How to do this Integration (in Orthonormal Family for Continuous Functions)

Question

Here is a common example in the discussion on orthonormal family. Let $\mathcal L = C[a, b]$. For $k \in \mathbb Z$, let $e_k \in \mathcal L$ be defined by $$e_k(\xi) := \frac{1}{\sqrt{b-a}} e^{\frac{2\pi i k(\xi - a)}{b-a}}, \forall \xi \in [a, b].$$ Then it is claimed that $\{e_k\}_{k=-\infty}^{+\infty}$ is an orthonormal family. It is clear to see that $$\langle e_k, e_l \rangle = \int_a^b e_k(\xi) \overline{e_l(\xi)} d\xi = \frac{1}{b-a} \int_a^b e^{\frac{2\pi i (k-l)(\xi - a)}{b-a}} d\xi.$$ From here I can see that the integral is $1$ when $k =l$. However, I could not figure out why it is $0$ when $k \neq l$. Basically, I could not do the integration. Could anyone explain how to integrate when $k \neq l$, please? Thank you!

Answer 1

A straightforward calculation reveals that when $k\neq l$ $$ \frac{1}{b-a}\int_a^b e^{\frac{2\pi i(k-l)(\xi-a)}{b-a}}\,d\xi =\frac{(b-a)}{(b-a)2\pi i (k-l)} \left[e^{\frac{2\pi i(k-l)(\xi-a)}{b-a}}\right]_{\xi=a}^{\xi=b}, $$ but $e^0 = e^{2\pi i n}= \cos(2\pi n) + i\sin(2\pi n) = 1$ for all $n\in \mathbb{Z}$, so the RHS is zero.

Answer 2

You can also use $e^{i\theta} = \cos \theta + \sin \theta$ then integrate both trigonometric functions.