解答 函数 $f(t)$ 对于 $-3\pi < t < 3\pi$ 的图像如图 12.3 所示。

图 12.3
例 12.2 中函数
$f(t)$ 的图像。

由 (12.15) 我们有

$$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) dt = \frac{1}{\pi} \int_{-\pi}^{\pi} (t^2 + t) dt = \frac{2}{3}\pi^2$$

以及

$$\begin{aligned} a_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \cos nt dt \quad (n = 1, 2, 3, \dots) \ &= \frac{1}{\pi} \int_{-\pi}^{\pi} (t^2 + t) \cos nt dt \end{aligned}$$

对其进行分部积分，得到

$$\begin{aligned} a_n &= \frac{1}{\pi} \left[ \frac{t^2}{n} \sin nt + \frac{2t}{n^2} \cos nt - \frac{2}{n^3} \sin nt + \frac{t}{n} \sin nt + \frac{1}{n^2} \cos nt \right]{-\pi}^{\pi} \ &= \frac{1}{\pi} \frac{4\pi}{n^2} \cos n\pi \quad \left( \text{since } \sin n\pi = 0 \text{ and } \left[ \frac{1}{n^2} \cos nt \right]{-\pi}^{\pi} = 0 \right) \ &= \frac{4}{n^2} (-1)^n \quad (\text{since } \cos n\pi = (-1)^n) \end{aligned}$$

由 (12.16)

$$\begin{aligned} b_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin nt dt \quad (n = 1, 2, 3, \dots) \ &= \frac{1}{\pi} \int_{-\pi}^{\pi} (t^2 + t) \sin nt dt \end{aligned}$$

对其进行分部积分，得到

$$\begin{aligned} b_n &= \frac{1}{\pi} \left[ -\frac{t^2}{n} \cos nt + \frac{2t}{n^2} \sin nt + \frac{2}{n^3} \cos nt - \frac{t}{n} \cos nt + \frac{1}{n^2} \sin nt \right]_{-\pi}^{\pi} \ &= -\frac{2}{n} \cos n\pi = -\frac{2}{n} (-1)^n \quad (\text{since } \cos n\pi = (-1)^n) \end{aligned}$$