229 lines
11 KiB
TeX
229 lines
11 KiB
TeX
\chapter{含参变量积分}
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\begin{definition}
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设二元函数$f(x, u)$在闭矩形$I = [a, b] \times [\alpha, \beta]$上连续，那么对于固定的$u \in [\alpha, \beta]$，函数$f(x, u)$对变量$x$在$[a,b]$上Riemann可积，这时称积分
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\[\int_a^b f(x, u) \dif x\]
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是含参变量$u$的常义积分。如果对于固定的$u$，$f(x, u)$是变量$x$在$[a, b]$中的无界函数，或者$[a, b]$是一个无限区间，则称相应的积分是含参变量$u$的反常积分。
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\end{definition}
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\section{含参变量的常义积分}
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\begin{theorem}
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如果函数$f(x, u)$在闭矩形$I = [a, b] \times [\alpha, \beta]$上连续，那么
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\[\varphi(u) = \int_a^b f(x, u) \dif x\]
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在区间$[\alpha, \beta]$上一致连续。
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\end{theorem}
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\begin{remark}
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这意味着
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\[\tolim{t}{t_0} \int_\alpha^\beta f(x, t) \dif x = \int_\alpha^\beta \tolim{t}{t_0} f(x, t) \dif x\]
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\end{remark}
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\begin{theorem}[含参积分的连续性]
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如果函数$f$及其偏导数$\dfrac{\partial f}{\partial u}$都在闭矩形$I = [a, b] \times [\alpha, \beta]$上连续，那么函数
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\[\varphi(u) = \int_a^b f(x, u) \dif x\eqper\]
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在$[\alpha, \beta]$上可微，而且
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\[\dfrac{\dif}{\dif u} \varphi(u) = \int_a^b \left(\frac{\partial}{\partial u} f(x, u)\right)\dif x\eqper\]
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\end{theorem}
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\begin{remark}
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这意味着当积分
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\[\varphi(u) = \int_a^b f(x, u) \dif x\]
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难以计算，而$\varphi(u_0)$容易计算时，可以先求出
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\[\deriv{\varphi(u)} = \int_a^b \frac{\partial f}{\partial u} (x, u) \dif x\]
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再对$u$积分：
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\[\varphi(u) = \varphi(u_0) + \int_{u_0}^u \deriv{\varphi}(t) \dif t\eqper\]
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\end{remark}
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\begin{theorem}[含参积分的可微性]
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设$g(t, x)$。$\dfrac{\partial g}{\partial t} \in C([a, b] \times [c, d])$，$\alpha(t), \beta(t)$在$[a, b]$上可导，且对任意的$t \in [a, b]$，有
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\[c \leq \alpha(t), \beta(t) \leq d\]
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则
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\[f(t) = \int_{\alpha(t)}^{\beta(t)} g(t, x) \dif x\]
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在区间$[a, b]$上可导，且
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\[\deriv{f}(t) = \dfrac{\dif}{\dif t}\int_{\alpha(t)}^{\beta(t)} g(t, x) \dif x = \int_{\alpha(t)}^{\beta(t)} \dfrac{\partial g}{\partial t}(t, x) \dif x + g(t, \beta(t)) \deriv{\beta}(t) - g(t, \alpha(t)) \deriv{\alpha}(t)\eqper\]
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\end{theorem}
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\begin{proof}
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令$J(t, \alpha, \beta) = \dint_\alpha^\beta g(t, x) \dif x$，由$g(t, x), \dfrac{\partial g}{\partial t}(t, x)$的连续性，
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\[\deriv{J_t} = \int_\alpha^\beta \dfrac{\partial g}{\partial t}(t, x) \dif x, \deriv{J_\alpha} = - g(t, \alpha), \deriv{J_\beta} = g(t, \beta)\]
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均在$(t, \alpha, \beta) \in D = [a, b] \times [c, d] \times [c, d]$上连续。因此$J(t, \alpha, \beta)$在$D$上可微，复合函数
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\[f(t) = \int_{\alpha(t)}^{\beta(t)} g(t, x) \dif x = J(t, \alpha(t), \beta(t))\]
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在$t \in [a, b]$上可微，且
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\begin{align*}
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\deriv{f}(t) & = \deriv{J_t} + \deriv{J_\alpha} \cdot \deriv{\alpha}(t) + \deriv{J_\beta} \cdot \deriv{\beta}(t)\\
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& = \int_{\alpha(t)}^{\beta(t)} \dfrac{\partial g}{\partial t}(t, x) \dif x + g(t, \beta(t)) \cdot \deriv{\beta}(t) - g(t, \alpha(t)) \cdot \deriv{\alpha}(t)\eqper
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\end{align*}
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\end{proof}
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\begin{theorem}[含参积分的可积性]
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设$g(t, x)$在$(t, x) \in D = [a, b] \times [\alpha, \beta]$上连续，则$\dint_\alpha^\beta g(t, x) \dif x$在$t \in [a, b]$上可积，$\dint_a^b g(t, x) \dif t$在$x \in [\alpha, \beta]$上可积，且
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\[\int_a^b \left(\int_\alpha^\beta g(t, x) \dif x\right) \dif t = \int_\alpha^\beta \left(\int_a^b g(t, x) \dif t\right)\dif x\]
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简记为
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\[\int_a^b \dif t \int_\alpha^\beta g(t, x) \dif x = \int_\alpha^\beta \dif x \int_\alpha^\beta g(t, x) \dif t\eqper\]
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\end{theorem}
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\section{含参反常积分的一致收敛}
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设$f(t, x)$在$D = [\alpha, \beta] \times [a, +\infty)$上连续，对任意的$t \in [\alpha, \beta]$，广义积分
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\[I(t) = \int_a^{+\infty} f(t, x) \dif x\]
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收敛。那么$I(t) \in C[\alpha, \beta]$吗？这个问题与先前遇到的函数项级数的连续性问题类似，都需要一致收敛的条件。
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\begin{definition}
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设对任意的$t \in \Omega \subset \realnum$，$\dint_a^{+\infty} f(t, x) \dif x$收敛。如果对于任意给定的$\varepsilon > 0$，总能找到只与$\varepsilon$有关的$A_0(\varepsilon)$，使得当$A > A_0$时
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\[\abs{\int_A^{+\infty} f(t, x) \dif x} < \varepsilon\]
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对任意的$t \in \Omega$成立，则称反常积分$\dint_a^{+\infty} f(t, x) \dif x$关于$t \in \Omega$一致收敛。
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\end{definition}
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记
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\[\eta(A) = \sup \limits_{u \in [\alpha, \beta]} \abs{\int_A^{+\infty} f(t, x) \dif x} \eqper\]
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\begin{theorem}
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积分$\dint_a^{+\infty} f(t, x) \dif x$在$\Omega$上一致收敛的充分必要条件是
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\[\tolim{A}{+\infty} \eta(A) = 0\eqper\]
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\end{theorem}
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\begin{theorem}[Cauchy收敛原理]
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积分$\dint_a^{+\infty} f(t, x) \dif x$在$\Omega$上一致收敛的充分必要条件是，对任意$\varepsilon > 0$，存在仅与$\varepsilon$有关的$A_0$当$A^\prime, A^{\prime \prime} > A_0$时，不等式
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\[\abs{\int_{A^\prime}^{A^{\prime \prime}} f(t, x) \dif x} < \varepsilon\]
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对任意$t \in \Omega$都成立。
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\end{theorem}
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\begin{theorem}[Weierstranss判别法]
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设$f(t, x)$对$x$在$[a, +\infty)$上连续。如果坐在$[a, +\infty)$上的连续函数$F$，使得$\dint_a^{+\infty} F(x) \dif x$收敛，且对一切充分大的$x$及$\Omega$上的一切$t$，都有
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\[\abs{f(t, x)} \leq F(x)\]
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那么积分$\dint_a^{+\infty} f(t, x) \dif x$在$\Omega$上一致收敛。
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\end{theorem}
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\begin{theorem}[Dirichlet判别法]
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如果$f, g$满足下面两个条件：
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\begin{enumerate}
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\item 当$A \to +\infty$时，积分$\dint_a^A f(t, x) \dif x$对$t \in \Omega$一致有界，即存在常数$M$，使得当$A$充分大时，对每个$t \in \Omega$有
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\[\abs{\int_a^A f(t, x) \dif x} \leq M\]
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\item $g(t, x)$是$x$的单调函数，且当$x \to +\infty$时关于$t$一致地趋于0。
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\end{enumerate}
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那么积分
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\[\int_a^{+\infty} f(t, x) g(t, x) \dif x\]
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在$\Omega$上一致收敛。
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\end{theorem}
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\begin{theorem}[Abel判别法]
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如果$f, g$满足下面两个条件：
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\begin{enumerate}
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\item 积分$\dint_a^{+\infty} f(t, x) \dif x$关于$t \in \Omega$一致收敛；
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\item $g(t, x)$对$x$单调，且关于$t$一致有界。
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\end{enumerate}
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那么积分
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\[\int_a^{+\infty} f(t, x) g(t, x) \dif x\]
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在$\Omega$上一致收敛。
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\end{theorem}
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\section{含参变量反常积分的性质}
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\begin{theorem}
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如果函数$f(t, x)$在$[\alpha, \beta] \times [a, +\infty)$上连续，而且积分
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\[I(t) = \int_a^{+\infty} f(t, x) \dif x\]
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关于$t \in [\alpha, \beta]$一致收敛，则$I(t) \in C[\alpha, \beta]$。
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\end{theorem}
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\begin{theorem}
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设函数$f(t, x), \dfrac{\partial f}{\partial t}(t, x) \in C([\alpha, \beta] \times [a, +\infty])$，且对任意的$t \in [\alpha, \beta]$，
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\[I(t) = \int_a^{+\infty} f(t, x) \dif x\]
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收敛，同时
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\[\int_a^{+\infty} \dfrac{\partial f}{\partial t}(t, x) \dif x\]
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关于$t \in [\alpha, \beta]$一致收敛，那么$I(t) \in C^1[\alpha, \beta]$且
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\[\deriv{I}(t) = \frac{\dif}{\dif t} \int_a^{+\infty} f(t, x) \dif x = \int_a^{+\infty} \frac{\partial f}{\partial t}(t, x) \dif x\eqper\]
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\end{theorem}
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\begin{theorem}
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设函数$f(x, y) \in C([a, +\infty) \times [\alpha, \beta])$，含参积分
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\[I(t) = \int_a^{+\infty} f(x, y) \dif x\]
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关于$y \in [\alpha, \beta]$一致收敛，则$I(y)$在$[\alpha, \beta]$上可积，且
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\[\int_\alpha^\beta I(y) \dif y = \int_a^{+\infty} \left(\int_\alpha^\beta f(x, y) \dif y\right)\dif x\]
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即
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\[\int_\alpha^\beta \dif y \int_a^{+\infty} f(x, y) \dif x = \int_a^{+\infty} \dif x \int_\alpha^\beta f(x, y) \dif y\eqper\]
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\end{theorem}
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\begin{theorem}
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如果函数$f(x, y)$满足下列条件：
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\begin{enumerate}
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\item $f$在$[a, +\infty) \times [\alpha, +\infty)$上连续；
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\item 对任意的$\beta > \alpha$，积分
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\[\int_a^{+\infty} f(x, y) \dif x\]
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在$y \in [\alpha, \beta]$上一致收敛；
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对任意的$b > a$，积分
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\[\int_\alpha^{+\infty} f(x, y) \dif y\]
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在$x \in [a, b]$上一致收敛；
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\item 积分
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\[\int_a^{+\infty} \left(\int_\alpha^{+\infty} \abs{f(x, u)} \dif u\right)\dif x, \quad \int_\alpha^{+\infty} \left(\int_a^{+\infty} \abs{f(x, u)} \dif x\right)\dif u\]
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中至少有一个存在，
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\end{enumerate}
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那么积分
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\[\int_a^{+\infty} \left(\int_\alpha^{+\infty} f(x, u) \dif u\right)\dif x, \quad \int_\alpha^{+\infty} \left(\int_a^{+\infty} f(x, u) \dif x\right)\dif u\]
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都存在且相等，即
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\[\int_a^{+\infty} \left(\int_\alpha^{+\infty} f(x, u) \dif u\right)\dif x = \int_\alpha^{+\infty} \left(\int_a^{+\infty} f(x, u) \dif x\right)\dif u\eqper\]
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\end{theorem}
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\section[Γ函数与B函数]{$\Gamma$函数与$\Beta$函数}
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含参变量的广义积分
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\begin{align*}
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\Gamma(s) = \int_0^{+\infty} t^{s - 1} e^{-t} \dif t, \quad (s > 0)\\
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\Beta(p, q) = \int_0^1 t^{p - 1} (1 - t)^{q - 1} \dif t, \quad (p > 0 , q > 0)
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\end{align*}
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分别称为$\Gamma$函数和$\Beta$函数。
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\subsection[Γ函数]{$\Gamma$函数}
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\begin{theorem}
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$\Gamma(s)$在$(0, +\infty)$上连续，且有各阶连续导数。
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\end{theorem}
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\begin{theorem}
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$\Gamma$函数具有下面三条性质：
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\begin{enumerate}
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\item 对任意$s > 0$，$\Gamma(s) > 0$，且$\Gamma(1) = 1$；
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\item $\Gamma(s + 1) = s \Gamma(s)$对任意$s > 0$成立；
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\item $\ln \Gamma(s)$是$(0, +\infty)$上的凸函数。
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\end{enumerate}
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\end{theorem}
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$\Gamma$函数可以看作是阶乘函数的推广：
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\[\Gamma(n + 1) = n!, n = 1, 2, 3, \dots\]
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\begin{theorem}[Bohr-Mollerup]
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如果$(0, +\infty)$上的函数$f$满足下面三个条件：
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\begin{enumerate}
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\item 对任意$x > 0$，$f(x) > 0$且$f(1) = 1$；
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\item $f(x + 1) = xf(x)$对任意$x > 0$成立；
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\item $\ln f$是$(0, +\infty)$上的凸函数，
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\end{enumerate}
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那么$f(x) = \Gamma(x)$对任意$x > 0$成立。
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\end{theorem}
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\begin{theorem}
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对任意$x > 0$，有
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\[\Gamma(x) = \tolim{n}{\infty} \frac{n^x n!}{x(x + 1) \dots (x + n)}\eqper\]
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\end{theorem}
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\begin{theorem}[$\Gamma$函数的余元公式]
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对任意的$x \in (0, 1)$，有
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\[\Gamma(x) \Gamma(1 - x) = \frac{\pi}{\sin \pi x}\eqper\]
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\end{theorem}
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\subsection[B函数]{$\Beta$函数}
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\begin{theorem}
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$\Beta$函数有下面的性质：
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\begin{enumerate}
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\item $\Beta(p, q) > 0$，且$\Beta(1, q) = \dfrac{1}{q}$；
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\item $\Beta(p + 1, q) = \dfrac{x}{p + q} \Beta(p, q)$；
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\item 给定$q > 0$，$\ln \Beta(p, q)$关于$x$在$(0, + \infty)$是凸函数。
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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对任意$p > 0$，$q > 0$，有
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\[\Beta(p, q) = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p + q)}\eqper\]
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\end{theorem}
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\begin{corollary}
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对任意的$p > 0, q > 0$，$\Beta$函数有如下性质：
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\begin{enumerate}
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\item $\Beta(p, q) = \Beta(q, p)$；
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\item $\Beta(p + 1, q + 1) = \dfrac{pq}{(p + q + 1)(p + q)}B(p, q)$。
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\end{enumerate}
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\end{corollary} |