$$\sum_{n = 1}^\infty u_n^\prime(x)$$

#### 含参变量无穷积分的换序（Fubini 定理）

$$\varphi(y) = \int_a^{+\infty} f(x, y) \, \mathrm{d} x$$

$$\int_D \varphi(y) \, \mathrm{d} y = \int_a^{+\infty} \, \mathrm{d} x \int_D f(x, y) \, \mathrm{d} y$$

$$\int_D \, \mathrm{d} y \int_a^{+\infty} f(x, y) \, \mathrm{d} x = \int_a^{+\infty} \, \mathrm{d} x \int_D f(x, y) \, \mathrm{d} y$$

$$\int_D \varphi(y) \, \mathrm{d} y = \int_D \, \mathrm{d} y \int_a^A f(x, y) \, \mathrm{d} x + \int_D \, \mathrm{d} y \int_A^{+\infty} f(x, y) \, \mathrm{d} x \\ = \int_a^A \, \mathrm{d} x \int_D x f(x, y) \, \mathrm{d} y + \int_D \, \mathrm{d} y \int_A^{+\infty} f(x, y) \, \mathrm{d} x$$

$$\left|\int_a^A \, \mathrm{d} x \int_D x f(x, y) \, \mathrm{d} y - \int_D \varphi(y) \, \mathrm{d} y \right| \le \int_D \left|\int_A^{+\infty} f(x, y) \, \mathrm{d} x\right| \, \mathrm{d} y$$

$$\varphi(y) = \int_a^{+\infty} f(x, y) \, \mathrm{d} x$$

$$\sup_{y \in D} \left|\int_A^{+\infty} f(x, y) \, \mathrm{d} x \right| < \frac{\varepsilon}{\mu(D)}$$ 代入前面的估计： $$\left|\int_a^A \, \mathrm{d} x \int_D x f(x, y) \, \mathrm{d} y - \int_D \varphi(y) \, \mathrm{d} y \right| \le \frac{\varepsilon}{\mu(D)} \int_D \, \mathrm{d} y = \varepsilon$$ 此即： $$\lim_{A \to +\infty} \int_a^A \, \mathrm{d} x \int_D f(x, y) \, \mathrm{d} y = \int_D \varphi(y) \, \mathrm{d} y = \int_D \, \mathrm{d} y \int_a^{+\infty} f(x, y) \, \mathrm{d} x \ \ \ \square$$ ：积分换序要求在 $D$ 整体一致连续。