From 3cbde21116de50c08d9241e8246590705933122b Mon Sep 17 00:00:00 2001 From: unlockable Date: Sat, 3 Dec 2022 11:35:32 +0800 Subject: [PATCH] =?UTF-8?q?=E7=AC=AC11=E5=91=A8=E3=80=82?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 05差值与逼近初步.tex | 6 - 05插值与逼近初步.tex | 6 + 07函数的积分.tex | 361 +++++++++++++++++++++++++++++++++++++++++++ 高等微积分.tex | 9 +- 4 files changed, 372 insertions(+), 10 deletions(-) delete mode 100644 05差值与逼近初步.tex create mode 100644 05插值与逼近初步.tex create mode 100644 07函数的积分.tex diff --git a/05差值与逼近初步.tex b/05差值与逼近初步.tex deleted file mode 100644 index c49ceb2..0000000 --- a/05差值与逼近初步.tex +++ /dev/null @@ -1,6 +0,0 @@ -\chapter{差值与逼近初步} -\section{Lagrange差值公式} -略。 - -\section{Bernstein多项式} -见定理\ref{Weierstrass第一逼近定理}。 \ No newline at end of file diff --git a/05插值与逼近初步.tex b/05插值与逼近初步.tex new file mode 100644 index 0000000..37af42e --- /dev/null +++ b/05插值与逼近初步.tex @@ -0,0 +1,6 @@ +\chapter{插值与逼近初步} +\section{Lagrange差值公式} +略。 + +\section{Bernstein多项式} +见定理\ref{Weierstrass第一逼近定理}。 \ No newline at end of file diff --git a/07函数的积分.tex b/07函数的积分.tex new file mode 100644 index 0000000..a18fa90 --- /dev/null +++ b/07函数的积分.tex @@ -0,0 +1,361 @@ +\chapter{函数的积分} +\section{积分的概念} +问题来源：求曲边梯形的面积。方法：将所求区域分割为小长方形、将面积叠加来逼近。 + +\begin{definition} + 设$f: [a,b] \to \realnum$，如果$\exists A \in \realnum$使得$\forall \varepsilon > 0, \exists \delta > 0$，对于区间$[a,b]$上任意有限分割$T: a = x_0 < x_1 < x_2 < \cdots < x_n = b$，记分割宽度$\Vert T \Vert = \max \limits_{i = 1, 2, \cdots, n} \Delta x_i, \Delta x_i = x_i - x_{i-1}, i = 1, 2, \cdots, n$。只要$\Vert T \Vert < \delta$，都有 + \[\left| \sum_{i = 1}^n f(c_i) \Delta x_i = A \right| < \varepsilon\] + 其中$c_i \in [x_{i-1}, x_i]$任取，$i = 1, 2, \cdots, n$。 + + 这时称$f$在$[a,b]$上Riemann可积，记为$f \in R[a,b]$，并记 + \[\tolim{\Vert T \Vert}{0} \sum_{i=1}^n f(c_i) \Delta x_i = A = \int_a^b f(x) \dif x\] + 称为$f$在$[a,b]$上的积分值，$a$称为积分下限，$b$称为积分上限。 +\end{definition} + +\begin{proposition}[积分的线性性质] + 设$f, g \in R[a, b]$，$\alpha, \beta \in \realnum$，则 + \[\alpha f + \beta g \in R[a,b]\] + 且 + \[\int_a^b [\alpha f(x) + \beta g(x)] \dif x = \alpha \int_a^b f(x) \dif x + \beta \int_a^b g(x) \dif x\eqper\] +\end{proposition} + +\begin{proof} + 任取$[a,b]$上有限分割$T: a = x_0 < x_1 < x_2 < \cdots < x_n = b$。令$\Delta x_i = x_i - x_{i-1}$，$\forall c_i \in [x_{i-1}, x_i], i = 1, 2, \cdots, n$，考虑相应的和式 + \[\sum_{i=1}^n [\alpha f(c_i) + \beta g(c_i)] \Delta x_i = \alpha \sum_{i=1}^n f(c_i) \Delta x_i + \beta \sum_{i=1}^n g(c_i) \Delta x_i\] + 设$\Vert T \Vert \max \limits_{i = 1, 2, \cdots, n} \Delta x_i$。令$\Vert T \Vert \to 0$。等式右侧有极限$\alpha \dint_a^b f(c_i) \Delta x_i + \beta \dint_a^b g(c_i) \Delta x_i$。因此左侧有相同的极限，依照定义$\alpha f + \beta g \in R[a,b]$且 + \[\int_a^b [\alpha f(x) + \beta g(x)] \dif x = \alpha \int_a^b f(x) \dif x + \beta \int_a^b g(x) \dif x\eqper\] +\end{proof} + +\begin{proposition}[积分的保号性质] + 设$f \in R[a,b]$，且$f \geq 0$，则 + \[\int_a^b f(x) \dif x \geq 0 \eqper\] +\end{proposition} + +\begin{corollary}[积分的保序性质] + 设$f, g \in R[a,b]$，且$f \geq g$，则 + \[\int_a^b f(x) \dif x \geq \int_a^b g(x) \dif x \eqper\] +\end{corollary} + +\begin{corollary} + 设$\vert f \vert, f \in R[a,b]$，则 + \[\left| \int_a^b f(x) \dif x \right| \leq \int_a^b \vert f(x) \vert \dif x \eqper\] +\end{corollary} + +\begin{proposition}[积分的区间可加性质] + 设$a < c < b$，则 + \[f \in R[a,b] \Leftrightarrow f \in R[a,c] \text{且} f \in R[c,b]\] + 这时有 + \[\int_a^b f(x) \dif x = \int_a^c f(x) \dif x + \int_c^b f(x) \dif x\eqper\] +\end{proposition} + +我们规定： +\[\int_a^a f(x) \dif x = 0,\quad \int_b^a f(x) \dif x = -\int_a^b f(x) \dif x\eqper\] + +\begin{corollary} + \[\int_a^b f(x) \dif x + \int_b^a f(x) \dif x = \int_a^a f(x) \dif x\eqper\] +\end{corollary} + +\begin{corollary} + \[\int_a^c f(x) \dif x = \int_a^b f(x) \dif x + \int_b^c f(x) \dif x\eqper\] +\end{corollary} + +\section{定积分的计算} +\begin{enumerate} + \item 利用定义：为简化计算通常取均匀分割 + \[T: \Delta x = \frac{b-a}{n}, x_i = a + i\Delta x, i = 0, 1, \cdots, n\] + 则可以取$c_i = x_i$，于是有 + \[\int_a^b f(x) \dif x = \tolim{n}{\infty} \sum_{i=1}^n f(x_i) \Delta x\] + \item 借助几何定义：$f(x)$在$x$轴上方围城的的面积$-$ $f(x)$在$x$轴下方围成的面积。 +\end{enumerate} + +\begin{example} + $f_1(x) \equiv c, x \in [a,b]$，求$\int_a^b f(x) \dif x$。 +\end{example} + +\begin{proof}[解] + 取分割$\Delta x = \dfrac{b-a}{n}$，$x_i = a + i \Delta x$。则分割宽度$\Vert T \Vert = \dfrac{b-a}{n}$。因此$\Vert T \Vert \to 0 \Leftrightarrow n \to \infty$。考察分割时函数的和式 + \[\sum_{i=1}^n f(x_i) \Delta x = \sum_{i=1}^n c \Delta x = c\sum_{i=1}^n \Delta x = c(b-a)\] + 因此 + \[\int_a^b f(x) \dif x = \tolim{n}{\infty} \sum_{i=1}^n f(x_i) \Delta x = c(b-a)\] + 即 + \[\int_a^b c \dif x = c(b-a) \eqper \qedhere\] +\end{proof} + +\begin{corollary}[积分的估值性质] + 设$f \in R[a,b]$且$m \leq f(x) \leq M$，则 + \[m(b-a) \leq \int_a^b f(x) \dif x \leq M(b-a)\eqper\] +\end{corollary} + +\begin{theorem}[Newton-Leibniz公式] + 设函数$f$在$[a,b]$上可积，且在$(a,b)$上有原函数$F$，如果$F$在$(a,b)$上连续，那么必有 + \[\int_a^b f(x) \dif x = F(b) - F(a)\eqper\] + 其中$F(b) - F(a)$也记为$\eval{F(x)}_a^b$。 +\end{theorem} + +上面的定理中的式子也可以写成： +\[\int_a^b \deriv{F}(x) \dif x = \int_a^b \dif F(x) = \eval{F(x)}_a^b\eqper\] + +\begin{proof} + 取$[a,b]$上均匀分割$T: a = x_0 < x_1 < x_2 < \cdots < x_n = b$，其中$x_i = a + i \Delta x, i = 0,1, \cdots, n, \Delta x = \dfrac{b - a}{n}$。这时有 + \[F(b) - F(a) = \sum_{i=1}^n [F(x_i) - F(x_{i-1})]\] + 在每个区间上应用Lagrange中值定理，得 + \[F(x_i) - F(x_{i-1}) = \deriv{F}(c_i) \Delta x, c_i \in (x_{i-1}, x_i)\] + 已知$\deriv{F} = f \in R[a,b]$，因此 + \[\tolim{n}{\infty} \sum_{i=1}^n f(c_i) \Delta x = \int_a^b f(x) \dif x\] + 这说明 + \[F(b) - F(a) = \int_a^b f(x) \dif x\eqper \qedhere\] +\end{proof} + +因此可以将求可积函数$f$的积分问题转化为了求$f$的原函数的问题。 + +\section{可积函数的性质} +目的：寻找可积函数的特性，寻找函数可积的判别条件。 + +\begin{proposition}[可积函数的有界性] + 若$f \in R[a,b]$，则$f$在$[a,b]$上有界。 +\end{proposition} + +\begin{proof} + 令$\int_a^b f(x) \dif x = A$，根据积分的定义，存在均匀分割 + \[T: x_i = a + i\Delta x, i = 0, 1, \cdots, n, \Delta x = \frac{b - a}{n}\] + 使得 + \[\left|\sum_{i=1}^n f(c_i) \Delta x - A\right| < 1, c_i \in [x_{i-1}, x_i]\text{任取}, i = 1, 2, \cdots, n\] + 由此导出 + \[\left|\sum_{i=1}^n f(c_i) \Delta x \right| < \left|\sum_{i=1}^n f(c_i) \Delta x - A\right| + \vert A \vert < 1 + \vert A \vert\] + 因此 + \[\left|\sum_{i=1}^n f(c_i) \right| \leq \frac{1 + \vert A \vert}{\Delta x}\] + 进一步有 + \[\vert f(c_i) \vert \leq \left| \sum_{i=1}^n f(c_i) \right| + \left| \sum_{i=2}^n f(c_i)\right| \leq \frac{1 + \vert A \vert}{\Delta x} + \left| \sum_{i=2}^n f(x_i) \right|\] + 即 + \[\vert f(x) \vert \leq \frac{1 + \vert A \vert}{\Delta x} + \left|\sum_{i\neq 1}^n f(x_i)\right| = M_1, \forall x \in [x_0, x_1]\] + 类似地可以得到 + \begin{align*} + \vert f(x) \vert \leq \frac{1 + \vert A \vert}{\Delta x} + \left|\sum_{i\neq 2}^n f(x_i)\right| & = M_2, \forall x \in [x_1, x_2]\\ + \vdots & \\ + \vert f(x) \vert \leq \frac{1 + \vert A \vert}{\Delta x} + \left|\sum_{i\neq n}^n f(x_i)\right| & = M_n, \forall x \in [x_{n-1}, x_n] + \end{align*} + 将这$n$个式子综合起来就可以得到 + \[\vert f(x) \vert \leq \max \{M_1, \cdots, M_n\}, \forall x \in [a,b] \eqper \qedhere\] +\end{proof} + +\begin{remark} + 有界是可积的必要条件，但不是充分条件。$f$在$[a,b]$上有界未必可积，例如Dirichlet函数$D(x)$。 +\end{remark} + +\begin{proposition}[连续函数的可积性] + $C[a,b] \subset R[a,b]$。 +\end{proposition} + +\begin{corollary}[积分中值定理] + 设$f,g \in C[a,b]$，且$g(x)$不变号，则$\exists c \in [a,b]$使 + \[\int_a^b f(x)g(x) \dif x = f(c) \int_a^b g(x) \dif x\eqper\] + + 特例：设$f \in C[a,b]$，则$\exists c_0 \in [a,b]$使得 + \[f(c_0) = \frac{1}{b-a} \int_a^b f(x) \dif x\] + 这称为$f$在区间$[a,b]$上的平均值。 +\end{corollary} + +\begin{proof} + 不妨设$g(x) \geq 0$，由保号性$\int_a^b g(x) \dif x \geq 0$。同时已知$f \in C[a,b]$，根据最值性质，存在$m = \min \limits_{a \leq x \leq b} f(x) , M = \max \limits_{a \leq x \leq b} f(x)$。这导出 + \[mg(x) \leq f(x)g(x) \leq Mg(x), x \in [a,b]\] + 应用积分的保序性与线性性质有 + \[m \int_a^b g(x) \dif x \leq \int_a^b f(x)g(x) \dif x \leq M \int_a^b g(x) \dif x\] + 不妨设$\dint_a^b g(x) \dif x > 0$，那么上式导出 + \[m \leq \frac{\dint_a^b f(x)g(x) \dif x}{\dint_a^b g(x) \dif x} \leq M\] + 由连续函数介质性质$\exists c \in [a,b]$使得 + \[f(c) = \frac{\dint_a^b f(x)g(x) \dif x}{\dint_a^b g(x) \dif x} \eqper \qedhere\] +\end{proof} + +\section{微积分基本定理} +变上限积分：设$f \in R[a,b]$，则$\forall x \in [a,b], f \in R[a,x]$。定义 +\[F(x) = \int_a^x f(t) \dif t\] +由此得到函数$F: [a,b] \to \realnum$。 + +记号约定： +\[C^n (a,b) = \{f \in C(a,b) \mid f^{(n)} \in C(a,b)\}\\ +C^n [a,b] = \{f \in C[a,b] \mid f^{(n)} \in C[a,b]\}\] + +\begin{theorem} + 设$f$在$[a,b]$上可积，那么带变动上限的积分$F(x) = \int_a^x f(t) \dif t$在$[a,b]$上连续。 +\end{theorem} +\begin{proof} + 任取$x, x + \Delta x \in [a,b], \Delta x > 0$ + \begin{align*} + F(x + \Delta x) - F(x) & = \int_a^{x+\Delta x} f(t) \dif t - \int_a^x f(t) \dif t\\ + & = \int_a^{x + \Delta x} f(t) \dif t + \int_x^a f(t) \dif t \\ + & = \int_x^{x + \Delta x} f(t) \dif t + \end{align*} + 已知可积函数有界，再由估值不等式得到 + \[\vert F(x + \Delta x) - F(x) \vert \leq \int_x^{x + \Delta x} \vert f(t) \vert \dif t \leq M \Delta x\] + 因此 + \[\tolim{\Delta x}{0} \vert F(x + \Delta x) - F(x) \vert = 0\] + 因此$F(x)$连续。 +\end{proof} + +\begin{theorem} + 设函数$f$在$[a,b]$上可积且连续，那么$F$可导且 + \[\deriv{F}(x) = f(x)\eqper\] +\end{theorem} + + +\begin{proof} + 已知 + \[F(x + \Delta x) - F(x) = \int_x^{x + \Delta x} f(t) \dif t\] + 若$f \in C[a,b]$，则可应用积分中值定理：$\exists c \in (x, x + \Delta x)$满足 + \[F(x + \Delta x) - F(x) = \int_x^{x + \Delta x} f(t) \dif t = f(c) \Delta x\] + 即 + \[\frac{F(x + \Delta x) - F(x)}{\Delta x} = f(c) \quad (\Delta x \neq 0)\] + 令$\Delta x \to 0$，则$c \to x$，进而$f(c) \to f(x)$。 +\end{proof} + +结论：设$f \in R[a,b]$，则$F(x) = \int_a^x f(t) \dif t \in C[a,b]$。又若$f \in C[a,b]$，则$F \in C^1 [a,b]$且$\deriv{F}(x) = f(x)$，$\forall x \in [a,b]$。这就是下面的 + +\begin{theorem}[微积分基本定理] + 设$f \in C[a,b]$，则 + \[\frac{\dif}{\dif x} \int_a^x f(t) \dif t = f(x), \forall x \in [a,b]\] + 或写成 + \[\dif \int_a^x f(t) \dif t = f(x) \dif x, \forall x \in [a,b] \eqper\] +\end{theorem} + +\begin{theorem}[Newton-Leibniz公式] + 设$F \in C^1 [a,b]$，则 + \[\int_a^b \deriv{F}(x) \dif x = \eval{F(x)}_a^b\] + 或写成 + \[\int_a^b \dif F(x) = \eval{F(x)}_a^b\eqper\] +\end{theorem} + +\begin{remark} + 考虑变下限积分的导数：若$f$连续， + \[\frac{\dif}{\dif x} \int_x^c f(t) \dif t = \frac{\dif }{\dif x} \left[-\int_c^x f(t) \dif t\right] = -f(x)\eqper\] +\end{remark} + +\begin{example} + 求$\tolim{x}{0} \dfrac{1}{x^3} \dint_0^x tf(t) \dif t$，其中$f$满足$f(0) = 0$，且$\deriv{f}(0)$存在。 +\end{example} + +\begin{proof}[解] + 考虑应用L'Hospital法则： + \[\frac{\dif}{\dif x} \int_0^x tf(t) \dif t = x f(x)\] + 因此 + \begin{align*} + \text{所求} & = \tolim{x}{0} \frac{1}{\deriv{\left(x^3\right)}} \deriv{\left(\int_0^x tf(t) \dif t\right)} = \tolim{x}{0} \frac{xf(x)}{3x^2}\\ + & = \frac{1}{3} \tolim{x}{0} \frac{f(x)}{x} = \frac{1}{3} \tolim{x}{0} \frac{f(x) - f(0)}{x - 0} = \frac{\deriv{f}(0)}{3} \qedhere + \end{align*} +\end{proof} + +\begin{example} + 设$f \in C[a,b], u(t), v(t)$可导且$a \leq u(t), v(t) \leq b$。计算导数 + \[\frac{\dif }{\dif t}\int_{v(t)}^{u(t)} f(x) \dif x\eqper\] +\end{example} + +\begin{proof}[解] + 记 + \[F(u) = \int_a^u f(x) \dif x\] + 则$\deriv{F}(u) = f(u)$。 + 因此有 + \begin{align*} + \int_v^u f(x) \dif x & = \int_v^a f(x) \dif x + \int_a^u f(x) \dif x\\ + & = \int_a^u f(x) \dif x - \int_a^v f(x) \dif x\\ + & = F(u) - F(v) + \intertext{应用链式法则} + \frac{\dif }{\dif t} \int_{v(t)}^{u(t)} f(x) \dif x & = \frac{\dif}{\dif t} [F(u) - F(v)]\\ + & = \deriv{F}(u)\deriv{u} - \deriv{F}(v) \deriv{v}\\ + & = f(u(t))\deriv{u}(t) - f(v(t))\deriv{v}(t) \eqper\qedhere + \end{align*} +\end{proof} + +\section{分部积分与换元} +\subsection{分部积分} +分部积分公式 +\[\int_a^b u(x) \deriv{v}(x) \dif x = \eval{u(x) v(x)}_a^b - \int_a^b v(x) \deriv{u}(x) \dif x\] +或者可以写成 +\[\int_a^b u(x) \dif [v(x)] = \eval{u(x) v(x)}_a^b - \int_a^b v(x) \dif [u(x)]\] + +\begin{example} + 求$I = \dint_0^\pi x^3 \sin x \dif x$。 +\end{example} + +\begin{proof}[解] + 考虑应用分部积分。取$u = x^3, v = -\cos x$。 + \begin{align*} + I & = \eval{-x^3 \cos x}_0^\pi + 3 \int_0^\pi x^2 \cos x \dif x = \pi^3 + 3 \int_0^\pi x^2 \cos x \dif x\\ + & = \pi^3 + 3\left(\eval{x^2 \sin x}_0^\pi - 2\int_0^\pi x \sin x \dif x\right)\\ + & = \pi^3 + 3\times 2 \int_0^\pi x \dif \cos x\\ + & = \pi^3 + 6 \left(\eval{x \cos x}_0^\pi - \int_0^\pi \cos x \dif x\right)\\ + & = \pi^3 + 6 \left(-\pi - \eval{\sin x}_0^\pi\right)\\ + & = \pi^3 - 6\pi \eqper\qedhere + \end{align*} +\end{proof} + +分部积分的应用：设$f$在$[a,b]$上所需的导数均存在。那么做分部积分 +\begin{equation*} + f(x) = f(a) + \int_a^x \deriv{f}(t) \dif t \tag{1} \label{积分Taylor公式1} +\end{equation*} +同时注意到 +\begin{align*} + \int_a^x f(t) \dif t & = \int_a^x \deriv{f}(t) \dif (t - x)\\ + & = \eval{(t - x) \deriv{f}(t)}_a^x - \int_a^x (t-x)f^{\prime \prime}(t) \dif t\\ + & = (x-a) \deriv{f}(a) - \int_a^x (t-x)f^{\prime \prime} (t) \dif t \tag{2} \label{积分Taylor公式2} +\end{align*} + +将\eqref{积分Taylor公式2}式带回\eqref{积分Taylor公式1}式得到 +\[f(x) = f(a) + \deriv{f}(a) (x-a) + \int_a^x (x-t)f^{\prime \prime} (t) \dif t\] + +类似地有 +\begin{align*} + \int_a^x (x-t)f^{\prime \prime} (t) \dif t & = -\frac{1}{2} \int_a^x f^{\prime \prime} (t) \dif \left[(t-x)^2\right]\\ + & = -\frac{1}{2} \left(\eval{\left((t-x)^2 f^{\prime \prime} (t)\right)}_a^x - \int_a^x (t-x)^2 f^{\prime \prime \prime} (t) \dif t\right)\\ + & = \frac{1}{2} (x-a)^2 f^{\prime \prime} (a) + \frac{1}{2} \int_a^x (t-x)^2 f^{\prime \prime \prime} (t) \dif t \tag{3} \label{积分Taylor公式3} +\end{align*} + +将\eqref{积分Taylor公式3}式带回\eqref{积分Taylor公式2}式中，有 +\[f(x) = f(a) + \deriv{f}(a) (x-a) + \frac{1}{2} (x-a)^2 f^{\prime \prime}(a) + \frac{1}{2}\int_a^x (x-t)^2 f^{\prime \prime \prime}(t) \dif t\] +不断如此重复，我们有 +\begin{theorem}[带积分余项的Taylor公式] + 设函数$f$在$(a,b)$上有直到$n+1$阶的连续导函数，那么对任意固定的$x_0 \in (a,b)$，我们有 + \[f(x) = f(x_0) + \frac{1}{1!}\deriv{f}(x_0) (x-x_0) + \cdots + \frac{1}{n!}f^{(n)}(x_0)(x-x_0)^n + R_n(x)\] + 其中 + \[R_n(x) = \frac{1}{n!}\int_{x_0}^x (x-t)^n f^{(n+1)}(t) \dif t, x \in (a, b)\eqper\] +\end{theorem} + +\subsection{定积分换元法} +\begin{theorem} + 设$f \in C(I)$，$I$为区间，$\varphi \in C^1 [\alpha, \beta]$，且$\varphi([\alpha, \beta]) \subset I$。令$\varphi(\alpha) = a, \varphi(\beta) = b$，则 + \[\int_a^b f(x) \dif x = \int_\alpha^\beta f(\varphi(t))\deriv{\varphi}(t) \dif t\eqper\] +\end{theorem} + +\begin{remark} + 不定积分换元公式中$x = \varphi(t)$要存在反函数，而定积分的换元$x = \varphi(t)$并不要求单调性或反函数的存在性，只要函数$\varphi$的值域落在$f$的连续区间内（不必完全在$[a,b]$内）。 +\end{remark} + +\begin{proof} + 记$F(u) = \dint_a^u f(x) \dif x$，则$\deriv{F}(u) = f(u) \in C(I)$。已知$\varphi \in C^1 [a,b]$，且$\varphi([\alpha, \beta]) \subset I$，应用链式法则求导公式 + \[\frac{\dif}{\dif t}[F(\varphi(t))] = \deriv{F}(\varphi(t))\deriv{\varphi}(t) \in C[\alpha, \beta]\] + 再应用Newton-Leibniz公式 + \[\int_\alpha^\beta f(\varphi(t))\deriv{\varphi}(t) \dif t = \eval{F(\varphi(t))}_\alpha^\beta = \int_{\varphi(\alpha)}^{\varphi(\beta)} f(x) \dif x = \int_a^b f(x) \dif x\eqper\qedhere\] +\end{proof} + +\subsubsection{对称区间积分} +设$f \in R[-a,a]$。由区间可加性 +\[\int_{-a}^a f(x) \dif x = \int_0^a f(x) \dif x + \int_{-a}^0 f(x) \dif x \tag{1} \label{对称区间积分1}\] +在区间$[-a,0]$的积分中引入换元$x = -t$，则 +\[\dif x = - \dif t, x(0) = 0, x(-a) = a\] +因此 +\[\int_{-a}^0 f(x) \dif x = - \int_a^0 f(-t) \dif t = \int_0^a f(-x) \dif x \tag{2} \label{对称区间积分2}\] +将\eqref{对称区间积分2}式带入\eqref{对称区间积分1}式，得到 +\[\int_{-a}^a f(x) \dif x = \int_0^a f(x) \dif x + \int_0^a f(-x) \dif x\eqper\] +因此对函数$f \in [-a,a]$ +\begin{enumerate} + \item 若$f$为奇函数，则$\dint_{-a}^a f(x) \dif x = 0$； + \item 若$f$为偶函数，则$\dint_{-a}^a f(x) \dif x = 2 \dint_0^a f(x) \dif x$。 +\end{enumerate} + +\subsubsection{周期函数积分} +设$f \in C(-\infty, + \infty)$有周期$T > 0$。由区间可加性 +\[\int_a^{a+T} f(x) \dif x = \int_a^0 f(x) \dif x + \int_0^T f(x) \dif x + \int_T^{a+T} f(x) \dif x \tag{1} \label{周期函数积分1}\] +引入换元$x = t + T$，那么 +\[\dif x = dif t, x(0) = T, x(a) = a + T\] +因此 +\[\int_T^{a+T} f(x) \dif x = \int_0^a f(t + T) \dif t = \int_0^a f(t) \dif t \tag{2} \label{周期函数积分2}\] +将\eqref{周期函数积分2}式带入\eqref{周期函数积分1}式中，得到 +\[\int_a^{a+T} f(x) \dif x = \int_a^0 f(x) \dif x + \int_0^T f(x) \dif x + \int_0^a f(x) \dif x = \int_0^T f(x) \dif x \text{对} \forall a \in \realnum\text{成立。}\] diff --git a/高等微积分.tex b/高等微积分.tex index 21cb5d9..eea07fc 100644 --- a/高等微积分.tex +++ b/高等微积分.tex @@ -14,6 +14,7 @@ \usepackage{multicol} \usepackage{float} \usepackage{extarrows} +\usepackage{physics} % \usepackage{mathptmx} \geometry{a4paper,scale=0.8} @@ -22,7 +23,7 @@ \ctexset{fontset=macnew} % \ctexset{fontset=windows} % On Windows -\allowdisplaybreaks[4] +\allowdisplaybreaks[3] \newtheorem{theorem}{定理}[section] \newtheorem{axiom}{公理}[section] @@ -51,14 +52,13 @@ \newcommand{\dint}{\displaystyle\int} \DeclareMathOperator{\sgn}{sgn} -\DeclareMathOperator{\arccot}{arccot} \title{{\Huge{\textbf{高等微积分}}}} \author{} \date{} % linespread{1.5} -% \includeonly{06求导的逆运算.tex} +% \includeonly{07函数的积分.tex} \begin{document} \maketitle @@ -73,6 +73,7 @@ \include{02函数及其连续性.tex} \include{03函数的导数.tex} \include{04微分与Taylor定理.tex} - \include{05差值与逼近初步.tex} + \include{05插值与逼近初步.tex} \include{06求导的逆运算.tex} + \include{07函数的积分.tex} \end{document} \ No newline at end of file