Given \(f\relax (x) \) which is periodic on \(0<x<L\), so period is \(L\), then Fourier series is\[ f\relax (x) \sim \frac {1}{\sqrt {L}}{\displaystyle \sum \limits _{n=-\infty }^{\infty }} c_{n}e^{in\frac {2\pi }{L}x}\] Where \begin {align*} c_{n} & =\langle n|f\rangle \\ & =\frac {1}{\sqrt {L}}\int _{0}^{L}f\relax (x) e^{-in\frac {2\pi }{L}x}dx \end {align*}
The basis are \(|n\rangle =\frac {1}{\sqrt {L}}e^{-in\frac {2\pi }{L}x}\) and \(L\) is the period.
Fourier transform for non periodic \(f\relax (x) \) is (sum above becomes integral)\begin {align*} f\relax (x) & =\frac {1}{2\pi }\int _{-\infty }^{\infty }c_{k}e^{ikx}dk\\ c_{k} & =\int _{-\infty }^{\infty }f\relax (x) e^{-ikx}dx \end {align*}
This gives rise to\[ \delta \left (x-x^{\prime }\right ) =\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{ik\left (x-x^{\prime }\right ) }dk \]