方企勤 第三章 一元函数积分学 第3题
📝 题目
例 3 设 $f\left( x\right)$ 是可积的且以 $T$ 为周期的周期函数. 求证:
$$
{\int }_{0}^{T}f\left( x\right) \mathrm{d}x = {\int }_{a}^{a + T}f\left( x\right) \mathrm{d}x\;\left( {\forall a \in \mathbf{R}}\right) .
$$
💡 答案解析
证 由题设条件得
$$
{\int }_{a}^{a + T}f\left( x\right) \mathrm{d}x = {\int }_{a}^{0}f\left( x\right) \mathrm{d}x + {\int }_{0}^{T}f\left( x\right) \mathrm{d}x + {\int }_{T}^{a + T}f\left( x\right) \mathrm{d}x
$$
$$
= - {\int }_{0}^{a}f\left( x\right) \mathrm{d}x + {\int }_{0}^{T}f\left( x\right) \mathrm{d}x + {\int }_{0}^{a}f\left( {x + T}\right) \mathrm{d}x
$$
$$
= {\int }_{0}^{a}\left( {f\left( {x + T}\right) - f\left( x\right) }\right) \mathrm{d}x + {\int }_{0}^{T}f\left( x\right) \mathrm{d}x
$$
$$
= {\int }_{0}^{T}f\left( x\right) \mathrm{d}x
$$
提问 本题如下证法对吗?
$F\left( a\right) \overset{\text{ 定义 }}{ = }{\int }_{a}^{a + T}f\left( x\right) \mathrm{d}x$ ,则
$$
{F}^{\prime }\left( a\right) = {\left\{ {\int }_{0}^{a + T}f\left( x\right) \mathrm{d}x - {\int }_{0}^{a}f\left( x\right) \mathrm{d}x\right\} }^{\prime }
$$
$$
= f\left( {a + T}\right) - f\left( a\right) \equiv 0,
$$
因此 $F\left( a\right) \equiv C$ (常数). 又 $F\left( 0\right) = {\int }_{0}^{T}f\left( x\right) \mathrm{d}x$ ,所以
$$
{\int }_{a}^{a + T}f\left( x\right) \mathrm{d}x = F\left( a\right) \equiv C = {\int }_{0}^{T}f\left( x\right) \mathrm{d}x.
$$
解答 这证明是错误的. 错误的原因在于: 本题条件只假定函数 $f\left( x\right)$ 是可积的,这未必能保证变上限积分 $\displaystyle{\int }_{0}^{a}f\left( x\right) \mathrm{d}x$ 对 $a$ 可导.