第十五周。

20含参变量积分.tex

@@ -17,9 +17,100 @@
     \[\tolim{t}{t_0} \int_\alpha^\beta f(x, t) \dif x = \int_\alpha^\beta \tolim{t}{t_0} f(x, t) \dif x\]
 \end{remark}
 
-\begin{theorem}
+\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
-    \[\varphi(u) = \int_a^b f(x, u) \dif x\]
+    \[\varphi(u) = \int_a^b f(x, u) \dif x\eqper\]
     在$[\alpha, \beta]$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]
+    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
+    \[c \leq \alpha(t), \beta(t) \leq d\]
+    则
+    \[f(t) = \int_{\alpha(t)}^{\beta(t)} g(t, x) \dif x\]
+    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}
+    令$J(t, \alpha, \beta) = \dint_\alpha^\beta g(t, x) \dif x$，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$J(t, \alpha, \beta)$在$D$上可微，复合函数
+    \[f(t) = \int_{\alpha(t)}^{\beta(t)} g(t, x) \dif x = J(t, \alpha(t), \beta(t))\]
+    在$t \in [a, b]$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]
+    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
+    \[\int_a^b \left(\int_\alpha^\beta g(t, x) \dif x\right) \dif t = \int_\alpha^\beta \left(\int_a^b g(t, x) \dif t\right)\dif x\]
+    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}
+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
+\[I(t) = \int_a^{+\infty} f(t, x) \dif x\]
+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$A > A_0$时
+    \[\abs{\int_A^{+\infty} f(t, x) \dif x} < \varepsilon\]
+    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}
+
+记
+\[\eta(A) = \sup \limits_{u \in [\alpha, \beta]} \abs{\int_A^{+\infty} f(t, x) \dif x} \eqper\]
+
+\begin{theorem}
+    积分$\dint_a^{+\infty} f(t, x) \dif x$在$\Omega$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$\dint_a^{+\infty} f(t, x) \dif x$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
+    \[\abs{\int_{A^\prime}^{A^{\prime \prime}} f(t, x) \dif x} < \varepsilon\]
+    对任意$t \in \Omega$都成立。
+\end{theorem}
+
+\begin{theorem}[Weierstranss判别法]
+    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
+    \[\abs{f(t, x)} \leq F(x)\]
+    那么积分$\dint_a^{+\infty} f(t, x) \dif x$在$\Omega$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 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
+        \[\abs{\int_a^A f(t, x) \dif x} \leq M\]
+        \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。
+    \end{enumerate}
+    那么积分
+    \[\int_a^{+\infty} f(t, x) g(t, x) \dif x\]
+    在$\Omega$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 $g(t, x)$对$x$单调，且关于$t$一致有界。
+    \end{enumerate}
+    那么积分
+    \[\int_a^{+\infty} f(t, x) g(t, x) \dif x\]
+    在$\Omega$上一致收敛。
 \end{theorem}