📝 题目
例 10 求证:
(1) $\mathop{\sum }\limits_{{n = 0}}^{\infty }{\int }_{0}^{x}{t}^{n}\sin {\pi t}\mathrm{\;d}t = {\int }_{0}^{x}\frac{\sin {\pi t}}{1 - t}\mathrm{\;d}t\left( {0 \leq x < 1}\right)$ ；
(2)级数 $\displaystyle{\mathop{\sum }\limits_{{n = 0}}^{\infty }{\int }_{0}^{x}{t}^{n}\sin {\pi t}\mathrm{\;d}t}$ 在 $x \in \left\lbrack {0,1}\right\rbrack$ 上一致收敛；
(3) $\displaystyle{\mathop{\sum }\limits_{{n = 0}}^{\infty }{\int }_{0}^{1}{t}^{n}\sin {\pi t}\mathrm{\;d}t = {\int }_{0}^{1}\frac{\sin {\pi t}}{t}\mathrm{\;d}t}$ .
💡 答案与解析
证 (1) 固定 $x < 1$ 时,因 $\left| {{t}^{n}\sin {\pi t}}\right| \leq {x}^{n}\left( {0 \leq t \leq x}\right)$ ,所以级数 $\displaystyle{\mathop{\sum }\limits_{{n = 0}}^{\infty }{t}^{n}\sin {\pi t}}$ 在 $\left\lbrack {0,x}\right\rbrack$ 上一致收敛. 由逐项积分定理得
$$ \mathop{\sum }\limits_{{n = 0}}^{\infty }{\int }_{0}^{x}{t}^{n}\sin {\pi t}\mathrm{\;d}t = {\int }_{0}^{x}\frac{\sin {\pi t}}{1 - t}\mathrm{\;d}t. $$
(2)当 $0 \leq x \leq 1$ 时,
$$ \left| {{\int }_{0}^{x}{t}^{n}\sin {\pi t}\mathrm{\;d}t}\right| \leq \left| {{\int }_{0}^{1}{t}^{n}\sin {\pi t}\mathrm{\;d}t}\right| = \left| {{\int }_{0}^{1}\sin {\pi t}\mathrm{\;d}\frac{{t}^{n + 1}}{n + 1}}\right| $$
$$ = {\left| \frac{{t}^{n + 1}}{n + 1}\sin \pi t\right| }_{0}^{1} - \frac{\pi }{n + 1}{\int }_{0}^{1}{t}^{n + 1}\cos {\pi t} \cdot \mathrm{d}t $$
$$ \leq \frac{\pi }{n + 1}{\int }_{0}^{1}\left| {{t}^{n + 1}\cos {\pi t}}\right| \mathrm{d}t $$
$$ \leq \frac{\pi }{n + 1}{\int }_{0}^{1}{t}^{n}\mathrm{\;d}t = \frac{\pi }{{\left( n + 1\right) }^{2}}. $$
由 $\mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{\pi }{{\left( n + 1\right) }^{2}} < + \infty$ 知级数 $\displaystyle{\mathop{\sum }\limits_{{n = 0}}^{\infty }{\int }_{0}^{t}{t}^{n}\sin {\pi t}\mathrm{\;d}t}$ 在 $\left\lbrack {0,1}\right\rbrack$ 上一致收敛.
(3)由连续性定理知级数和 $\mathop{\sum }\limits_{{n = 0}}^{\infty }{\int }_{0}^{x}{t}^{n}\sin {\pi t}\mathrm{\;d}t = S\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续,又由 (1) 知在 $\lbrack 0,1)$ 上 $S\left( x\right) = {\int }_{0}^{x}\frac{\sin {\pi t}}{1 - t}\mathrm{\;d}t$ . 令 $x \rightarrow 1$ ,故上式在 $x = 1$ 时也成立,即
$$ \mathop{\sum }\limits_{{n = 0}}^{\infty }{\int }_{0}^{1}{t}^{n}\sin {\pi tdt} = S\left( 1\right) = {\int }_{0}^{1}\frac{\sin {\pi t}}{1 - t}\overset{\text{ 令 }x = 1 - t}{ = }{\int }_{0}^{1}\frac{\sin {\pi x}}{x}\mathrm{\;d}x, $$
所以
$$ \mathop{\sum }\limits_{{n = 0}}^{\infty }{\int }_{0}^{1}{x}^{n}\sin {\pi x}\mathrm{\;d}x = {\int }_{0}^{1}\frac{\sin {\pi x}}{x}\mathrm{\;d}x. $$
评注 若能在闭区间上直接应用逐项积分定理当然更好, 否则先在缩小的区间上应用逐项积分定理, 然后通过连续性定理把结论推广到整个区间.