📝 题目
例 8 将函数 $f\left( x\right) = x\cos x\left( {-\frac{\pi }{2} \leq x \leq \frac{\pi }{2}}\right)$ 展开为傅氏级数.
💡 答案与解析
解 记 $l = \pi /2$ ,因为 $f\left( x\right)$ 是奇函数,所以 ${a}_{n} = 0\left( {n \geq 0}\right)$ ,且
$$ {b}_{n} = \frac{2}{l}{\int }_{0}^{l}x \cdot \cos x \cdot \sin \left( {2nx}\right) \mathrm{d}x $$
$$ = \frac{1}{l}{\int }_{0}^{l}x\left\lbrack {\sin \left( {{2n} - 1}\right) x + \sin \left( {{2n} + 1}\right) x}\right\rbrack \mathrm{d}x $$
$$ = \frac{1}{l}{\left\lbrack -\frac{1}{\left( 2n - 1\right) }x\cos \left( 2n - 1\right) x\right| }_{0}^{l} $$
$$ \left. {+\frac{1}{{2n} - 1}{\int }_{0}^{l}\cos \left( {{2n} - 1}\right) x\mathrm{\;d}x}\right\rbrack $$
$$ + \frac{1}{l}{\left\lbrack -\frac{1}{\left( 2n + 1\right) } \times x\cos \left( 2n + 1\right) x\right| }_{0}^{l} $$
$$ \left. {+\frac{1}{{2n} + 1}{\int }_{0}^{l}\cos \left( {{2n} + 1}\right) x\mathrm{\;d}x}\right\rbrack $$
$$ = \frac{1}{l}\left\lbrack {\frac{1}{{\left( 2n - 1\right) }^{2}}\sin \left( {{2n} - 1}\right) l + \frac{1}{{\left( 2n + 1\right) }^{2}}\sin \left( {{2n} + 1}\right) l}\right\rbrack $$
$$ = \frac{{\left( -1\right) }^{n - 1}}{l}\left\lbrack {\frac{1}{{\left( 2n - 1\right) }^{2}} - \frac{1}{2{\left( n + 1\right) }^{2}}}\right\rbrack $$
$$ = \frac{{\left( -1\right) }^{n - 1}}{\pi } \cdot \frac{16n}{{\left( 4{n}^{2} - 1\right) }^{2}}, $$
即得
$$ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{\pi } \cdot \frac{16n}{{\left( 4{n}^{2} - 1\right) }^{2}}\sin {2nx} = x\cos x\;\left( {\left| x\right| \leq \frac{\pi }{2}}\right) . $$