T1

设 $f(x)$ 在 $[0, +\infty)$ 有定义，且在任意 $[0,a]$ 上可积，$\lim\limits_{x \to +\infty} f(x) = A$

证明：
$$
\boldsymbol{\lim\limits_{x \to +\infty} \frac{1}{x}\int_{0}^{x} f(t),\mathrm{d}t = A}
$$

证：
$$
\begin{align}
\left| \frac{1}{x}\int_{0}^{x} f(t),\mathrm{d}t - A \right| &= \left| \frac{1}{x}\int_{0}^{x} \left( f(t) - A \right) ,\mathrm{d}t \right|\
&= \left| \frac{1}{x}\int_{0}^{M} \left( f(t) - A \right) ,\mathrm{d}t + \frac{1}{x}\int_{M}^{x} \left( f(t) - A \right) ,\mathrm{d}t \right|\
&\leq \left| \frac{1}{x}\int_{0}^{M} \left( f(t) - A \right) ,\mathrm{d}t \right| + \left| \frac{1}{x}\int_{M}^{x} \left( f(t) - A \right) ,\mathrm{d}t \right|
\end{align}
$$
由于
$$
\lim\limits_{x \to +\infty} \frac{1}{x} = 0 \quad \lim\limits_{x \to +\infty} f(x) = A
$$
则对 $\forall \varepsilon > 0$，$\exists M > 0$，当 $x \geq M$ 时，有
$$
|f(x) - A| < \frac{\varepsilon}{2}
$$
由于 $\displaystyle \left| \int_{0}^{M} \left( f(t) - A \right) ,\mathrm{d}t \right|$ 有界，则有
$$
\lim\limits_{x \to +\infty} \left| \frac{1}{x}\int_{0}^{M} \left( f(t) - A \right) ,\mathrm{d}t \right| = 0
$$
从而对 $\varepsilon > 0$，$\exists N > M > 0$，当 $x \geq N$ 时，有
$$
\left| \frac{1}{x}\int_{0}^{M} \left( f(t) - A \right) ,\mathrm{d}t \right| < \frac{\varepsilon}{2}
$$
则当 $x \geq N$ 时，有
$$
\begin{align}
\left| \frac{1}{x}\int_{0}^{x} f(t),\mathrm{d}t - A \right| &\leq \left| \frac{1}{x}\int_{0}^{M} \left( f(t) - A \right) ,\mathrm{d}t \right| + \left| \frac{1}{x}\int_{M}^{x} \left( f(t) - A \right) ,\mathrm{d}t \right|\
&< \frac{\varepsilon}{2} + \frac{x - M}{x} \cdot \frac{\varepsilon}{2}\
&< \varepsilon
\end{align}
$$
故
$$
\lim\limits_{x \to +\infty} \frac{1}{x}\int_{0}^{x} f(t),\mathrm{d}t = A
$$

T2

设 $f$ 是定义在 $(-\infty, +\infty)$ 上的一个连续周期函数，周期为 $p$

证明：
$$
\boldsymbol{\lim\limits_{x \to +\infty} \frac{1}{x}\int_{0}^{x} f(t),\mathrm{d}t = \frac{1}{p}\int_{0}^{p} f(t),\mathrm{d}t}
$$

证：
设 $[0, x)$ 上包含 $n$ 个周期，即 $\displaystyle n = \left[ \frac{x}{p} \right]$
$$
\frac{1}{x}\int_{0}^{x} f(t),\mathrm{d}t = \frac{1}{x}\int_{0}^{np} f(t),\mathrm{d}t + \frac{1}{x}\int_{np}^{x} f(t),\mathrm{d}t
$$

由于 $f(t)$ 周期性，$\displaystyle \int_{0}^{np} f(t),\mathrm{d}t = n\int_{0}^{p} f(t),\mathrm{d}t$，故：

$$
\frac{1}{x}\int_{0}^{x} f(t),\mathrm{d}t = \frac{n}{x}\int_{0}^{p} f(t),\mathrm{d}t + \frac{1}{x}\int_{np}^{x} f(t),\mathrm{d}t
$$

放缩得：

$$
\begin{align}
\frac{1}{x}\int_{0}^{x} f(t),\mathrm{d}t &\leq \frac{n}{x}\int_{0}^{p} f(t),\mathrm{d}t + \frac{1}{x}\int_{np}^{(n + 1)p} f(t),\mathrm{d}t\ &= \frac{n + 1}{x}\int_{0}^{p} f(t),\mathrm{d}t
\end{align}
$$

且

$$
\frac{1}{x}\int_{0}^{x} f(t),\mathrm{d}t \geq \frac{n}{x}\int_{0}^{p} f(t),\mathrm{d}t
$$

则：

$$
\left| \frac{1}{x}\int_{0}^{x} f(t),\mathrm{d}t - \frac{1}{p}\int_{0}^{p} f(t),\mathrm{d}t \right| \leq \left| \left( \frac{n + 1}{x} - \frac{1}{p} \right)\int_{0}^{p} f(t),\mathrm{d}t \right|
$$

由
$$
\frac{x}{p} - 1 \leq n \leq \frac{x}{p}
$$
可得
$$
\frac{1}{p} \leq \frac{n + 1}{x} \leq \frac{1}{p} + \frac{1}{x}
$$
根据夹逼定理得
$$
\lim\limits_{x \to +\infty} \frac{n + 1}{x} = \frac{1}{p}
$$
对 $\forall \varepsilon > 0$，$\exists M > 0$，当 $x > M$ 时：

$$
\left| \frac{n + 1}{x} - \frac{1}{p} \right| < \frac{\varepsilon}{\displaystyle \left| \int_{0}^{p} f(t),\mathrm{d}t \right|}
$$

故当 $x > M$ 时：

$$
\left| \frac{1}{x}\int_{0}^{x} f(t),\mathrm{d}t - \frac{1}{p}\int_{0}^{p} f(t),\mathrm{d}t \right| < \varepsilon
$$

即：

$$
\lim\limits_{x \to +\infty} \frac{1}{x}\int_{0}^{x} f(t),\mathrm{d}t = \frac{1}{p}\int_{0}^{p} f(t),\mathrm{d}t
$$
此题可以作为结论，引申出许多题目，例如下面这题

T3

求极限：
$$
\boldsymbol{\lim\limits_{x \to +\infty} \frac{\displaystyle \int_{0}^{x} |\sin t| ,\mathrm{d}t}{x}}
$$

解：

已知 $|\sin x|$ 为 $[0, +\infty)$ 上的连续周期函数，周期为 $\pi$，则存在 $n \in \mathbb{N}$，使得
$$
n\pi \leq x < (n + 1)\pi
$$
则有：
$$
\int_{0}^{n\pi} |\sin t| ,\mathrm{d}t \leq \int_{0}^{x} |\sin t| ,\mathrm{d}t \leq \int_{0}^{(n + 1)\pi} |\sin t| ,\mathrm{d}t
$$
因 $|\sin t|$ 周期为 $\pi$，且
$$
\int_{0}^{\pi} |\sin t| ,\mathrm{d}t = \int_{0}^{\pi} \sin t ,\mathrm{d}t = 2
$$
故：
$$
\begin{align}
\int_{0}^{n\pi} |\sin t| ,\mathrm{d}t &= 2n\
\int_{0}^{(n + 1)\pi} |\sin t| ,\mathrm{d}t &= 2(n + 1)
\end{align}
$$
结合 $n\pi \leq x < (n + 1)\pi$，对原式放缩得：
$$
\frac{2n}{(n + 1)\pi} \leq \frac{\displaystyle \int_{0}^{x} |\sin t| ,\mathrm{d}t}{x} \leq \frac{2(n + 1)}{n\pi}
$$
当$x \to +\infty$时，$n \to +\infty$，计算两端极限：
$$
\begin{align}
\lim\limits_{n \to +\infty} \frac{2n}{(n + 1)\pi} &= \lim\limits_{n \to +\infty} \frac{2}{\left(1 + \frac{1}{n}\right)\pi} = \frac{2}{\pi}\
\lim\limits_{n \to +\infty} \frac{2(n + 1)}{n\pi} &= \lim\limits_{n \to +\infty} \frac{2\left(1 + \frac{1}{n}\right)}{\pi} = \frac{2}{\pi}
\end{align}
$$
由夹逼定理，得：
$$
\lim\limits_{x \to +\infty} \frac{\displaystyle \int_{0}^{x} |\sin t| ,\mathrm{d}t}{x} = \frac{2}{\pi}
$$

T4

若 $f$ 在 $[0,a]$ 上连续可微，且 $f(0) = 0$

证明：
$$
\boldsymbol{\int_{0}^{a} |f(x)f’(x)| , \mathrm{d}x \leq \frac{a}{2}\int_{0}^{a} [f’(x)]^2 ,\mathrm{d}x}
$$

证：

绝对值不太好处理，不妨设 $\displaystyle F(x)=\int_{0}^{x} |f’(t)| ,\mathrm{d}t$

易知：
$$
\begin{align}
F(x)=\int_{0}^{x} |f’(t)|,\mathrm{d}t &\geq \left| \int_{0}^{x} f’(t) ,\mathrm{d}t \right| = |f(x)|\
F’(x) &=|f’(x)|
\end{align}
$$
则原式化为
$$
\begin{align}
\int_{0}^{a} |f(x)||f’(x)| ,\mathrm{d}x &\leq \int_{0}^{a} F(x) \cdot F’(x) ,\mathrm{d}x \
&= \frac{1}{2}F^2(x)\big|{0}^{a} = \frac{1}{2}F^2(a)\
& = \frac{1}{2}\left( \int{0}^{a} |f’(t)| ,\mathrm{d}t \right)^2 \
\end{align}
$$
最后由 Cauchy Schwartz 不等式可得
$$
\left( \int_{0}^{a} |f’(t)| \cdot 1 ,\mathrm{d}t \right)^2 \leq \int_{0}^{a} [f’(t)]^2 ,\mathrm{d}t \cdot \int_{0}^{a} 1^2 ,\mathrm{d}t = a\int_{0}^{a} [f’(t)]^2 ,\mathrm{d}t
$$
即
$$
\int_{0}^{a} |f(x)f’(x)| ,\mathrm{d}x \leq \frac{a}{2}\int_{0}^{a} [f’(x)]^2 ,\mathrm{d}x
$$

T5

证明：设 $f$ 在 $[a,b]$ 上可积，且处处有 $f(x)>0$，则 $\displaystyle \int_{a}^{b} f(x),\mathrm{d}x > 0$

**证：**由 $f$ 在 $[a,b]$ 上可积，故 $\exists\ x_0 \in [a,b]$，$f(x)$ 在 $x_0$ 连续且 $f(x_0)>0$

于是 $\exists\ [\alpha, \beta] \subseteq [a,b]$，$s.t. \ f(x) \geq \displaystyle \frac{f(x_0)}{2},\ x \in [\alpha, \beta]$，则有
$$
\int_{a}^{b} f(x),\mathrm{d}x \geq \int_{\alpha}^{\beta} f(x),\mathrm{d}x \geq \frac{f(x_0)}{2} (\beta - \alpha) > 0
$$