The set $\{ e^{inx}: n\in \mathbb{Z} \}$ forms an orthonormal basis of $(L_{2}(-\pi,\pi),\langle \cdot , \cdot \rangle)$ where $\langle f, g\rangle = \int_{-\pi}^{\pi} f\bar{g}$. That is, $\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{ikx}e^{-ilx} dx = \begin{cases} 1, & \text{if }k=l, \\ 0, & \text{otherwise}, \end{cases}\quad \quad k,l \in \mathbb{Z}$or equivalently, > [!NOTE] > > $\frac{1}{2\pi}\int_{-\pi}^\pi e^{ikx} = \begin{cases} > 1, &k= 0, \\ > 0, & k\neq 0 > \end{cases}, \quad k\in \mathbb{Z}.$ ###### Proof This follows from [[Cauchy-Goursat Theorem]], since $\int_{-\pi}^{\pi} e^{inx} \, dx = \frac{1}{i} \int_{|z|=1} z^{n-1} \, dz $and $z\mapsto z^{n-1}$ is entire except when $n= 0$. ###### Proof If then and the assertion follows immediately. If $k \neq l$ then we make use of $e^{i\pi}=e^{-i\pi}=-1$: $\begin{align} \int_{-\pi}^{\pi}e^{\mathrm{i}kx}e^{-\mathrm{i}lx}dx& =\int_{-\pi}^{\pi}e^{\mathrm{i}(k-l)x}dx \\ &=\left.\frac1{\mathrm{i}(k-l)}e^{\mathrm{i}(k-1)x}\right|_{-\pi}^\pi \\ &=\frac1{\mathrm{i}(k-l)}\big(e^{\mathrm{i}(k-1)\pi}-e^{-\mathrm{i}(k-l)\pi}\big) \\ &=\frac1{\mathrm{i}(k-l)}\big((-1)^{(k-l)}-(-1)^{(k-1)}\big) \\ &=0. \end{align}$