且

$$\begin{aligned} a_n &= \frac{1}{\pi} \int_0^{2\pi} f(t) \cos nt , dt \quad (n = 1, 2, 3, \dots) \ &= \frac{1}{\pi} \left[ \int_0^{\pi/2} t \cos nt , dt + \int_{\pi/2}^{\pi} \frac{1}{2} \pi \cos nt , dt + \int_{\pi}^{2\pi} \left( \pi - \frac{1}{2}t \right) \cos nt , dt \right] \ &= \frac{1}{\pi} \left( \left[ \frac{t}{n} \sin nt + \frac{\cos nt}{n^2} \right]0^{\pi/2} + \left[ \frac{\pi}{2n} \sin nt \right]{\pi/2}^{\pi} + \left[ \frac{2\pi - t}{2} \frac{\sin nt}{n} - \frac{\cos nt}{2n^2} \right]_{\pi}^{2\pi} \right) \ &= \frac{1}{\pi} \left( \frac{\pi}{2n} \sin \frac{1}{2}n\pi + \frac{1}{n^2} \cos \frac{1}{2}n\pi - \frac{1}{n^2} - \frac{\pi}{2n} \sin \frac{1}{2}n\pi - \frac{1}{2n^2} + \frac{1}{2n^2} \cos n\pi \right) \ &= \frac{1}{2\pi n^2} (2 \cos \frac{1}{2}n\pi - 3 + \cos n\pi) \end{aligned}$$

也就是说，

$$a_n = \begin{cases} \frac{1}{\pi n^2} [(-1)^{n/2} - 1] & (\text{even } n) \ -\frac{2}{\pi n^2} & (\text{odd } n) \end{cases}$$

由 (12.16)，

$$\begin{aligned} b_n &= \frac{1}{\pi} \int_0^{2\pi} f(t) \sin nt , dt \quad (n = 1, 2, 3, \dots) \ &= \frac{1}{\pi} \left[ \int_0^{\pi/2} t \sin nt , dt + \int_{\pi/2}^{\pi} \frac{1}{2} \pi \sin nt , dt + \int_{\pi}^{2\pi} \left( \pi - \frac{1}{2}t \right) \sin nt , dt \right] \ &= \frac{1}{\pi} \left( \left[ -\frac{t}{n} \cos nt + \frac{1}{n^2} \sin nt \right]0^{\pi/2} + \left[ -\frac{\pi}{2n} \cos nt \right]{\pi/2}^{\pi} \right. \ &\quad \left. + \left[ \frac{t - 2\pi}{2n} \cos nt - \frac{1}{2n^2} \sin nt \right]_{\pi}^{2\pi} \right) \ &= \frac{1}{\pi} \left( -\frac{\pi}{2n} \cos \frac{1}{2}n\pi + \frac{1}{n^2} \sin \frac{1}{2}n\pi - \frac{\pi}{2n} \cos n\pi + \frac{\pi}{2n} \cos \frac{1}{2}n\pi + \frac{\pi}{2n} \cos n\pi \right) \ &= \frac{1}{\pi n^2} \sin \frac{1}{2}n\pi \ &= \begin{cases} 0 & (\text{even } n) \ \frac{(-1)^{(n-1)/2}}{\pi n^2} & (\text{odd } n) \end{cases} \end{aligned}$$