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MathematicalAnalysis/20含参变量积分.tex
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\chapter{含参变量积分}
\begin{definition}
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$$[a,b]$上Riemann可积这时称积分
\[\int_a^b f(x, u) \dif x\]
是含参变量$u$的常义积分。如果对于固定的$u$$f(x, u)$是变量$x$$[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$的反常积分。
\end{definition}
\section{含参变量的常义积分}
\begin{theorem}
如果函数$f(x, u)$在闭矩形$I = [a, b] \times [\alpha, \beta]$上连续,那么
\[\varphi(u) = \int_a^b f(x, u) \dif x\]
在区间$[\alpha, \beta]$上一致连续。
\end{theorem}
\begin{remark}
这意味着
\[\tolim{t}{t_0} \int_\alpha^\beta f(x, t) \dif x = \int_\alpha^\beta \tolim{t}{t_0} f(x, t) \dif x\]
\end{remark}
\begin{theorem}[含参积分的连续性]
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
\[\varphi(u) = \int_a^b f(x, u) \dif x\eqper\]
$[\alpha, \beta]$上可微,而且
\[\dfrac{\dif}{\dif u} \varphi(u) = \int_a^b \left(\frac{\partial}{\partial u} f(x, u)\right)\dif x\eqper\]
\end{theorem}
\begin{remark}
这意味着当积分
\[\varphi(u) = \int_a^b f(x, u) \dif x\]
难以计算,而$\varphi(u_0)$容易计算时,可以先求出
\[\deriv{\varphi(u)} = \int_a^b \frac{\partial f}{\partial u} (x, u) \dif x\]
再对$u$积分:
\[\varphi(u) = \varphi(u_0) + \int_{u_0}^u \deriv{\varphi}(t) \dif t\eqper\]
\end{remark}
\begin{theorem}[含参积分的可微性]
$g(t, x)$$\dfrac{\partial g}{\partial t} \in C([a, b] \times [c, d])$$\alpha(t), \beta(t)$$[a, b]$上可导,且对任意的$t \in [a, b]$,有
\[c \leq \alpha(t), \beta(t) \leq d\]
\[f(t) = \int_{\alpha(t)}^{\beta(t)} g(t, x) \dif x\]
在区间$[a, b]$上可导,且
\[\deriv{f}(t) = \dfrac{\dif}{\dif t}\int_{\alpha(t)}^{\beta(t)} g(t, x) \dif x = \int_{\alpha(t)}^{\beta(t)} \dfrac{\partial g}{\partial t}(t, x) \dif x + g(t, \beta(t)) \deriv{\beta}(t) - g(t, \alpha(t)) \deriv{\alpha}(t)\eqper\]
\end{theorem}
\begin{proof}
$J(t, \alpha, \beta) = \dint_\alpha^\beta g(t, x) \dif x$,由$g(t, x), \deriv{g_t}(t, x)$的连续性,
\[\deriv{J_t} = \int_\alpha^\beta \dfrac{\partial g}{\partial t}(t, x) \dif x, \deriv{J_\alpha} = - g(t, \alpha), \deriv{J_\beta} = g(t, \beta)\]
均在$(t, \alpha, \beta) \in D = [a, b] \times [c, d] \times [c, d]$上连续。因此$J(t, \alpha, \beta)$$D$上可微,复合函数
\[f(t) = \int_{\alpha(t)}^{\beta(t)} g(t, x) \dif x = J(t, \alpha(t), \beta(t))\]
$t \in [a, b]$上可微,且
\begin{align*}
\deriv{f}(t) & = \deriv{J_t} + \deriv{J_\alpha} \cdot \deriv{\alpha}(t) + \deriv{J_\beta} \cdot \deriv{\beta}(t)\\
& = \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
\end{align*}
\end{proof}
\begin{theorem}[含参积分的可积性]
$g(t, x)$$(t, x) \in D = [a, b] \times [\alpha, \beta]$上连续,则$\dint_\alpha^\beta g(t, x) \dif x$$t \in [a, b]$上可积,$\dint_a^b g(t, x) \dif t$$x \in [\alpha, \beta]$上可积,且
\[\int_a^b \left(\int_\alpha^\beta g(t, x) \dif x\right) \dif t = \int_\alpha^\beta \left(\int_a^b g(t, x) \dif t\right)\dif x\]
简记为
\[\int_a^b \dif t \int_\alpha^\beta g(t, x) \dif x = \int_\alpha^\beta \dif x \int_\alpha^\beta g(t, x) \dif t\eqper\]
\end{theorem}
\section{含参反常积分的一致收敛}
$f(t, x)$$D = [\alpha, \beta] \times [a, +\infty)$上连续,对任意的$t \in [\alpha, \beta]$,广义积分
\[I(t) = \int_a^{+\infty} f(t, x) \dif x\]
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。
\begin{definition}
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$
\[\abs{\int_A^{+\infty} f(t, x) \dif x} < \varepsilon\]
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。
\end{definition}
\[\eta(A) = \sup \limits_{u \in [\alpha, \beta]} \abs{\int_A^{+\infty} f(t, x) \dif x} \eqper\]
\begin{theorem}
积分$\dint_a^{+\infty} f(t, x) \dif x$$\Omega$上一致收敛的充分必要条件是
\[\tolim{A}{+\infty} \eta(A) = 0\eqper\]
\end{theorem}
\begin{theorem}[Cauchy收敛原理]
积分$\dint_a^{+\infty} f(t, x) \dif x$$\Omega$上一致收敛的充分必要条件是,对任意$\varepsilon > 0$,存在仅与$\varepsilon$有关的$A_0$$A^\prime, A^{\prime \prime} > A_0$时,不等式
\[\abs{\int_{A^\prime}^{A^{\prime \prime}} f(t, x) \dif x} < \varepsilon\]
对任意$t \in \Omega$都成立。
\end{theorem}
\begin{theorem}[Weierstranss判别法]
$f(t, x)$$x$$[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$$\Omega$上的一切$t$,都有
\[\abs{f(t, x)} \leq F(x)\]
那么积分$\dint_a^{+\infty} f(t, x) \dif x$$\Omega$上一致收敛。
\end{theorem}
\begin{theorem}[Dirichlet判别法]
如果$f, g$满足下面两个条件:
\begin{enumerate}
\item$A \to +\infty$时,积分$\dint_a^A f(t, x) \dif x$$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$
\[\abs{\int_a^A f(t, x) \dif x} \leq M\]
\item $g(t, x)$$x$的单调函数,且当$x \to +\infty$时关于$t$一致地趋于0。
\end{enumerate}
那么积分
\[\int_a^{+\infty} f(t, x) g(t, x) \dif x\]
$\Omega$上一致收敛。
\end{theorem}
\begin{theorem}[Abel判别法]
如果$f, g$满足下面两个条件:
\begin{enumerate}
\item 积分$\dint_a^{+\infty} f(t, x) \dif x$关于$t \in \Omega$一致收敛;
\item $g(t, x)$$x$单调,且关于$t$一致有界。
\end{enumerate}
那么积分
\[\int_a^{+\infty} f(t, x) g(t, x) \dif x\]
$\Omega$上一致收敛。
\end{theorem}