取 $T = 2$ 在 (12.11) 和 (12.12) 中，我们得到

$$\begin{aligned} a_0 &= \frac{2}{2} \int_0^2 f(t) dt = \int_0^1 3t dt + \int_1^2 3 dt = \frac{9}{2} \ a_n &= \frac{2}{2} \int_0^2 f(t) \cos \frac{n\pi t}{1} dt \quad (n = 1, 2, 3, \dots) \ &= \int_0^1 3t \cos n\pi t dt + \int_1^2 3 \cos n\pi t dt \ &= \left[ \frac{3t \sin n\pi t}{n\pi} + \frac{3 \cos n\pi t}{(n\pi)^2} \right]_0^1 + \left[ \frac{3 \sin n\pi t}{n\pi} \right]_1^2 \ &= \frac{3}{(n\pi)^2} (\cos n\pi - 1) \ &= \begin{cases} 0 & (\text{even } n) \ -6/(n\pi)^2 & (\text{odd } n) \end{cases} \end{aligned}$$

以及

$$\begin{aligned} b_n &= \frac{2}{2} \int_0^2 f(t) \sin \frac{n\pi t}{1} dt \quad (n = 1, 2, 3, \dots) \ &= \int_0^1 3t \sin n\pi t dt + \int_1^2 3 \sin n\pi t dt \ &= \left[ -\frac{3 \cos n\pi t}{n\pi} + \frac{3 \sin n\pi t}{(n\pi)^2} \right]_0^1 + \left[ -\frac{3 \cos n\pi t}{n\pi} \right]_1^2 = -\frac{3}{n\pi} \cos 2n\pi = -\frac{3}{n\pi} \end{aligned}$$

因此，根据 (12.10)， $f(t)$ 的傅里叶级数展开为

$$\begin{aligned} f(t) &= \frac{9}{4} - \frac{6}{\pi^2} (\cos \pi t + \frac{1}{9} \cos 3\pi t + \frac{1}{25} \cos 5\pi t + \dots) \ &\quad - \frac{3}{\pi} (\sin \pi t + \frac{1}{2} \sin 2\pi t + \frac{1}{3} \sin 3\pi t + \dots) \ &= \frac{9}{4} - \frac{6}{\pi^2} \sum_{n=1}^{\infty} \frac{\cos(2n-1)\pi t}{(2n-1)^2} - \frac{3}{\pi} \sum_{n=1}^{\infty} \frac{\sin n\pi t}{n} \end{aligned}$$