$${f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\left( \frac{n \pi t}{L} \right) + \sum_{n=1}^{\infty} b_n \sin\left( \frac{n \pi t}{L} \right) }$$
$${a_0 = \frac{1}{L} \int_{-L}^{L} f(t) \ dt }$$
$${a_n = \frac{1}{L} \int_{-L}^{L} f(t) \cos\left( \frac{n \pi t}{L} \right) \ dt }$$
$${b_n = \frac{1}{L} \int_{-L}^{L} f(t) \sin\left( \frac{n \pi t}{L} \right) \ dt }$$
$${f(t) = \sum_{n=- \infty}^{\infty} C_n e^{\frac{in \pi t}{L}} }$$
$${C_n = \frac{1}{2L} \int_{-L}^{L} f(t) e^{\frac{-in \pi t}{L}} \ dt }$$
$${\frac{1}{\pi} \int_{-\pi}^{\pi} \left(f(t) \right)^2 \ dt = 2 (a_0)^2 + \sum_{n=1}^{\infty} \left( a_n^2 + b_n^2 \right) }$$
The series converges to f(x) on the interval (-L, L) when continuous and to the midpoint of the two one-sided limits at the discontinuities.