傅里叶变换推导
傅里叶级数
设三角级数
$$ \frac{a_0}{2} + \sum_{n = 1}^{\infty} \left( a_n \cos n x + b_n \sin n x \right) $$
假设它在区间 $[-\pi, \pi]$ 上收敛于 $f(x)$, 即
$$ f(x) = \frac{a_0}{2} + \sum_{n = 1}^{\infty} \left( a_n \cos n x + b_n \sin n x \right) \tag{1.1} $$
将等式 $(1.1)$ 的两端在 $[-\pi, \pi]$ 上逐项积分, 得
$$ \begin{align} \int_{-\pi}^{\pi} f(x) dx &= \int_{-\pi}^{\pi} \left( \frac{a_0}{2} + \sum_{n = 1}^{\infty} \left( a_n \cos n x + b_n \sin n x \right) \right) dx \\ &= \frac{a_0}{2} \int_{-\pi}^{\pi} dx + \sum_{n = 1}^{\infty} \left( a_n \int_{-\pi}^{\pi} \cos n x dx + b_n \int_{-\pi}^{\pi} \sin n x dx \right) \\ &= \pi a_0 + \sum_{n = 1}^{\infty} \left( \frac{a_n}{n} \int_{-\pi}^{\pi} \cos n x dnx + \frac{b_n}{n} \int_{-\pi}^{\pi} \sin n x dnx \right) \\ &= \pi a_0 + \sum_{n = 1}^{\infty} \left( \frac{a_n}{n} \left[ \sin(n \pi) - \sin(- n \pi) \right] + \frac{b_n}{n} \left[ \cos(-n \pi) - \cos(n \pi) \right] \right) \\ &= \pi a_0 \end{align} \tag{.} $$