If x(t) = x(t + T) for all t (x(t) periodic with $ \omega_0 = 2\pi/T $)
$ \begin{align} x(t) &= \Sigma_{k=-\infty}^{\infty} a_k e^{jk (2\pi/T) t} \\ \\ where \\ a_k &= 1/T \int_{T}^{} x(t) e^{-jk (2\pi/T) t} dt \end{align} $
