(1) $\frac{1}{\alpha}+2 \alpha \sum_{n=1}^{\infty} \frac{(-1)^n}{\alpha^2-n^2}=\frac{\pi}{\sin (\alpha \pi)}$.
(2) $\sum_{n=1}^{\infty} \frac{1}{\left(4 n^2-1\right)^2}=\frac{\pi^2-8}{16}$.
【答案】 $a_0=\frac{1}{\pi} \int_{-\pi}^\pi \cos \alpha x \mathrm{~d} x=\frac{2 \sin \alpha \pi}{\alpha \pi}, b_n=0$
\begin{aligned} a_n & =\frac{2}{\pi} \int_0^\pi \cos \alpha x \cos n x \mathrm{~d} x \\ & =\frac{1}{\pi} \int_0^\pi[\cos (\alpha x+n x)+\cos (\alpha x-n x)] \mathrm{d} x \\ & =\frac{1}{\pi}\left[\frac{1}{\alpha+n} \sin (\alpha+n) x+\frac{1}{\alpha-n} \sin (\alpha-n) x\right]_0^\pi \\ & =\frac{1}{\pi}\left[\frac{(-1)^n \sin (\alpha x)}{\alpha+n}+\frac{(-1)^n \sin (\alpha x)}{\alpha-n}\right] \\ & =\frac{(-1)^n}{\pi}\left(\frac{2 \alpha \sin (\alpha \pi)}{\alpha^2-n^2}\right) \\ \Rightarrow & \cos \alpha x=\frac{\sin \alpha \pi}{\alpha \pi}+\frac{2 \alpha \sin \alpha \pi}{\pi} \sum_{n=1}^{+\infty} \frac{(-1)^n \cos n x}{\alpha^2-n^2} \end{aligned}

(1) 在 $\cos \alpha x$ 的 Fourier 级数展开中令 $x=0$ ，
$$\Rightarrow \frac{\pi}{\sin \alpha \pi}=\frac{1}{\alpha}+\sum_{n=1}^{\infty}(-1)^a \frac{2 \alpha}{\alpha^2-n^2}$$
(2) 由帕赛瓦尔等式，有:
\begin{aligned} & \int_{-\pi}^\pi \cos ^2 \frac{x}{2} d x=\pi\left(\frac{8}{\pi^2}+\frac{16}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{\left(4 n^2-1\right)^2}\right) \\ & \Rightarrow \pi=\pi\left(\frac{8}{\pi^2}+\frac{16}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{\left(4 n^2-1\right)^2}\right) \\ & \Rightarrow \frac{\pi^2-8}{16}=\sum_{n=1}^{\infty} \frac{1}{\left(4 n^2-1\right)^2} \end{aligned}