362 lines
17 KiB
TeX
362 lines
17 KiB
TeX
\chapter{函数的积分}
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\section{积分的概念}
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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。
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\begin{definition}
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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
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\[\left| \sum_{i = 1}^n f(c_i) \Delta x_i = A \right| < \varepsilon\]
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其中$c_i \in [x_{i-1}, x_i]$任取，$i = 1, 2, \cdots, n$。
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这时称$f$在$[a,b]$上Riemann可积，记为$f \in R[a,b]$，并记
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\[\tolim{\Vert T \Vert}{0} \sum_{i=1}^n f(c_i) \Delta x_i = A = \int_a^b f(x) \dif x\]
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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。
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\end{definition}
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\begin{proposition}[积分的线性性质]
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设$f, g \in R[a, b]$，$\alpha, \beta \in \realnum$，则
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\[\alpha f + \beta g \in R[a,b]\]
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且
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\[\int_a^b [\alpha f(x) + \beta g(x)] \dif x = \alpha \int_a^b f(x) \dif x + \beta \int_a^b g(x) \dif x\eqper\]
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\end{proposition}
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\begin{proof}
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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
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\[\sum_{i=1}^n [\alpha f(c_i) + \beta g(c_i)] \Delta x_i = \alpha \sum_{i=1}^n f(c_i) \Delta x_i + \beta \sum_{i=1}^n g(c_i) \Delta x_i\]
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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
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\[\int_a^b [\alpha f(x) + \beta g(x)] \dif x = \alpha \int_a^b f(x) \dif x + \beta \int_a^b g(x) \dif x\eqper\]
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\end{proof}
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\begin{proposition}[积分的保号性质]
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设$f \in R[a,b]$，且$f \geq 0$，则
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\[\int_a^b f(x) \dif x \geq 0 \eqper\]
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\end{proposition}
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\begin{corollary}[积分的保序性质]
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设$f, g \in R[a,b]$，且$f \geq g$，则
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\[\int_a^b f(x) \dif x \geq \int_a^b g(x) \dif x \eqper\]
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\end{corollary}
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\begin{corollary}
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设$\vert f \vert, f \in R[a,b]$，则
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\[\left| \int_a^b f(x) \dif x \right| \leq \int_a^b \vert f(x) \vert \dif x \eqper\]
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\end{corollary}
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\begin{proposition}[积分的区间可加性质]
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设$a < c < b$，则
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\[f \in R[a,b] \Leftrightarrow f \in R[a,c] \text{且} f \in R[c,b]\]
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这时有
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\[\int_a^b f(x) \dif x = \int_a^c f(x) \dif x + \int_c^b f(x) \dif x\eqper\]
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\end{proposition}
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我们规定：
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\[\int_a^a f(x) \dif x = 0,\quad \int_b^a f(x) \dif x = -\int_a^b f(x) \dif x\eqper\]
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\begin{corollary}
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\[\int_a^b f(x) \dif x + \int_b^a f(x) \dif x = \int_a^a f(x) \dif x\eqper\]
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\end{corollary}
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\begin{corollary}
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\[\int_a^c f(x) \dif x = \int_a^b f(x) \dif x + \int_b^c f(x) \dif x\eqper\]
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\end{corollary}
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\section{定积分的计算}
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\begin{enumerate}
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\item 利用定义：为简化计算通常取均匀分割
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\[T: \Delta x = \frac{b-a}{n}, x_i = a + i\Delta x, i = 0, 1, \cdots, n\]
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则可以取$c_i = x_i$，于是有
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\[\int_a^b f(x) \dif x = \tolim{n}{\infty} \sum_{i=1}^n f(x_i) \Delta x\]
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\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。
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\end{enumerate}
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\begin{example}
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$f_1(x) \equiv c, x \in [a,b]$，求$\int_a^b f(x) \dif x$。
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\end{example}
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\begin{proof}[解]
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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
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\[\sum_{i=1}^n f(x_i) \Delta x = \sum_{i=1}^n c \Delta x = c\sum_{i=1}^n \Delta x = c(b-a)\]
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因此
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\[\int_a^b f(x) \dif x = \tolim{n}{\infty} \sum_{i=1}^n f(x_i) \Delta x = c(b-a)\]
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即
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\[\int_a^b c \dif x = c(b-a) \eqper \qedhere\]
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\end{proof}
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\begin{corollary}[积分的估值性质]
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设$f \in R[a,b]$且$m \leq f(x) \leq M$，则
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\[m(b-a) \leq \int_a^b f(x) \dif x \leq M(b-a)\eqper\]
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\end{corollary}
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\begin{theorem}[Newton-Leibniz公式]
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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
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\[\int_a^b f(x) \dif x = F(b) - F(a)\eqper\]
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其中$F(b) - F(a)$也记为$\eval{F(x)}_a^b$。
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\end{theorem}
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上面的定理中的式子也可以写成：
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\[\int_a^b \deriv{F}(x) \dif x = \int_a^b \dif F(x) = \eval{F(x)}_a^b\eqper\]
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\begin{proof}
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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
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\[F(b) - F(a) = \sum_{i=1}^n [F(x_i) - F(x_{i-1})]\]
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在每个区间上应用Lagrange中值定理，得
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\[F(x_i) - F(x_{i-1}) = \deriv{F}(c_i) \Delta x, c_i \in (x_{i-1}, x_i)\]
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已知$\deriv{F} = f \in R[a,b]$，因此
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\[\tolim{n}{\infty} \sum_{i=1}^n f(c_i) \Delta x = \int_a^b f(x) \dif x\]
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这说明
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\[F(b) - F(a) = \int_a^b f(x) \dif x\eqper \qedhere\]
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\end{proof}
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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。
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\section{可积函数的性质}
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目的：寻找可积函数的特性，寻找函数可积的判别条件。
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\begin{proposition}[可积函数的有界性]
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若$f \in R[a,b]$，则$f$在$[a,b]$上有界。
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\end{proposition}
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\begin{proof}
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令$\int_a^b f(x) \dif x = A$，根据积分的定义，存在均匀分割
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\[T: x_i = a + i\Delta x, i = 0, 1, \cdots, n, \Delta x = \frac{b - a}{n}\]
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使得
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\[\left|\sum_{i=1}^n f(c_i) \Delta x - A\right| < 1, c_i \in [x_{i-1}, x_i]\text{任取}, i = 1, 2, \cdots, n\]
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由此导出
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\[\left|\sum_{i=1}^n f(c_i) \Delta x \right| < \left|\sum_{i=1}^n f(c_i) \Delta x - A\right| + \vert A \vert < 1 + \vert A \vert\]
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因此
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\[\left|\sum_{i=1}^n f(c_i) \right| \leq \frac{1 + \vert A \vert}{\Delta x}\]
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进一步有
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\[\vert f(c_i) \vert \leq \left| \sum_{i=1}^n f(c_i) \right| + \left| \sum_{i=2}^n f(c_i)\right| \leq \frac{1 + \vert A \vert}{\Delta x} + \left| \sum_{i=2}^n f(x_i) \right|\]
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即
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\[\vert f(x) \vert \leq \frac{1 + \vert A \vert}{\Delta x} + \left|\sum_{i\neq 1}^n f(x_i)\right| = M_1, \forall x \in [x_0, x_1]\]
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类似地可以得到
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\begin{align*}
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\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]\\
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\vdots & \\
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\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]
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\end{align*}
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将这$n$个式子综合起来就可以得到
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\[\vert f(x) \vert \leq \max \{M_1, \cdots, M_n\}, \forall x \in [a,b] \eqper \qedhere\]
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\end{proof}
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\begin{remark}
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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$D(x)$。
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\end{remark}
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\begin{proposition}[连续函数的可积性]
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$C[a,b] \subset R[a,b]$。
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\end{proposition}
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\begin{corollary}[积分中值定理]
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设$f,g \in C[a,b]$，且$g(x)$不变号，则$\exists c \in [a,b]$使
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\[\int_a^b f(x)g(x) \dif x = f(c) \int_a^b g(x) \dif x\eqper\]
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特例：设$f \in C[a,b]$，则$\exists c_0 \in [a,b]$使得
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\[f(c_0) = \frac{1}{b-a} \int_a^b f(x) \dif x\]
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这称为$f$在区间$[a,b]$上的平均值。
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\end{corollary}
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\begin{proof}
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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
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\[mg(x) \leq f(x)g(x) \leq Mg(x), x \in [a,b]\]
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应用积分的保序性与线性性质有
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\[m \int_a^b g(x) \dif x \leq \int_a^b f(x)g(x) \dif x \leq M \int_a^b g(x) \dif x\]
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不妨设$\dint_a^b g(x) \dif x > 0$，那么上式导出
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\[m \leq \frac{\dint_a^b f(x)g(x) \dif x}{\dint_a^b g(x) \dif x} \leq M\]
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由连续函数介质性质$\exists c \in [a,b]$使得
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\[f(c) = \frac{\dint_a^b f(x)g(x) \dif x}{\dint_a^b g(x) \dif x} \eqper \qedhere\]
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\end{proof}
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\section{微积分基本定理}
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变上限积分：设$f \in R[a,b]$，则$\forall x \in [a,b], f \in R[a,x]$。定义
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\[F(x) = \int_a^x f(t) \dif t\]
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由此得到函数$F: [a,b] \to \realnum$。
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记号约定：
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\[C^n (a,b) = \{f \in C(a,b) \mid f^{(n)} \in C(a,b)\}\\
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C^n [a,b] = \{f \in C[a,b] \mid f^{(n)} \in C[a,b]\}\]
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\begin{theorem}
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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。
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\end{theorem}
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\begin{proof}
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任取$x, x + \Delta x \in [a,b], \Delta x > 0$
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\begin{align*}
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F(x + \Delta x) - F(x) & = \int_a^{x+\Delta x} f(t) \dif t - \int_a^x f(t) \dif t\\
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& = \int_a^{x + \Delta x} f(t) \dif t + \int_x^a f(t) \dif t \\
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& = \int_x^{x + \Delta x} f(t) \dif t
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\end{align*}
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已知可积函数有界，再由估值不等式得到
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\[\vert F(x + \Delta x) - F(x) \vert \leq \int_x^{x + \Delta x} \vert f(t) \vert \dif t \leq M \Delta x\]
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因此
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\[\tolim{\Delta x}{0} \vert F(x + \Delta x) - F(x) \vert = 0\]
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因此$F(x)$连续。
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\end{proof}
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\begin{theorem}
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设函数$f$在$[a,b]$上可积且连续，那么$F$可导且
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\[\deriv{F}(x) = f(x)\eqper\]
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\end{theorem}
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\begin{proof}
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已知
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\[F(x + \Delta x) - F(x) = \int_x^{x + \Delta x} f(t) \dif t\]
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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
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\[F(x + \Delta x) - F(x) = \int_x^{x + \Delta x} f(t) \dif t = f(c) \Delta x\]
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即
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\[\frac{F(x + \Delta x) - F(x)}{\Delta x} = f(c) \quad (\Delta x \neq 0)\]
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令$\Delta x \to 0$，则$c \to x$，进而$f(c) \to f(x)$。
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\end{proof}
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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
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\begin{theorem}[微积分基本定理]
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设$f \in C[a,b]$，则
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\[\frac{\dif}{\dif x} \int_a^x f(t) \dif t = f(x), \forall x \in [a,b]\]
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或写成
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\[\dif \int_a^x f(t) \dif t = f(x) \dif x, \forall x \in [a,b] \eqper\]
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\end{theorem}
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\begin{theorem}[Newton-Leibniz公式]
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设$F \in C^1 [a,b]$，则
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\[\int_a^b \deriv{F}(x) \dif x = \eval{F(x)}_a^b\]
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或写成
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\[\int_a^b \dif F(x) = \eval{F(x)}_a^b\eqper\]
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\end{theorem}
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\begin{remark}
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考虑变下限积分的导数：若$f$连续，
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\[\frac{\dif}{\dif x} \int_x^c f(t) \dif t = \frac{\dif }{\dif x} \left[-\int_c^x f(t) \dif t\right] = -f(x)\eqper\]
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\end{remark}
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\begin{example}
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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。
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\end{example}
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\begin{proof}[解]
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考虑应用L'Hospital法则：
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\[\frac{\dif}{\dif x} \int_0^x tf(t) \dif t = x f(x)\]
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因此
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\begin{align*}
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\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}\\
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& = \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
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\end{align*}
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\end{proof}
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\begin{example}
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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
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\[\frac{\dif }{\dif t}\int_{v(t)}^{u(t)} f(x) \dif x\eqper\]
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\end{example}
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\begin{proof}[解]
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记
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\[F(u) = \int_a^u f(x) \dif x\]
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则$\deriv{F}(u) = f(u)$。
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因此有
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\begin{align*}
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\int_v^u f(x) \dif x & = \int_v^a f(x) \dif x + \int_a^u f(x) \dif x\\
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& = \int_a^u f(x) \dif x - \int_a^v f(x) \dif x\\
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& = F(u) - F(v)
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\intertext{应用链式法则}
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\frac{\dif }{\dif t} \int_{v(t)}^{u(t)} f(x) \dif x & = \frac{\dif}{\dif t} [F(u) - F(v)]\\
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& = \deriv{F}(u)\deriv{u} - \deriv{F}(v) \deriv{v}\\
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& = f(u(t))\deriv{u}(t) - f(v(t))\deriv{v}(t) \eqper\qedhere
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\end{align*}
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\end{proof}
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\section{分部积分与换元}
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\subsection{分部积分}
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分部积分公式
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\[\int_a^b u(x) \deriv{v}(x) \dif x = \eval{u(x) v(x)}_a^b - \int_a^b v(x) \deriv{u}(x) \dif x\]
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或者可以写成
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\[\int_a^b u(x) \dif [v(x)] = \eval{u(x) v(x)}_a^b - \int_a^b v(x) \dif [u(x)]\]
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\begin{example}
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求$I = \dint_0^\pi x^3 \sin x \dif x$。
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\end{example}
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\begin{proof}[解]
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考虑应用分部积分。取$u = x^3, v = -\cos x$。
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\begin{align*}
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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\\
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& = \pi^3 + 3\left(\eval{x^2 \sin x}_0^\pi - 2\int_0^\pi x \sin x \dif x\right)\\
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& = \pi^3 + 3\times 2 \int_0^\pi x \dif \cos x\\
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& = \pi^3 + 6 \left(\eval{x \cos x}_0^\pi - \int_0^\pi \cos x \dif x\right)\\
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& = \pi^3 + 6 \left(-\pi - \eval{\sin x}_0^\pi\right)\\
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& = \pi^3 - 6\pi \eqper\qedhere
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\end{align*}
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\end{proof}
|
||
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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
|
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\begin{equation*}
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f(x) = f(a) + \int_a^x \deriv{f}(t) \dif t \tag{1} \label{积分Taylor公式1}
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\end{equation*}
|
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同时注意到
|
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\begin{align*}
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\int_a^x f(t) \dif t & = \int_a^x \deriv{f}(t) \dif (t - x)\\
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& = \eval{(t - x) \deriv{f}(t)}_a^x - \int_a^x (t-x)f^{\prime \prime}(t) \dif t\\
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& = (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}
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\end{align*}
|
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|
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将\eqref{积分Taylor公式2}式带回\eqref{积分Taylor公式1}式得到
|
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\[f(x) = f(a) + \deriv{f}(a) (x-a) + \int_a^x (x-t)f^{\prime \prime} (t) \dif t\]
|
||
|
||
类似地有
|
||
\begin{align*}
|
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\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]\\
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& = -\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)\\
|
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& = \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
|
||
\[\int_a^{a+T} f(x) \dif x = \int_a^0 f(x) \dif x + \int_0^T f(x) \dif x + \int_0^a f(x) \dif x = \int_0^T f(x) \dif x \text{对} \forall a \in \realnum\text{成立。}\]
|